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Please send them to linear@axler.net. The print version of this book is published by Springer. \n\nOpen Access This book is licensed under the terms of the Creative Commons Attribution-NonCommercial 4.0 International License (https://creativecommons.org/licenses/by-nc/4.0), which permits any noncommercial use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to original author and source, provide a link to the Creative Commons license, and indicate if changes were made. \n\nThe images or other third party material in this book are included in the books Creative Commons license, unless indicated otherwise in a credit line to the material. If material is not included in the books Creative Commons license and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. \n\n# = 1\n\n# 51 + 52 \n\n \n\n# 1 52 \n\n \n\n# About the Author \n\nSheldon Axler received his undergraduate degree from Princeton University, followed by a PhD in mathematics from the University of California at Berkeley. As a postdoctoral Moore Instructor at MIT, Axler received a university-wide teaching award. He was then an assistant professor, associate professor, and professor at Michigan State University, where he received the first J. Sutherland Frame Teaching Award and the Distinguished Faculty Award. Axler received the Lester R. Ford Award for expository writing from the Math-ematical Association of America in 1996, for a paper that eventually expanded into this book. In addition to publishing numerous research papers, he is the author of six mathematics textbooks, ranging from freshman to graduate level. Previous editions of this book have been adopted as a textbook at over 375 universities and colleges and have been translated into three languages. Axler has served as Editor-in-Chief of the Mathematical Intelligencer and Associate Editor of the American Mathematical Monthly . He has been a member of the Council of the American Mathematical Society and of the Board of Trustees of the Mathematical Sciences Research Institute. He has also served on the editorial board of Springers series Undergraduate Texts in Mathematics, Graduate Texts in Mathematics, Universitext, and Springer Monographs in Mathematics. Axler is a Fellow of the American Mathematical Society and has been a recipient of numerous grants from the National Science Foundation. Axler joined San Francisco State University as chair of the Mathematics Department in 1997. He served as dean of the College of Science & Engineering from 2002 to 2015, when he returned to a regular faculty appointment as a professor in the Mathematics Department. \n\n> Carrie Heeter, Bishnu Sarangi\n\nThe author and his cat Moon. \n\nCover equation : Formula for the th Fibonacci number. Exercise 21 in Section 5D uses linear algebra to derive this formula. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n> v\n\n# Contents \n\n## About the Author v\n\n## Preface for Students xii \n\n## Preface for Instructors xiii \n\n## Acknowledgments xvii \n\nChapter 1 \n\n## Vector Spaces 1\n\n1A and 2Complex Numbers 2Lists 5\n\n 6Digression on Fields 10 Exercises 1A 10 \n\n1B Definition of Vector Space 12 Exercises 1B 16 \n\n1C Subspaces 18 Sums of Subspaces 19 Direct Sums 21 Exercises 1C 24 \n\nChapter 2 \n\n## Finite-Dimensional Vector Spaces 27 \n\n2A Span and Linear Independence 28 Linear Combinations and Span 28 Linear Independence 31 Exercises 2A 37 \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n> vi Contents vii\n\n2B Bases 39 Exercises 2B 42 \n\n2C Dimension 44 Exercises 2C 48 \n\n> Chapter 3\n\n## Linear Maps 51 \n\n3A Vector Space of Linear Maps 52 Definition and Examples of Linear Maps 52 Algebraic Operations on ( , ) 55 Exercises 3A 57 \n\n3B Null Spaces and Ranges 59 Null Space and Injectivity 59 Range and Surjectivity 61 Fundamental Theorem of Linear Maps 62 Exercises 3B 66 \n\n3C Matrices 69 Representing a Linear Map by a Matrix 69 Addition and Scalar Multiplication of Matrices 71 Matrix Multiplication 72 ColumnRow Factorization and Rank of a Matrix 77 Exercises 3C 79 \n\n3D Invertibility and Isomorphisms 82 Invertible Linear Maps 82 Isomorphic Vector Spaces 86 Linear Maps Thought of as Matrix Multiplication 88 Change of Basis 90 Exercises 3D 93 \n\n3E Products and Quotients of Vector Spaces 96 Products of Vector Spaces 96 Quotient Spaces 98 Exercises 3E 103 \n\n3F Duality 105 Dual Space and Dual Map 105 Null Space and Range of Dual of Linear Map 109 \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler viii Contents\n\nMatrix of Dual of Linear Map 113 Exercises 3F 115 \n\n> Chapter 4\n\n## Polynomials 119 \n\nZeros of Polynomials 122 Division Algorithm for Polynomials 123 Factorization of Polynomials over 124 Factorization of Polynomials over 127 Exercises 4 129 \n\n> Chapter 5\n\n## Eigenvalues and Eigenvectors 132 \n\n5A Invariant Subspaces 133 Eigenvalues 133 Polynomials Applied to Operators 137 Exercises 5A 139 \n\n5B The Minimal Polynomial 143 Existence of Eigenvalues on Complex Vector Spaces 143 Eigenvalues and the Minimal Polynomial 144 Eigenvalues on Odd-Dimensional Real Vector Spaces 149 Exercises 5B 150 \n\n5C Upper-Triangular Matrices 154 Exercises 5C 160 \n\n5D Diagonalizable Operators 163 Diagonal Matrices 163 Conditions for Diagonalizability 165 Gershgorin Disk Theorem 170 Exercises 5D 172 \n\n5E Commuting Operators 175 Exercises 5E 179 \n\n> Chapter 6\n\n## Inner Product Spaces 181 \n\n6A Inner Products and Norms 182 Inner Products 182 \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Contents ix\n\nNorms 186 Exercises 6A 191 \n\n6B Orthonormal Bases 197 Orthonormal Lists and the GramSchmidt Procedure 197 Linear Functionals on Inner Product Spaces 204 Exercises 6B 207 \n\n6C Orthogonal Complements and Minimization Problems 211 Orthogonal Complements 211 Minimization Problems 217 Pseudoinverse 220 Exercises 6C 224 \n\n> Chapter 7\n\n## Operators on Inner Product Spaces 227 \n\n7A Self-Adjoint and Normal Operators 228 Adjoints 228 Self-Adjoint Operators 233 Normal Operators 235 Exercises 7A 239 \n\n7B Spectral Theorem 243 Real Spectral Theorem 243 Complex Spectral Theorem 246 Exercises 7B 247 \n\n7C Positive Operators 251 Exercises 7C 255 \n\n7D Isometries, Unitary Operators, and Matrix Factorization 258 Isometries 258 Unitary Operators 260 QR Factorization 263 Cholesky Factorization 266 Exercises 7D 268 \n\n7E Singular Value Decomposition 270 Singular Values 270 SVD for Linear Maps and for Matrices 273 Exercises 7E 278 \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler xContents\n\n7F Consequences of Singular Value Decomposition 280 Norms of Linear Maps 280 Approximation by Linear Maps with Lower-Dimensional Range 283 Polar Decomposition 285 Operators Applied to Ellipsoids and Parallelepipeds 287 Volume via Singular Values 291 Properties of an Operator as Determined by Its Eigenvalues 293 Exercises 7F 294 \n\n> Chapter 8\n\n## Operators on Complex Vector Spaces 297 \n\n8A Generalized Eigenvectors and Nilpotent Operators 298 Null Spaces of Powers of an Operator 298 Generalized Eigenvectors 300 Nilpotent Operators 303 Exercises 8A 306 \n\n8B Generalized Eigenspace Decomposition 308 Generalized Eigenspaces 308 Multiplicity of an Eigenvalue 310 Block Diagonal Matrices 314 Exercises 8B 316 \n\n8C Consequences of Generalized Eigenspace Decomposition 319 Square Roots of Operators 319 Jordan Form 321 Exercises 8C 324 \n\n8D Trace: A Connection Between Matrices and Operators 326 Exercises 8D 330 \n\n> Chapter 9\n\n## Multilinear Algebra and Determinants 332 \n\n9A Bilinear Forms and Quadratic Forms 333 Bilinear Forms 333 Symmetric Bilinear Forms 337 Quadratic Forms 341 Exercises 9A 344 \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Contents xi\n\n9B Alternating Multilinear Forms 346 Multilinear Forms 346 Alternating Multilinear Forms and Permutations 348 Exercises 9B 352 \n\n9C Determinants 354 Defining the Determinant 354 Properties of Determinants 357 Exercises 9C 367 \n\n9D Tensor Products 370 Tensor Product of Two Vector Spaces 370 Tensor Product of Inner Product Spaces 376 Tensor Product of Multiple Vector Spaces 378 Exercises 9D 380 \n\n## Photo Credits 383 \n\n## Symbol Index 384 \n\n## Index 385 \n\n## Colophon: Notes on Typesetting 390 \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n# Preface for Students \n\nYou are probably about to begin your second exposure to linear algebra. Unlike your first brush with the subject, which probably emphasized Euclidean spaces and matrices, this encounter will focus on abstract vector spaces and linear maps. These terms will be defined later, so dont worry if you do not know what they mean. This book starts from the beginning of the subject, assuming no knowledge of linear algebra. The key point is that you are about to immerse yourself in serious mathematics, with an emphasis on attaining a deep understanding of the definitions, theorems, and proofs. You cannot read mathematics the way you read a novel. If you zip through a page in less than an hour, you are probably going too fast. When you encounter the phrase as you should verify, you should indeed do the verification, which will usually require some writing on your part. When steps are left out, you need to supply the missing pieces. You should ponder and internalize each definition. For each theorem, you should seek examples to show why each hypothesis is necessary. Discussions with other students should help. As a visual aid, definitions are in yellow boxes and theorems are in blue boxes (in color versions of the book). Each theorem has an informal descriptive name. Please check the website below for additional information about the book, including a link to videos that are freely available to accompany the book. Your suggestions, comments, and corrections are most welcome. Best wishes for success and enjoyment in learning linear algebra! Sheldon Axler San Francisco State University website: https://linear.axler.net e-mail: linear@axler.net \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n> xii\n\n# Preface for Instructors \n\nYou are about to teach a course that will probably give students their second exposure to linear algebra. During their first brush with the subject, your students probably worked with Euclidean spaces and matrices. In contrast, this course will emphasize abstract vector spaces and linear maps. The title of this book deserves an explanation. Most linear algebra textbooks use determinants to prove that every linear operator on a finite-dimensional com-plex vector space has an eigenvalue. Determinants are difficult, nonintuitive, and often defined without motivation. To prove the theorem about existence of eigenvalues on complex vector spaces, most books must define determinants, prove that a linear operator is not invertible if and only if its determinant equals 0,and then define the characteristic polynomial. This tortuous (torturous?) path gives students little feeling for why eigenvalues exist. In contrast, the simple determinant-free proofs presented here (for example, see 5.19) offer more insight. Once determinants have been moved to the end of the book, a new route opens to the main goal of linear algebraunderstanding the structure of linear operators. This book starts at the beginning of the subject, with no prerequisites other than the usual demand for suitable mathematical maturity. A few examples and exercises involve calculus concepts such as continuity, differentiation, and integration. You can easily skip those examples and exercises if your students have not had calculus. If your students have had calculus, then those examples and exercises can enrich their experience by showing connections between different parts of mathematics. Even if your students have already seen some of the material in the first few chapters, they may be unaccustomed to working exercises of the type presented here, most of which require an understanding of proofs. Here is a chapter-by-chapter summary of the highlights of the book: \n\n Chapter 1: Vector spaces are defined in this chapter, and their basic properties are developed. \n\n Chapter 2: Linear independence, span, basis, and dimension are defined in this chapter, which presents the basic theory of finite-dimensional vector spaces. \n\n Chapter 3: This chapter introduces linear maps. The key result here is the fundamental theorem of linear maps: if is a linear map on , then dim = \n\ndim null + dim range . Quotient spaces and duality are topics in this chapter at a higher level of abstraction than most of the book; these topics can be skipped (except that duality is needed for tensor products in Section 9D). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n> xiii xiv Preface for Instructors\n> \n\nChapter 4: The part of the theory of polynomials that will be needed to un-derstand linear operators is presented in this chapter. This chapter contains no linear algebra. It can be covered quickly, especially if your students are already familiar with these results. \n\n> \n\nChapter 5: The idea of studying a linear operator by restricting it to small sub-spaces leads to eigenvectors in the early part of this chapter. The highlight of this chapter is a simple proof that on complex vector spaces, eigenvalues always ex-ist. This result is then used to show that each linear operator on a complex vector space has an upper-triangular matrix with respect to some basis. The minimal polynomial plays an important role here and later in the book. For example, this chapter gives a characterization of the diagonalizable operators in terms of the minimal polynomial. Section 5E can be skipped if you want to save some time. \n\n> \n\nChapter 6: Inner product spaces are defined in this chapter, and their basic properties are developed along with tools such as orthonormal bases and the GramSchmidt procedure. This chapter also shows how orthogonal projections can be used to solve certain minimization problems. The pseudoinverse is then introduced as a useful tool when the inverse does not exist. The material on the pseudoinverse can be skipped if you want to save some time. \n\n> \n\nChapter 7: The spectral theorem, which characterizes the linear operators for which there exists an orthonormal basis consisting of eigenvectors, is one of the highlights of this book. The work in earlier chapters pays off here with espe-cially simple proofs. This chapter also deals with positive operators, isometries, unitary operators, matrix factorizations, and especially the singular value de-composition, which leads to the polar decomposition and norms of linear maps. \n\n> \n\nChapter 8: This chapter shows that for each operator on a complex vector space, there is a basis of the vector space consisting of generalized eigenvectors of the operator. Then the generalized eigenspace decomposition describes a linear operator on a complex vector space. The multiplicity of an eigenvalue is defined as the dimension of the corresponding generalized eigenspace. These tools are used to prove that every invertible linear operator on a complex vector space has a square root. Then the chapter gives a proof that every linear operator on a complex vector space can be put into Jordan form. The chapter concludes with an investigation of the trace of operators. \n\n> \n\nChapter 9: This chapter begins by looking at bilinear forms and showing that the vector space of bilinear forms is the direct sum of the subspaces of symmetric bilinear forms and alternating bilinear forms. Then quadratic forms are diag-onalized. Moving to multilinear forms, the chapter shows that the subspace of alternating -linear forms on an -dimensional vector space has dimension one. This result leads to a clean basis-free definition of the determinant of an opera-tor. For complex vector spaces, the determinant turns out to equal the product of the eigenvalues, with each eigenvalue included in the product as many times as its multiplicity. The chapter concludes with an introduction to tensor products. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Preface for Instructors xv\n\nThis book usually develops linear algebra simultaneously for real and complex vector spaces by letting denote either the real or the complex numbers. If you and your students prefer to think of as an arbitrary field, then see the comments at the end of Section 1A. I prefer avoiding arbitrary fields at this level because they intro-duce extra abstraction without leading to any new linear algebra. Also, students are more comfortable thinking of polynomials as functions instead of the more formal objects needed for polynomials with coefficients in finite fields. Finally, even if the beginning part of the theory were developed with arbitrary fields, inner product spaces would push consideration back to just real and complex vector spaces. You probably cannot cover everything in this book in one semester. Going through all the material in the first seven or eight chapters during a one-semester course may require a rapid pace. If you must reach Chapter 9, then consider skipping the material on quotient spaces in Section 3E, skipping Section 3F on duality (unless you intend to cover tensor products in Section 9D), covering Chapter 4 on polynomials in a half hour, skipping Section 5E on commuting operators, and skipping the subsection in Section 6C on the pseudoinverse. A goal more important than teaching any particular theorem is to develop in students the ability to understand and manipulate the objects of linear algebra. Mathematics can be learned only by doing. Fortunately, linear algebra has many good homework exercises. When teaching this course, during each class I usually assign as homework several of the exercises, due the next class. Going over the homework might take up significant time in a typical class. Some of the exercises are intended to lead curious students into important topics beyond what might usually be included in a basic second course in linear algebra. \n\nThe authors top ten \n\nListed below are the authors ten favorite results in the book, in order of their appearance in the book. Students who leave your course with a good understanding of these crucial results will have an excellent foundation in linear algebra. \n\n> \n\nany two bases of a vector space have the same length (2.34) \n\n> \n\nfundamental theorem of linear maps (3.21) \n\n> \n\nexistence of eigenvalues if = (5.19) \n\n> \n\nupper-triangular form always exists if = (5.47) \n\n> \n\nCauchySchwarz inequality (6.14) \n\n> \n\nGramSchmidt procedure (6.32) \n\n> \n\nspectral theorem (7.29 and 7.31) \n\n> \n\nsingular value decomposition (7.70) \n\n> \n\ngeneralized eigenspace decomposition theorem when = (8.22) \n\n> \n\ndimension of alternating -linear forms on is 1 if dim = (9.37) \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler xvi Preface for Instructors\n\nMajor improvements and additions for the fourth edition \n\n> \n\nOver 250 new exercises and over 70 new examples. \n\n> \n\nIncreasing use of the minimal polynomial to provide cleaner proofs of multiple results, including necessary and sufficient conditions for an operator to have an upper-triangular matrix with respect to some basis (see Section 5C), necessary and sufficient conditions for diagonalizability (see Section 5D), and the real spectral theorem (see Section 7B). \n\n> \n\nNew section on commuting operators (see Section 5E). \n\n> \n\nNew subsection on pseudoinverse (see Section 6C). \n\n> \n\nNew subsections on QR factorization/Cholesky factorization (see Section 7D). \n\n> \n\nSingular value decomposition now done for linear maps from an inner product space to another (possibly different) inner product space, rather than only deal-ing with linear operators from an inner product space to itself (see Section 7E). \n\n> \n\nPolar decomposition now proved from singular value decomposition, rather than in the opposite order; this has led to cleaner proofs of both the singular value decomposition (see Section 7E) and the polar decomposition (see Section 7F). \n\n> \n\nNew subsection on norms of linear maps on finite-dimensional inner prod-uct spaces, using the singular value decomposition to avoid even mentioning supremum in the definition of the norm of a linear map (see Section 7F). \n\n> \n\nNew subsection on approximation by linear maps with lower-dimensional range (see Section 7F). \n\n> \n\nNew elementary proof of the important result that if is an operator on a finite-dimensional complex vector space , then there exists a basis of consisting of generalized eigenvectors of (see 8.9). \n\n> \n\nNew Chapter 9 on multilinear algebra, including bilinear forms, quadratic forms, multilinear forms, and tensor products. Determinants now are defined using a basis-free approach via alternating multilinear forms. \n\n> \n\nNew formatting to improve the student-friendly appearance of the book. For example, the definition and result boxes now have rounded corners instead of right-angle corners, for a gentler look. The main font size has been reduced from 11 point to 10.5 point. Please check the website below for additional links and information about the book. Your suggestions, comments, and corrections are most welcome. Best wishes for teaching a successful linear algebra class! \n\n> Contact the author, or Springer if the author is not available, for permission for translations or other commercial reuse of the contents of this book.\n\nSheldon Axler San Francisco State University website: https://linear.axler.net e-mail: linear@axler.net \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n# Acknowledgments \n\nI owe a huge intellectual debt to all the mathematicians who created linear algebra over the past two centuries. The results in this book belong to the common heritage of mathematics. A special case of a theorem may first have been proved long ago, then sharpened and improved by many mathematicians in different time periods. Bestowing proper credit on all contributors would be a difficult task that I have not undertaken. In no case should the reader assume that any result presented here represents my original contribution. Many people helped make this a better book. The three previous editions of this book were used as a textbook at over 375 universities and colleges around the world. I received thousands of suggestions and comments from faculty and students who used the book. Many of those suggestions led to improvements in this edition. The manuscript for this fourth edition was class tested at 30 universities. I am extremely grateful for the useful feedback that I received from faculty and students during this class testing. The long list of people who should be thanked for their suggestions would fill up many pages. Lists are boring to read. Thus to represent all contributors to this edition, I will mention only Noel Hinton, a graduate student at Australian National University, who sent me more suggestions and corrections for this fourth edition than anyone else. To everyone who contributed suggestions, let me say how truly grateful I am to all of you. Many many thanks! I thank Springer for providing me with help when I needed it and for allowing me the freedom to make the final decisions about the content and appearance of this book. Special thanks to the two terrific mathematics editors at Springer who worked with me on this projectLoretta Bartolini during the first half of my work on the fourth edition, and Elizabeth Loew during the second half of my work on the fourth edition. I am deeply indebted to David Kramer, who did a magnificent job of copyediting and prevented me from making many mistakes. Extra special thanks to my fantastic partner Carrie Heeter. Her understanding and encouragement enabled me to work intensely on this new edition. Our won-derful cat Moon, whose picture appears on the About the Author page, provided sweet breaks throughout the writing process. Moon died suddenly due to a blood clot as this book was being finished. We are grateful for five precious years with him. Sheldon Axler \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n> xvii\n\n# Chapter 1 \n\n# Vector Spaces \n\nLinear algebra is the study of linear maps on finite-dimensional vector spaces. Eventually we will learn what all these terms mean. In this chapter we will define vector spaces and discuss their elementary properties. In linear algebra, better theorems and more insight emerge if complex numbers are investigated along with real numbers. Thus we will begin by introducing the complex numbers and their basic properties. We will generalize the examples of a plane and of ordinary space to and \n\n , which we then will generalize to the notion of a vector space. As we will see, a vector space is a set with operations of addition and scalar multiplication that satisfy natural algebraic properties. Then our next topic will be subspaces, which play a role for vector spaces analogous to the role played by subsets for sets. Finally, we will look at sums of subspaces (analogous to unions of subsets) and direct sums of subspaces (analogous to unions of disjoint sets). \n\n> Pierre Louis Dumesnil, Nils Forsberg\n\nRen Descartes explaining his work to Queen Christina of Sweden. Vector spaces are a generalization of the description of a plane using two coordinates, as published by Descartes in 1637. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n> 12Chapter 1 Vector Spaces\n\n# 1A and \n\n# Complex Numbers \n\nYou should already be familiar with basic properties of the set of real numbers. Complex numbers were invented so that we can take square roots of negative numbers. The idea is to assume we have a square root of 1 , denoted by , that obeys the usual rules of arithmetic. Here are the formal definitions. \n\n> 1.1 definition:\n\ncomplex numbers, A complex number is an ordered pair ( , ) , where , , but we will write this as + .\n\n The set of all complex numbers is denoted by :\n\n = { + , }. Addition and multiplication on are defined by \n\n( + ) + ( + ) = ( + ) + ( + ) ,\n\n( + )( + ) = ( ) + ( + ) ;\n\nhere , , , .If , we identify + 0 with the real number . Thus we think of as a subset of . We usually write 0 + as just , and we usually write 0 + 1 as just . \n\n> The symbol was first used to denote\n> 1 by Leonhard Euler in 1777.\n\nTo motivate the definition of complex multiplication given above, pretend that we knew that 2 = 1 and then use the usual rules of arithmetic to derive the formula above for the product of two complex numbers. Then use that formula to verify that we indeed have \n\n 2 = 1. \n\nDo not memorize the formula for the product of two complex numbersyou can always rederive it by recalling that 2 = 1 and then using the usual rules of arithmetic (as given by 1.3). The next example illustrates this procedure. \n\n> 1.2 example:\n\ncomplex arithmetic \n\nThe product (2 + 3)(4 + 5) can be evaluated by applying the distributive and commutative properties from 1.3: \n\n(2 + 3)(4 + 5) = 2 (4 + 5) + (3)(4 + 5) = 2 4 + 2 5 + 3 4 + (3)(5) = 8 + 10 + 12 15 = 7 + 22. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 1A and 3\n\nOur first result states that complex addition and complex multiplication have the familiar properties that we expect. \n\n> 1.3\n\nproperties of complex arithmetic \n\ncommutativity \n\n + = + and = for all , .\n\nassociativity \n\n( + ) + = + ( + ) and () = () for all , , .\n\nidentities \n\n + 0 = and 1 = for all .\n\nadditive inverse \n\nFor every , there exists a unique such that + = 0 .\n\nmultiplicative inverse \n\nFor every with 0 , there exists a unique such that = 1 .\n\ndistributive property \n\n( + ) = + for all , , .The properties above are proved using the familiar properties of real numbers and the definitions of complex addition and multiplication. The next example shows how commutativity of complex multiplication is proved. Proofs of the other properties above are left as exercises. \n\n> 1.4 example:\n\ncommutativity of complex multiplication \n\nTo show that = for all , , suppose \n\n = + and = + ,\n\nwhere , , , . Then the definition of multiplication of complex numbers shows that \n\n = ( + )( + ) = ( ) + ( + ) \n\nand \n\n = ( + )( + ) = ( ) + ( + ). \n\nThe equations above and the commutativity of multiplication and addition of real numbers show that = .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n4 Chapter 1 Vector Spaces \n\nNext, we define the additive and multiplicative inverses of complex numbers, and then use those inverses to define subtraction and division operations with complex numbers. \n\n> 1.5 definition:\n\n , subtraction, 1/ , division \n\nSuppose , .\n\n Let denote the additive inverse of . Thus is the unique complex number such that \n\n + () = 0. Subtraction on is defined by \n\n = + (). For 0 , let 1/ and 1 denote the multiplicative inverse of . Thus 1/ is the unique complex number such that \n\n(1/) = 1. For 0 , division by is defined by \n\n/ = (1/). \n\nSo that we can conveniently make definitions and prove theorems that apply to both real and complex numbers, we adopt the following notation. \n\n> 1.6 notation:\n\n \n\nThroughout this book, stands for either or .\n\nThe letter is used because and \n\nare examples of what are called fields .\n\nThus if we prove a theorem involving \n\n , we will know that it holds when is replaced with and when is replaced with .Elements of are called scalars . The word scalar (which is just a fancy word for number) is often used when we want to emphasize that an object is a number, as opposed to a vector (vectors will be defined soon). For and a positive integer, we define to denote the product of \n\nwith itself times: \n\n = \n\n> times\n\n.\n\nThis definition implies that \n\n( ) = and () = \n\nfor all , and all positive integers , . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 1A and 5\n\n# Lists \n\nBefore defining and , we look at two important examples. \n\n> 1.7 example:\n\n 2 and 3\n\n The set 2, which you can think of as a plane, is the set of all ordered pairs of real numbers: \n\n 2 = {( , ) , }. The set 3, which you can think of as ordinary space, is the set of all ordered triples of real numbers: \n\n 3 = {( , , ) , , }. \n\nTo generalize 2 and 3 to higher dimensions, we first need to discuss the concept of lists. \n\n> 1.8 definition:\n\nlist, length \n\n Suppose is a nonnegative integer. A list of length is an ordered collec-tion of elements (which might be numbers, other lists, or more abstract objects). \n\n Two lists are equal if and only if they have the same length and the same elements in the same order. \n\n> Many mathematicians call a list of length an -tuple .\n\nLists are often written as elements separated by commas and surrounded by parentheses. Thus a list of length two is an ordered pair that might be written as ( , ) . A list of length three is an ordered triple that might be written as ( , , ) . A list of length might look like this: \n\n( 1, , ). \n\nSometimes we will use the word list without specifying its length. Remember, however, that by definition each list has a finite length that is a nonnegative integer. Thus an object that looks like ( 1, 2, ) , which might be said to have infinite length, is not a list. A list of length 0 looks like this: ( ) . We consider such an object to be a list so that some of our theorems will not have trivial exceptions. Lists differ from finite sets in two ways: in lists, order matters and repetitions have meaning; in sets, order and repetitions are irrelevant. \n\n> 1.9 example:\n\nlists versus sets \n\n The lists (3 , 5) and (5 , 3) are not equal, but the sets {3 , 5} and {5 , 3} are equal. \n\n The lists (4 , 4) and (4 , 4, 4) are not equal (they do not have the same length), although the sets {4 , 4} and {4 , 4, 4} both equal the set {4} .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n6 Chapter 1 Vector Spaces \n\n## \n\nTo define the higher-dimensional analogues of 2 and 3, we will simply replace \n\n with (which equals or ) and replace the 2 or 3 with an arbitrary positive integer. \n\n> 1.10 notation:\n\n \n\nFix a positive integer for the rest of this chapter. \n\n> 1.11 definition:\n\n , coordinate \n\n is the set of all lists of length of elements of :\n\n = {( 1, , ) for = 1 , , }. \n\nFor ( 1, , ) and {1 , , } , we say that is the th coordinate \n\nof ( 1, , ).If = and equals 2 or 3, then the definition above of agrees with our previous notions of 2 and 3. \n\n> 1.12 example:\n\n 4\n\n 4 is the set of all lists of four complex numbers: \n\n 4 = {( 1, 2, 3, 4) 1, 2, 3, 4 }. \n\nRead Flatland: A Romance of Many Dimensions , by Edwin A. Abbott, for an amusing account of how 3 would be perceived by creatures living in 2.This novel, published in 1884, may help you imagine a physical space of four or more dimensions. \n\nIf 4 , we cannot visualize as a physical object. Similarly, 1 can be thought of as a plane, but for 2 , the human brain cannot provide a full image of . However, even if is large, we can perform algebraic manipulations in \n\n as easily as in 2 or 3. For example, addition in is defined as follows. \n\n> 1.13 definition:\n\naddition in \n\nAddition in is defined by adding corresponding coordinates: \n\n( 1, , ) + ( 1, , ) = ( 1 + 1, , + ). \n\nOften the mathematics of becomes cleaner if we use a single letter to denote a list of numbers, without explicitly writing the coordinates. For example, the next result is stated with and in even though the proof requires the more cumbersome notation of ( 1, , ) and ( 1, , ).\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 1A and 7 \n\n> 1.14\n\ncommutativity of addition in \n\nIf , , then + = + .\n\nProof Suppose = ( 1, , ) and = ( 1, , ) . Then \n\n + = ( 1, , ) + ( 1, , )= ( 1 + 1, , + )= ( 1 + 1, , + )= ( 1, , ) + ( 1, , )= + ,\n\nwhere the second and fourth equalities above hold because of the definition of addition in and the third equality holds because of the usual commutativity of addition in .\n\nThe symbol means end of proof . If a single letter is used to denote an element of , then the same letter with appropriate subscripts is often used when coordinates must be displayed. For example, if , then letting equal \n\n( 1, , ) is good notation, as shown in the proof above. Even better, work with just and avoid explicit coordinates when possible. \n\n> 1.15 notation:\n\n0\n\nLet 0 denote the list of length whose coordinates are all 0:\n\n0 = (0 , , 0). \n\nHere we are using the symbol 0 in two different wayson the left side of the equation above, the symbol 0 denotes a list of length , which is an element of ,whereas on the right side, each 0 denotes a number. This potentially confusing practice actually causes no problems because the context should always make clear which 0 is intended. \n\n> 1.16 example:\n\ncontext determines which 0 is intended \n\nConsider the statement that 0 is an additive identity for :\n\n + 0 = for all .\n\nHere the 0 above is the list defined in 1.15, not the number 0, because we have not defined the sum of an element of (namely, ) and the number 0. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 8Chapter 1 Vector Spaces\n> Elements of 2can be thought of as points or as vectors.\n\nA picture can aid our intuition. We will draw pictures in 2 because we can sketch this space on two-dimensional surfaces such as paper and computer screens. A typical element of 2 is a point = ( , ) .Sometimes we think of not as a point but as an arrow starting at the origin and ending at ( , ) , as shown here. When we think of an element of 2 as an arrow, we refer to it as a vector .\n\n> A vector.\n\nWhen we think of vectors in 2 as arrows, we can move an arrow parallel to itself (not changing its length or direction) and still think of it as the same vector. With that viewpoint, you will often gain better understanding by dispensing with the coordinate axes and the explicit coordinates and just thinking of the vector, as shown in the figure here. The two arrows shown here have the same length and same direction, so we think of them as the same vector. \n\n> Mathematical models of the economy can have thousands of variables, say\n> 1,, 5000 , which means that we must work in 5000 . Such a space cannot be dealt with geometrically. However, the algebraic approach works well. Thus our subject is called linear algebra .\n\nWhenever we use pictures in 2 or use the somewhat vague language of points and vectors, remember that these are just aids to our understanding, not sub-stitutes for the actual mathematics that we will develop. Although we cannot draw good pictures in high-dimensional spaces, the elements of these spaces are as rigorously defined as elements of 2.For example, (2 , 3 , 17 , , 2) is an element of 5, and we may casually refer to it as a point in 5 or a vector in 5 without worrying about whether the geometry of 5 has any physical meaning. Recall that we defined the sum of two elements of to be the element of \n\nobtained by adding corresponding coordinates; see 1.13. As we will now see, addition has a simple geometric interpretation in the special case of 2.\n\n> The sum of two vectors.\n\nSuppose we have two vectors and in 2\n\nthat we want to add. Move the vector parallel to itself so that its initial point coincides with the end point of the vector , as shown here. The sum + then equals the vector whose initial point equals the initial point of and whose end point equals the end point of the vector , as shown here. In the next definition, the 0 on the right side of the displayed equation is the list 0 .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 1A and 9 \n\n> 1.17 definition:\n\nadditive inverse in , \n\nFor , the additive inverse of , denoted by , is the vector \n\nsuch that \n\n + () = 0. \n\nThus if = ( 1, , ), then = ( 1, , ).\n\nA vector and its additive inverse. \n\nThe additive inverse of a vector in 2 is the vector with the same length but pointing in the opposite direction. The figure here illustrates this way of thinking about the additive inverse in 2. As you can see, the vector labeled has the same length as the vector labeled but points in the opposite direction. Having dealt with addition in , we now turn to multiplication. We could define a multiplication in in a similar fashion, starting with two elements of \n\n and getting another element of by multiplying corresponding coordinates. Experience shows that this definition is not useful for our purposes. Another type of multiplication, called scalar multiplication, will be central to our subject. Specifically, we need to define what it means to multiply an element of by an element of . \n\n> 1.18 definition:\n\nscalar multiplication in \n\nThe product of a number and a vector in is computed by multiplying each coordinate of the vector by :\n\n( 1, , ) = ( 1, , );\n\nhere and ( 1, , ) .\n\nScalar multiplication in multiplies together a scalar and a vector, getting a vector. In contrast, the dot product in \n\n> 2\n\nor 3 multiplies together two vec-tors and gets a scalar. Generalizations of the dot product will become impor-tant in Chapter 6. \n\nScalar multiplication has a nice geo-metric interpretation in 2. If > 0 and \n\n 2, then is the vector that points in the same direction as and whose length is times the length of . In other words, to get , we shrink or stretch \n\nby a factor of , depending on whether \n\n < 1 or > 1 .\n\nScalar multiplication. \n\nIf < 0 and 2, then is the vector that points in the direction opposite to that of and whose length is || times the length of , as shown here. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 10 Chapter 1 Vector Spaces\n\n# Digression on Fields \n\nA field is a set containing at least two distinct elements called 0 and 1, along with operations of addition and multiplication satisfying all properties listed in 1.3. Thus and are fields, as is the set of rational numbers along with the usual operations of addition and multiplication. Another example of a field is the set \n\n{0 , 1} with the usual operations of addition and multiplication except that 1 + 1 is defined to equal 0.In this book we will not deal with fields other than and . However, many of the definitions, theorems, and proofs in linear algebra that work for the fields \n\n and also work without change for arbitrary fields. If you prefer to do so, throughout much of this book (except for Chapters 6 and 7, which deal with inner product spaces) you can think of as denoting an arbitrary field instead of \n\nor . For results (except in the inner product chapters) that have as a hypothesis that is , you can probably replace that hypothesis with the hypothesis that \n\nis an algebraically closed field, which means that every nonconstant polynomial with coefficients in has a zero. A few results, such as Exercise 13 in Section 1C, require the hypothesis on that 1 + 1 0 .\n\n# Exercises 1A \n\n1 Show that + = + for all , .\n\n2 Show that ( + ) + = + ( + ) for all , , .\n\n3 Show that () = () for all , , .\n\n4 Show that ( + ) = + for all , , .\n\n5 Show that for every , there exists a unique such that + = 0 .\n\n6 Show that for every with 0 , there exists a unique such that = 1 .\n\n7 Show that \n\n1 + 3 2\n\nis a cube root of 1 (meaning that its cube equals 1). \n\n8 Find two distinct square roots of .\n\n9 Find 4 such that \n\n(4 , 3 , 1, 7) + 2 = (5 , 9, 6 , 8). \n\n10 Explain why there does not exist such that \n\n(2 3 , 5 + 4 , 6 + 7) = (12 5 , 7 + 22 , 32 9). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 1A and 11 \n\n11 Show that ( + ) + = + ( + ) for all , , .\n\n12 Show that () = () for all and all , .\n\n13 Show that 1 = for all .\n\n14 Show that ( + ) = + for all and all , .\n\n15 Show that ( + ) = + for all , and all .Can you do addition? the White Queen asked. Whats one and one and one and one and one and one and one and one and one and one? I dont know, said Alice. I lost count. \n\nThrough the Looking Glass , Lewis Carroll \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 12 Chapter 1 Vector Spaces\n\n# 1B Definition of Vector Space \n\nThe motivation for the definition of a vector space comes from properties of addition and scalar multiplication in : Addition is commutative, associative, and has an identity. Every element has an additive inverse. Scalar multiplication is associative. Scalar multiplication by 1 acts as expected. Addition and scalar multiplication are connected by distributive properties. We will define a vector space to be a set with an addition and a scalar multiplication on that satisfy the properties in the paragraph above. \n\n> 1.19 definition:\n\naddition, scalar multiplication \n\n An addition on a set is a function that assigns an element + \n\nto each pair of elements , .\n\n A scalar multiplication on a set is a function that assigns an element \n\n to each and each .Now we are ready to give the formal definition of a vector space. \n\n> 1.20 definition:\n\nvector space \n\nA vector space is a set along with an addition on and a scalar multiplication on such that the following properties hold. \n\ncommutativity \n\n + = + for all , .\n\nassociativity \n\n( + ) + = + ( + ) and () = () for all , , and for all , .\n\nadditive identity \n\nThere exists an element 0 such that + 0 = for all .\n\nadditive inverse \n\nFor every , there exists such that + = 0 .\n\nmultiplicative identity \n\n1 = for all .\n\ndistributive properties \n\n( + ) = + and ( + ) = + for all , and all , .The following geometric language sometimes aids our intuition. \n\n> 1.21 definition:\n\nvector, point \n\nElements of a vector space are called vectors or points .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 1B Definition of Vector Space 13 \n\nThe scalar multiplication in a vector space depends on . Thus when we need to be precise, we will say that is a vector space over instead of saying simply that is a vector space. For example, is a vector space over , and is a vector space over . \n\n> 1.22 definition:\n\nreal vector space, complex vector space \n\n A vector space over is called a real vector space .\n\n A vector space over is called a complex vector space .Usually the choice of is either clear from the context or irrelevant. Thus we often assume that is lurking in the background without specifically mentioning it. \n\nThe simplest vector space is {0} , which contains only one point. \n\nWith the usual operations of addition and scalar multiplication, is a vector space over , as you should verify. The example of motivated our definition of vector space. \n\n> 1.23 example:\n\n \n\n is defined to be the set of all sequences of elements of :\n\n = {( 1, 2, ) for = 1 , 2, }. \n\nAddition and scalar multiplication on are defined as expected: \n\n( 1, 2, ) + ( 1, 2, ) = ( 1 + 1, 2 + 2, ) ,\n\n( 1, 2, ) = ( 1, 2, ). \n\nWith these definitions, becomes a vector space over , as you should verify. The additive identity in this vector space is the sequence of all 0s. Our next example of a vector space involves a set of functions. \n\n> 1.24 notation:\n\n \n\n If is a set, then denotes the set of functions from to .\n\n For , , the sum + is the function defined by \n\n( + )() = () + () \n\nfor all .\n\n For and , the product is the function defined by \n\n( )() = () \n\nfor all .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n14 Chapter 1 Vector Spaces \n\nAs an example of the notation above, if is the interval [0 , 1] and = , then \n\n [0 ,1] is the set of real-valued functions on the interval [0 , 1] .You should verify all three bullet points in the next example. \n\n> 1.25 example:\n\n is a vector space \n\n If is a nonempty set, then (with the operations of addition and scalar multiplication as defined above) is a vector space over .\n\n The additive identity of is the function 0 defined by \n\n0() = 0 \n\nfor all .\n\n For , the additive inverse of is the function defined by \n\n( )() = () \n\nfor all .\n\nThe elements of the vector space [0 ,1] \n\nare real-valued functions on [0 , 1] , not lists. In general, a vector space is an abstract entity whose elements might be lists, functions, or weird objects. \n\nThe vector space is a special case of the vector space because each \n\n( 1, , ) can be thought of as a function from the set {1 , 2, , } to \n\n by writing () instead of for the th \n\ncoordinate of ( 1, , ). In other words, we can think of as {1 ,2, ,} . Similarly, we can think of as {1 ,2, } .Soon we will see further examples of vector spaces, but first we need to develop some of the elementary properties of vector spaces. The definition of a vector space requires it to have an additive identity. The next result states that this identity is unique. \n\n> 1.26\n\nunique additive identity \n\nA vector space has a unique additive identity. \n\nProof Suppose 0 and 0 are both additive identities for some vector space .Then \n\n0 = 0 + 0 = 0 + 0 = 0 ,\n\nwhere the first equality holds because 0 is an additive identity, the second equality comes from commutativity, and the third equality holds because 0 is an additive identity. Thus 0 = 0 , proving that has only one additive identity. Each element in a vector space has an additive inverse, an element in the vector space such that + = 0 . The next result shows that each element in a vector space has only one additive inverse. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 1B Definition of Vector Space 15\n> 1.27\n\nunique additive inverse \n\nEvery element in a vector space has a unique additive inverse. \n\n> Proof\n\nSuppose is a vector space. Let . Suppose and are additive inverses of . Then \n\n = + 0 = + ( + ) = ( + ) + = 0 + = .\n\nThus = , as desired. Because additive inverses are unique, the following notation now makes sense. \n\n> 1.28 notation:\n\n , \n\nLet , . Then \n\n denotes the additive inverse of ;\n\n is defined to be + () .Almost all results in this book involve some vector space. To avoid having to restate frequently that is a vector space, we now make the necessary declaration once and for all. \n\n> 1.29 notation:\n\n \n\nFor the rest of this book, denotes a vector space over .In the next result, 0 denotes a scalar (the number 0 ) on the left side of the equation and a vector (the additive identity of ) on the right side of the equation. \n\n> 1.30\n\nthe number 0 times a vector \n\n0 = 0 for every . \n\n> The result in 1.30 involves the additive identity of and scalar multiplication. The only part of the definition of a vec-tor space that connects vector addition and scalar multiplication is the dis-tributive property. Thus the distribu-tive property must be used in the proof of 1.30.\n> Proof\n\nFor , we have \n\n0 = (0 + 0) = 0 + 0. \n\nAdding the additive inverse of 0 to both sides of the equation above gives 0 = 0 ,as desired. In the next result, 0 denotes the addi-tive identity of . Although their proofs are similar, 1.30 and 1.31 are not identical. More precisely, 1.30 states that the product of the scalar 0 and any vector equals the vector 0, whereas 1.31 states that the product of any scalar and the vector 0 equals the vector 0. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 16 Chapter 1 Vector Spaces\n> 1.31\n\na number times the vector 00 = 0 for every . \n\n> Proof\n\nFor , we have \n\n0 = (0 + 0) = 0 + 0. \n\nAdding the additive inverse of 0 to both sides of the equation above gives 0 = 0 ,as desired. Now we show that if an element of is multiplied by the scalar 1 , then the result is the additive inverse of the element of . \n\n> 1.32\n\nthe number 1 times a vector \n\n(1) = for every . \n\n> Proof\n\nFor , we have \n\n + (1) = 1 + (1) = (1 + (1)) = 0 = 0. \n\nThis equation says that (1) , when added to , gives 0. Thus (1) is the additive inverse of , as desired. \n\n# Exercises 1B \n\n1 Prove that () = for every .\n\n2 Suppose , , and = 0 . Prove that = 0 or = 0 .\n\n3 Suppose , . Explain why there exists a unique such that \n\n + 3 = .\n\n4 The empty set is not a vector space. The empty set fails to satisfy only one of the requirements listed in the definition of a vector space (1.20). Which one? \n\n5 Show that in the definition of a vector space (1.20), the additive inverse condition can be replaced with the condition that \n\n0 = 0 for all . \n\nHere the 0 on the left side is the number 0, and the 0 on the right side is the additive identity of .\n\n> The phrase a condition can be replaced in a definition means that the collection of objects satisfying the definition is unchanged if the original condition is replaced with the new condition.\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 1B Definition of Vector Space 17 \n\n6 Let and denote two distinct objects, neither of which is in . Define an addition and scalar multiplication on { , } as you could guess from the notation. Specifically, the sum and product of two real numbers is as usual, and for define \n\n = {{{{ if < 0 ,\n\n0 if = 0 ,\n\n if > 0 ,\n\n() = {{{{ if < 0 ,\n\n0 if = 0 ,\n\n if > 0 ,\n\nand \n\n + = + = + = ,\n\n + () = () + = () + () = ,\n\n + () = () + = 0. \n\nWith these operations of addition and scalar multiplication, is { , } \n\na vector space over ? Explain. \n\n7 Suppose is a nonempty set. Let denote the set of functions from to .Define a natural addition and scalar multiplication on , and show that \n\nis a vector space with these definitions. \n\n8 Suppose is a real vector space. \n\n The complexification of , denoted by , equals . An element of \n\n is an ordered pair ( , ) , where , , but we write this as + .\n\n Addition on is defined by \n\n( 1 + 1) + ( 2 + 2) = ( 1 + 2) + ( 1 + 2)\n\nfor all 1, 1, 2, 2 .\n\n Complex scalar multiplication on is defined by \n\n( + )( + ) = ( ) + ( + ) \n\nfor all , and all , .Prove that with the definitions of addition and scalar multiplication as above, \n\n is a complex vector space. \n\nThink of as a subset of by identifying with + 0 . The construc-tion of from can then be thought of as generalizing the construction of from . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 18 Chapter 1 Vector Spaces\n\n# 1C Subspaces \n\nBy considering subspaces, we can greatly expand our examples of vector spaces. \n\n> 1.33 definition:\n\nsubspace \n\nA subset of is called a subspace of if is also a vector space with the same additive identity, addition, and scalar multiplication as on . \n\n> Some people use the terminology\n> linear subspace ,which means the same as subspace.\n\nThe next result gives the easiest way to check whether a subset of a vector space is a subspace. \n\n> 1.34\n\nconditions for a subspace \n\nA subset of is a subspace of if and only if satisfies the following three conditions. \n\nadditive identity \n\n0 .\n\nclosed under addition \n\n , implies + .\n\nclosed under scalar multiplication \n\n and implies . \n\n> The additive identity condition above could be replaced with the condition that is nonempty (because then tak-ing and multiplying it by 0\n> would imply that 0 ). However, if a subset of is indeed a sub-space, then usually the quickest way to show that is nonempty is to show that 0 .\n> Proof\n\nIf is a subspace of , then \n\nsatisfies the three conditions above by the definition of vector space. Conversely, suppose satisfies the three conditions above. The first condi-tion ensures that the additive identity of \n\n is in . The second condition ensures that addition makes sense on . The third condition ensures that scalar multiplica-tion makes sense on .If , then [ which equals (1) by 1.32 ] is also in by the third condition above. Hence every element of has an additive inverse in .The other parts of the definition of a vector space, such as associativity and commutativity, are automatically satisfied for because they hold on the larger space . Thus is a vector space and hence is a subspace of .The three conditions in the result above usually enable us to determine quickly whether a given subset of is a subspace of . You should verify all assertions in the next example. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 1C Subspaces 19 \n\n> 1.35 example:\n\nsubspaces \n\n(a) If , then \n\n{( 1, 2, 3, 4) 4 3 = 5 4 + } \n\nis a subspace of 4 if and only if = 0 .(b) The set of continuous real-valued functions on the interval [0 , 1] is a subspace of [0 ,1] .(c) The set of differentiable real-valued functions on is a subspace of .(d) The set of differentiable real-valued functions on the interval (0 , 3) such that (2) = is a subspace of (0 ,3) if and only if = 0 .(e) The set of all sequences of complex numbers with limit 0 is a subspace of .\n\nThe set {0} is the smallest subspace of \n\n> \n\n, and itself is the largest subspace of . The empty set is not a subspace of because a subspace must be a vector space and hence must contain at least one element, namely, an additive identity. \n\nVerifying some of the items above shows the linear structure underlying parts of calculus. For example, (b) above requires the result that the sum of two continuous functions is continuous. As another example, (d) above requires the result that for a constant , the derivative of equals times the derivative of .The subspaces of 2 are precisely {0} , all lines in 2 containing the origin, and 2. The subspaces of 3 are precisely {0} , all lines in 3 containing the origin, all planes in 3 containing the origin, and 3. To prove that all these objects are indeed subspaces is straightforwardthe hard part is to show that they are the only subspaces of 2 and 3. That task will be easier after we introduce some additional tools in the next chapter. \n\n# Sums of Subspaces \n\nThe union of subspaces is rarely a sub-space (see Exercise 12 ), which is why we usually work with sums rather than unions. \n\nWhen dealing with vector spaces, we are usually interested only in subspaces , as opposed to arbitrary subsets. The notion of the sum of subspaces will be useful. \n\n> 1.36 definition:\n\nsum of subspaces \n\nSuppose 1, , are subspaces of . The sum of 1, , , denoted by \n\n 1 + + , is the set of all possible sums of elements of 1, , . More precisely, \n\n 1 + + = { 1 + + 1 1, , }. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 20 Chapter 1 Vector Spaces\n\nLets look at some examples of sums of subspaces. \n\n> 1.37 example:\n\na sum of subspaces of 3\n\nSuppose is the set of all elements of 3 whose second and third coordinates equal 0, and is the set of all elements of 3 whose first and third coordinates equal 0:\n\n = {( , 0, 0) 3 } and = {(0 , , 0) 3 }. \n\nThen \n\n + = {( , , 0) 3 , } ,\n\nas you should verify. \n\n> 1.38 example:\n\na sum of subspaces of 4\n\nSuppose \n\n = {( , , , ) 4 , } and = {( , , , ) 4 , }. \n\nUsing words rather than symbols, we could say that is the set of elements of 4 whose first two coordinates equal each other and whose third and fourth coordinates equal each other. Similarly, is the set of elements of 4 whose first three coordinates equal each other. To find a description of + , consider a typical element ( , , , ) of and a typical element ( , , , ) of , where , , , . We have \n\n( , , , ) + ( , , , ) = ( + , + , + , + ) ,\n\nwhich shows that every element of + has its first two coordinates equal to each other. Thus \n\n> 1.39\n\n + {( , , , ) 4 , , }. \n\nTo prove the inclusion in the other direction, suppose , , . Then \n\n( , , , ) = ( , , , ) + (0 , 0, 0, ) ,\n\nwhere the first vector on the right is in and the second vector on the right is in . Thus ( , , , ) + , showing that the inclusion 1.39 also holds in the opposite direction. Hence \n\n + = {( , , , ) 4 , , } ,\n\nwhich shows that + is the set of elements of 4 whose first two coordinates equal each other. The next result states that the sum of subspaces is a subspace, and is in fact the smallest subspace containing all the summands (which means that every subspace containing all the summands also contains the sum). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 1C Subspaces 21 \n\n> 1.40\n\nsum of subspaces is the smallest containing subspace \n\nSuppose 1, , are subspaces of . Then 1 + + is the smallest subspace of containing 1, , .\n\nProof The reader can verify that 1 + + contains the additive identity 0\n\nand is closed under addition and scalar multiplication. Thus 1.34 implies that \n\n 1 + + is a subspace of .\n\nSums of subspaces in the theory of vec-tor spaces are analogous to unions of subsets in set theory. Given two sub-spaces of a vector space, the smallest subspace containing them is their sum. Analogously, given two subsets of a set, the smallest subset containing them is their union. \n\nThe subspaces 1, , are all con-tained in 1 + + (to see this, con-sider sums 1 + + where all except one of the s are 0). Conversely, every subspace of containing 1, , con-tains 1 + + (because subspaces must contain all finite sums of their ele-ments). Thus 1 + + is the smallest subspace of containing 1, , .\n\n# Direct Sums \n\nSuppose 1, , are subspaces of . Every element of 1 + + can be written in the form \n\n 1 + + ,\n\nwhere each . Of special interest are cases in which each vector in \n\n 1 + + can be represented in the form above in only one way. This situation is so important that it gets a special name (direct sum) and a special symbol ( ). \n\n> 1.41 definition:\n\ndirect sum, \n\nSuppose 1, , are subspaces of .\n\n The sum 1 + + is called a direct sum if each element of 1 + + \n\ncan be written in only one way as a sum 1 + + , where each .\n\n If 1 + + is a direct sum, then 1 denotes 1 + + ,with the notation serving as an indication that this is a direct sum. \n\n> 1.42 example:\n\na direct sum of two subspaces \n\nSuppose is the subspace of 3 of those vectors whose last coordinate equals 0,and is the subspace of 3 of those vectors whose first two coordinates equal 0:\n\n = {( , , 0) 3 , } and = {(0 , 0, ) 3 }. \n\nThen 3 = , as you should verify. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 7\n\nChapter 1 Vector Spaces \n\n> 1.43 example:\n\na direct sum of multiple subspaces \n\nTo produce in TEX, type \\oplus .Suppose is the subspace of of those vectors whose coordinates are all 0,except possibly in the th slot; for example, 2 = {(0 , , 0, , 0) } .Then \n\n = 1 ,\n\nas you should verify. Sometimes nonexamples add to our understanding as much as examples. \n\n> 1.44 example:\n\na sum that is not a direct sum \n\nSuppose \n\n 1 = {( , , 0) 3 , } ,\n\n 2 = {(0 , 0, ) 3 } ,\n\n 3 = {(0 , , ) 3 }. \n\nThen 3 = 1 + 2 + 3 because every vector ( , , ) 3 can be written as \n\n( , , ) = ( , , 0) + (0 , 0, ) + (0 , 0, 0) ,\n\nwhere the first vector on the right side is in 1, the second vector is in 2, and the third vector is in 3.However, 3 does not equal the direct sum of 1, 2, 3, because the vector \n\n(0 , 0, 0) can be written in more than one way as a sum 1 + 2 + 3, with each \n\n . Specifically, we have \n\n(0 , 0, 0) = (0 , 1, 0) + (0 , 0, 1) + (0 , 1 , 1) \n\nand, of course, \n\n(0 , 0, 0) = (0 , 0, 0) + (0 , 0, 0) + (0 , 0, 0) ,\n\nwhere the first vector on the right side of each equation above is in 1, the second vector is in 2, and the third vector is in 3. Thus the sum 1 + 2 + 3 is not a direct sum. \n\nThe symbol , which is a plus sign inside a circle, reminds us that we are dealing with a special type of sum of subspaceseach element in the direct sum can be represented in only one way as a sum of elements from the specified subspaces. \n\nThe definition of direct sum requires every vector in the sum to have a unique representation as an appropriate sum. The next result shows that when deciding whether a sum of subspaces is a direct sum, we only need to consider whether 0\n\ncan be uniquely written as an appropriate sum. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 1C Subspaces 23 \n\n> 1.45\n\ncondition for a direct sum \n\nSuppose 1, , are subspaces of . Then 1 + + is a direct sum if and only if the only way to write 0 as a sum 1 + + , where each ,is by taking each equal to 0.\n\nProof First suppose 1 + + is a direct sum. Then the definition of direct sum implies that the only way to write 0 as a sum 1 ++ , where each ,is by taking each equal to 0.Now suppose that the only way to write 0 as a sum 1 + + , where each \n\n , is by taking each equal to 0. To show that 1 + + is a direct sum, let 1 + + . We can write \n\n = 1 + + \n\nfor some 1 1, , . To show that this representation is unique, suppose we also have \n\n = 1 + + ,\n\nwhere 1 1, , . Subtracting these two equations, we have \n\n0 = ( 1 1) + + ( ). \n\nBecause 1 1 1, , , the equation above implies that each \n\n equals 0. Thus 1 = 1, , = , as desired. \n\nThe symbol used below means if and only if ; this symbol could also be read to mean is equivalent to. \n\nThe next result gives a simple con-dition for testing whether a sum of two subspaces is a direct sum. \n\n> 1.46\n\ndirect sum of two subspaces \n\nSuppose and are subspaces of . Then \n\n + is a direct sum = {0}. \n\nProof First suppose that + is a direct sum. If , then 0 = + () ,where and . By the unique representation of 0 as the sum of a vector in and a vector in , we have = 0 . Thus = {0} , completing the proof in one direction. To prove the other direction, now suppose = {0} . To prove that + \n\nis a direct sum, suppose , , and \n\n0 = + . \n\nTo complete the proof, we only need to show that = = 0 (by 1.45). The equation above implies that = . Thus . Hence = 0 , which by the equation above implies that = 0 , completing the proof. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n24 Chapter 1 Vector Spaces \n\nSums of subspaces are analogous to unions of subsets. Similarly, direct sums of subspaces are analogous to disjoint unions of subsets. No two sub-spaces of a vector space can be disjoint, because both contain 0. So disjoint-ness is replaced, at least in the case of two subspaces, with the requirement that the intersection equal {0} .\n\nThe result above deals only with the case of two subspaces. When ask-ing about a possible direct sum with more than two subspaces, it is not enough to test that each pair of the subspaces intersect only at 0. To see this, consider Example 1.44. In that nonexample of a direct sum, we have \n\n 1 2 = 1 3 = 2 3 = {0} .\n\n# Exercises 1C \n\n1 For each of the following subsets of 3, determine whether it is a subspace of 3.(a) {( 1, 2, 3) 3 1 + 2 2 + 3 3 = 0} \n\n(b) {( 1, 2, 3) 3 1 + 2 2 + 3 3 = 4} \n\n(c) {( 1, 2, 3) 3 1 2 3 = 0} \n\n(d) {( 1, 2, 3) 3 1 = 5 3}\n\n2 Verify all assertions about subspaces in Example 1.35. \n\n3 Show that the set of differentiable real-valued functions on the interval \n\n(4 , 4) such that (1) = 3 (2) is a subspace of (4 ,4) .\n\n4 Suppose . Show that the set of continuous real-valued functions on the interval [0 , 1] such that 10 = is a subspace of [0 ,1] if and only if \n\n = 0 .\n\n5 Is 2 a subspace of the complex vector space 2?\n\n6 (a) Is {( , , ) 3 3 = 3} a subspace of 3?(b) Is {( , , ) 3 3 = 3} a subspace of 3?\n\n7 Prove or give a counterexample: If is a nonempty subset of 2 such that \n\n is closed under addition and under taking additive inverses (meaning \n\n whenever ), then is a subspace of 2.\n\n8 Give an example of a nonempty subset of 2 such that is closed under scalar multiplication, but is not a subspace of 2.\n\n9 A function is called periodic if there exists a positive number \n\nsuch that () = ( + ) for all . Is the set of periodic functions from to a subspace of ? Explain. \n\n10 Suppose 1 and 2 are subspaces of . Prove that the intersection 1 2\n\nis a subspace of .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 1C Subspaces 25 \n\n11 Prove that the intersection of every collection of subspaces of is a subspace of .\n\n12 Prove that the union of two subspaces of is a subspace of if and only if one of the subspaces is contained in the other. \n\n13 Prove that the union of three subspaces of is a subspace of if and only if one of the subspaces contains the other two. \n\nThis exercise is surprisingly harder than Exercise 12, possibly because this exercise is not true if we replace with a field containing only two elements. \n\n14 Suppose \n\n = {( , , 2) 3 } and = {( , , 2) 3 }. \n\nDescribe + using symbols, and also give a description of + that uses no symbols. \n\n15 Suppose is a subspace of . What is + ?\n\n16 Is the operation of addition on the subspaces of commutative? In other words, if and are subspaces of , is + = + ?\n\n17 Is the operation of addition on the subspaces of associative? In other words, if 1, 2, 3 are subspaces of , is \n\n( 1 + 2) + 3 = 1 + ( 2 + 3)? \n\n18 Does the operation of addition on the subspaces of have an additive identity? Which subspaces have additive inverses? \n\n19 Prove or give a counterexample: If 1, 2, are subspaces of such that \n\n 1 + = 2 + ,\n\nthen 1 = 2.\n\n20 Suppose \n\n = {( , , , ) 4 , }. \n\nFind a subspace of 4 such that 4 = .\n\n21 Suppose \n\n = {( , , + , , 2) 5 , }. \n\nFind a subspace of 5 such that 5 = .\n\n22 Suppose \n\n = {( , , + , , 2) 5 , }. \n\nFind three subspaces 1, 2, 3 of 5, none of which equals {0} , such that \n\n 5 = 1 2 3.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n26 Chapter 1 Vector Spaces \n\n23 Prove or give a counterexample: If 1, 2, are subspaces of such that \n\n = 1 and = 2 ,\n\nthen 1 = 2.\n\nHint: When trying to discover whether a conjecture in linear algebra is true or false, it is often useful to start by experimenting in 2.\n\n24 A function is called even if \n\n () = () \n\nfor all . A function is called odd if \n\n () = () \n\nfor all . Let e denote the set of real-valued even functions on \n\nand let o denote the set of real-valued odd functions on . Show that \n\n = e o.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n# Chapter 2 \n\n# Finite-Dimensional Vector Spaces \n\nIn the last chapter we learned about vector spaces. Linear algebra focuses not on arbitrary vector spaces, but on finite-dimensional vector spaces, which we introduce in this chapter. We begin this chapter by considering linear combinations of lists of vectors. This leads us to the crucial concept of linear independence. The linear dependence lemma will become one of our most useful tools. A list of vectors in a vector space that is small enough to be linearly independent and big enough so the linear combinations of the list fill up the vector space is called a basis of the vector space. We will see that every basis of a vector space has the same length, which will allow us to define the dimension of a vector space. This chapter ends with a formula for the dimension of the sum of two subspaces. \n\nstanding assumptions for this chapter \n\n denotes or .\n\n denotes a vector space over .\n\nThe main building of the Institute for Advanced Study, in Princeton, New Jersey. Paul Halmos (19162006 ) wrote the first modern linear algebra book in this building. Halmoss linear algebra book was published in 1942 (second edition published in 1958 ).The title of Halmoss book was the same as the title of this chapter. \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler \n\n27 28 Chapter 2 Finite-Dimensional Vector Spaces \n\n# 2A Span and Linear Independence \n\nWe have been writing lists of numbers surrounded by parentheses, and we will continue to do so for elements of ; for example, (2 , 7 , 8) 3. However, now we need to consider lists of vectors (which may be elements of or of other vector spaces). To avoid confusion, we will usually write lists of vectors without surrounding parentheses. For example, (4 , 1, 6) , (9 , 5, 7) is a list of length two of vectors in 3. \n\n> 2.1 notation:\n\nlist of vectors \n\nWe will usually write lists of vectors without surrounding parentheses. \n\n# Linear Combinations and Span \n\nA sum of scalar multiples of the vectors in a list is called a linear combination of the list. Here is the formal definition. \n\n> 2.2 definition:\n\nlinear combination \n\nA linear combination of a list 1, , of vectors in is a vector of the form \n\n 1 1 + + ,\n\nwhere 1, , . \n\n> 2.3 example:\n\nlinear combinations in 3\n\n (17 , 4 , 2) is a linear combination of (2 , 1, 3) , (1 , 2 , 4) , which is a list of length two of vectors in 3, because \n\n(17 , 4 , 2) = 6(2 , 1, 3) + 5(1 , 2 , 4). (17 , 4 , 5) is not a linear combination of (2 , 1, 3) , (1 , 2 , 4) , which is a list of length two of vectors in 3, because there do not exist numbers 1, 2 \n\nsuch that \n\n(17 , 4 , 5) = 1(2 , 1, 3) + 2(1 , 2 , 4). \n\nIn other words, the system of equations \n\n17 = 2 1 + 2\n\n4 = 1 2 2\n\n5 = 3 1 + 4 2\n\nhas no solutions (as you should verify). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 2A Span and Linear Independence 29 \n\n> 2.4 definition:\n\nspan \n\nThe set of all linear combinations of a list of vectors 1, , in is called the span of 1, , , denoted by span ( 1, , ). In other words, \n\nspan ( 1, , ) = { 1 1 + + 1, , }. \n\nThe span of the empty list ( ) is defined to be {0} . \n\n> 2.5 example:\n\nspan \n\nThe previous example shows that in 3,\n\n (17 , 4 , 2) span ((2 , 1, 3) , (1 , 2 , 4)) ;\n\n (17 , 4 , 5) span ((2 , 1, 3) , (1 , 2 , 4)) .Some mathematicians use the term linear span , which means the same as span. \n\n> 2.6\n\nspan is the smallest containing subspace \n\nThe span of a list of vectors in is the smallest subspace of containing all vectors in the list. \n\nProof Suppose 1, , is a list of vectors in .First we show that span ( 1, , ) is a subspace of . The additive identity is in span ( 1, , ) because \n\n0 = 0 1 + + 0 .\n\nAlso, span ( 1, , ) is closed under addition because \n\n( 1 1 + + ) + ( 1 1 + + ) = ( 1 + 1) 1 + + ( + ) .\n\nFurthermore, span ( 1, , ) is closed under scalar multiplication because \n\n( 1 1 + + ) = 1 1 + + .\n\nThus span ( 1, , ) is a subspace of (by 1.34). Each is a linear combination of 1, , (to show this, set = 1 and let the other s in 2.2 equal 0). Thus span ( 1, , ) contains each . Conversely, because subspaces are closed under scalar multiplication and addition, every sub-space of that contains each contains span ( 1, , ). Thus span ( 1, , )\n\nis the smallest subspace of containing all the vectors 1, , . \n\n> 2.7 definition:\n\nspans \n\nIf span ( 1, , ) equals , we say that the list 1, , spans .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n30 Chapter 2 Finite-Dimensional Vector Spaces \n\n> 2.8 example:\n\na list that spans \n\nSuppose is a positive integer. We want to show that \n\n(1 , 0, , 0) , (0 , 1, 0, , 0) , , (0 , , 0, 1) \n\nspans . Here the th vector in the list above has 1 in the th slot and 0 in all other slots. Suppose ( 1, , ) . Then \n\n( 1, , ) = 1(1 , 0, , 0) + 2(0 , 1, 0, , 0) + + (0 , , 0, 1). \n\nThus ( 1, , ) span ((1 , 0, , 0) , (0 , 1, 0, , 0) , , (0 , , 0, 1)) , as desired. Now we can make one of the key definitions in linear algebra. \n\n> 2.9 definition:\n\nfinite-dimensional vector space \n\nA vector space is called finite-dimensional if some list of vectors in it spans the space. \n\nRecall that by definition every list has finite length. \n\nExample 2.8 above shows that is a finite-dimensional vector space for every positive integer .The definition of a polynomial is no doubt already familiar to you. \n\n> 2.10 definition:\n\npolynomial, () A function is called a polynomial with coefficients in if there exist 0, , such that \n\n() = 0 + 1 + 2 2 + + \n\nfor all .\n\n () is the set of all polynomials with coefficients in .With the usual operations of addition and scalar multiplication, () is a vector space over , as you should verify. Hence () is a subspace of , the vector space of functions from to .If a polynomial (thought of as a function from to ) is represented by two sets of coefficients, then subtracting one representation of the polynomial from the other produces a polynomial that is identically zero as a function on and hence has all zero coefficients (if you are unfamiliar with this fact, just believe it for now; we will prove it latersee 4.8). Conclusion: the coefficients of a polynomial are uniquely determined by the polynomial. Thus the next definition uniquely defines the degree of a polynomial. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 2A Span and Linear Independence 31 \n\n> 2.11 definition:\n\ndegree of a polynomial, deg A polynomial () is said to have degree if there exist scalars \n\n 0, 1, , with 0 such that for every , we have \n\n() = 0 + 1 + + . The polynomial that is identically 0 is said to have degree .\n\n The degree of a polynomial is denoted by deg .In the next definition, we use the convention that < , which means that the polynomial 0 is in () . \n\n> 2.12 notation:\n\n () \n\nFor a nonnegative integer, () denotes the set of all polynomials with coefficients in and degree at most .If is a nonnegative integer, then () = span (1 , , , ) [ here we slightly abuse notation by letting denote a function ]. Thus () is a finite-dimensional vector space for each nonnegative integer . \n\n> 2.13 definition:\n\ninfinite-dimensional vector space \n\nA vector space is called infinite-dimensional if it is not finite-dimensional. \n\n> 2.14 example:\n\n () is infinite-dimensional. \n\nConsider any list of elements of () . Let denote the highest degree of the polynomials in this list. Then every polynomial in the span of this list has degree at most . Thus + 1 is not in the span of our list. Hence no list spans () .Thus () is infinite-dimensional. \n\n# Linear Independence \n\nSuppose 1, , and span ( 1, , ). By the definition of span, there exist 1, , such that \n\n = 1 1 + + .\n\nConsider the question of whether the choice of scalars in the equation above is unique. Suppose 1, , is another set of scalars such that \n\n = 1 1 + + .\n\nSubtracting the last two equations, we have \n\n0 = ( 1 1) 1 + + ( ) .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n32 Chapter 2 Finite-Dimensional Vector Spaces \n\nThus we have written 0 as a linear combination of ( 1, , ). If the only way to do this is by using 0 for all the scalars in the linear combination, then each \n\n equals 0, which means that each equals (and thus the choice of scalars was indeed unique). This situation is so important that we give it a special namelinear independencewhich we now define. \n\n> 2.15 definition:\n\nlinearly independent \n\n A list 1, , of vectors in is called linearly independent if the only choice of 1, , that makes \n\n 1 1 + + = 0 \n\nis 1 = = = 0 .\n\n The empty list ( ) is also declared to be linearly independent. The reasoning above shows that 1, , is linearly independent if and only if each vector in span ( 1, , ) has only one representation as a linear combination of 1, , . \n\n> 2.16 example:\n\nlinearly independent lists \n\n(a) To see that the list (1 , 0, 0, 0) , (0 , 1, 0, 0) , (0 , 0, 1, 0) is linearly independent in \n\n 4, suppose 1, 2, 3 and \n\n 1(1 , 0, 0, 0) + 2(0 , 1, 0, 0) + 3(0 , 0, 1, 0) = (0 , 0, 0, 0). \n\nThus \n\n( 1, 2, 3, 0) = (0 , 0, 0, 0). \n\nHence 1 = 2 = 3 = 0 . Thus the list (1 , 0, 0, 0) , (0 , 1, 0, 0) , (0 , 0, 1, 0) is linearly independent in 4.(b) Suppose is a nonnegative integer. To see that the list 1, , , is linearly independent in () , suppose 0, 1, , and \n\n 0 + 1 + + = 0 ,\n\nwhere we think of both sides as elements of () . Then \n\n 0 + 1 + + = 0 \n\nfor all . As discussed earlier (and as follows from 4.8), this implies that \n\n 0 = 1 = = = 0 . Thus 1, , , is a linearly independent list in \n\n () .(c) A list of length one in a vector space is linearly independent if and only if the vector in the list is not 0.(d) A list of length two in a vector space is linearly independent if and only if neither of the two vectors in the list is a scalar multiple of the other. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 2A Span and Linear Independence 33 \n\nIf some vectors are removed from a linearly independent list, the remaining list is also linearly independent, as you should verify. \n\n> 2.17 definition:\n\nlinearly dependent \n\n A list of vectors in is called linearly dependent if it is not linearly inde-pendent. \n\n In other words, a list 1, , of vectors in is linearly dependent if there exist 1, , , not all 0, such that 1 1 + + = 0 . \n\n> 2.18 example:\n\nlinearly dependent lists \n\n (2 , 3, 1) , (1 , 1 , 2) , (7 , 3, 8) is linearly dependent in 3 because \n\n2(2 , 3, 1) + 3(1 , 1 , 2) + (1)(7 , 3, 8) = (0 , 0, 0). The list (2 , 3, 1) , (1 , 1 , 2) , (7 , 3, ) is linearly dependent in 3 if and only if \n\n = 8 , as you should verify. \n\n If some vector in a list of vectors in is a linear combination of the other vectors, then the list is linearly dependent. (Proof: After writing one vector in the list as equal to a linear combination of the other vectors, move that vector to the other side of the equation, where it will be multiplied by 1 .) \n\n Every list of vectors in containing the 0 vector is linearly dependent. (This is a special case of the previous bullet point.) The next lemma is a terrific tool. It states that given a linearly dependent list of vectors, one of the vectors is in the span of the previous ones. Furthermore, we can throw out that vector without changing the span of the original list. \n\n> 2.19\n\nlinear dependence lemma \n\nSuppose 1, , is a linearly dependent list in . Then there exists \n\n {1 , 2, , } such that \n\n span ( 1, , 1 ). \n\nFurthermore, if satisfies the condition above and the th term is removed from 1, , , then the span of the remaining list equals span ( 1, , ).\n\nProof Because the list 1, , is linearly dependent, there exist numbers \n\n 1, , , not all 0, such that \n\n 1 1 + + = 0. \n\nLet be the largest element of {1 , , } such that 0 . Then \n\n = 1\n\n \n\n 1 1 \n\n \n\n 1 ,\n\nwhich proves that span ( 1, , 1 ), as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 34 Chapter 2 Finite-Dimensional Vector Spaces\n\nNow suppose is any element of {1 , , } such that span ( 1, , 1 ).Let 1, , 1 be such that \n\n> 2.20\n\n = 1 1 + + 1 1 .\n\nSuppose span ( 1, , ). Then there exist 1, , such that \n\n = 1 1 + + .\n\nIn the equation above, we can replace with the right side of 2.20, which shows that is in the span of the list obtained by removing the th term from 1, , .Thus removing the th term of the list 1, , does not change the span of the list. If = 1 in the linear dependence lemma, then span ( 1, , 1 ) means that 1 = 0 , because span ( ) = {0} . Note also that parts of the proof of the linear dependence lemma need to be modified if = 1 . In general, the proofs in the rest of the book will not call attention to special cases that must be considered involving lists of length 0, the subspace {0} , or other trivial cases for which the result is true but needs a slightly different proof. Be sure to check these special cases yourself. \n\n> 2.21 example:\n\nsmallest in linear dependence lemma \n\nConsider the list \n\n(1 , 2, 3) , (6 , 5, 4) , (15 , 16 , 17) , (8 , 9, 7) \n\nin 3. This list of length four is linearly dependent, as we will soon see. Thus the linear dependence lemma implies that there exists {1 , 2, 3, 4} such that the th \n\nvector in this list is a linear combination of the previous vectors in the list. Lets see how to find the smallest value of that works. Taking = 1 in the linear dependence lemma works if and only if the first vector in the list equals 0. Because (1 , 2, 3) is not the 0 vector, we cannot take \n\n = 1 for this list. Taking = 2 in the linear dependence lemma works if and only if the second vector in the list is a scalar multiple of the first vector. However, there does not exist such that (6 , 5, 4) = (1 , 2, 3) . Thus we cannot take = 2 for this list. Taking = 3 in the linear dependence lemma works if and only if the third vector in the list is a linear combination of the first two vectors. Thus for the list in this example, we want to know whether there exist , such that \n\n(15 , 16 , 17) = (1 , 2, 3) + (6 , 5, 4). \n\nThe equation above is equivalent to a system of three linear equations in the two unknowns , . Using Gaussian elimination or appropriate software, we find that \n\n = 3 , = 2 is a solution of the equation above, as you can verify. Thus for the list in this example, taking = 3 is the smallest value of that works in the linear dependence lemma. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 2A Span and Linear Independence 35\n\nNow we come to a key result. It says that no linearly independent list in is longer than a spanning list in . \n\n> 2.22\n\nlength of linearly independent list length of spanning list \n\nIn a finite-dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors. \n\n> Proof\n\nSuppose that 1, , is linearly independent in . Suppose also that \n\n 1, , spans . We need to prove that . We do so through the process described below with steps; note that in each step we add one of the s and remove one of the s. \n\nStep 1 \n\nLet be the list 1, , , which spans . Adjoining 1 at the beginning of this list produces a linearly dependent list (because 1 can be written as a linear combination of 1, , ). In other words, the list \n\n 1, 1, , \n\nis linearly dependent. Thus by the linear dependence lemma (2.19), one of the vectors in the list above is a linear combination of the previous vectors in the list. We know that 1 0 \n\nbecause the list 1, , is linearly independent. Thus 1 is not in the span of the previous vectors in the list above (because 1 is not in {0} , which is the span of the empty list). Hence the linear dependence lemma implies that we can remove one of the s so that the new list (of length ) consisting of 1\n\nand the remaining s spans .\n\nStep k, for k = 2, , m\n\nThe list (of length ) from step 1 spans . In particular, is in the span of the list . Thus the list of length ( +1) obtained by adjoining to , placing it just after 1, , 1 , is linearly dependent. By the linear dependence lemma (2.19), one of the vectors in this list is in the span of the previous ones, and because 1, , is linearly independent, this vector cannot be one of the s. Hence there still must be at least one remaining at this step. We can remove from our new list (after adjoining in the proper place) a that is a linear combination of the previous vectors in the list, so that the new list (of length \n\n ) consisting of 1, , and the remaining s spans .After step , we have added all the s and the process stops. At each step as we add a to , the linear dependence lemma implies that there is some to remove. Thus there are at least as many s as s. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 36 Chapter 2 Finite-Dimensional Vector Spaces\n\nThe next two examples show how the result above can be used to show, without any computations, that certain lists are not linearly independent and that certain lists do not span a given vector space. \n\n> 2.23 example:\n\nno list of length 4 is linearly independent in 3\n\nThe list (1 , 0, 0) , (0 , 1, 0) , (0 , 0, 1) , which has length three, spans 3. Thus no list of length larger than three is linearly independent in 3.For example, we now know that (1 , 2, 3) , (4 , 5, 8) , (9 , 6, 7) , (3 , 2, 8) , which is a list of length four, is not linearly independent in 3. \n\n> 2.24 example:\n\nno list of length 3 spans 4\n\nThe list (1 , 0, 0, 0) , (0 , 1, 0, 0) , (0 , 0, 1, 0) , (0 , 0, 0, 1) , which has length four, is linearly independent in 4. Thus no list of length less than four spans 4.For example, we now know that (1 , 2, 3, 5) , (4 , 5, 8, 3) , (9 , 6, 7, 1) , which is a list of length three, does not span 4.Our intuition suggests that every subspace of a finite-dimensional vector space should also be finite-dimensional. We now prove that this intuition is correct. \n\n> 2.25\n\nfinite-dimensional subspaces \n\nEvery subspace of a finite-dimensional vector space is finite-dimensional. \n\n> Proof\n\nSuppose is finite-dimensional and is a subspace of . We need to prove that is finite-dimensional. We do this through the following multistep construction. \n\nStep 1 \n\nIf = {0} , then is finite-dimensional and we are done. If {0} , then choose a nonzero vector 1 .\n\nStep k\n\nIf = span ( 1, , 1 ), then is finite-dimensional and we are done. If \n\n span ( 1, , 1 ), then choose a vector such that \n\n span ( 1, , 1 ). \n\nAfter each step, as long as the process continues, we have constructed a list of vectors such that no vector in this list is in the span of the previous vectors. Thus after each step we have constructed a linearly independent list, by the linear dependence lemma (2.19). This linearly independent list cannot be longer than any spanning list of (by 2.22). Thus the process eventually terminates, which means that is finite-dimensional. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 2A Span and Linear Independence 37 \n\n# Exercises 2A \n\n1 Find a list of four distinct vectors in 3 whose span equals \n\n{( , , ) 3 + + = 0}. \n\n2 Prove or give a counterexample: If 1, 2, 3, 4 spans , then the list \n\n 1 2, 2 3, 3 4, 4\n\nalso spans .\n\n3 Suppose 1, , is a list of vectors in . For {1 , , } , let \n\n = 1 + + .\n\nShow that span ( 1, , ) = span ( 1, , ).\n\n4 (a) Show that a list of length one in a vector space is linearly independent if and only if the vector in the list is not 0.(b) Show that a list of length two in a vector space is linearly independent if and only if neither of the two vectors in the list is a scalar multiple of the other. \n\n5 Find a number such that \n\n(3 , 1, 4) , (2 , 3 , 5) , (5 , 9, ) \n\nis not linearly independent in 3.\n\n6 Show that the list (2 , 3, 1) , (1 , 1 , 2) , (7 , 3, ) is linearly dependent in 3 if and only if = 8 .\n\n7 (a) Show that if we think of as a vector space over , then the list \n\n1 + , 1 is linearly independent. (b) Show that if we think of as a vector space over , then the list \n\n1 + , 1 is linearly dependent. \n\n8 Suppose 1, 2, 3, 4 is linearly independent in . Prove that the list \n\n 1 2, 2 3, 3 4, 4\n\nis also linearly independent. \n\n9 Prove or give a counterexample: If 1, 2, , is a linearly independent list of vectors in , then \n\n5 1 4 2, 2, 3, , \n\nis linearly independent. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n38 Chapter 2 Finite-Dimensional Vector Spaces \n\n10 Prove or give a counterexample: If 1, 2, , is a linearly independent list of vectors in and with 0 , then 1, 2, , is linearly independent. \n\n11 Prove or give a counterexample: If 1, , and 1, , are linearly independent lists of vectors in , then the list 1 + 1, , + is linearly independent. \n\n12 Suppose 1, , is linearly independent in and . Prove that if \n\n 1 + , , + is linearly dependent, then span ( 1, , ).\n\n13 Suppose 1, , is linearly independent in and . Show that \n\n 1, , , is linearly independent span ( 1, , ). \n\n14 Suppose 1, , is a list of vectors in . For {1 , , } , let \n\n = 1 + + .\n\nShow that the list 1, , is linearly independent if and only if the list \n\n 1, , is linearly independent. \n\n15 Explain why there does not exist a list of six polynomials that is linearly independent in 4() .\n\n16 Explain why no list of four polynomials spans 4() .\n\n17 Prove that is infinite-dimensional if and only if there is a sequence 1, 2, \n\nof vectors in such that 1, , is linearly independent for every positive integer .\n\n18 Prove that is infinite-dimensional. \n\n19 Prove that the real vector space of all continuous real-valued functions on the interval [0 , 1] is infinite-dimensional. \n\n20 Suppose 0, 1, , are polynomials in () such that (2) = 0 for each {0 , , } . Prove that 0, 1, , is not linearly independent in \n\n () .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 2B Bases 39 \n\n# 2B Bases \n\nIn the previous section, we discussed linearly independent lists and we also discussed spanning lists. Now we bring these concepts together by considering lists that have both properties. \n\n> 2.26 definition:\n\nbasis \n\nA basis of is a list of vectors in that is linearly independent and spans . \n\n> 2.27 example:\n\nbases \n\n(a) The list (1 , 0, , 0) , (0 , 1, 0, , 0) , , (0 , , 0, 1) is a basis of , called the \n\nstandard basis of .(b) The list (1 , 2) , (3 , 5) is a basis of 2. Note that this list has length two, which is the same as the length of the standard basis of 2. In the next section, we will see that this is not a coincidence. (c) The list (1 , 2, 4) , (7 , 5 , 6) is linearly independent in 3 but is not a basis of 3 because it does not span 3.(d) The list (1 , 2) , (3 , 5) , (4 , 13) spans 2 but is not a basis of 2 because it is not linearly independent. (e) The list (1 , 1, 0) , (0 , 0, 1) is a basis of {( , , ) 3 , } .(f) The list (1 , 1 , 0) , (1 , 0, 1) is a basis of \n\n{( , , ) 3 + + = 0}. \n\n(g) The list 1, , , is a basis of () , called the standard basis of () .In addition to the standard basis, has many other bases. For example, \n\n(7 , 5) , (4 , 9) and (1 , 2) , (3 , 5) \n\nare both bases of 2.The next result helps explain why bases are useful. Recall that uniquely means in only one way. \n\n> 2.28\n\ncriterion for basis \n\nA list 1, , of vectors in is a basis of if and only if every can be written uniquely in the form \n\n> 2.29\n\n = 1 1 + + ,\n\nwhere 1, , .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n40 Chapter 2 Finite-Dimensional Vector Spaces \n\nThis proof is essentially a repetition of the ideas that led us to the definition of linear independence. \n\nProof First suppose that 1, , is a basis of . Let . Because 1, , \n\nspans , there exist 1, , such that 2.29 holds. To show that the repre-sentation in 2.29 is unique, suppose 1, , are scalars such that we also have \n\n = 1 1 + + .\n\nSubtracting the last equation from 2.29, we get \n\n0 = ( 1 1) 1 + + ( ) .\n\nThis implies that each equals 0 (because 1, , is linearly independent). Hence 1 = 1, , = . We have the desired uniqueness, completing the proof in one direction. For the other direction, suppose every can be written uniquely in the form given by 2.29. This implies that the list 1, , spans . To show that \n\n 1, , is linearly independent, suppose 1, , are such that \n\n0 = 1 1 + + .\n\nThe uniqueness of the representation 2.29 (taking = 0 ) now implies that \n\n 1 = = = 0 . Thus 1, , is linearly independent and hence is a basis of .A spanning list in a vector space may not be a basis because it is not linearly independent. Our next result says that given any spanning list, some (possibly none) of the vectors in it can be discarded so that the remaining list is linearly independent and still spans the vector space. As an example in the vector space 2, if the procedure in the proof below is applied to the list (1 , 2) , (3 , 6) , (4 , 7) , (5 , 9) , then the second and fourth vectors will be removed. This leaves (1 , 2) , (4 , 7) , which is a basis of 2. \n\n> 2.30\n\nevery spanning list contains a basis \n\nEvery spanning list in a vector space can be reduced to a basis of the vector space. \n\nProof Suppose 1, , spans . We want to remove some of the vectors from \n\n 1, , so that the remaining vectors form a basis of . We do this through the multistep process described below. Start with equal to the list 1, , .\n\nStep 1 \n\nIf 1 = 0 , then delete 1 from . If 1 0 , then leave unchanged. \n\nStep k\n\nIf is in span ( 1, , 1 ), then delete from the list . If is not in \n\nspan ( 1, , 1 ), then leave unchanged. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 2B Bases 41\n\nStop the process after step , getting a list . This list spans because our original list spanned and we have discarded only vectors that were already in the span of the previous vectors. The process ensures that no vector in is in the span of the previous ones. Thus is linearly independent, by the linear dependence lemma (2.19). Hence is a basis of .We now come to an important corollary of the previous result. \n\n> 2.31\n\nbasis of finite-dimensional vector space \n\nEvery finite-dimensional vector space has a basis. \n\n> Proof\n\nBy definition, a finite-dimensional vector space has a spanning list. The previous result tells us that each spanning list can be reduced to a basis. Our next result is in some sense a dual of 2.30, which said that every spanning list can be reduced to a basis. Now we show that given any linearly independent list, we can adjoin some additional vectors (this includes the possibility of adjoining no additional vectors) so that the extended list is still linearly independent but also spans the space. \n\n> 2.32\n\nevery linearly independent list extends to a basis \n\nEvery linearly independent list of vectors in a finite-dimensional vector space can be extended to a basis of the vector space. \n\n> Proof\n\nSuppose 1, , is linearly independent in a finite-dimensional vector space . Let 1, , be a list of vectors in that spans . Thus the list \n\n 1, , , 1, , \n\nspans . Applying the procedure of the proof of 2.30 to reduce this list to a basis of produces a basis consisting of the vectors 1, , and some of the \n\n s (none of the s get deleted in this procedure because 1, , is linearly independent). As an example in 3, suppose we start with the linearly independent list \n\n(2 , 3, 4) , (9 , 6, 8) . If we take 1, 2, 3 to be the standard basis of 3, then applying the procedure in the proof above produces the list \n\n(2 , 3, 4) , (9 , 6, 8) , (0 , 1, 0) ,\n\nwhich is a basis of 3. \n\n> Using the same ideas but more ad-vanced tools, the next result can be proved without the hypothesis that is finite-dimensional.\n\nAs an application of the result above, we now show that every subspace of a finite-dimensional vector space can be paired with another subspace to form a direct sum of the whole space. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n42 Chapter 2 Finite-Dimensional Vector Spaces \n\n> 2.33\n\nevery subspace of is part of a direct sum equal to \n\nSuppose is finite-dimensional and is a subspace of . Then there is a subspace of such that = .\n\nProof Because is finite-dimensional, so is (see 2.25). Thus there is a basis \n\n 1, , of (by 2.31). Of course 1, , is a linearly independent list of vectors in . Hence this list can be extended to a basis 1, , , 1, , of \n\n(by 2.32). Let = span ( 1, , ).To prove that = , by 1.46 we only need to show that \n\n = + and = {0}. \n\nTo prove the first equation above, suppose . Then, because the list \n\n 1, , , 1, , spans , there exist 1, , , 1, , such that \n\n = 1 1 + + \n\n> \n\n+ 1 1 + + \n\n> \n\n.\n\nWe have = + , where and are defined as above. Thus \n\n + , completing the proof that = + .To show that = {0} , suppose . Then there exist scalars \n\n 1, , , 1, , such that \n\n = 1 1 + + = 1 1 + + .\n\nThus \n\n 1 1 + + 1 1 = 0. \n\nBecause 1, , , 1, , is linearly independent, this implies that \n\n 1 = = = 1 = = = 0. \n\nThus = 0 , completing the proof that = {0} .\n\n# Exercises 2B \n\n1 Find all vector spaces that have exactly one basis. \n\n2 Verify all assertions in Example 2.27. \n\n3 (a) Let be the subspace of 5 defined by \n\n = {( 1, 2, 3, 4, 5) 5 1 = 3 2 and 3 = 7 4}. \n\nFind a basis of .(b) Extend the basis in (a) to a basis of 5.(c) Find a subspace of 5 such that 5 = .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 2B Bases 43 \n\n4 (a) Let be the subspace of 5 defined by \n\n = {( 1, 2, 3, 4, 5) 5 6 1 = 2 and 3 + 2 4 + 3 5 = 0}. \n\nFind a basis of .(b) Extend the basis in (a) to a basis of 5.(c) Find a subspace of 5 such that 5 = .\n\n5 Suppose is finite-dimensional and , are subspaces of such that \n\n = + . Prove that there exists a basis of consisting of vectors in \n\n .\n\n6 Prove or give a counterexample: If 0, 1, 2, 3 is a list in 3() such that none of the polynomials 0, 1, 2, 3 has degree 2, then 0, 1, 2, 3 is not a basis of 3() .\n\n7 Suppose 1, 2, 3, 4 is a basis of . Prove that \n\n 1 + 2, 2 + 3, 3 + 4, 4\n\nis also a basis of .\n\n8 Prove or give a counterexample: If 1, 2, 3, 4 is a basis of and is a subspace of such that 1, 2 and 3 and 4 , then 1, 2 is a basis of .\n\n9 Suppose 1, , is a list of vectors in . For {1 , , } , let \n\n = 1 + + .\n\nShow that 1, , is a basis of if and only if 1, , is a basis of .\n\n10 Suppose and are subspaces of such that = . Suppose also that 1, , is a basis of and 1, , is a basis of . Prove that \n\n 1, , , 1, , \n\nis a basis of .\n\n11 Suppose is a real vector space. Show that if 1, , is a basis of (as a real vector space), then 1, , is also a basis of the complexification \n\n(as a complex vector space). \n\nSee Exercise 8 in Section 1B for the definition of the complexification . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 44 Chapter 2 Finite-Dimensional Vector Spaces\n\n# 2C Dimension \n\nAlthough we have been discussing finite-dimensional vector spaces, we have not yet defined the dimension of such an object. How should dimension be defined? A reasonable definition should force the dimension of to equal . Notice that the standard basis \n\n(1 , 0, , 0) , (0 , 1, 0, , 0) , , (0 , , 0, 1) \n\nof has length . Thus we are tempted to define the dimension as the length of a basis. However, a finite-dimensional vector space in general has many different bases, and our attempted definition makes sense only if all bases in a given vector space have the same length. Fortunately that turns out to be the case, as we now show. \n\n> 2.34\n\nbasis length does not depend on basis \n\nAny two bases of a finite-dimensional vector space have the same length. \n\n> Proof\n\nSuppose is finite-dimensional. Let 1 and 2 be two bases of . Then \n\n 1 is linearly independent in and 2 spans , so the length of 1 is at most the length of 2 (by 2.22). Interchanging the roles of 1 and 2, we also see that the length of 2 is at most the length of 1. Thus the length of 1 equals the length of 2, as desired. Now that we know that any two bases of a finite-dimensional vector space have the same length, we can formally define the dimension of such spaces. \n\n> 2.35 definition:\n\ndimension, dim The dimension of a finite-dimensional vector space is the length of any basis of the vector space. \n\n The dimension of a finite-dimensional vector space is denoted by dim . \n\n> 2.36 example:\n\ndimensions \n\n dim = because the standard basis of has length .\n\n dim () = + 1 because the standard basis 1, , , of () has length \n\n + 1.\n\n If = {( , , ) 3 , } , then dim = 2 because (1 , 1, 0) , (0 , 0, 1) is a basis of .\n\n If = {( , , ) 3 + + = 0} , then dim = 2 because the list \n\n(1 , 1 , 0) , (1 , 0, 1) is a basis of . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 2C Dimension 45\n\nEvery subspace of a finite-dimensional vector space is finite-dimensional (by 2.25) and so has a dimension. The next result gives the expected inequality about the dimension of a subspace. \n\n> 2.37\n\ndimension of a subspace \n\nIf is finite-dimensional and is a subspace of , then dim dim . \n\n> Proof\n\nSuppose is finite-dimensional and is a subspace of . Think of a basis of as a linearly independent list in , and think of a basis of as a spanning list in . Now use 2.22 to conclude that dim dim . \n\n> The real vector space 2has dimen-sion two; the complex vector space \n> has dimension one. As sets, 2can be identified with (and addition is the same on both spaces, as is scalar multiplication by real numbers ). Thus when we talk about the dimension of a vector space, the role played by the choice of cannot be neglected.\n\nTo check that a list of vectors in \n\nis a basis of , we must, according to the definition, show that the list in ques-tion satisfies two properties: it must be linearly independent and it must span .The next two results show that if the list in question has the right length, then we only need to check that it satisfies one of the two required properties. First we prove that every linearly independent list of the right length is a basis. \n\n> 2.38\n\nlinearly independent list of the right length is a basis \n\nSuppose is finite-dimensional. Then every linearly independent list of vectors in of length dim is a basis of . \n\n> Proof\n\nSuppose dim = and 1, , is linearly independent in . The list \n\n 1, , can be extended to a basis of (by 2.32). However, every basis of has length , so in this case the extension is the trivial one, meaning that no elements are adjoined to 1, , . Thus 1, , is a basis of , as desired. The next result is a useful consequence of the previous result. \n\n> 2.39\n\nsubspace of full dimension equals the whole space \n\nSuppose that is finite-dimensional and is a subspace of such that \n\ndim = dim . Then = . \n\n> Proof\n\nLet 1, , be a basis of . Thus = dim , and by hypothesis we also have = dim . Thus 1, , is a linearly independent list of vectors in \n\n(because it is a basis of ) of length dim . From 2.38, we see that 1, , is a basis of . In particular every vector in is a linear combination of 1, , .Thus = . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 46 Chapter 2 Finite-Dimensional Vector Spaces\n> 2.40 example:\n\na basis of 2\n\nConsider the list (5 , 7) , (4 , 3) of vectors in 2. This list of length two is linearly independent in 2 (because neither vector is a scalar multiple of the other). Note that 2 has dimension two. Thus 2.38 implies that the linearly independent list \n\n(5 , 7) , (4 , 3) of length two is a basis of 2 (we do not need to bother checking that it spans 2). \n\n> 2.41 example:\n\na basis of a subspace of 3() \n\nLet be the subspace of 3() defined by \n\n = { 3() (5) = 0}. \n\nTo find a basis of , first note that each of the polynomials 1, ( 5) 2, and ( 5) 3\n\nis in .Suppose , , and \n\n + ( 5) 2 + ( 5) 3 = 0 \n\nfor every . Without explicitly expanding the left side of the equation above, we can see that the left side has a 3 term. Because the right side has no 3\n\nterm, this implies that = 0 . Because = 0 , we see that the left side has a 2\n\nterm, which implies that = 0 . Because = = 0 , we can also conclude that \n\n = 0 . Thus the equation above implies that = = = 0 . Hence the list \n\n1, ( 5) 2, ( 5) 3 is linearly independent in . Thus 3 dim . Hence \n\n3 dim dim 3() = 4 ,\n\nwhere we have used 2.37. The polynomial is not in because its derivative is the constant function 1.Thus 3() . Hence dim 4 (by 2.39). The inequality above now implies that dim = 3 . Thus the linearly independent list 1, ( 5) 2, ( 5) 3 in has length dim and hence is a basis of (by 2.38). Now we prove that a spanning list of the right length is a basis. \n\n> 2.42\n\nspanning list of the right length is a basis \n\nSuppose is finite-dimensional. Then every spanning list of vectors in of length dim is a basis of . \n\n> Proof\n\nSuppose dim = and 1, , spans . The list 1, , can be reduced to a basis of (by 2.30). However, every basis of has length , so in this case the reduction is the trivial one, meaning that no elements are deleted from 1, , . Thus 1, , is a basis of , as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 2C Dimension 47 \n\nThe next result gives a formula for the dimension of the sum of two subspaces of a finite-dimensional vector space. This formula is analogous to a familiar counting formula: the number of elements in the union of two finite sets equals the number of elements in the first set, plus the number of elements in the second set, minus the number of elements in the intersection of the two sets. \n\n> 2.43\n\ndimension of a sum \n\nIf 1 and 2 are subspaces of a finite-dimensional vector space, then \n\ndim ( 1 + 2) = dim 1 + dim 2 dim ( 1 2). \n\nProof Let 1, , be a basis of 1 2; thus dim ( 1 2) = . Because \n\n 1, , is a basis of 1 2, it is linearly independent in 1. Hence this list can be extended to a basis 1, , , 1, , of 1 (by 2.32). Thus dim 1 = + .Also extend 1, , to a basis 1, , , 1, , of 2; thus dim 2 = + .We will show that \n\n> 2.44\n\n 1, , , 1, , , 1, , \n\nis a basis of 1 + 2. This will complete the proof, because then we will have \n\ndim ( 1 + 2) = + + = ( + ) + ( + ) = dim 1 + dim 2 dim ( 1 2). \n\nThe list 2.44 is contained in 1 2 and thus is contained in 1 + 2. The span of this list contains 1 and contains 2 and hence is equal to 1 + 2. Thus to show that 2.44 is a basis of 1 + 2 we only need to show that it is linearly independent. To prove that 2.44 is linearly independent, suppose \n\n 1 1 + + + 1 1 + + + 1 1 + + = 0 ,\n\nwhere all the s, s, and s are scalars. We need to prove that all the s, s, and s equal 0. The equation above can be rewritten as \n\n> 2.45\n\n 1 1 + + = 1 1 1 1 ,\n\nwhich shows that 1 1 + + 1. All the s are in 2, so this implies that \n\n 1 1 + + 1 2. Because 1, , is a basis of 1 2, we have \n\n 1 1 + + = 1 1 + + \n\nfor some scalars 1, , . But 1, , , 1, , is linearly independent, so the last equation implies that all the s (and s) equal 0. Thus 2.45 becomes the equation \n\n 1 1 + + + 1 1 + + = 0. \n\nBecause the list 1, , , 1, , is linearly independent, this equation implies that all the s and s are 0, completing the proof. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n48 Chapter 2 Finite-Dimensional Vector Spaces \n\nFor a finite set, let # denote the number of elements of . The table below compares finite sets with finite-dimensional vector spaces, showing the analogy between # (for sets) and dim (for vector spaces), as well as the analogy between unions of subsets (in the context of sets) and sums of subspaces (in the context of vector spaces). \n\nsets vector spaces \n\n is a finite set is a finite-dimensional vector space \n\n# dim \n\nfor subsets 1, 2 of , the union 1 2\n\nis the smallest subset of containing 1\n\nand 2\n\nfor subspaces 1, 2 of , the sum 1 + 2\n\nis the smallest subspace of containing \n\n 1 and 2\n\n#( 1 2) dim ( 1 + 2)= # 1 + # 2 #( 1 2) = dim 1 + dim 2 dim ( 1 2)#( 1 2) = # 1 + # 2 dim ( 1 + 2) = dim 1 + dim 2\n\n 1 2 = 1 2 = {0} 1 is a disjoint union #( 1 ) = # 1 + + # \n\n 1 + + is a direct sum \n\ndim ( 1 + + )= dim 1 + + dim \n\nThe last row above focuses on the analogy between disjoint unions (for sets) and direct sums (for vector spaces). The proof of the result in the last box above will be given in 3.94. You should be able to find results about sets that correspond, via analogy, to the results about vector spaces in Exercises 12 through 18. \n\n# Exercises 2C \n\n1 Show that the subspaces of 2 are precisely {0} , all lines in 2 containing the origin, and 2.\n\n2 Show that the subspaces of 3 are precisely {0} , all lines in 3 containing the origin, all planes in 3 containing the origin, and 3.\n\n3 (a) Let = { 4() (6) = 0} . Find a basis of .(b) Extend the basis in (a) to a basis of 4() .(c) Find a subspace of 4() such that 4() = .\n\n4 (a) Let = { 4() (6) = 0} . Find a basis of .(b) Extend the basis in (a) to a basis of 4() .(c) Find a subspace of 4() such that 4() = .\n\n5 (a) Let = { 4() (2) = (5)} . Find a basis of .(b) Extend the basis in (a) to a basis of 4() .(c) Find a subspace of 4() such that 4() = .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 2C Dimension 49 \n\n6 (a) Let = { 4() (2) = (5) = (6)} . Find a basis of .(b) Extend the basis in (a) to a basis of 4() .(c) Find a subspace of 4() such that 4() = .\n\n7 (a) Let = { 4() 11 = 0} . Find a basis of .(b) Extend the basis in (a) to a basis of 4() .(c) Find a subspace of 4() such that 4() = .\n\n8 Suppose 1, , is linearly independent in and . Prove that \n\ndim span ( 1 + , , + ) 1. \n\n9 Suppose is a positive integer and 0, 1, , () are such that each \n\n has degree . Prove that 0, 1, , is a basis of () .\n\n10 Suppose is a positive integer. For 0 , let \n\n () = (1 ) .\n\nShow that 0, , is a basis of () .\n\nThe basis in this exercise leads to what are called Bernstein polynomials .You can do a web search to learn how Bernstein polynomials are used to approximate continuous functions on [0 , 1] .\n\n11 Suppose and are both four-dimensional subspaces of 6. Prove that there exist two vectors in such that neither of these vectors is a scalar multiple of the other. \n\n12 Suppose that and are subspaces of 8 such that dim = 3 , dim = 5 ,and + = 8. Prove that 8 = .\n\n13 Suppose and are both five-dimensional subspaces of 9. Prove that \n\n {0} .\n\n14 Suppose is a ten-dimensional vector space and 1, 2, 3 are subspaces of with dim 1 = dim 2 = dim 3 = 7 . Prove that 1 2 3 {0} .\n\n15 Suppose is finite-dimensional and 1, 2, 3 are subspaces of with \n\ndim 1 + dim 2 + dim 3 > 2 dim . Prove that 1 2 3 {0} .\n\n16 Suppose is finite-dimensional and is a subspace of with . Let \n\n = dim and = dim . Prove that there exist subspaces of ,each of dimension 1 , whose intersection equals .\n\n17 Suppose that 1, , are finite-dimensional subspaces of . Prove that \n\n 1 + + is finite-dimensional and \n\ndim ( 1 + + ) dim 1 + + dim .\n\nThe inequality above is an equality if and only if 1 + + is a direct sum, as will be shown in 3.94. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n50 Chapter 2 Finite-Dimensional Vector Spaces \n\n18 Suppose is finite-dimensional, with dim = 1 . Prove that there exist one-dimensional subspaces 1, , of such that \n\n = 1 .\n\n19 Explain why you might guess, motivated by analogy with the formula for the number of elements in the union of three finite sets, that if 1, 2, 3 are subspaces of a finite-dimensional vector space, then \n\ndim ( 1 + 2 + 3)= dim 1 + dim 2 + dim 3\n\n dim ( 1 2) dim ( 1 3) dim ( 2 3)\n\n+ dim ( 1 2 3). \n\nThen either prove the formula above or give a counterexample. \n\n20 Prove that if 1, 2, and 3 are subspaces of a finite-dimensional vector space, then \n\ndim ( 1 + 2 + 3)= dim 1 + dim 2 + dim 3\n\n dim ( 1 2) + dim ( 1 3) + dim ( 2 3)3 dim (( 1 + 2) 3) + dim (( 1 + 3) 2) + dim (( 2 + 3) 1)3 .\n\nThe formula above may seem strange because the right side does not look like an integer. \n\nI at once gave up my former occupations, set down natural history and all its progeny as a deformed and abortive creation, and entertained the greatest disdain for a would-be science which could never even step within the threshold of real knowledge. In this mood I betook myself to the mathematics and the branches of study appertaining to that science as being built upon secure foundations, and so worthy of my consideration. \n\nFrankenstein , Mary Wollstonecraft Shelley \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n# Chapter 3 \n\n# Linear Maps \n\nSo far our attention has focused on vector spaces. No one gets excited about vector spaces. The interesting part of linear algebra is the subject to which we now turnlinear maps. We will frequently use the powerful fundamental theorem of linear maps, which states that the dimension of the domain of a linear map equals the dimension of the subspace that gets sent to 0 plus the dimension of the range. This will imply the striking result that a linear map from a finite-dimensional vector space to itself is one-to-one if and only if its range is the whole space. A major concept that we will introduce in this chapter is the matrix associated with a linear map and with a basis of the domain space and a basis of the target space. This correspondence between linear maps and matrices provides much insight into key aspects of linear algebra. This chapter concludes by introducing product, quotient, and dual spaces. In this chapter we will need additional vector spaces, which we call and ,in addition to . Thus our standing assumptions are now as follows. \n\nstanding assumptions for this chapter \n\n denotes or .\n\n , , and denote vector spaces over .\n\n> Stefan Schfer CC BY-SA\n\nThe twelfth-century Dankwarderode Castle in Brunswick (Braunschweig ), where Carl Friedrich Gauss (17771855 ) was born and grew up. In 1809 Gauss published a method for solving systems of linear equations. This method, now called Gaussian elimination, was used in a Chinese book written over 1600 years earlier. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n> 51 52 Chapter 3 Linear Maps\n\n# 3A Vector Space of Linear Maps \n\n# Definition and Examples of Linear Maps \n\nNow we are ready for one of the key definitions in linear algebra. \n\n> 3.1 definition:\n\nlinear map \n\nA linear map from to is a function with the following properties. \n\nadditivity \n\n( + ) = + for all , .\n\nhomogeneity \n\n() = () for all and all .\n\n> Some mathematicians use the phrase\n> linear transformation , which means the same as linear map.\n\nNote that for linear maps we often use the notation as well as the usual function notation () . \n\n> 3.2 notation:\n\n( , ) , () The set of linear maps from to is denoted by ( , ) .\n\n The set of linear maps from to is denoted by () . In other words, \n\n() = ( , ) .Lets look at some examples of linear maps. Make sure you verify that each of the functions defined in the next example is indeed a linear map: \n\n> 3.3 example:\n\nlinear maps \n\nzero \n\nIn addition to its other uses, we let the symbol 0 denote the linear map that takes every element of some vector space to the additive identity of another (or possibly the same) vector space. To be specific, 0 ( , ) is defined by \n\n0 = 0. \n\nThe 0 on the left side of the equation above is a function from to , whereas the 0 on the right side is the additive identity in . As usual, the context should allow you to distinguish between the many uses of the symbol 0.\n\nidentity operator \n\nThe identity operator , denoted by , is the linear map on some vector space that takes each element to itself. To be specific, () is defined by \n\n = . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3A Vector Space of Linear Maps 53 \n\ndifferentiation \n\nDefine ( ()) by \n\n = .\n\nThe assertion that this function is a linear map is another way of stating a basic result about differentiation: ( + ) = + and ( ) = whenever , are differentiable and is a constant. \n\nintegration \n\nDefine ( () , ) by \n\n = 10\n\n. \n\nThe assertion that this function is linear is another way of stating a basic result about integration: the integral of the sum of two functions equals the sum of the integrals, and the integral of a constant times a function equals the constant times the integral of the function. \n\nmultiplication by 2\n\nDefine a linear map ( ()) by \n\n()() = 2() \n\nfor each .\n\nbackward shift \n\nRecall that denotes the vector space of all sequences of elements of . Define a linear map ( ) by \n\n( 1, 2, 3, ) = ( 2, 3, ). \n\nfrom 3 to 2\n\nDefine a linear map ( 3, 2) by \n\n( , , ) = (2 + 3 , 7 + 5 6). \n\nfrom to \n\nTo generalize the previous example, let and be positive integers, let , \n\nfor each = 1 , , and each = 1 , , , and define a linear map ( , )\n\nby \n\n( 1, , ) = ( 1,1 1 + + 1, , , ,1 1 + + , ). \n\nActually every linear map from to is of this form. \n\ncomposition \n\nFix a polynomial () . Define a linear map ( ()) by \n\n()() = (()). \n\nThe existence part of the next result means that we can find a linear map that takes on whatever values we wish on the vectors in a basis. The uniqueness part of the next result means that a linear map is completely determined by its values on a basis. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n54 Chapter 3 Linear Maps \n\n> 3.4\n\nlinear map lemma \n\nSuppose 1, , is a basis of and 1, , . Then there exists a unique linear map such that \n\n = \n\nfor each = 1 , , .\n\nProof First we show the existence of a linear map with the desired property. Define by \n\n( 1 1 + + ) = 1 1 + + ,\n\nwhere 1, , are arbitrary elements of . The list 1, , is a basis of . Thus the equation above does indeed define a function from to (because each element of can be uniquely written in the form 1 1 + + ). For each , taking = 1 and the other s equal to 0 in the equation above shows that = .If , with = 1 1 + + and = 1 1 + + , then \n\n( + ) = (( 1 + 1) 1 + + ( + ) )= ( 1 + 1) 1 + + ( + ) \n\n= ( 1 1 + + ) + ( 1 1 + + )= + . \n\nSimilarly, if and = 1 1 + + , then \n\n() = ( 1 1 + + )= 1 1 + + \n\n= ( 1 1 + + )= . \n\nThus is a linear map from to .To prove uniqueness, now suppose that ( , ) and that = for each = 1 , , . Let 1, , . Then the homogeneity of implies that \n\n( ) = for each = 1 , , . The additivity of now implies that \n\n( 1 1 + + ) = 1 1 + + .\n\nThus is uniquely determined on span ( 1, , ) by the equation above. Because \n\n 1, , is a basis of , this implies that is uniquely determined on , as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 3A Vector Space of Linear Maps 55\n\n# Algebraic Operations on ( , ) \n\nWe begin by defining addition and scalar multiplication on ( , ) . \n\n> 3.5 definition:\n\naddition and scalar multiplication on ( , ) \n\nSuppose , ( , ) and . The sum + and the product are the linear maps from to defined by \n\n( + )() = + and ()() = () \n\nfor all . \n\n> Linear maps are pervasive throughout mathematics. However, they are not as ubiquitous as imagined by people who seem to think cos is a linear map from\n> to when they incorrectly write that\n> cos ( +) equals cos +cos and that\n> cos 2 equals 2cos .\n\nYou should verify that + and \n\nas defined above are indeed linear maps. In other words, if , ( , ) and \n\n , then + ( , ) and \n\n( , ) .Because we took the trouble to de-fine addition and scalar multiplication on \n\n( , ) , the next result should not be a surprise. \n\n> 3.6\n\n( , ) is a vector space \n\nWith the operations of addition and scalar multiplication as defined above, \n\n( , ) is a vector space. The routine proof of the result above is left to the reader. Note that the additive identity of ( , ) is the zero linear map defined in Example 3.3. Usually it makes no sense to multiply together two elements of a vector space, but for some pairs of linear maps a useful product exists, as in the next definition. \n\n> 3.7 definition:\n\nproduct of linear maps \n\nIf ( , ) and ( , ) , then the product ( , ) is defined by \n\n()() = () \n\nfor all .Thus is just the usual composition of two functions, but when both functions are linear, we usually write instead of . The product notation \n\n helps make the distributive properties (see next result) seem natural. Note that is defined only when maps into the domain of . You should verify that is indeed a linear map from to whenever ( , ) and \n\n ( , ) . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 56 Chapter 3 Linear Maps\n> 3.8\n\nalgebraic properties of products of linear maps \n\nassociativity \n\n( 1 2) 3 = 1( 2 3) whenever 1, 2, and 3 are linear maps such that the products make sense (meaning 3 maps into the domain of 2, and 2\n\nmaps into the domain of 1). \n\nidentity \n\n = = whenever ( , ) ; here the first is the identity operator on , and the second is the identity operator on .\n\ndistributive properties \n\n( 1 + 2) = 1 + 2 and ( 1 + 2) = 1 + 2 whenever \n\n , 1, 2 ( , ) and , 1, 2 ( , ) .The routine proof of the result above is left to the reader. Multiplication of linear maps is not commutative. In other words, it is not necessarily true that = , even if both sides of the equation make sense. \n\n> 3.9 example:\n\ntwo noncommuting linear maps from () to () \n\nSuppose ( ()) is the differentiation map defined in Example 3.3 and ( ()) is the multiplication by 2 map defined earlier in this section. Then \n\n(())() = 2 () but (())() = 2 () + 2(). \n\nThus differentiating and then multiplying by 2 is not the same as multiplying by 2 and then differentiating. \n\n> 3.10\n\nlinear maps take 0 to 0\n\nSuppose is a linear map from to . Then (0) = 0 . \n\n> Proof\n\nBy additivity, we have \n\n(0) = (0 + 0) = (0) + (0). \n\nAdd the additive inverse of (0) to each side of the equation above to conclude that (0) = 0 .Suppose , . The function defined by \n\n () = + \n\nis a linear map if and only if = 0 (use 3.10). Thus the linear functions of high school algebra are not the same as linear maps in the context of linear algebra. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3A Vector Space of Linear Maps 57 \n\n# Exercises 3A \n\n1 Suppose , . Define 3 2 by \n\n( , , ) = (2 4 + 3 + , 6 + ). \n\nShow that is linear if and only if = = 0 .\n\n2 Suppose , . Define () 2 by \n\n = (3(4) + 5 (6) + (1)(2) , 21 \n\n 3() + sin (0)). \n\nShow that is linear if and only if = = 0 .\n\n3 Suppose that ( , ). Show that there exist scalars , for \n\n = 1 , , and = 1 , , such that \n\n( 1, , ) = ( 1,1 1 + + 1, , , ,1 1 + + , )\n\nfor every ( 1, , ) .\n\nThis exercise shows that the linear map has the form promised in the second to last item of Example 3.3. \n\n4 Suppose ( , ) and 1, , is a list of vectors in such that \n\n 1, , is a linearly independent list in . Prove that 1, , is linearly independent. \n\n5 Prove that ( , ) is a vector space, as was asserted in 3.6. \n\n6 Prove that multiplication of linear maps has the associative, identity, and distributive properties asserted in 3.8. \n\n7 Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if dim = 1 and \n\n () , then there exists such that = for all .\n\n8 Give an example of a function 2 such that \n\n() = () \n\nfor all and all 2 but is not linear. \n\nThis exercise and the next exercise show that neither homogeneity nor additivity alone is enough to imply that a function is a linear map. \n\n9 Give an example of a function such that \n\n( + ) = () + () \n\nfor all , but is not linear. (Here is thought of as a complex vector space.) \n\nThere also exists a function such that satisfies the additivity condition above but is not linear. However, showing the existence of such a function involves considerably more advanced tools. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n58 Chapter 3 Linear Maps \n\n10 Prove or give a counterexample: If () and () () is defined by = , then is a linear map. \n\nThe function defined here differs from the function defined in the last bullet point of 3.3 by the order of the functions in the compositions. \n\n11 Suppose is finite-dimensional and () . Prove that is a scalar multiple of the identity if and only if = for every () .\n\n12 Suppose is a subspace of with . Suppose ( , ) and \n\n 0 (which means that 0 for some ). Define by \n\n = {{ if ,\n\n0 if and . \n\nProve that is not a linear map on .\n\n13 Suppose is finite-dimensional. Prove that every linear map on a subspace of can be extended to a linear map on . In other words, show that if is a subspace of and ( , ) , then there exists ( , ) such that \n\n = for all .\n\nThe result in this exercise is used in the proof of 3.125. \n\n14 Suppose is finite-dimensional with dim > 0 , and suppose is infinite-dimensional. Prove that ( , ) is infinite-dimensional. \n\n15 Suppose 1, , is a linearly dependent list of vectors in . Suppose also that {0} . Prove that there exist 1, , such that no \n\n ( , ) satisfies = for each = 1 , , .\n\n16 Suppose is finite-dimensional with dim > 1 . Prove that there exist \n\n , () such that .\n\n17 Suppose is finite-dimensional. Show that the only two-sided ideals of \n\n() are {0} and () .\n\nA subspace of () is called a two-sided ideal of () if and \n\n> \n\nfor all and all () .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3B Null Spaces and Ranges 59 \n\n# 3B Null Spaces and Ranges \n\n# Null Space and Injectivity \n\nIn this section we will learn about two subspaces that are intimately connected with each linear map. We begin with the set of vectors that get mapped to 0. \n\n> 3.11 definition:\n\nnull space, null \n\nFor ( , ) , the null space of , denoted by null , is the subset of \n\nconsisting of those vectors that maps to 0:\n\nnull = { = 0}. \n\n> 3.12 example:\n\nnull space \n\n If is the zero map from to , meaning that = 0 for every , then \n\nnull = .\n\n Suppose ( 3, ) is defined by ( 1, 2, 3) = 1 + 2 2 + 3 3. Then \n\nnull equals {( 1, 2, 3) 3 1 + 2 2 + 3 3 = 0} , which is a subspace of the domain of . We will soon see that the null space of each linear map is a subspace of its domain. \n\n The word null means zero. Thus the term null spaceshould remind you of the connection to 0. Some mathe-maticians use the term kernel instead of null space. \n\nSuppose ( ()) is the dif-ferentiation map defined by = .The only functions whose derivative equals the zero function are the con-stant functions. Thus the null space of \n\n equals the set of constant functions. \n\n Suppose that ( ()) is the multiplication by 2 map defined by \n\n()() = 2() . The only polynomial such that 2() = 0 for all \n\nis the 0 polynomial. Thus null = {0} .\n\n Suppose ( ) is the backward shift defined by \n\n( 1, 2, 3, ) = ( 2, 3, ). \n\nThen ( 1, 2, 3, ) equals 0 if and only if the numbers 2, 3, are all 0.Thus null = {( , 0, 0, ) } .The next result shows that the null space of each linear map is a subspace of the domain. In particular, 0 is in the null space of every linear map. \n\n> 3.13\n\nthe null space is a subspace \n\nSuppose ( , ) . Then null is a subspace of . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 60 Chapter 3 Linear Maps Proof\n\nBecause is a linear map, (0) = 0 (by 3.10). Thus 0 null .Suppose , null . Then \n\n( + ) = + = 0 + 0 = 0. \n\nHence + null . Thus null is closed under addition. Suppose null and . Then \n\n() = = 0 = 0. \n\nHence null . Thus null is closed under scalar multiplication. We have shown that null contains 0 and is closed under addition and scalar multiplication. Thus null is a subspace of (by 1.34). As we will soon see, for a linear map the next definition is closely connected to the null space. \n\n> 3.14 definition:\n\ninjective \n\nA function is called injective if = implies = . \n\n> The term one-to-one means the same as injective.\n\nWe could rephrase the definition above to say that is injective if \n\nimplies that . Thus is injective if and only if it maps distinct inputs to distinct outputs. The next result says that we can check whether a linear map is injective by checking whether 0 is the only vector that gets mapped to 0. As a simple application of this result, we see that of the linear maps whose null spaces we computed in 3.12, only multiplication by 2 is injective (except that the zero map is injective in the special case = {0} ). \n\n> 3.15\n\ninjectivity null space equals {0} \n\nLet ( , ) . Then is injective if and only if null = {0} . \n\n> Proof\n\nFirst suppose is injective. We want to prove that null = {0} . We already know that {0} null (by 3.10). To prove the inclusion in the other direction, suppose null . Then \n\n() = 0 = (0). \n\nBecause is injective, the equation above implies that = 0 . Thus we can conclude that null = {0} , as desired. To prove the implication in the other direction, now suppose null = {0} . We want to prove that is injective. To do this, suppose , and = . Then \n\n0 = = ( ). \n\nThus is in null , which equals {0} . Hence = 0 , which implies that \n\n = . Hence is injective, as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 3B Null Spaces and Ranges 61\n\n# Range and Surjectivity \n\nNow we give a name to the set of outputs of a linear map. \n\n> 3.16 definition:\n\nrange \n\nFor ( , ) , the range of is the subset of consisting of those vectors that are equal to for some :\n\nrange = { }. \n\n> 3.17 example:\n\nrange \n\n If is the zero map from to , meaning that = 0 for every , then \n\nrange = {0} .\n\n Suppose ( 2, 3) is defined by ( , ) = (2 , 5 , + ) . Then \n\nrange = {(2 , 5 , + ) , }. \n\nNote that range is a subspace of 3. We will soon see that the range of each element of ( , ) is a subspace of .\n\n Suppose ( ()) is the differentiation map defined by = . Because for every polynomial () there exists a polynomial () such that \n\n = , the range of is () .The next result shows that the range of each linear map is a subspace of the vector space into which it is being mapped. \n\n> 3.18\n\nthe range is a subspace \n\nIf ( , ) , then range is a subspace of . \n\n> Proof\n\nSuppose ( , ) . Then (0) = 0 (by 3.10), which implies that \n\n0 range .If 1, 2 range , then there exist 1, 2 such that 1 = 1 and \n\n 2 = 2. Thus \n\n( 1 + 2) = 1 + 2 = 1 + 2.\n\nHence 1 + 2 range . Thus range is closed under addition. If range and , then there exists such that = . Thus \n\n() = = . \n\nHence range . Thus range is closed under scalar multiplication. We have shown that range contains 0 and is closed under addition and scalar multiplication. Thus range is a subspace of (by 1.34). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n62 Chapter 3 Linear Maps \n\n> 3.19 definition:\n\nsurjective \n\nA function is called surjective if its range equals .To illustrate the definition above, note that of the ranges we computed in 3.17, only the differentiation map is surjective (except that the zero map is surjective in the special case = {0} ). \n\nSome people use the term onto , which means the same as surjective. \n\nWhether a linear map is surjective de-pends on what we are thinking of as the vector space into which it maps. \n\n> 3.20 example:\n\nsurjectivity depends on the target space \n\nThe differentiation map ( 5()) defined by = is not surjective, because the polynomial 5 is not in the range of . However, the differentiation map ( 5() , 4()) defined by = is surjective, because its range equals 4() , which is the vector space into which maps. \n\n# Fundamental Theorem of Linear Maps \n\nThe next result is so important that it gets a dramatic name. \n\n> 3.21\n\nfundamental theorem of linear maps \n\nSuppose is finite-dimensional and ( , ) . Then range is finite-dimensional and \n\ndim = dim null + dim range . \n\nProof Let 1, , be a basis of null ; thus dim null = . The linearly independent list 1, , can be extended to a basis \n\n 1, , , 1, , \n\nof (by 2.32). Thus dim = + . To complete the proof, we need to show that \n\nrange is finite-dimensional and dim range = . We will do this by proving that 1, , is a basis of range .Let . Because 1, , , 1, , spans , we can write \n\n = 1 1 + + + 1 1 + + ,\n\nwhere the s and s are in . Applying to both sides of this equation, we get \n\n = 1 1 + + ,\n\nwhere the terms of the form disappeared because each is in null . The last equation implies that the list 1, , spans range . In particular, range is finite-dimensional. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3B Null Spaces and Ranges 63 \n\nTo show 1, , is linearly independent, suppose 1, , and \n\n 1 1 + + = 0. \n\nThen \n\n( 1 1 + + ) = 0. \n\nHence \n\n 1 1 + + null . \n\nBecause 1, , spans null , we can write \n\n 1 1 + + = 1 1 + + ,\n\nwhere the s are in . This equation implies that all the s (and s) are 0 (be-cause 1, , , 1, , is linearly independent). Thus 1, , is linearly independent and hence is a basis of range , as desired. Now we can show that no linear map from a finite-dimensional vector space to a smaller vector space can be injective, where smaller is measured by dimension. \n\n> 3.22\n\nlinear map to a lower-dimensional space is not injective \n\nSuppose and are finite-dimensional vector spaces such that \n\ndim > dim . Then no linear map from to is injective. \n\nProof Let ( , ) . Then \n\ndim null = dim dim range dim dim > 0 ,\n\nwhere the first line above comes from the fundamental theorem of linear maps (3.21) and the second line follows from 2.37. The inequality above states that \n\ndim null > 0 . This means that null contains vectors other than 0. Thus is not injective (by 3.15). \n\n> 3.23 example:\n\nlinear map from 4 to 3 is not injective \n\nDefine a linear map 4 3 by \n\n( 1, 2, 3, 4) = ( 7 1 + 2 + 4, 97 1 + 3 2 + 2 3, 2 + 6 3 + 7 4). \n\nBecause dim 4 > dim 3, we can use 3.22 to assert that is not injective, without doing any calculations. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n64 Chapter 3 Linear Maps \n\nThe next result shows that no linear map from a finite-dimensional vector space to a bigger vector space can be surjective, where bigger is measured by dimension. \n\n> 3.24\n\nlinear map to a higher-dimensional space is not surjective \n\nSuppose and are finite-dimensional vector spaces such that \n\ndim < dim . Then no linear map from to is surjective. \n\nProof Let ( , ) . Then \n\ndim range = dim dim null dim < dim ,\n\nwhere the equality above comes from the fundamental theorem of linear maps (3.21). The inequality above states that dim range < dim . This means that \n\nrange cannot equal . Thus is not surjective. As we will soon see, 3.22 and 3.24 have important consequences in the theory of linear equations. The idea is to express questions about systems of linear equations in terms of linear maps. Lets begin by rephrasing in terms of linear maps the question of whether a homogeneous system of linear equations has a nonzero solution. \n\nHomogeneous , in this context, means that the constant term on the right side of each equation below is 0.\n\nFix positive integers and , and let \n\n , for = 1 , , and = 1 , , .Consider the homogeneous system of lin-ear equations \n\n> \n> \n> = 1\n\n 1, = 0 \n\n> \n> \n> = 1\n\n , = 0. \n\nClearly 1 = = = 0 is a solution of the system of equations above; the question here is whether any other solutions exist. Define by \n\n> 3.25\n\n( 1, , ) = ( \n\n> \n> \n> = 1\n\n 1, , , \n\n> \n> = 1\n\n , ). \n\nThe equation ( 1, , ) = 0 (the 0 here is the additive identity in , namely, the list of length of all 0s) is the same as the homogeneous system of linear equations above. Thus we want to know if null is strictly bigger than {0} , which is equivalent to not being injective (by 3.15). The next result gives an important condition for ensuring that is not injective. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3B Null Spaces and Ranges 65 \n\n> 3.26\n\nhomogeneous system of linear equations \n\nA homogeneous system of linear equations with more variables than equations has nonzero solutions. \n\nProof Use the notation and result from the discussion above. Thus is a linear map from to , and we have a homogeneous system of linear equations with variables 1, , . From 3.22 we see that is not injective if > .Example of the result above: a homogeneous system of four linear equations with five variables has nonzero solutions. Now we consider the question of whether a system of linear equations has no solutions for some choice of the constant terms. To rephrase this question in terms of a linear map, fix positive integers and , and let , for all = 1 , , \n\nand all = 1 , , . For 1, , , consider the system of linear equations \n\n> \n> \n> = 1\n\n 1, = 1\n\n3.27 \n\n> \n> \n> = 1\n\n , = .\n\nWith this notation, the question here is whether there is some choice of the constant terms 1, , such that no solution exists to the system above. \n\nThe results 3.26 and 3.28, which com-pare the number of variables and the number of equations, can also be proved using Gaussian elimina-tion. The abstract approach taken here seems to provide cleaner proofs. \n\nDefine as in 3.25. The equation ( 1, , ) = ( 1, , ) is the same as the system of equations 3.27. Thus we want to know if range .Hence we can rephrase our question about not having a solution for some choice of 1, , as follows: What condition ensures that is not surjective? The next result gives one such condition. \n\n> 3.28\n\nsystem of linear equations with more equations than variables \n\nA system of linear equations with more equations than variables has no solution for some choice of the constant terms. \n\nProof Use the notation from the discussion above. Thus is a linear map from \n\n to , and we have a system of equations with variables 1, , ; see 3.27. If < , then 3.24 implies that is not surjective. As discussed above, this shows that if we have more equations than variables in a system of linear equations, then there is no solution for some choice of the constant terms. Example of the result above: a system of five linear equations with four variables has no solution for some choice of the constant terms. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n66 Chapter 3 Linear Maps \n\n# Exercises 3B \n\n1 Give an example of a linear map with dim null = 3 and dim range = 2 .\n\n2 Suppose , () are such that range null . Prove that () 2 = 0 .\n\n3 Suppose 1, , is a list of vectors in . Define ( , ) by \n\n( 1, , ) = 1 1 + + .\n\n(a) What property of corresponds to 1, , spanning ?(b) What property of corresponds to the list 1, , being linearly independent? \n\n4 Show that { ( 5, 4) dim null > 2} is not a subspace of ( 5, 4).\n\n5 Give an example of ( 4) such that range = null .\n\n6 Prove that there does not exist ( 5) such that range = null .\n\n7 Suppose and are finite-dimensional with 2 dim dim . Show that { ( , ) is not injective } is not a subspace of ( , ) .\n\n8 Suppose and are finite-dimensional with dim dim 2 . Show that { ( , ) is not surjective } is not a subspace of ( , ) .\n\n9 Suppose ( , ) is injective and 1, , is linearly independent in .Prove that 1, , is linearly independent in .\n\n10 Suppose 1, , spans and ( , ) . Show that 1, , spans \n\nrange .\n\n11 Suppose that is finite-dimensional and that ( , ) . Prove that there exists a subspace of such that \n\n null = {0} and range = { }. \n\n12 Suppose is a linear map from 4 to 2 such that \n\nnull = {( 1, 2, 3, 4) 4 1 = 5 2 and 3 = 7 4}. \n\nProve that is surjective. \n\n13 Suppose is a three-dimensional subspace of 8 and that is a linear map from 8 to 5 such that null = . Prove that is surjective. \n\n14 Prove that there does not exist a linear map from 5 to 2 whose null space equals {( 1, 2, 3, 4, 5) 5 1 = 3 2 and 3 = 4 = 5}.\n\n15 Suppose there exists a linear map on whose null space and range are both finite-dimensional. Prove that is finite-dimensional. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 3B Null Spaces and Ranges 67\n\n16 Suppose and are both finite-dimensional. Prove that there exists an injective linear map from to if and only if dim dim .\n\n17 Suppose and are both finite-dimensional. Prove that there exists a surjective linear map from onto if and only if dim dim .\n\n18 Suppose and are finite-dimensional and that is a subspace of .Prove that there exists ( , ) such that null = if and only if \n\ndim dim dim .\n\n19 Suppose is finite-dimensional and ( , ) . Prove that is injective if and only if there exists ( , ) such that is the identity operator on .\n\n20 Suppose is finite-dimensional and ( , ) . Prove that is surjective if and only if there exists ( , ) such that is the identity operator on .\n\n21 Suppose is finite-dimensional, ( , ) , and is a subspace of .Prove that { } is a subspace of and \n\ndim { } = dim null + dim ( range ). \n\n22 Suppose and are finite-dimensional vector spaces and ( , ) and \n\n ( , ) . Prove that \n\ndim null dim null + dim null . \n\n23 Suppose and are finite-dimensional vector spaces and ( , ) and \n\n ( , ) . Prove that \n\ndim range min {dim range , dim range }. \n\n24 (a) Suppose dim = 5 and , () are such that = 0 . Prove that \n\ndim range 2 .(b) Give an example of , ( 5) with = 0 and dim range = 2 .\n\n25 Suppose that is finite-dimensional and , ( , ) . Prove that \n\nnull null if and only if there exists () such that = .\n\n26 Suppose that is finite-dimensional and , ( , ) . Prove that \n\nrange range if and only if there exists () such that = .\n\n27 Suppose () and 2 = . Prove that = null range .\n\n28 Suppose ( ()) is such that deg = ( deg ) 1 for every non-constant polynomial () . Prove that is surjective. \n\n> The notation is used above to remind you of the differentiation map that sends a polynomial to .\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n68 Chapter 3 Linear Maps \n\n29 Suppose () . Prove that there exists a polynomial () such that 5 + 3 = .\n\nThis exercise can be done without linear algebra, but its more fun to do it using linear algebra. \n\n30 Suppose ( , ) and 0 . Suppose is not in null . Prove that \n\n = null { }. \n\n31 Suppose is finite-dimensional, is a subspace of , and is a finite-dimensional subspace of . Prove that there exists ( , ) such that \n\nnull = and range = if and only if dim + dim = dim .\n\n32 Suppose is finite-dimensional with dim > 1 . Show that if () \n\nis a linear map such that () = ()() for all , () , then \n\n = 0 .\n\nHint: The description of the two-sided ideals of () given by Exercise 17 in Section 3A might be useful. \n\n33 Suppose that and are real vector spaces and ( , ) . Define \n\n by \n\n ( + ) = + \n\nfor all , .(a) Show that is a (complex) linear map from to .(b) Show that is injective if and only if is injective. (c) Show that range = if and only if range = .\n\nSee Exercise 8 in Section 1B for the definition of the complexification .The linear map is called the complexification of the linear map .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3C Matrices 69 \n\n# 3C Matrices \n\n# Representing a Linear Map by a Matrix \n\nWe know that if 1, , is a basis of and is linear, then the values of 1, , determine the values of on arbitrary vectors in see the linear map lemma (3.4). As we will soon see, matrices provide an efficient method of recording the values of the s in terms of a basis of . \n\n> 3.29 definition:\n\nmatrix, , \n\nSuppose and are nonnegative integers. An -by- matrix is a rectangular array of elements of with rows and columns: \n\n = 1,1 1, \n\n ,1 , \n\n.\n\nThe notation , denotes the entry in row , column of . \n\n> 3.30 example:\n\n , equals entry in row , column of \n\nWhen dealing with matrices, the first index refers to the row number; the sec-ond index refers to the column number. \n\nSuppose = 8 4 5 3 1 9 7 .Thus 2,3 refers to the entry in the sec-ond row, third column of , which means that 2,3 = 7 .Now we come to the key definition in this section. \n\n> 3.31 definition:\n\nmatrix of a linear map, () \n\nSuppose ( , ) and 1, , is a basis of and 1, , is a basis of . The matrix of with respect to these bases is the -by- matrix () \n\nwhose entries , are defined by \n\n = 1, 1 + + , .\n\nIf the bases 1, , and 1, , are not clear from the context, then the notation ( , ( 1, , ), ( 1, , )) is used. The matrix () of a linear map ( , ) depends on the basis 1, , \n\nof and the basis 1, , of , as well as on . However, the bases should be clear from the context, and thus they are often not included in the notation. To remember how () is constructed from , you might write across the top of the matrix the basis vectors 1, , for the domain and along the left the basis vectors 1, , for the vector space into which maps, as follows: \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n70 Chapter 3 Linear Maps \n\n 1 \n\n 1\n\n() = \n\n 1, \n\n , \n\n.\n\nThe th column of () consists of the scalars needed to write as a linear combination of 1, , : \n\n> =\n> \n> \n> = 1\n> , .\n\nIn the matrix above only the th col-umn is shown. Thus the second index of each displayed entry of the matrix above is . The picture above should remind you that can be computed from () by multiplying each entry in the th column by the corresponding from the left col-umn, and then adding up the resulting vectors. \n\nIf is a linear map from an \n\n> \n\n-dimensional vector space to an \n\n> \n\n-dimensional vector space, then \n\n() is an -by- matrix. \n\nIf is a linear map from to ,then unless stated otherwise, assume the bases in question are the standard ones (where the th basis vector is 1 in the th \n\nslot and 0 in all other slots). If you think of elements of as columns of numbers, then you can think of the th column of () as applied to the th standard basis vector. \n\n> 3.32 example:\n\nthe matrix of a linear map from 2 to 3\n\nSuppose ( 2, 3) is defined by \n\n( , ) = ( + 3 , 2 + 5 , 7 + 9). \n\nBecause (1 , 0) = (1 , 2, 7) and (0 , 1) = (3 , 5, 9) , the matrix of with respect to the standard bases is the 3-by-2 matrix below: \n\n() = 1 32 57 9.\n\nWhen working with () , use the standard basis 1, , 2, , unless the context indicates otherwise. \n\n> 3.33 example:\n\nmatrix of the differentiation map from 3() to 2() \n\nSuppose ( 3() , 2()) is the differentiation map defined by = .Because ( ) = 1 , the matrix of with respect to the standard bases is the \n\n3-by-4 matrix below: \n\n() = 0 1 0 00 0 2 00 0 0 3.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3C Matrices 71 \n\n# Addition and Scalar Multiplication of Matrices \n\nFor the rest of this section, assume that , , and are finite-dimensional and that a basis has been chosen for each of these vector spaces. Thus for each linear map from to , we can talk about its matrix (with respect to the chosen bases). Is the matrix of the sum of two linear maps equal to the sum of the matrices of the two maps? Right now this question does not yet make sense because although we have defined the sum of two linear maps, we have not defined the sum of two matrices. Fortunately, the natural definition of the sum of two matrices has the right properties. Specifically, we make the following definition. \n\n> 3.34 definition:\n\nmatrix addition \n\nThe sum of two matrices of the same size is the matrix obtained by adding corresponding entries in the matrices: \n\n 1,1 1, \n\n ,1 , \n\n\n\n+ 1,1 1, \n\n ,1 , \n\n= 1,1 + 1,1 1, + 1, \n\n ,1 + ,1 , + , \n\n.\n\nIn the next result, the assumption is that the same bases are used for all three linear maps + , , and . \n\n> 3.35\n\nmatrix of the sum of linear maps \n\nSuppose , ( , ) . Then ( + ) = () + () .The verification of the result above follows from the definitions and is left to the reader. Still assuming that we have some bases in mind, is the matrix of a scalar times a linear map equal to the scalar times the matrix of the linear map? Again, the question does not yet make sense because we have not defined scalar multiplication on matrices. Fortunately, the natural definition again has the right properties. \n\n> 3.36 definition:\n\nscalar multiplication of a matrix \n\nThe product of a scalar and a matrix is the matrix obtained by multiplying each entry in the matrix by the scalar: \n\n 1,1 1, \n\n ,1 , \n\n= 1,1 1, \n\n ,1 , \n\n.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n72 Chapter 3 Linear Maps \n\n> 3.37 example:\n\naddition and scalar multiplication of matrices \n\n23 11 5 + 4 21 6 = 6 22 10 + 4 21 6 = 10 41 16 \n\nIn the next result, the assumption is that the same bases are used for both the linear maps and . \n\n> 3.38\n\nthe matrix of a scalar times a linear map \n\nSuppose and ( , ) . Then () = () .The verification of the result above is also left to the reader. Because addition and scalar multiplication have now been defined for matrices, you should not be surprised that a vector space is about to appear. First we introduce a bit of notation so that this new vector space has a name, and then we find the dimension of this new vector space. \n\n> 3.39 notation:\n\n , \n\nFor and positive integers, the set of all -by- matrices with entries in \n\nis denoted by , . \n\n> 3.40\n\ndim , = \n\nSuppose and are positive integers. With addition and scalar multiplication defined as above, , is a vector space of dimension .\n\nProof The verification that , is a vector space is left to the reader. Note that the additive identity of , is the -by- matrix all of whose entries equal 0.The reader should also verify that the list of distinct -by- matrices that have \n\n0 in all entries except for a 1 in one entry is a basis of , . There are such matrices, so the dimension of , equals .\n\n# Matrix Multiplication \n\nSuppose, as previously, that 1, , is a basis of and 1, , is a basis of . Suppose also that 1, , is a basis of .Consider linear maps and . The composition is a linear map from to . Does () equal () () ? This question does not yet make sense because we have not defined the product of two matrices. We will choose a definition of matrix multiplication that forces this question to have a positive answer. Lets see how to do this. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3C Matrices 73 \n\nSuppose () = and () = . For 1 , we have \n\n() = ( \n\n> \n\n\n\n> = 1\n\n , )=\n\n> \n\n\n\n> = 1\n\n , \n\n=\n\n> \n\n\n\n> = 1\n\n , \n\n\n\n> = 1\n\n , \n\n=\n\n> \n\n\n\n> = 1\n\n(\n\n> \n\n\n\n> = 1\n\n , , ) .\n\nThus () is the -by- matrix whose entry in row , column , equals \n\n> \n\n\n\n> = 1\n\n , , .\n\nNow we see how to define matrix multiplication so that the desired equation \n\n() = () () holds. \n\n> 3.41 definition:\n\nmatrix multiplication \n\nSuppose is an -by- matrix and is an -by- matrix. Then is defined to be the -by- matrix whose entry in row , column , is given by the equation \n\n() , =\n\n> \n\n\n\n> = 1\n\n , , .\n\nThus the entry in row , column , of is computed by taking row of and column of , multiplying together corresponding entries, and then summing. \n\nYou may have learned this definition of matrix multiplication in an earlier course, although you may not have seen this motivation for it. \n\nNote that we define the product of two matrices only when the number of columns of the first matrix equals the number of rows of the second matrix. \n\n> 3.42 example:\n\nmatrix multiplication \n\nHere we multiply together a 3-by-2 matrix and a 2-by-4 matrix, obtaining a \n\n3-by-4 matrix: \n\n1 23 45 66 5 4 32 1 0 1 = 10 7 4 126 19 12 542 31 20 9.\n\nMatrix multiplication is not commutative is not necessarily equal to \n\n even if both products are defined (see Exercise 10). Matrix multiplication is distributive and associative (see Exercises 11 and 12). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n74 Chapter 3 Linear Maps \n\nIn the next result, we assume that the same basis of is used in considering \n\n ( , ) and ( , ) , the same basis of is used in considering \n\n ( , ) and ( , ) , and the same basis of is used in considering \n\n ( , ) and ( , ) . \n\n> 3.43\n\nmatrix of product of linear maps \n\nIf ( , ) and ( , ) , then () = () () .The proof of the result above is the calculation that was done as motivation before the definition of matrix multiplication. In the next piece of notation, note that as usual the first index refers to a row and the second index refers to a column, with a vertically centered dot used as a placeholder. \n\n> 3.44 notation:\n\n , , , \n\nSuppose is an -by- matrix. \n\n If 1 , then , denotes the 1-by- matrix consisting of row of .\n\n If 1 , then , denotes the -by-1 matrix consisting of column \n\nof . \n\n> 3.45 example:\n\n , equals th row of and , equals th column of \n\nThe notation 2, denotes the second row of and ,2 denotes the second column of . Thus if = 8 4 51 9 7 , then \n\n 2, = ( 1 9 7 ) and ,2 = 49 .\n\nThe product of a 1-by- matrix and an -by-1 matrix is a 1-by-1 matrix. How-ever, we will frequently identify a 1-by-1 matrix with its entry. For example, \n\n( 3 4 ) 62 = ( 26 ) \n\nbecause 3 6 + 4 2 = 26 . However, we can identify ( 26 ) with 26 , writing \n\n( 3 4 ) 62 = 26 .The next result uses the convention discussed in the paragraph above to give another way to think of matrix multiplication. For example, the next result and the calculation in the paragraph above explain why the entry in row 2, column 1,of the product in Example 3.42 equals 26 .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3C Matrices 75 \n\n> 3.46\n\nentry of matrix product equals row times column \n\nSuppose is an -by- matrix and is an -by- matrix. Then \n\n() , = , , \n\nif 1 and 1 . In other words, the entry in row , column , of \n\n equals (row of ) times (column of ). \n\nProof Suppose 1 and 1 . The definition of matrix multiplication states that \n\n> 3.47\n\n() , = ,1 1, + + , , .\n\nThe definition of matrix multiplication also implies that the product of the 1-by- \n\nmatrix , and the -by-1 matrix , is the 1-by-1 matrix whose entry is the number on the right side of the equation above. Thus the entry in row , column ,of equals (row of ) times (column of ). The next result gives yet another way to think of matrix multiplication. In the result below, () , is column of the -by- matrix . Thus () , is an \n\n -by-1 matrix. Also, , is an -by-1 matrix because it is the product of an \n\n -by- matrix and an -by-1 matrix. Thus the two sides of the equation in the result below have the same size, making it reasonable that they might be equal. \n\n> 3.48\n\ncolumn of matrix product equals matrix times column \n\nSuppose is an -by- matrix and is an -by- matrix. Then \n\n() , = , \n\nif 1 . In other words, column of equals times column of .\n\nProof As discussed above, () , and , are both -by-1 matrices. If 1 , then the entry in row of () , is the left side of 3.47 and the entry in row of , is the right side of 3.47. Thus () , = , .Our next result will give another way of thinking about the product of an \n\n -by- matrix and an -by-1 matrix, motivated by the next example. \n\n> 3.49 example:\n\nproduct of a 3-by-2 matrix and a 2-by-1 matrix \n\nUse our definitions and basic arithmetic to verify that \n\n1 23 45 651 = 719 31 = 5 135\n\n+ 1246.\n\nThus in this example, the product of a 3-by-2 matrix and a 2-by-1 matrix is a linear combination of the columns of the 3-by-2 matrix, with the scalars ( 5 and 1)that multiply the columns coming from the 2-by-1 matrix. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 76 Chapter 3 Linear Maps\n\nThe next result generalizes the example above. \n\n> 3.50\n\nlinear combination of columns \n\nSuppose is an -by- matrix and = 1\n\n \n\n\n\nis an -by-1 matrix. Then \n\n = 1 ,1 + + , .\n\nIn other words, is a linear combination of the columns of , with the scalars that multiply the columns coming from . \n\n> Proof\n\nIf {1 , , } , then the definition of matrix multiplication implies that the entry in row of the -by-1 matrix is \n\n ,1 1 + + , .\n\nThe entry in row of 1 ,1 + + , also equals the number displayed above. Because and 1 ,1 + + , have the same entry in row for each \n\n {1 , , } , we conclude that = 1 ,1 + + , .Our two previous results focus on the columns of a matrix. Analogous results hold for the rows of a matrix. Specifically, see Exercises 8 and 9, which can be proved using appropriate modifications of the proofs of 3.48 and 3.50. The next result is the main tool used in the next subsection to prove the columnrow factorization (3.56) and to prove that the column rank of a matrix equals the row rank (3.57). To be consistent with the notation often used with the columnrow factorization, including in the next subsection, the matrices in the next result are called and instead of and . \n\n> 3.51\n\nmatrix multiplication as linear combinations of columns or rows \n\nSuppose is an -by- matrix and is a -by- matrix. (a) If {1 , , } , then column of is a linear combination of the columns of , with the coefficients of this linear combination coming from column of .(b) If {1 , , } , then row of is a linear combination of the rows of \n\n , with the coefficients of this linear combination coming from row of \n\n . \n\n> Proof\n\nSuppose {1 , , } . Then column of equals , (by 3.48), which equals the linear combination of the columns of with coefficients coming from , (by 3.50). Thus (a) holds. To prove (b), follow the pattern of the proof of (a) but use rows instead of columns and use Exercises 8 and 9 instead of 3.48 and 3.50. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3C Matrices 77 \n\n# ColumnRow Factorization and Rank of a Matrix \n\nWe begin by defining two nonnegative integers associated with each matrix. \n\n> 3.52 definition:\n\ncolumn rank, row rank \n\nSuppose is an -by- matrix with entries in .\n\n The column rank of is the dimension of the span of the columns of \n\nin ,1.\n\n The row rank of is the dimension of the span of the rows of in 1, .If is an -by- matrix, then the column rank of is at most (because has \n\n columns) and the column rank of is also at most (because dim ,1 = ). Similarly, the row rank of is also at most min { , } . \n\n> 3.53 example:\n\ncolumn rank and row rank of a 2-by-4 matrix \n\nSuppose \n\n = 4 7 1 83 5 2 9 .\n\nThe column rank of is the dimension of \n\nspan 43 , 75 , 12 , 89 \n\nin 2,1. Neither of the first two vectors listed above in 2,1 is a scalar multiple of the other. Thus the span of this list of length four has dimension at least two. The span of this list of vectors in 2,1 cannot have dimension larger than two because \n\ndim 2,1 = 2 . Thus the span of this list has dimension two, which means that the column rank of is two. The row rank of is the dimension of \n\nspan (( 4 7 1 8 ) , ( 3 5 2 9 )) \n\nin 1,4. Neither of the two vectors listed above in 1,4 is a scalar multiple of the other. Thus the span of this list of length two has dimension two, which means that the row rank of is two. We now define the transpose of a matrix. \n\n> 3.54 definition:\n\ntranspose, t\n\nThe transpose of a matrix , denoted by t, is the matrix obtained from by interchanging rows and columns. Specifically, if is an -by- matrix, then \n\n t is the -by- matrix whose entries are given by the equation \n\n( t) , = , . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 78 Chapter 3 Linear Maps\n> 3.55 example:\n\ntranspose of a matrix \n\nIf = 5 7 3 84 2\n\n, then t = 5 3 4 7 8 2 .Note that here is a 3-by-2 matrix and t is a 2-by-3 matrix. The transpose has nice algebraic properties: ( + ) t = t + t, () t = t,and () t = t t for all -by- matrices , , all , and all -by- matrices \n\n (see Exercises 14 and 15). The next result will be the main tool used to prove that the column rank equals the row rank (see 3.57). \n\n> 3.56\n\ncolumnrow factorization \n\nSuppose is an -by- matrix with entries in and column rank 1 . Then there exist an -by- matrix and a -by- matrix , both with entries in ,such that = . \n\n> Proof\n\nEach column of is an -by-1 matrix. The list ,1, , , of columns of can be reduced to a basis of the span of the columns of (by 2.30). This basis has length , by the definition of the column rank. The columns in this basis can be put together to form an -by- matrix .If {1 , , } , then column of is a linear combination of the columns of . Make the coefficients of this linear combination into column of a -by- \n\nmatrix that we call . Then = , as follows from 3.51(a). In Example 3.53, the column rank and row rank turned out to equal each other. The next result states that this happens for all matrices. \n\n> 3.57\n\ncolumn rank equals row rank \n\nSuppose , . Then the column rank of equals the row rank of . \n\n> Proof\n\nLet denote the column rank of . Let = be the columnrow factorization of given by 3.56, where is an -by- matrix and is a -by- \n\nmatrix. Then 3.51(b) tells us that every row of is a linear combination of the rows of . Because has rows, this implies that the row rank of is less than or equal to the column rank of .To prove the inequality in the other direction, apply the result in the previous paragraph to t, getting column rank of = row rank of t\n\n column rank of t\n\n= row rank of . \n\nThus the column rank of equals the row rank of .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3C Matrices 79 \n\nBecause the column rank equals the row rank, the last result allows us to dispense with the terms column rank and row rank and just use the simpler term rank. \n\n> 3.58 definition:\n\nrank \n\nThe rank of a matrix , is the column rank of .See 3.133 and Exercise 8 in Section 7A for alternative proofs that the column rank equals the row rank. \n\n# Exercises 3C \n\n1 Suppose ( , ) . Show that with respect to each choice of bases of \n\nand , the matrix of has at least dim range nonzero entries. \n\n2 Suppose ( , ) , where and are finite-dimensional and nonzero. Prove that dim range = 1 if and only if there exist a basis of and a basis of such that with respect to these bases, all entries of () equal 1.\n\n3 Suppose 1, , is a basis of and 1, , is a basis of .(a) Show that if , ( , ) , then ( + ) = () + () .(b) Show that if and ( , ) , then () = () .\n\nThis exercise asks you to verify 3.35 and 3.38. \n\n4 Suppose that ( 3() , 2()) is the differentiation map defined by \n\n = . Find a basis of 3() and a basis of 2() such that the matrix of \n\n with respect to these bases is \n\n1 0 0 00 1 0 00 0 1 0.\n\nCompare with Example 3.33. The next exercise generalizes this exercise. \n\n5 Suppose and are finite-dimensional and ( , ) . Prove that there exist a basis of and a basis of such that with respect to these bases, all entries of () are 0 except that the entries in row , column , equal 1 if \n\n1 dim range .\n\n6 Suppose 1, , is a basis of and is finite-dimensional. Suppose \n\n ( , ) . Prove that there exists a basis 1, , of such that all entries in the first column of () [with respect to the bases 1, , and \n\n 1, , ] are 0 except for possibly a 1 in the first row, first column. \n\nIn this exercise, unlike Exercise 5, you are given the basis of instead of being able to choose a basis of .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n80 Chapter 3 Linear Maps \n\n7 Suppose 1, , is a basis of and is finite-dimensional. Suppose \n\n ( , ) . Prove that there exists a basis 1, , of such that all entries in the first row of () [with respect to the bases 1, , and \n\n 1, , ] are 0 except for possibly a 1 in the first row, first column. \n\nIn this exercise, unlike Exercise 5, you are given the basis of instead of being able to choose a basis of .\n\n8 Suppose is an -by- matrix and is an -by- matrix. Prove that \n\n() , = , \n\nfor each 1 . In other words, show that row of equals (row of )times .\n\nThis exercise gives the row version of 3.48. \n\n9 Suppose = ( 1 ) is a 1-by- matrix and is an -by- matrix. Prove that \n\n = 1 1, + + , .\n\nIn other words, show that is a linear combination of the rows of , with the scalars that multiply the rows coming from .\n\nThis exercise gives the row version of 3.50. \n\n10 Give an example of 2-by-2 matrices and such that .\n\n11 Prove that the distributive property holds for matrix addition and matrix multiplication. In other words, suppose , , , , , and are matrices whose sizes are such that ( + ) and ( + ) make sense. Explain why \n\n + and + both make sense and prove that \n\n( + ) = + and ( + ) = + . \n\n12 Prove that matrix multiplication is associative. In other words, suppose , ,and are matrices whose sizes are such that () makes sense. Explain why () makes sense and prove that \n\n() = (). \n\nTry to find a clean proof that illustrates the following quote from Emil Artin: It is my experience that proofs involving matrices can be shortened by 50% if one throws the matrices out. \n\n13 Suppose is an -by- matrix and 1 , . Show that the entry in row , column , of 3 (which is defined to mean ) is \n\n> \n> \n> = 1 \n> \n> = 1\n\n , , , .\n\n14 Suppose and are positive integers. Prove that the function t is a linear map from , to , .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3C Matrices 81 \n\n15 Prove that if is an -by- matrix and is an -by- matrix, then \n\n() t = t t.\n\nThis exercise shows that the transpose of the product of two matrices is the product of the transposes in the opposite order. \n\n16 Suppose is an -by- matrix with 0 . Prove that the rank of is 1\n\nif and only if there exist ( 1, , ) and ( 1, , ) such that \n\n , = for every = 1 , , and every = 1 , , .\n\n17 Suppose () , and 1, , and 1, , are bases of . Prove that the following are equivalent. (a) is injective. (b) The columns of () are linearly independent in ,1.(c) The columns of () span ,1.(d) The rows of () span 1, .(e) The rows of () are linearly independent in 1, .Here () means ( , ( 1, , ), ( 1, , )) .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n82 Chapter 3 Linear Maps \n\n# 3D Invertibility and Isomorphisms \n\n# Invertible Linear Maps \n\nWe begin this section by defining the notions of invertible and inverse in the context of linear maps. \n\n> 3.59 definition:\n\ninvertible, inverse \n\n A linear map ( , ) is called invertible if there exists a linear map \n\n ( , ) such that equals the identity operator on and equals the identity operator on .\n\n A linear map ( , ) satisfying = and = is called an \n\ninverse of (note that the first is the identity operator on and the second \n\n is the identity operator on ). The definition above mentions an inverse. However, the next result shows that we can change this terminology to the inverse. \n\n> 3.60\n\ninverse is unique \n\nAn invertible linear map has a unique inverse. \n\nProof Suppose ( , ) is invertible and 1 and 2 are inverses of . Then \n\n 1 = 1 = 1( 2) = ( 1) 2 = 2 = 2.\n\nThus 1 = 2.Now that we know that the inverse is unique, we can give it a notation. \n\n> 3.61 notation:\n\n 1 \n\nIf is invertible, then its inverse is denoted by 1 . In other words, if \n\n ( , ) is invertible, then 1 is the unique element of ( , ) such that 1 = and 1 = . \n\n> 3.62 example:\n\ninverse of a linear map from 3 to 3\n\nSuppose ( 3) is defined by ( , , ) = ( , , 4) . Thus is a counterclockwise rotation by 90 in the -plane and a stretch by a factor of 4 in the direction of the -axis. Hence the inverse map 1 ( 3) is the clockwise rotation by 90 in the \n\n -plane and a stretch by a factor of 14 in the direction of the -axis: \n\n 1 ( , , ) = ( , , 14 ). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 3D Invertibility and Isomorphisms 83\n\nThe next result shows that a linear map is invertible if and only if it is one-to-one and onto. \n\n> 3.63\n\ninvertibility injectivity and surjectivity \n\nA linear map is invertible if and only if it is injective and surjective. \n\n> Proof\n\nSuppose ( , ) . We need to show that is invertible if and only if it is injective and surjective. First suppose is invertible. To show that is injective, suppose , \n\nand = . Then \n\n = 1 () = 1 () = ,\n\nso = . Hence is injective. We are still assuming that is invertible. Now we want to prove that is surjective. To do this, let . Then = ( 1 ) , which shows that is in the range of . Thus range = . Hence is surjective, completing this direction of the proof. Now suppose is injective and surjective. We want to prove that is invertible. For each , define () to be the unique element of such that (()) = (the existence and uniqueness of such an element follow from the surjectivity and injectivity of ). The definition of implies that equals the identity operator on .To prove that equals the identity operator on , let . Then \n\n(( )) = ( )() = () = . \n\nThis equation implies that ( ) = (because is injective). Thus equals the identity operator on .To complete the proof, we need to show that is linear. To do this, suppose \n\n 1, 2 . Then \n\n(( 1) + ( 2)) = (( 1)) + (( 2)) = 1 + 2.\n\nThus ( 1) + ( 2) is the unique element of that maps to 1 + 2. By the definition of , this implies that ( 1 + 2) = ( 1) + ( 2). Hence satisfies the additive property required for linearity. The proof of homogeneity is similar. Specifically, if and , then \n\n(()) = (()) = . \n\nThus () is the unique element of that maps to . By the definition of ,this implies that () = () . Hence is linear, as desired. For a linear map from a vector space to itself, you might wonder whether injectivity alone, or surjectivity alone, is enough to imply invertibility. On infinite-dimensional vector spaces, neither condition alone implies invertibility, as illus-trated by the next example, which uses two familiar linear maps from Example 3.3. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 84 Chapter 3 Linear Maps\n> 3.64 example:\n\nneither injectivity nor surjectivity implies invertibility \n\n The multiplication by 2 linear map from () to () (see 3.3) is injective but it is not invertible because it is not surjective (the polynomial 1 is not in the range). \n\n The backward shift linear map from to (see 3.3) is surjective but it is not invertible because it is not injective [the vector (1 , 0, 0, 0, ) is in the null space ].In view of the example above, the next result is remarkableit states that for a linear map from a finite-dimensional vector space to a vector space of the same dimension, either injectivity or surjectivity alone implies the other condition. Note that the hypothesis below that dim = dim is automatically satisfied in the important special case where is finite-dimensional and = . \n\n> 3.65\n\ninjectivity is equivalent to surjectivity (if dim = dim < )Suppose that and are finite-dimensional vector spaces, dim = dim ,and ( , ) . Then \n\n is invertible is injective is surjective . \n\n> Proof\n\nThe fundamental theorem of linear maps (3.21) states that \n\n> 3.66\n\ndim = dim null + dim range . \n\nIf is injective (which by 3.15 is equivalent to the condition dim null = 0 ), then the equation above implies that \n\ndim range = dim dim null = dim = dim ,\n\nwhich implies that is surjective (by 2.39). Conversely, if is surjective, then 3.66 implies that \n\ndim null = dim dim range = dim dim = 0 ,\n\nwhich implies that is injective. Thus we have shown that is injective if and only if is surjective. Thus if \n\n is either injective or surjective, then is both injective and surjective, which implies that is invertible. Hence is invertible if and only if is injective if and only if is surjective. The next example illustrates the power of the previous result. Although it is possible to prove the result in the example below without using linear algebra, the proof using linear algebra is cleaner and easier. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 3D Invertibility and Isomorphisms 85\n> 3.67 example:\n\nthere exists a polynomial such that (( 2 + 5 + 7)) = \n\nThe linear map \n\n (( 2 + 5 + 7)) \n\nfrom () to itself is injective, as you can show. Thus we are tempted to use 3.65 to show that this map is surjective. However, Example 3.64 shows that the magic of 3.65 does not apply to the infinite-dimensional vector space () . We will get around this problem by restricting attention to the finite-dimensional vector space () .Suppose () . There exists a nonnegative integer such that () .Define () () by \n\n = (( 2 + 5 + 7)) .\n\nMultiplying a nonzero polynomial by ( 2 + 5 + 7) increases the degree by 2, and then differentiating twice reduces the degree by 2. Thus is indeed a linear map from () to itself. Every polynomial whose second derivative equals 0 is of the form + ,where , . Thus null = {0} . Hence is injective. Thus is surjective (by 3.65), which means that there exists a polynomial \n\n () such that (( 2 + 5 + 7)) = , as claimed in the title of this example. Exercise 35 in Section 6A gives a similar but more spectacular example of using 3.65. The hypothesis in the result below that dim = dim holds in the important special case in which is finite-dimensional and = . Thus in that case, the equation = implies that = , even though we do not have multiplicative commutativity of arbitrary linear maps from to . \n\n> 3.68\n\n = = (on vector spaces of the same dimension )Suppose and are finite-dimensional vector spaces of the same dimension, \n\n ( , ) , and ( , ) . Then = if and only if = . \n\n> Proof\n\nFirst suppose = . If and = 0 , then \n\n = = () = () = (0) = 0. \n\nThus is injective (by 3.15). Because and have the same dimension, this implies that is invertible (by 3.65). Now multiply both sides of the equation = by 1 on the right, getting \n\n = 1 .\n\nThus = 1 = , as desired. To prove the implication in the other direction, simply reverse the roles of \n\nand (and and ) in the direction we have already proved, showing that if \n\n = , then = . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 86 Chapter 3 Linear Maps\n\n# Isomorphic Vector Spaces \n\nThe next definition captures the idea of two vector spaces that are essentially the same, except for the names of their elements. \n\n> 3.69 definition:\n\nisomorphism, isomorphic \n\n An isomorphism is an invertible linear map. \n\n Two vector spaces are called isomorphic if there is an isomorphism from one vector space onto the other one. Think of an isomorphism as relabeling as . This viewpoint explains why two isomorphic vector spaces have the same vector space properties. The terms isomorphism and invertible linear map mean the same thing. Use isomorphism when you want to emphasize that the two spaces are essentially the same. It can be difficult to determine whether two mathematical structures (such as groups or topological spaces) are essentially the same, differing only in the names of the elements of underlying sets. However, the next result shows that we need to look at only a single number (the dimension) to determine whether two vector spaces are isomorphic. \n\n> 3.70\n\ndimension shows whether vector spaces are isomorphic \n\nTwo finite-dimensional vector spaces over are isomorphic if and only if they have the same dimension. \n\n> Proof\n\nFirst suppose and are isomorphic finite-dimensional vector spaces. Thus there exists an isomorphism from onto . Because is invertible, we have null = {0} and range = . Thus \n\ndim null = 0 and dim range = dim . \n\nThe formula \n\ndim = dim null + dim range \n\n(the fundamental theorem of linear maps, which is 3.21) thus becomes the equation \n\ndim = dim , completing the proof in one direction. To prove the other direction, suppose and are finite-dimensional vector spaces of the same dimension. Let 1, , be a basis of and 1, , be a basis of . Let ( , ) be defined by \n\n( 1 1 + + ) = 1 1 + + .\n\nThen is a well-defined linear map because 1, , is a basis of . Also, \n\nis surjective because 1, , spans . Furthermore, null = {0} because \n\n 1, , is linearly independent. Thus is injective. Because is injective and surjective, it is an isomorphism (see 3.63). Hence and are isomorphic. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3D Invertibility and Isomorphisms 87 \n\nEvery finite-dimensional vector space is isomorphic to some . Thus why not just study instead of more general vector spaces? To answer this ques-tion, note that an investigation of \n\nwould soon lead to other vector spaces. For example, we would encounter the null space and range of linear maps. Although each of these vector spaces is isomorphic to some , thinking of them that way often adds complexity but no new insight. \n\nThe previous result implies that each finite-dimensional vector space is iso-morphic to , where = dim . For example, if is a nonnegative integer, then () is isomorphic to + 1.Recall that the notation , denotes the vector space of -by- matrices with entries in . If 1, , is a basis of \n\nand 1, , is a basis of , then for each ( , ) , we have a matrix \n\n() , . Thus once bases have been fixed for and , becomes a function from ( , ) to , . Notice that 3.35 and 3.38 show that is a lin-ear map. This linear map is actually an isomorphism, as we now show. \n\n> 3.71\n\n( , ) and , are isomorphic \n\nSuppose 1, , is a basis of and 1, , is a basis of . Then is an isomorphism between ( , ) and , .\n\nProof We already noted that is linear. We need to prove that is injective and surjective. We begin with injectivity. If ( , ) and () = 0 , then = 0 for each = 1 , , . Because 1, , is a basis of , this implies = 0 . Thus \n\nis injective (by 3.15). To prove that is surjective, suppose , . By the linear map lemma (3.4), there exists ( , ) such that \n\n =\n\n> \n> \n> = 1\n\n , \n\nfor each = 1 , , . Because () equals , the range of equals , , as desired. Now we can determine the dimension of the vector space of linear maps from one finite-dimensional vector space to another. \n\n> 3.72\n\ndim ( , ) = ( dim )( dim ) \n\nSuppose and are finite-dimensional. Then ( , ) is finite-dimensional and \n\ndim ( , ) = ( dim )( dim ). \n\nProof The desired result follows from 3.71, 3.70, and 3.40. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n88 Chapter 3 Linear Maps \n\n# Linear Maps Thought of as Matrix Multiplication \n\nPreviously we defined the matrix of a linear map. Now we define the matrix of a vector. \n\n> 3.73 definition:\n\nmatrix of a vector, () \n\nSuppose and 1, , is a basis of . The matrix of with respect to this basis is the -by-1 matrix \n\n() = 1\n\n \n\n\n\n,\n\nwhere 1, , are the scalars such that \n\n = 1 1 + + .\n\nThe matrix () of a vector depends on the basis 1, , of , as well as on . However, the basis should be clear from the context and thus it is not included in the notation. \n\n> 3.74 example:\n\nmatrix of a vector \n\n The matrix of the polynomial 2 7 + 5 3 + 4 with respect to the standard basis of 4() is \n\n27 051. The matrix of a vector with respect to the standard basis is obtained by writing the coordinates of as the entries in an -by-1 matrix. In other words, if = ( 1, , ) , then \n\n() = 1\n\n \n\n.\n\nOccasionally we want to think of elements of as relabeled to be -by-1\n\nmatrices. Once a basis 1, , is chosen, the function that takes to \n\n() is an isomorphism of onto ,1 that implements this relabeling. Recall that if is an -by- matrix, then , denotes the th column of ,thought of as an -by-1 matrix. In the next result, ( ) is computed with respect to the basis 1, , of .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3D Invertibility and Isomorphisms 89 \n\n> 3.75\n\n() , = ( ).\n\nSuppose ( , ) and 1, , is a basis of and 1, , is a basis of . Let 1 . Then the th column of () , which is denoted by \n\n() , , equals ( ).\n\nProof The desired result follows immediately from the definitions of () and \n\n( ).The next result shows how the notions of the matrix of a linear map, the matrix of a vector, and matrix multiplication fit together. \n\n> 3.76\n\nlinear maps act like matrix multiplication \n\nSuppose ( , ) and . Suppose 1, , is a basis of and \n\n 1, , is a basis of . Then \n\n() = () (). \n\nProof Suppose = 1 1 + + , where 1, , . Thus \n\n> 3.77\n\n = 1 1 + + .\n\nHence \n\n() = 1( 1) + + ( )= 1() ,1 + + () , \n\n= () () ,\n\nwhere the first equality follows from 3.77 and the linearity of , the second equality comes from 3.75, and the last equality comes from 3.50. Each -by- matrix induces a linear map from ,1 to ,1, namely the matrix multiplication function that takes ,1 to ,1. The result above can be used to think of every linear map (from a finite-dimensional vector space to another finite-dimensional vector space) as a matrix multiplication map after suitable relabeling via the isomorphisms given by . Specifically, if ( , ) \n\nand we identify with () ,1, then the result above says that we can identify with () () .Because the result above allows us to think (via isomorphisms) of each linear map as multiplication on ,1 by some matrix , keep in mind that the specific matrix depends not only on the linear map but also on the choice of bases. One of the themes of many of the most important results in later chapters will be the choice of a basis that makes the matrix as simple as possible. In this book, we concentrate on linear maps rather than on matrices. However, sometimes thinking of linear maps as matrices (or thinking of matrices as linear maps) gives important insights that we will find useful. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n90 Chapter 3 Linear Maps \n\nNotice that no bases are in sight in the statement of the next result. Although \n\n() in the next result depends on a choice of bases of and , the next result shows that the column rank of () is the same for all such choices (because \n\nrange does not depend on a choice of basis). \n\n> 3.78\n\ndimension of range equals column rank of () \n\nSuppose and are finite-dimensional and ( , ) . Then dim range \n\nequals the column rank of () .\n\nProof Suppose 1, , is a basis of and 1, , is a basis of . The linear map that takes to () is an isomorphism from onto the space ,1\n\nof -by-1 column vectors. The restriction of this isomorphism to range [ which equals span ( 1, , ) by Exercise 10 in Section 3B ] is an isomorphism from \n\nrange onto span (( 1), , ( )) . For each {1 , , } , the -by-1\n\nmatrix ( ) equals column of () . Thus \n\ndim range = the column rank of () ,\n\nas desired. \n\n# Change of Basis \n\nIn Section 3C we defined the matrix \n\n( , ( 1, , ), ( 1, , )) \n\nof a linear map from to a possibly different vector space , where 1, , \n\nis a basis of and 1, , is a basis of . For linear maps from a vector space to itself, we usually use the same basis for both the domain vector space and the target vector space. When using a single basis in both capacities, we often write the basis only once. In other words, if () and 1, , is a basis of ,then the notation ( , ( 1, , )) is defined by the equation \n\n( , ( 1, , )) = ( , ( 1, , ), ( 1, , )). \n\nIf the basis 1, , is clear from the context, then we can write just () . \n\n> 3.79 definition:\n\nidentity matrix, I \n\nSuppose is a positive integer. The -by- matrix \n\n1 00 1\n\nwith 1s on the diagonal (the entries where the row number equals the column number) and 0s elsewhere is called the identity matrix and is denoted by .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3D Invertibility and Isomorphisms 91 \n\nIn the definition above, the 0 in the lower left corner of the matrix indicates that all entries below the diagonal are 0, and the 0 in the upper right corner indicates that all entries above the diagonal are 0.With respect to each basis of , the matrix of the identity operator () \n\nis the identity matrix . Note that the symbol is used to denote both the identity operator and the identity matrix. The context indicates which meaning of is intended. For example, consider the equation () = ; on the left side denotes the identity operator, and on the right side denotes the identity matrix. If is a square matrix (meaning it has the same number of rows as columns) with the same size as , then = = , as you should verify. \n\n> 3.80 definition:\n\ninvertible, inverse, 1 \n\nA square matrix is called invertible if there is a square matrix of the same size such that = = ; we call the inverse of and denote it by 1 .\n\nSome mathematicians use the terms \n\nnonsingular and singular , which mean the same as invertible and non-invertible. \n\nThe same proof as used in 3.60 shows that if is an invertible square matrix, then there is a unique matrix such that \n\n = = (and thus the notation \n\n = 1 is justified). If is an invertible matrix, then ( 1 )1 = because \n\n 1 = 1 = . \n\nAlso, if and are invertible square matrices of the same size, then is invertible and () 1 = 1 1 because \n\n()( 1 1 ) = ( 1 ) 1 \n\n= 1 \n\n= 1 \n\n= ,\n\nand similarly ( 1 1 )() = .The next result holds because we defined matrix multiplication to make it truesee 3.43 and the material preceding it. Now we are just being more explicit about the bases involved. \n\n> 3.81\n\nmatrix of product of linear maps \n\nSuppose ( , ) and ( , ) . If 1, , is a basis of ,\n\n 1, , is a basis of , and 1, , is a basis of , then \n\n( , ( 1, , ), ( 1, , )) = \n\n( , ( 1, , ), ( 1, , )) ( , ( 1, , ), ( 1, , )). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n92 Chapter 3 Linear Maps \n\nThe next result deals with the matrix of the identity operator with respect to two different bases. Note that the th column of ( , ( 1, , ), ( 1, , )) \n\nconsists of the scalars needed to write as a linear combination of the basis \n\n 1, , .In the statement of the next result, denotes the identity operator from to .In the proof, also denotes the -by- identity matrix. \n\n> 3.82\n\nmatrix of identity operator with respect to two bases \n\nSuppose that 1, , and 1, , are bases of . Then the matrices \n\n( , ( 1, , ), ( 1, , )) and ( , ( 1, , ), ( 1, , )) \n\nare invertible, and each is the inverse of the other. \n\nProof In 3.81, replace with , and replace and with , getting \n\n = ( , ( 1, , ), ( 1, , )) ( , ( 1, , ), ( 1, , )). \n\nNow interchange the roles of the s and s, getting \n\n = ( , ( 1, , ), ( 1, , )) ( , ( 1, , ), ( 1, , )). \n\nThese two equations above give the desired result. \n\n> 3.83 example:\n\nmatrix of identity operator on 2 with respect to two bases \n\nConsider the bases (4 , 2) , (5 , 3) and (1 , 0) , (0 , 1) of 2. Because (4 , 2) = 4(1 , 0) + 2(0 , 1) and (5 , 3) = 5(1 , 0) + 3(0 , 1) , we have \n\n( , ((4 , 2) , (5 , 3)) , ((1 , 0) , (0 , 1))) = 4 52 3 .\n\nThe inverse of the matrix above is \n\n \n\n> 32\n\n 52\n\n1 2\n\n,\n\nas you should verify. Thus 3.82 implies that \n\n( , ((1 , 0) , (0 , 1)) , ((4 , 2) , (5 , 3))) = \n\n> 32\n\n 52\n\n1 2.\n\nOur next result shows how the matrix of changes when we change bases. In the next result, we have two different bases of , each of which is used as a basis for the domain space and as a basis for the target space. Recall our shorthand notation that allows us to display a basis only once when it is used in both capacities: \n\n( , ( 1, , )) = ( , ( 1, , ), ( 1, , )). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3D Invertibility and Isomorphisms 93 \n\n> 3.84\n\nchange-of-basis formula \n\nSuppose () . Suppose 1, , and 1, , are bases of . Let \n\n = ( , ( 1, , )) and = ( , ( 1, , )) \n\nand = ( , ( 1, , ), ( 1, , )) . Then \n\n = 1 . \n\nProof In 3.81, replace with and replace with , getting \n\n> 3.85\n\n = 1 ( , ( 1, , ), ( 1, , )) ,\n\nwhere we have used 3.82. Again use 3.81, this time replacing with . Also replace with and replace with , getting \n\n( , ( 1, , ), ( 1, , )) = . \n\nSubstituting the equation above into 3.85 gives the equation = 1 .The proof of the next result is left as an exercise. \n\n> 3.86\n\nmatrix of inverse equals inverse of matrix \n\nSuppose that 1, , is a basis of and () is invertible. Then \n\n( 1 ) = ( ()) 1 , where both matrices are with respect to the basis \n\n 1, , .\n\n# Exercises 3D \n\n1 Suppose ( , ) is invertible. Show that 1 is invertible and \n\n( 1 )1 = . \n\n2 Suppose ( , ) and ( , ) are both invertible linear maps. Prove that ( , ) is invertible and that () 1 = 1 1 .\n\n3 Suppose is finite-dimensional and () . Prove that the following are equivalent. (a) is invertible. (b) 1, , is a basis of for every basis 1, , of .(c) 1, , is a basis of for some basis 1, , of .\n\n4 Suppose is finite-dimensional and dim > 1 . Prove that the set of noninvertible linear maps from to itself is not a subspace of () .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n94 Chapter 3 Linear Maps \n\n5 Suppose is finite-dimensional, is a subspace of , and ( , ) .Prove that there exists an invertible linear map from to itself such that \n\n = for every if and only if is injective. \n\n6 Suppose that is finite-dimensional and , ( , ) . Prove that \n\nnull = null if and only if there exists an invertible () such that \n\n = .\n\n7 Suppose that is finite-dimensional and , ( , ) . Prove that \n\nrange = range if and only if there exists an invertible () such that = .\n\n8 Suppose and are finite-dimensional and , ( , ) . Prove that there exist invertible 1 () and 2 () such that = 2 1 if and only if dim null = dim null .\n\n9 Suppose is finite-dimensional and is a surjective linear map of onto . Prove that there is a subspace of such that | is an isomorphism of onto .\n\nHere | means the function restricted to . Thus | is the function whose domain is , with | defined by | () = for every .\n\n10 Suppose and are finite-dimensional and is a subspace of . Let \n\n = { ( , ) null }. \n\n(a) Show that is a subspace of ( , ) .(b) Find a formula for dim in terms of dim , dim , and dim .\n\nHint: Define ( , ) ( , ) by () = | . What is null ?What is range ?\n\n11 Suppose is finite-dimensional and , () . Prove that \n\n is invertible and are invertible .\n\n12 Suppose is finite-dimensional and , , () and = . Show that is invertible and that 1 = .\n\n13 Show that the result in Exercise 12 can fail without the hypothesis that is finite-dimensional. \n\n14 Prove or give a counterexample: If is a finite-dimensional vector space and , , () are such that is surjective, then is injective. \n\n15 Suppose () and 1, , is a list in such that 1, , \n\nspans . Prove that 1, , spans .\n\n16 Prove that every linear map from ,1 to ,1 is given by a matrix multipli-cation. In other words, prove that if ( ,1, ,1), then there exists an \n\n -by- matrix such that = for every ,1.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3D Invertibility and Isomorphisms 95 \n\n17 Suppose is finite-dimensional and () . Define (()) by \n\n () = \n\nfor () .(a) Show that dim null = ( dim )( dim null ) .(b) Show that dim range = ( dim )( dim range ) .\n\n18 Show that and ( , ) are isomorphic vector spaces. \n\n19 Suppose is finite-dimensional and () . Prove that has the same matrix with respect to every basis of if and only if is a scalar multiple of the identity operator. \n\n20 Suppose () . Prove that there exists a polynomial () such that \n\n() = ( 2 + ) () + 2 () + (3) \n\nfor all .\n\n21 Suppose is a positive integer and , for all , = 1 , , . Prove that the following are equivalent (note that in both parts below, the number of equations equals the number of variables). (a) The trivial solution 1 = = = 0 is the only solution to the homogeneous system of equations \n\n> \n> \n> = 1\n\n 1, = 0 \n\n> \n> \n> = 1\n\n , = 0. \n\n(b) For every 1, , , there exists a solution to the system of equations \n\n> \n> \n> = 1\n\n 1, = 1\n\n\n\n> \n> \n> = 1\n\n , = .\n\n22 Suppose () and 1, , is a basis of . Prove that \n\n( , ( 1, , )) is invertible is invertible .\n\n23 Suppose that 1, , and 1, , are bases of . Let () be such that = for each = 1 , , . Prove that \n\n( , ( 1, , )) = ( , ( 1, , ), ( 1, , )). \n\n24 Suppose and are square matrices of the same size and = . Prove that = .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n96 Chapter 3 Linear Maps \n\n# 3E Products and Quotients of Vector Spaces \n\n# Products of Vector Spaces \n\nAs usual when dealing with more than one vector space, all vector spaces in use should be over the same field. \n\n> 3.87 definition:\n\nproduct of vector spaces \n\nSuppose 1, , are vector spaces over .\n\n The product 1 is defined by \n\n 1 = {( 1, , ) 1 1, , }. Addition on 1 is defined by \n\n( 1, , ) + ( 1, , ) = ( 1 + 1, , + ). Scalar multiplication on 1 is defined by \n\n( 1, , ) = ( 1, , ). \n\n> 3.88 example:\n\nproduct of the vector spaces 5() and 3\n\nElements of 5() 3 are lists of length two, with the first item in the list an element of 5() and the second item in the list an element of 3.For example, (5 6 + 4 2, (3 , 8, 7)) and ( + 9 5, (2 , 2, 2)) are elements of \n\n 5() 3. Their sum is defined by \n\n(5 6 + 4 2, (3 , 8, 7)) + ( + 9 5, (2 , 2, 2)) = (5 5 + 4 2 + 9 5, (5 , 10 , 9)). \n\nAlso, 2(5 6 + 4 2, (3 , 8, 7)) = (10 12 + 8 2, (6 , 16 , 14)) .The next result should be interpreted to mean that the product of vector spaces is a vector space with the operations of addition and scalar multiplication as defined by 3.87. \n\n> 3.89\n\nproduct of vector spaces is a vector space \n\nSuppose 1, , are vector spaces over . Then 1 is a vector space over .The proof of the result above is left to the reader. Note that the additive identity of 1 is (0 , , 0) , where the 0 in the th slot is the additive identity of .The additive inverse of ( 1, , ) 1 is ( 1, , ).\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3E Products and Quotients of Vector Spaces 97 \n\n> 3.90 example:\n\n 2 3 5 but 2 3 is isomorphic to 5\n\nElements of the vector space 2 3 are lists \n\n(( 1, 2), ( 3, 4, 5)) ,\n\nwhere 1, 2, 3, 4, 5 . Elements of 5 are lists \n\n( 1, 2, 3, 4, 5),\n\nwhere 1, 2, 3, 4, 5 .Although elements of 2 3 and 5 look similar, they are not the same kind of object. Elements of 2 3 are lists of length two (with the first item itself a list of length two and the second item a list of length three), and elements of 5\n\nare lists of length five. Thus 2 3 does not equal 5.\n\nThis isomorphism is so natural that we should think of it as a relabel-ing. Some people informally say that \n\n> 2 3\n\nequals 5, which is not techni-cally correct but which captures the spirit of identification via relabeling. \n\nThe linear map \n\n(( 1, 2), ( 3, 4, 5)) ( 1, 2, 3, 4, 5)\n\nis an isomorphism of the vector space \n\n 2 3 onto the vector space 5. Thus these two vector spaces are isomorphic, al-though they are not equal. The next example illustrates the idea that we will use in the proof of 3.92. \n\n> 3.91 example:\n\na basis of 2() 2\n\nConsider this list of length five of elements of 2() 2:\n\n(1 , (0 , 0)) , ( , (0 , 0)) , ( 2, (0 , 0)) , (0 , (1 , 0)) , (0 , (0 , 1)). \n\nThe list above is linearly independent and it spans 2() 2. Thus it is a basis of 2() 2. \n\n> 3.92\n\ndimension of a product is the sum of dimensions \n\nSuppose 1, , are finite-dimensional vector spaces. Then 1 \n\nis finite-dimensional and \n\ndim ( 1 ) = dim 1 + + dim .\n\nProof Choose a basis of each . For each basis vector of each , consider the element of 1 that equals the basis vector in the th slot and 0 in the other slots. The list of all such vectors is linearly independent and spans 1 .Thus it is a basis of 1 . The length of this basis is dim 1 + + dim ,as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n98 Chapter 3 Linear Maps \n\nIn the next result, the map is surjective by the definition of 1 ++ . Thus the last word in the result below could be changed from injective to invertible. \n\n> 3.93\n\nproducts and direct sums \n\nSuppose that 1, , are subspaces of . Define a linear map \n\n 1 1 + + by \n\n( 1, , ) = 1 + + .\n\nThen 1 + + is a direct sum if and only if is injective. \n\nProof By 3.15, is injective if and only if the only way to write 0 as a sum \n\n 1 + + , where each is in , is by taking each equal to 0. Thus 1.45 shows that is injective if and only if 1 + + is a direct sum, as desired. \n\n> 3.94\n\na sum is a direct sum if and only if dimensions add up \n\nSuppose is finite-dimensional and 1, , are subspaces of . Then \n\n 1 + + is a direct sum if and only if \n\ndim ( 1 + + ) = dim 1 + + dim .\n\nProof The map in 3.93 is surjective. Thus by the fundamental theorem of linear maps (3.21), is injective if and only if \n\ndim ( 1 + + ) = dim ( 1 ). \n\nCombining 3.93 and 3.92 now shows that 1 + + is a direct sum if and only if \n\ndim ( 1 + + ) = dim 1 + + dim ,\n\nas desired. In the special case = 2 , an alternative proof that 1 + 2 is a direct sum if and only if dim ( 1 + 2) = dim 1 + dim 2 can be obtained by combining 1.46 and 2.43. \n\n# Quotient Spaces \n\nWe begin our approach to quotient spaces by defining the sum of a vector and a subset. \n\n> 3.95 notation:\n\n + \n\nSuppose and . Then + is the subset of defined by \n\n + = { + }. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3E Products and Quotients of Vector Spaces 99 \n\n> 3.96 example:\n\nsum of a vector and a one-dimensional subspace of 2\n\n> (17\n\n, 20) + is parallel to the subspace .\n\nSuppose \n\n = {( , 2) 2 }. \n\nHence is the line in 2 through the origin with slope 2. Thus \n\n(17 , 20) + \n\nis the line in 2 that contains the point (17 , 20) \n\nand has slope 2.Because \n\n(10 , 20) and (17 , 20) (17 , 20) + ,\n\nwe see that (17 , 20) + is obtained by moving \n\nto the right by 7 units. \n\n> 3.97 definition:\n\ntranslate \n\nFor and a subset of , the set + is said to be a translate of . \n\n> 3.98 example:\n\ntranslates \n\n If is the line in 2 defined by = {( , 2) 2 } , then all lines in \n\n 2 with slope 2 are translates of . See Example 3.96 above for a drawing of \n\n and one of its translates. \n\n More generally, if is a line in 2, then the set of all translates of is the set of all lines in 2 that are parallel to .\n\n If = {( , , 0) 3 , } , then the translates of are the planes in \n\n 3 that are parallel to the -plane .\n\n More generally, if is a plane in 3, then the set of all translates of is the set of all planes in 3 that are parallel to (see, for example, Exercise 7). \n\n> 3.99 definition:\n\nquotient space, / \n\nSuppose is a subspace of . Then the quotient space / is the set of all translates of . Thus \n\n/ = { + }. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 100 Chapter 3 Linear Maps\n> 3.100 example:\n\nquotient spaces \n\n If = {( , 2) 2 } , then 2/ is the set of all lines in 2 that have slope 2.\n\n If is a line in 3 containing the origin, then 3/ is the set of all lines in 3\n\nparallel to .\n\n If is a plane in 3 containing the origin, then 3/ is the set of all planes in \n\n 3 parallel to .Our next goal is to make / into a vector space. To do this, we will need the next result. \n\n> 3.101\n\ntwo translates of a subspace are equal or disjoint \n\nSuppose is a subspace of and , . Then \n\n + = + ( + ) ( + ) . \n\n> Proof\n\nFirst suppose . If , then \n\n + = + (( ) + ) + . \n\nThus + + . Similarly, + + . Thus + = + , completing the proof that implies + = + .The equation + = + implies that ( + ) ( + ) .Now suppose ( + ) ( + ) . Thus there exist 1, 2 such that \n\n + 1 = + 2.\n\nThus = 2 1. Hence , showing that ( + ) ( + ) \n\nimplies , which completes the proof. Now we can define addition and scalar multiplication on / . \n\n> 3.102 definition:\n\naddition and scalar multiplication on / \n\nSuppose is a subspace of . Then addition and scalar multiplication are defined on / by \n\n( + ) + ( + ) = ( + ) + ( + ) = () + \n\nfor all , and all .As part of the proof of the next result, we will show that the definitions above make sense. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 3E Products and Quotients of Vector Spaces 101\n> 3.103\n\nquotient space is a vector space \n\nSuppose is a subspace of . Then / , with the operations of addition and scalar multiplication as defined above, is a vector space. \n\n> Proof\n\nThe potential problem with the definitions above of addition and scalar multiplication on / is that the representation of a translate of is not unique. Specifically, suppose 1, 2, 1, 2 are such that \n\n 1 + = 2 + and 1 + = 2 + . \n\nTo show that the definition of addition on / given above makes sense, we must show that ( 1 + 1) + = ( 2 + 2) + .By 3.101, we have \n\n 1 2 and 1 2 . \n\nBecause is a subspace of and thus is closed under addition, this implies that \n\n( 1 2) + ( 1 2) . Thus ( 1 + 1) ( 2 + 2) . Using 3.101 again, we see that \n\n( 1 + 1) + = ( 2 + 2) + ,\n\nas desired. Thus the definition of addition on / makes sense. Similarly, suppose . We are still assuming that 1 + = 2 + .Because is a subspace of and thus is closed under scalar multiplication, we have ( 1 2) . Thus 1 2 . Hence 3.101 implies that \n\n( 1) + = ( 2) + . \n\nThus the definition of scalar multiplication on / makes sense. Now that addition and scalar multiplication have been defined on / , the verification that these operations make / into a vector space is straightforward and is left to the reader. Note that the additive identity of / is 0 + (which equals ) and that the additive inverse of + is () + .The next concept will lead to a computation of the dimension of / . \n\n> 3.104 definition:\n\nquotient map, \n\nSuppose is a subspace of . The quotient map / is the linear map defined by \n\n() = + \n\nfor each .The reader should verify that is indeed a linear map. Although depends on as well as , these spaces are left out of the notation because they should be clear from the context. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 102 Chapter 3 Linear Maps\n> 3.105\n\ndimension of quotient space \n\nSuppose is finite-dimensional and is a subspace of . Then \n\ndim / = dim dim . \n\n> Proof\n\nLet denote the quotient map from to / . If , then + = 0 + \n\nif and only if (by 3.101), which implies that null = . The definition of \n\n implies range = / . The fundamental theorem of linear maps (3.21) now implies dim = dim + dim / , which gives the desired result. Each linear map on induces a linear map on /( null ) , as defined below. \n\n> 3.106 notation:\n\n \n\nSuppose ( , ) . Define /( null ) by ( + null ) = . \n\nTo show that the definition of makes sense, suppose , are such that \n\n + null = + null . By 3.101, we have null . Thus ( ) = 0 .Hence = . Thus the definition of indeed makes sense. The routine verification that is a linear map from /( null ) to is left to the reader. The next result shows that we can think of as a modified version of , with a domain that produces a one-to-one map. \n\n> 3.107\n\nnull space and range of \n\nSuppose ( , ) . Then (a) = , where is the quotient map of onto /( null ) ;(b) is injective; (c) range = range ;(d) /( null ) and range are isomorphic vector spaces. \n\n> Proof\n\n(a) If , then ( )() = (()) = ( + null ) = , as desired. (b) Suppose and ( + null ) = 0 . Then = 0 . Thus null .Hence 3.101 implies that + null = 0 + null . This implies that null = {0 + null } . Hence is injective, as desired. (c) The definition of shows that range = range .(d) Now (b) and (c) imply that if we think of as mapping into range , then \n\nis an isomorphism from /( null ) onto range .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3E Products and Quotients of Vector Spaces 103 \n\n# Exercises 3E \n\n1 Suppose is a function from to . The graph of is the subset of \n\ndefined by graph of = {( , ) }. \n\nProve that is a linear map if and only if the graph of is a subspace of \n\n .\n\nFormally, a function from to is a subset of such that for each , there exists exactly one element ( , ) . In other words, formally a function is what is called above its graph. We do not usually think of functions in this formal manner. However, if we do become formal, then this exercise could be rephrased as follows: Prove that a function \n\nfrom to is a linear map if and only if is a subspace of .\n\n2 Suppose that 1, , are vector spaces such that 1 is finite-dimensional. Prove that is finite-dimensional for each = 1 , , .\n\n3 Suppose 1, , are vector spaces. Prove that ( 1 , ) and \n\n( 1, ) ( , ) are isomorphic vector spaces. \n\nThere is no assumption in the exercise above or in the two following exercises that the vector spaces are finite-dimensional. \n\n4 Suppose 1, , are vector spaces. Prove that ( , 1 ) and \n\n( , 1) ( , ) are isomorphic vector spaces. \n\n5 For a positive integer, define by \n\n = \n\n> times\n\n.\n\nProve that and ( , ) are isomorphic vector spaces. \n\n6 Suppose that , are vectors in and that , are subspaces of such that + = + . Prove that = .\n\n7 Let = {( , , ) 3 2 + 3 + 5 = 0} . Suppose 3. Prove that \n\n is a translate of if and only if there exists such that \n\n = {( , , ) 3 2 + 3 + 5 = }. \n\n8 (a) Suppose ( , ) and . Prove that { = } is either the empty set or is a translate of null .(b) Explain why the set of solutions to a system of linear equations such as 3.27 is either the empty set or is a translate of some subspace of .\n\n9 Prove that a nonempty subset of is a translate of some subspace of if and only if + (1 ) for all , and all .\n\n10 Suppose 1 = + 1 and 2 = + 2 for some , and some subspaces 1, 2 of . Prove that the intersection 1 2 is either a translate of some subspace of or is the empty set. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n104 Chapter 3 Linear Maps \n\n11 Suppose = {( 1, 2, ) 0 for only finitely many } .(a) Show that is a subspace of .(b) Prove that / is infinite-dimensional. \n\n12 Suppose 1, , . Let \n\n = { 1 1 + + 1, , and 1 + + = 1}. \n\n(a) Prove that is a translate of some subspace of .(b) Prove that if is a translate of some subspace of and { 1, , } ,then .(c) Prove that is a translate of some subspace of of dimension less than .\n\n13 Suppose is a subspace of such that / is finite-dimensional. Prove that is isomorphic to (/) .\n\n14 Suppose and are subspaces of and = . Suppose 1, , \n\nis a basis of . Prove that 1 + , , + is a basis of / .\n\n15 Suppose is a subspace of and 1 + , , + is a basis of / and \n\n 1, , is a basis of . Prove that 1, , , 1, , is a basis of .\n\n16 Suppose ( , ) and 0 . Prove that dim /( null ) = 1 .\n\n17 Suppose is a subspace of such that dim / = 1 . Prove that there exists \n\n ( , ) such that null = .\n\n18 Suppose that is a subspace of such that / is finite-dimensional. (a) Show that if is a finite-dimensional subspace of and = + ,then dim dim / .(b) Prove that there exists a finite-dimensional subspace of such that \n\ndim = dim / and = .\n\n19 Suppose ( , ) and is a subspace of . Let denote the quotient map from onto / . Prove that there exists (/ , ) such that \n\n = if and only if null .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3F Duality 105 \n\n# 3F Duality \n\n# Dual Space and Dual Map \n\nLinear maps into the scalar field play a special role in linear algebra, and thus they get a special name. \n\n> 3.108 definition:\n\nlinear functional \n\nA linear functional on is a linear map from to . In other words, a linear functional is an element of ( , ) . \n\n> 3.109 example:\n\nlinear functionals \n\n Define 3 by ( , , ) = 4 5 + 2 . Then is a linear functional on 3.\n\n Fix ( 1, , ) . Define by ( 1, , ) = 1 1 + + .Then is a linear functional on .\n\n Define () by \n\n() = 3 (5) + 7(4). \n\nThen is a linear functional on () .\n\n Define () by \n\n() = 10\n\n \n\nfor each () . Then is a linear functional on () .The vector space ( , ) also gets a special name and special notation. \n\n> 3.110 definition:\n\ndual space, \n\nThe dual space of , denoted by , is the vector space of all linear functionals on . In other words, = ( , ) . \n\n> 3.111\n\ndim = dim \n\nSuppose is finite-dimensional. Then is also finite-dimensional and \n\ndim = dim . \n\nProof By 3.72 we have \n\ndim = dim ( , ) = ( dim )( dim ) = dim ,\n\nas desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n106 Chapter 3 Linear Maps \n\nIn the following definition, the linear map lemma (3.4) implies that each is well defined. \n\n> 3.112 definition:\n\ndual basis \n\nIf 1, , is a basis of , then the dual basis of 1, , is the list 1, , \n\nof elements of , where each is the linear functional on such that \n\n ( ) = {{1 if = ,\n\n0 if . \n\n> 3.113 example:\n\nthe dual basis of the standard basis of \n\nSuppose is a positive integer. For 1 , define to be the linear functional on that selects the th coordinate of a vector in . Thus \n\n ( 1, , ) = \n\nfor each ( 1, , ) .Let 1, , be the standard basis of . Then \n\n ( ) = {{1 if = ,\n\n0 if . \n\nThus 1, , is the dual basis of the standard basis 1, , of .The next result shows that the dual basis of a basis of consists of the linear functionals on that give the coefficients for expressing a vector in as a linear combination of the basis vectors. \n\n> 3.114\n\ndual basis gives coefficients for linear combination \n\nSuppose 1, , is a basis of and 1, , is the dual basis. Then \n\n = 1() 1 + + () \n\nfor each .\n\nProof Suppose . Then there exist 1, , such that \n\n> 3.115\n\n = 1 1 + + .\n\nIf {1 , , } , then applying to both sides of the equation above gives \n\n () = .\n\nSubstituting the values for 1, , given by the equation above into 3.115 shows that = 1() 1 + + () .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3F Duality 107 \n\nThe next result shows that the dual basis is indeed a basis of the dual space. Thus the terminology dual basis is justified. \n\n> 3.116\n\ndual basis is a basis of the dual space \n\nSuppose is finite-dimensional. Then the dual basis of a basis of is a basis of .\n\nProof Suppose 1, , is a basis of . Let 1, , denote the dual basis. To show that 1, , is a linearly independent list of elements of , suppose \n\n 1, , are such that \n\n> 3.117\n\n 1 1 + + = 0. \n\nNow \n\n( 1 1 + + )( ) = \n\nfor each = 1 , , . Thus 3.117 shows that 1 = = = 0 . Hence 1, , \n\nis linearly independent. Because 1, , is a linearly independent list in whose length equals \n\ndim (by 3.111), we can conclude that 1, , is a basis of (see 2.38). In the definition below, note that if is a linear map from to then is a linear map from to . \n\n> 3.118 definition:\n\ndual map, \n\nSuppose ( , ) . The dual map of is the linear map ( , )\n\ndefined for each by \n\n () = . \n\nIf ( , ) and , then () is defined above to be the composition of the linear maps and . Thus () is indeed a linear map from to ; in other words, () .The following two bullet points show that is a linear map from to .\n\n If , , then \n\n ( + ) = ( + ) = + = () + (). If and , then \n\n () = () = ( ) = (). \n\nThe prime notation appears with two unrelated meanings in the next example: \n\n denotes the dual of the linear map , and denotes the derivative of a polynomial .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n108 Chapter 3 Linear Maps \n\n> 3.119 example:\n\ndual map of the differentiation linear map \n\nDefine () () by = .\n\n Suppose is the linear functional on () defined by () = (3) . Then \n\n () is the linear functional on () given by \n\n( ())() = ( )() = () = ( ) = (3). \n\nThus () is the linear functional on () taking to (3) .\n\n Suppose is the linear functional on () defined by () = 10 . Then \n\n () is the linear functional on () given by \n\n( ())() = ( )() = () = ( )= 10\n\n \n\n= (1) (0). \n\nThus () is the linear functional on () taking to (1) (0) .In the next result, (a) and (b) imply that the function that takes to is a linear map from ( , ) to ( , ).In (c) below, note the reversal of order from on the left to on the right. \n\n> 3.120\n\nalgebraic properties of dual maps \n\nSuppose ( , ) . Then (a) ( + ) = + for all ( , ) ;(b) () = for all ;(c) () = for all ( , ) .\n\nProof The proofs of (a) and (b) are left to the reader. To prove (c), suppose . Then \n\n() () = () = ( ) = ( ) = ( ()) = ( )() ,\n\nSome books use the notation and \n\n> \n\nfor duality instead of and .However, here we reserve the notation \n\n> \n\nfor the adjoint, which will be intro-duced when we study linear maps on inner product spaces in Chapter 7. \n\nwhere the first, third, and fourth equal-ities above hold because of the defini-tion of the dual map, the second equality holds because composition of functions is associative, and the last equality fol-lows from the definition of composition. The equation above shows that \n\n() () = ( )() for all .Thus () = .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3F Duality 109 \n\n# Null Space and Range of Dual of Linear Map \n\nOur goal in this subsection is to describe null and range in terms of range \n\nand null . To do this, we will need the next definition. \n\n> 3.121 definition:\n\nannihilator, 0\n\nFor , the annihilator of , denoted by 0, is defined by \n\n 0 = { () = 0 for all }. \n\n> 3.122 example:\n\nelement of an annihilator \n\nSuppose is the subspace of () consisting of polynomial multiples of 2.If is the linear functional on () defined by () = (0) , then 0.For , the annihilator 0 is a subset of the dual space . Thus 0\n\ndepends on the vector space containing , so a notation such as 0 would be more precise. However, the containing vector space will always be clear from the context, so we will use the simpler notation 0. \n\n> 3.123 example:\n\nthe annihilator of a two-dimensional subspace of 5\n\nLet 1, 2, 3, 4, 5 denote the standard basis of 5; let 1, 2, 3, 4, 5 ( 5) denote the dual basis of 1, 2, 3, 4, 5. Suppose \n\n = span ( 1, 2) = {( 1, 2, 0, 0, 0) 5 1, 2 }. \n\nWe want to show that 0 = span ( 3, 4, 5).Recall (see 3.113) that is the linear functional on 5 that selects the th \n\ncoordinate: ( 1, 2, 3, 4, 5) = .First suppose span ( 3, 4, 5). Then there exist 3, 4, 5 such that \n\n = 3 3 + 4 4 + 5 5. If ( 1, 2, 0, 0, 0) , then \n\n( 1, 2, 0, 0, 0) = ( 3 3 + 4 4 + 5 5)( 1, 2, 0, 0, 0) = 0. \n\nThus 0. Hence we have shown that span ( 3, 4, 5) 0.To show the inclusion in the other direction, suppose that 0. Be-cause the dual basis is a basis of ( 5), there exist 1, 2, 3, 4, 5 such that \n\n = 1 1 + 2 2 + 3 3 + 4 4 + 5 5. Because 1 and 0, we have \n\n0 = ( 1) = ( 1 1 + 2 2 + 3 3 + 4 4 + 5 5)( 1) = 1.\n\nSimilarly, 2 and thus 2 = 0 . Hence = 3 3 + 4 4 + 5 5. Thus \n\n span ( 3, 4, 5), which shows that 0 span ( 3, 4, 5).Thus 0 = span ( 3, 4, 5).\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n110 Chapter 3 Linear Maps \n\n> 3.124\n\nthe annihilator is a subspace \n\nSuppose . Then 0 is a subspace of .\n\nProof Note that 0 0 (here 0 is the zero linear functional on ) because the zero linear functional applied to every vector in equals 0 .Suppose , 0. Thus , and () = () = 0 for every .If , then \n\n( + )() = () + () = 0 + 0 = 0. \n\nThus + 0.Similarly, 0 is closed under scalar multiplication. Thus 1.34 implies that 0\n\nis a subspace of .The next result shows that dim 0 is the difference of dim and dim . For example, this shows that if is a two-dimensional subspace of 5, then 0 is a three-dimensional subspace of ( 5), as in Example 3.123. The next result can be proved following the pattern of Example 3.123: choose a basis 1, , of , extend to a basis 1, , , , of , let 1, , , , \n\nbe the dual basis of , and then show that + 1, , is a basis of 0, which implies the desired result. You should construct the proof just outlined, even though a slicker proof is presented here. \n\n> 3.125\n\ndimension of the annihilator \n\nSuppose is finite-dimensional and is a subspace of . Then \n\ndim 0 = dim dim . \n\nProof Let ( , ) be the inclusion map defined by () = for each .Thus is a linear map from to . The fundamental theorem of linear maps (3.21) applied to shows that \n\ndim range + dim null = dim .\n\nHowever, null = 0 (as can be seen by thinking about the definitions) and \n\ndim = dim (by 3.111), so we can rewrite the equation above as \n\n> 3.126\n\ndim range + dim 0 = dim . \n\nIf , then can be extended to a linear functional on (see, for example, Exercise 13 in Section 3A). The definition of shows that () = .Thus range , which implies that range = . Hence \n\ndim range = dim = dim ,\n\nand then 3.126 becomes the equation dim + dim 0 = dim , as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 3F Duality 111\n\nThe next result can be a useful tool to show that a subspace is as big as possiblesee (a)or to show that a subspace is as small as possiblesee (b). \n\n> 3.127\n\ncondition for the annihilator to equal {0} or the whole space \n\nSuppose is finite-dimensional and is a subspace of . Then (a) 0 = {0} = ;(b) 0 = = {0} . \n\n> Proof\n\nTo prove (a), we have \n\n 0 = {0} dim 0 = 0 dim = dim = ,\n\nwhere the second equivalence follows from 3.125 and the third equivalence follows from 2.39. Similarly, to prove (b) we have \n\n 0 = dim 0 = dim \n\n dim 0 = dim dim = 0 = {0} ,\n\nwhere one direction of the first equivalence follows from 2.39, the second equiva-lence follows from 3.111, and the third equivalence follows from 3.125. The proof of (a) in the next result does not use the hypothesis that and \n\nare finite-dimensional. \n\n> 3.128\n\nthe null space of \n\nSuppose and are finite-dimensional and ( , ) . Then (a) null = ( range ) 0;(b) dim null = dim null + dim dim .\n\n> Proof\n\n(a) First suppose null . Thus 0 = () = . Hence \n\n0 = ( )() = () for every . \n\nThus ( range ) 0. This implies that null ( range ) 0.To prove the inclusion in the opposite direction, now suppose ( range ) 0.Thus () = 0 for every vector . Hence 0 = = () . In other words, null , which shows that (range ) 0 null , completing the proof of (a). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 112 Chapter 3 Linear Maps\n\n(b) We have \n\ndim null = dim (range ) 0\n\n= dim dim range = dim ( dim dim null ) = dim null + dim dim ,\n\nwhere the first equality comes from (a), the second equality comes from 3.125, and the third equality comes from the fundamental theorem of linear maps (3.21). The next result can be useful because sometimes it is easier to verify that \n\nis injective than to show directly that is surjective. \n\n> 3.129\n\n surjective is equivalent to injective \n\nSuppose and are finite-dimensional and ( , ) . Then \n\n is surjective is injective . \n\n> Proof\n\nWe have \n\n ( , ) is surjective range = ( range ) 0 = {0} null = {0} is injective ,\n\nwhere the second equivalence comes from 3.127(a) and the third equivalence comes from 3.128(a). \n\n> 3.130\n\nthe range of \n\nSuppose and are finite-dimensional and ( , ) . Then (a) dim range = dim range ;(b) range = ( null ) 0.\n\n> Proof\n\n(a) We have \n\ndim range = dim dim null \n\n= dim dim (range ) 0\n\n= dim range ,\n\nwhere the first equality comes from 3.21, the second equality comes from 3.111 and 3.128(a), and the third equality comes from 3.125. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3F Duality 113 \n\n(b) First suppose range . Thus there exists such that = () .If null , then \n\n() = ( ()) = ( )() = () = (0) = 0. \n\nHence ( null ) 0. This implies that range ( null ) 0.We will complete the proof by showing that range and (null ) 0 have the same dimension. To do this, note that \n\ndim range = dim range = dim dim null = dim (null ) 0,\n\nwhere the first equality comes from (a), the second equality comes from 3.21, and the third equality comes from 3.125. The next result should be compared to 3.129. \n\n> 3.131\n\n injective is equivalent to surjective \n\nSuppose and are finite-dimensional and ( , ) . Then \n\n is injective is surjective .\n\nProof We have \n\n is injective null = {0} ( null ) 0 = \n\n range = ,\n\nwhere the second equivalence follows from 3.127(b) and the third equivalence follows from 3.130(b). \n\n# Matrix of Dual of Linear Map \n\nThe setting for the next result is the assumption that we have a basis 1, , \n\nof , along with its dual basis 1, , of . We also have a basis 1, , \n\nof , along with its dual basis 1, , of . Thus () is computed with respect to the bases just mentioned of and , and ( ) is computed with respect to the dual bases just mentioned of and . Using these bases gives the following pretty result. \n\n> 3.132\n\nmatrix of is transpose of matrix of \n\nSuppose and are finite-dimensional and ( , ) . Then \n\n( ) = ( ()) t.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n114 Chapter 3 Linear Maps Proof Let = () and = ( ). Suppose 1 and 1 .From the definition of ( ) we have \n\n ( ) = \n\n> \n> \n> = 1\n\n , .\n\nThe left side of the equation above equals . Thus applying both sides of the equation above to gives \n\n( )( ) = \n\n> \n> \n> = 1\n\n , ( )= , .\n\nWe also have \n\n( )( ) = ( )= (\n\n> \n> \n> = 1\n\n , )=\n\n> \n> \n> = 1\n\n , ( )= , .\n\nComparing the last line of the last two sets of equations, we have , = , .Thus = t. In other words, ( ) = ( ()) t, as desired. Now we use duality to give an alternative proof that the column rank of a matrix equals the row rank of the matrix. This result was previously proved using different toolssee 3.57. \n\n> 3.133\n\ncolumn rank equals row rank \n\nSuppose , . Then the column rank of equals the row rank of .\n\nProof Define ,1 ,1 by = . Thus () = , where () is computed with respect to the standard bases of ,1 and ,1. Now column rank of = column rank of () = dim range = dim range \n\n= column rank of ( )= column rank of t\n\n= row rank of ,\n\nwhere the second equality comes from 3.78, the third equality comes from 3.130(a), the fourth equality comes from 3.78, the fifth equality comes from 3.132, and the last equality follows from the definitions of row and column rank. See Exercise 8 in Section 7A for another alternative proof of the result above. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3F Duality 115 \n\n# Exercises 3F \n\n1 Explain why each linear functional is surjective or is the zero map. \n\n2 Give three distinct examples of linear functionals on [0 ,1] .\n\n3 Suppose is finite-dimensional and with 0 . Prove that there exists such that () = 1 .\n\n4 Suppose is finite-dimensional and is a subspace of such that .Prove that there exists such that () = 0 for every but 0 .\n\n5 Suppose ( , ) and 1, , is a basis of range . Hence for each \n\n , there exist unique numbers 1() , , () such that \n\n = 1() 1 + + () ,\n\nthus defining functions 1, , from to . Show that each of the functions 1, , is a linear functional on .\n\n6 Suppose , . Prove that null null if and only if there exists \n\n such that = .\n\n7 Suppose that 1, , are vector spaces. Prove that ( 1 ) and \n\n 1 are isomorphic vector spaces. \n\n8 Suppose 1, , is a basis of and 1, , is the dual basis of . Define \n\n and by \n\n() = ( 1() , , ()) and ( 1, , ) = 1 1 + + .\n\nExplain why and are inverses of each other. \n\n9 Suppose is a positive integer. Show that the dual basis of the basis \n\n1, , , of () is 0, 1, , , where \n\n () = () (0) ! .\n\nHere () denotes the th derivative of , with the understanding that the 0th \n\nderivative of is .\n\n10 Suppose is a positive integer. (a) Show that 1, 5 , , ( 5) is a basis of () .(b) What is the dual basis of the basis in (a)? \n\n11 Suppose 1, , is a basis of and 1, , is the corresponding dual basis of . Suppose . Prove that \n\n = ( 1) 1 + + ( ) .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n116 Chapter 3 Linear Maps \n\n12 Suppose , ( , ) .(a) Prove that ( + ) = + .(b) Prove that () = for all .\n\nThis exercise asks you to verify (a) and (b) in 3.120. \n\n13 Show that the dual map of the identity operator on is the identity operator on .\n\n14 Define 3 2 by \n\n( , , ) = (4 + 5 + 6 , 7 + 8 + 9). \n\nSuppose 1, 2 denotes the dual basis of the standard basis of 2 and \n\n 1, 2, 3 denotes the dual basis of the standard basis of 3.(a) Describe the linear functionals ( 1) and ( 2).(b) Write ( 1) and ( 2) as linear combinations of 1, 2, 3.\n\n15 Define () () by \n\n()() = 2() + () \n\nfor each .(a) Suppose () is defined by () = (4) . Describe the linear functional () on () .(b) Suppose () is defined by () = 10 . Evaluate ( ())( 3).\n\n16 Suppose is finite-dimensional and ( , ) . Prove that \n\n = 0 = 0. \n\n17 Suppose and are finite-dimensional and ( , ) . Prove that is invertible if and only if ( , ) is invertible. \n\n18 Suppose and are finite-dimensional. Prove that the map that takes \n\n ( , ) to ( , ) is an isomorphism of ( , ) onto \n\n( , ).\n\n19 Suppose . Explain why \n\n 0 = { null }. \n\n20 Suppose is finite-dimensional and is a subspace of . Show that \n\n = { () = 0 for every 0}. \n\n21 Suppose is finite-dimensional and and are subspaces of .(a) Prove that 0 0 if and only if .(b) Prove that 0 = 0 if and only if = .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 3F Duality 117 \n\n22 Suppose is finite-dimensional and and are subspaces of .(a) Show that ( + ) 0 = 0 0.(b) Show that ( ) 0 = 0 + 0.\n\n23 Suppose is finite-dimensional and 1, , . Prove that the follow-ing three sets are equal to each other. (a) span ( 1, , )\n\n(b) (( null 1) ( null )) 0\n\n(c) { (null 1) ( null ) null } \n\n24 Suppose is finite-dimensional and 1, , . Define a linear map \n\n by () = (( 1), , ( )) .(a) Prove that 1, , spans if and only if is injective. (b) Prove that 1, , is linearly independent if and only if is surjective. \n\n25 Suppose is finite-dimensional and 1, , . Define a linear map \n\n by () = ( 1() , , ()) .(a) Prove that 1, , spans if and only if is injective. (b) Prove that 1, , is linearly independent if and only if is surjective. \n\n26 Suppose is finite-dimensional and is a subspace of . Prove that \n\n = { () = 0 for every } 0.\n\n27 Suppose ( 5()) and null = span () , where is the linear functional on 5() defined by () = (8) . Prove that \n\nrange = { 5() (8) = 0}. \n\n28 Suppose is finite-dimensional and 1, , is a linearly independent list in . Prove that \n\ndim (( null 1) ( null )) = ( dim ) . \n\n29 Suppose and are finite-dimensional and ( , ) .(a) Prove that if and null = span () , then range = null .(b) Prove that if and range = span () , then null = null .\n\n30 Suppose is finite-dimensional and 1, , is a basis of . Show that there exists a basis of whose dual basis is 1, , .\n\n31 Suppose is a subspace of . Let be the inclusion map defined by () = . Thus ( , ).(a) Show that null = 0.(b) Prove that if is finite-dimensional, then range = .(c) Prove that if is finite-dimensional, then is an isomorphism from \n\n / 0 onto .\n\nThe isomorphism in (c) is natural in that it does not depend on a choice of basis in either vector space. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n118 Chapter 3 Linear Maps \n\n32 The double dual space of , denoted by , is defined to be the dual space of . In other words, = ( ). Define by \n\n()() = () \n\nfor each and each .(a) Show that is a linear map from to .(b) Show that if () , then = , where = ( ).(c) Show that if is finite-dimensional, then is an isomorphism from \n\nonto .\n\nSuppose is finite-dimensional. Then and are isomorphic, but finding an isomorphism from onto generally requires choosing a basis of .In contrast, the isomorphism from onto does not require a choice of basis and thus is considered more natural. \n\n33 Suppose is a subspace of . Let / be the usual quotient map. Thus ((/) , ).(a) Show that is injective. (b) Show that range = 0.(c) Conclude that is an isomorphism from (/) onto 0.\n\nThe isomorphism in (c) is natural in that it does not depend on a choice of basis in either vector space. In fact, there is no assumption here that any of these vector spaces are finite-dimensional. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n# Chapter 4 \n\n# Polynomials \n\nThis chapter contains material on polynomials that we will use to investigate linear maps from a vector space to itself. Many results in this chapter will already be familiar to you from other courses; they are included here for completeness. Because this chapter is not about linear algebra, your instructor may go through it rapidly. You may not be asked to scrutinize all the proofs. Make sure, however, that you at least read and understand the statements of all results in this chapter they will be used in later chapters. This chapter begins with a brief discussion of algebraic properties of the complex numbers. Then we prove that a nonconstant polynomial cannot have more zeros than its degree. We also give a linear-algebra-based proof of the division algorithm for polynomials, which is worth reading even if you are already familiar with a proof that does not use linear algebra. As we will see, the fundamental theorem of algebra leads to a factorization of every polynomial into degree-one factors if the scalar field is or to factors of degree at most two if the scalar field is .\n\nstanding assumption for this chapter \n\n denotes or .\n\n> Alireza Javaheri CC BY\n\nStatue of mathematician and poet Omar Khayyam (10481131 ), whose algebra book written in 1070 contained the first serious study of cubic polynomials. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n> 119 120 Chapter 4 Polynomials\n\nBefore discussing polynomials with complex or real coefficients, we need to learn a bit more about the complex numbers. \n\n> 4.1 definition:\n\nreal part, Re , imaginary part, Im \n\nSuppose = + , where and are real numbers. \n\n The real part of , denoted by Re , is defined by Re = .\n\n The imaginary part of , denoted by Im , is defined by Im = .Thus for every complex number , we have \n\n = Re + (Im ). \n\n> 4.2 definition:\n\ncomplex conjugate, , absolute value, || \n\nSuppose .\n\n The complex conjugate of , denoted by , is defined by \n\n = Re ( Im ). The absolute value of a complex number , denoted by || , is defined by \n\n|| = (Re ) 2 + (Im ) 2. \n\n> 4.3 example:\n\nreal and imaginary part, complex conjugate, absolute value \n\nSuppose = 3 + 2 . Then \n\n Re = 3 and Im = 2 ;\n\n = 3 2 ;\n\n || = 32 + 22 = 13 .Identifying a complex number with the ordered pair (Re , Im ) 2\n\nidentifies with 2. Note that is a one-dimensional complex vector space, but we can also think of (identified with 2) as a two-dimensional real vector space. The absolute value of each complex number is a nonnegative number. Specif-ically, if , then || equals the distance from the origin in 2 to the point \n\n(Re , Im ) 2. \n\n> You should verify that = if and only if is a real number.\n\nThe real and imaginary parts, com-plex conjugate, and absolute value have the properties listed in the following multipart result. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nChapter 4 Polynomials 121 \n\n> 4.4\n\nproperties of complex numbers \n\nSuppose , . Then the following equalities and inequalities hold. \n\nsum of and + = 2 Re .\n\ndifference of and = 2( Im ) .\n\nproduct of and = || 2.\n\nadditivity and multiplicativity of complex conjugate \n\n + = + and = .\n\ndouble complex conjugate \n\n = .\n\nreal and imaginary parts are bounded by || | Re | || and | Im | || .\n\nabsolute value of the complex conjugate \n\n = || .\n\nmultiplicativity of absolute value \n\n|| = || || .\n\ntriangle inequality \n\n| + | || + || .\n\nGeometric interpretation of triangle in-equality: The length of each side of a triangle is less than or equal to the sum of the lengths of the two other sides. \n\nProof Except for the last item above, the routine verifications of the assertions above are left to the reader. To verify the triangle inequality, we have \n\n| + | 2 = ( + )( + ) = + + + = || 2 + || 2 + + = || 2 + || 2 + 2 Re () || 2 + || 2 + 2 = || 2 + || 2 + 2|| || = (|| + ||) 2.\n\nSee Exercise 2 for the reverse triangle inequality. \n\nTaking square roots now gives the desired inequality | + | || + || .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n122 Chapter 4 Polynomials \n\n# Zeros of Polynomials \n\nRecall that a function is called a polynomial of degree if there exist \n\n 0, , with 0 such that \n\n() = 0 + 1 + + \n\nfor all . A polynomial could have more than one degree if the representation of in the form above were not unique. Our first task is to show that this cannot happen. The solutions to the equation () = 0 play a crucial role in the study of a polynomial () . Thus these solutions have a special name. \n\n> 4.5 definition:\n\nzero of a polynomial \n\nA number is called a zero (or root ) of a polynomial () if \n\n() = 0. \n\nThe next result is the key tool that we will use to show that the degree of a polynomial is unique. \n\n> 4.6\n\neach zero of a polynomial corresponds to a degree-one factor \n\nSuppose is a positive integer and () is a polynomial of degree .Suppose . Then () = 0 if and only if there exists a polynomial \n\n () of degree 1 such that \n\n() = ( )() \n\nfor every .\n\nProof First suppose () = 0 . Let 0, 1, , be such that \n\n() = 0 + 1 + + \n\nfor all . Then \n\n> 4.7\n\n() = () () = 1( ) + + ( )\n\nfor all . For each {1 , , } , the equation \n\n = ( ) \n\n> \n> \n> = 1\n\n 1 \n\nshows that equals times some polynomial of degree 1 . Thus 4.7 shows that equals times some polynomial of degree 1 , as desired. To prove the implication in the other direction, now suppose that there is a polynomial () such that () = ( )() for every . Then \n\n() = ( )() = 0 , as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Chapter 4 Polynomials 123\n\nNow we can prove that polynomials do not have too many zeros. \n\n> 4.8\n\ndegree implies at most zeros \n\nSuppose is a positive integer and () is a polynomial of degree .Then has at most zeros in . \n\n> Proof\n\nWe will use induction on . The desired result holds if = 1 because if 1 0 then the polynomial 0 + 1 has only one zero (which equals 0/ 1). Thus assume that > 1 and the desired result holds for 1 .If has no zeros in , then the desired result holds and we are done. Thus suppose has a zero . By 4.6, there is polynomial () of degree \n\n 1 such that \n\n() = ( )() \n\nfor every . Our induction hypothesis implies that has at most 1 zeros in . The equation above shows that the zeros of in are exactly the zeros of \n\nin along with . Thus has at most zeros in .The result above implies that the coefficients of a polynomial are uniquely determined (because if a polynomial had two different sets of coefficients, then subtracting the two representations of the polynomial would give a polynomial with some nonzero coefficients but infinitely many zeros). In particular, the degree of a polynomial is uniquely defined. \n\n> The 0polynomial is declared to have degree so that exceptions are not needed for various reasonable results such as deg () = deg +deg .\n\nRecall that the degree of the 0 poly-nomial is defined to be . When necessary, use the expected arithmetic with . For example, < and \n\n + = for every integer .\n\n# Division Algorithm for Polynomials \n\nIf and are nonnegative integers, with 0 , then there exist nonnegative integers and such that \n\n = + \n\nand < . Think of dividing by , getting quotient with remainder . Our next result gives an analogous result for polynomials. Thus the next result is often called the division algorithm for polynomials, although as stated here it is not really an algorithm, just a useful result. \n\n> Think of the division algorithm for poly-nomials as giving a remainder polyno-mial when the polynomial is divided by the polynomial .\n\nThe division algorithm for polynomi-als could be proved without using any linear algebra. However, as is appropri-ate for a linear algebra textbook, the proof given here uses linear algebra techniques and makes nice use of a basis of () , which is the ( + 1) -dimensional vector space of polynomials with coefficients in and of degree at most .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n124 Chapter 4 Polynomials \n\n> 4.9\n\ndivision algorithm for polynomials \n\nSuppose that , () , with 0 . Then there exist unique polynomials \n\n , () such that \n\n = + \n\nand deg < deg .\n\nProof Let = deg and let = deg . If < , then take = 0 and = to get the desired equation = + with deg < deg . Thus we now assume that \n\n .The list \n\n> 4.10\n\n1, , , 1 , , , , \n\nis linearly independent in () because each polynomial in this list has a different degree. Also, the list 4.10 has length + 1, which equals dim () . Hence 4.10 is a basis of () [by 2.38]. Because () and 4.10 is a basis of () , there exist unique constants \n\n 0, 1, , 1 and 0, 1, , such that \n\n = 0 + 1 + + 1 1 + 0 + 1 + + 4.11 \n\n= 0 + 1 + + 1 1 \n\n \n\n> \n\n+ ( 0 + 1 + + \n\n \n\n> \n\n). \n\nWith and as defined above, we see that can be written as = + with \n\ndeg < deg , as desired. The uniqueness of , () satisfying these conditions follows from the uniqueness of the constants 0, 1, , 1 and 0, 1, , satisfy-ing 4.11. \n\n# Factorization of Polynomials over \n\nThe fundamental theorem of algebra is an existence theorem. Its proof does not lead to a method for finding zeros. The quadratic formula gives the zeros explicitly for polynomials of degree 2.Similar but more complicated formulas exist for polynomials of degree 3 and 4.No such formulas exist for polynomials of degree 5 and above. \n\nWe have been handling polynomials with complex coefficients and polynomials with real coefficients simultaneously, let-ting denote or . Now we will see differences between these two cases. First we treat polynomials with complex coefficients. Then we will use those re-sults to prove corresponding results for polynomials with real coefficients. Our proof of the fundamental theorem of algebra implicitly uses the result that a continuous real-valued function on a closed disk in 2 attains a minimum value. A web search can lead you to several \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nChapter 4 Polynomials 125 \n\nother proofs of the fundamental theorem of algebra. The proof using Liouvilles theorem is particularly nice if you are comfortable with analytic functions. All proofs of the fundamental theorem of algebra need to use some analysis, because the result is not true if is replaced, for example, with the set of numbers of the form + where , are rational numbers. \n\n> 4.12\n\nfundamental theorem of algebra, first version \n\nEvery nonconstant polynomial with complex coefficients has a zero in .\n\nProof De Moivres theorem, which you can prove using induction on and the addition formulas for cosine and sine, states that if is a positive integer and \n\n , then \n\n(cos + sin ) = cos + sin . \n\nSuppose and is a positive integer. Using polar coordinates, we know that there exist 0 and such that \n\n( cos + sin ) = . \n\nDe Moivres theorem implies that \n\n( 1/ (cos + sin )) \n\n> \n\n= . \n\nThus every complex number has a th root, a fact that we will soon use. Suppose is a nonconstant polynomial with complex coefficients and highest-order nonzero term . Then |()| as || (because |()|/ | |\n\nas || ). Thus the continuous function |()| has a global minimum at some point . To show that ( ) = 0 , suppose that ( ) 0 .Define a new polynomial by \n\n() = ( + ) ( ) .\n\nThe function |()| has a global minimum value of 1 at = 0 . Write \n\n() = 1 + + + ,\n\nwhere is the smallest positive integer such that the coefficient of is nonzero; in other words, 0 .Let be such that = 1 \n\n. There is a constant > 1 such that if \n\n (0 , 1) , then \n\n|()| 1 + + + 1 = 1 (1 ). \n\nThus taking to be 1/(2) in the inequality above, we have |()| < 1 , which contradicts the assumption that the global minimum of |()| is 1. This contradiction implies that ( ) = 0 , showing that has a zero, as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 126 Chapter 4 Polynomials\n\nComputers can use clever numerical methods to find good approximations to the zeros of any polynomial, even when exact zeros cannot be found. For example, no one will ever give an exact formula for a zero of the polynomial defined by \n\n() = 5 5 4 6 3 + 17 2 + 4 7. \n\nHowever, a computer can find that the zeros of are approximately the five numbers 1.87 , 0.74 , 0.62 , 1.47 , 5.51 .The first version of the fundamental theorem of algebra leads to the following factorization result for polynomials with complex coefficients. Note that in this factorization, the zeros of are the numbers 1, , , which are the only values of for which the right side of the equation in the next result equals 0. \n\n> 4.13\n\nfundamental theorem of algebra, second version \n\nIf () is a nonconstant polynomial, then has a unique factorization (except for the order of the factors) of the form \n\n() = ( 1)( ),\n\nwhere , 1, , . \n\n> Proof\n\nLet () and let = deg . We will use induction on . If = 1 ,then the desired factorization exists and is unique. So assume that > 1 and that the desired factorization exists and is unique for all polynomials of degree 1 .First we will show that the desired factorization of exists. By the first version of the fundamental theorem of algebra (4.12), has a zero . By 4.6, there is a polynomial of degree 1 such that \n\n() = ( )() \n\nfor all . Our induction hypothesis implies that has the desired factorization, which when plugged into the equation above gives the desired factorization of .Now we turn to the question of uniqueness. The number is uniquely deter-mined as the coefficient of in . So we only need to show that except for the order, there is only one way to choose 1, , . If \n\n( 1)( ) = ( 1)( )\n\nfor all , then because the left side of the equation above equals 0 when \n\n = 1, one of the s on the right side equals 1. Relabeling, we can assume that 1 = 1. Now if 1, we can divide both sides of the equation above by \n\n 1, getting \n\n( 2)( ) = ( 2)( )\n\nfor all except possibly = 1. Actually the equation above holds for all \n\n , because otherwise by subtracting the right side from the left side we would get a nonzero polynomial that has infinitely many zeros. The equation above and our induction hypothesis imply that except for the order, the s are the same as the s, completing the proof of uniqueness. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nChapter 4 Polynomials 127 \n\n# Factorization of Polynomials over \n\nThe failure of the fundamental theorem of algebra for accounts for the differ-ences between linear algebra on real and complex vector spaces, as we will see in later chapters. \n\nA polynomial with real coefficients may have no real zeros. For example, the poly-nomial 1 + 2 has no real zeros. To obtain a factorization theorem over \n\n , we will use our factorization theorem over . We begin with the next result. \n\n> 4.14\n\npolynomials with real coefficients have nonreal zeros in pairs \n\nSuppose () is a polynomial with real coefficients. If is a zero of , then so is .\n\nProof Let \n\n() = 0 + 1 + + ,\n\nwhere 0, , are real numbers. Suppose is a zero of . Then \n\n 0 + 1 + + = 0. \n\nTake the complex conjugate of both sides of this equation, obtaining \n\n 0 + 1 + + = 0 ,\n\nwhere we have used basic properties of the complex conjugate (see 4.4). The equation above shows that is a zero of .\n\nThink about the quadratic formula in connection with the result below. \n\nWe want a factorization theorem for polynomials with real coefficients. We begin with the following result. \n\n> 4.15\n\nfactorization of a quadratic polynomial \n\nSuppose , . Then there is a polynomial factorization of the form \n\n 2 + + = ( 1)( 2)\n\nwith 1, 2 if and only if 2 4 .\n\nProof Notice that \n\n 2 + + = ( + 2 )\n\n> 2\n\n+ ( 2\n\n4 ). \n\nThe equation above is the basis of the technique called completing the square. \n\nFirst suppose 2 < 4 . Then the right side of the equation above is positive for every . Hence the polynomial \n\n 2 + + has no real zeros and thus cannot be factored in the form ( 1)( 2) with 1, 2 . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 128 Chapter 4 Polynomials\n\nConversely, now suppose 2 4 . Then there is a real number such that \n\n 2 = 2 \n\n> 4\n\n . From the displayed equation above, we have \n\n 2 + + = ( + 2 )\n\n> 2\n\n 2\n\n= ( + 2 + )( + 2 ) ,\n\nwhich gives the desired factorization. The next result gives a factorization of a polynomial over . The idea of the proof is to use the second version of the fundamental theorem of algebra (4.13), which gives a factorization of as a polynomial with complex coefficients. Complex but nonreal zeros of come in pairs; see 4.14. Thus if the factorization of as an element of () includes terms of the form ( ) with a nonreal complex number, then ( ) is also a term in the factorization. Multiplying together these two terms, we get \n\n 2 2( Re ) + || 2,\n\nwhich is a quadratic term of the required form. The idea sketched in the paragraph above almost provides a proof of the existence of our desired factorization. However, we need to be careful about one point. Suppose is a nonreal complex number and ( ) is a term in the factorization of as an element of () . We are guaranteed by 4.14 that ( ) \n\nalso appears as a term in the factorization, but 4.14 does not state that these two factors appear the same number of times, as needed to make the idea above work. However, the proof works around this point. In the next result, either or may equal 0. The numbers 1, , are precisely the real zeros of , for these are the only real values of for which the right side of the equation in the next result equals 0. \n\n> 4.16\n\nfactorization of a polynomial over \n\nSuppose () is a nonconstant polynomial. Then has a unique factor-ization (except for the order of the factors) of the form \n\n() = ( 1)( )( 2 + 1 + 1)( 2 + + ),\n\nwhere , 1, , , 1, , , 1, , , with 2 < 4 for each . \n\n> Proof\n\nFirst we will prove that the desired factorization exists, and after that we will prove the uniqueness. Think of as an element of () . If all (complex) zeros of are real, then we have the desired factorization by 4.13. Thus suppose has a zero with \n\n . By 4.14, is a zero of . Thus we can write \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nChapter 4 Polynomials 129 \n\n() = ( )( )() = ( 2 2( Re ) + || 2)() \n\nfor some polynomial () of degree two less than the degree of . If we can prove that has real coefficients, then using induction on the degree of \n\ncompletes the proof of the existence part of this result. To prove that has real coefficients, we solve the equation above for , getting \n\n() = () 2 2( Re ) + || 2\n\nfor all . The equation above implies that () for all . Writing \n\n() = 0 + 1 + + 2 2 ,\n\nwhere = deg and 0, , 2 , we thus have \n\n0 = Im () = ( Im 0) + (Im 1) + + (Im 2 ) 2 \n\nfor all . This implies that Im 0, , Im 2 all equal 0 (by 4.8). Thus all coefficients of are real, as desired. Hence the desired factorization exists. Now we turn to the question of uniqueness of our factorization. A factor of \n\nof the form 2 + + with 2 < 4 can be uniquely written as ( )( )\n\nwith . A moments thought shows that two different factorizations of as an element of () would lead to two different factorizations of as an element of () , contradicting 4.13. \n\n# Exercises 4 \n\n1 Suppose , . Verify the following equalities and inequalities. (a) + = 2 Re \n\n(b) = 2( Im ) \n\n(c) = || 2\n\n(d) + = + and = \n\n(e) = \n\n(f) | Re | || and | Im | || \n\n(g) = || \n\n(h) || = || || \n\nThe results above are the parts of 4.4 that were left to the reader. \n\n2 Prove that if , , then || || | | .\n\nThe inequality above is called the reverse triangle inequality .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n130 Chapter 4 Polynomials \n\n3 Suppose is a complex vector space and . Define by \n\n() = Re () for each . Show that \n\n() = () () \n\nfor all .\n\n4 Suppose is a positive integer. Is the set \n\n{0} { () deg = } \n\na subspace of () ?\n\n5 Is the set \n\n{0} { () deg is even }\n\na subspace of () ?\n\n6 Suppose that and are positive integers with , and suppose \n\n 1, , . Prove that there exists a polynomial () with \n\ndeg = such that 0 = ( 1) = = ( ) and such that has no other zeros. \n\n7 Suppose that is a nonnegative integer, 1, , + 1 are distinct elements of , and 1, , + 1 . Prove that there exists a unique polynomial \n\n () such that \n\n( ) = \n\nfor each = 1 , , + 1.\n\nThis result can be proved without using linear algebra. However, try to find the clearer, shorter proof that uses some linear algebra. \n\n8 Suppose () has degree . Prove that has distinct zeros if and only if and its derivative have no zeros in common. \n\n9 Prove that every polynomial of odd degree with real coefficients has a real zero. \n\n10 For () , define by \n\n()() = {{{{() (3) 3 if 3 ,\n\n (3) if = 3 \n\nfor each . Show that () for every polynomial () and also show that () () is a linear map. \n\n11 Suppose () . Define by \n\n() = () (). \n\nProve that is a polynomial with real coefficients. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nChapter 4 Polynomials 131 \n\n12 Suppose is a nonnegative integer and () is such that there are distinct real numbers 0, 1, , with ( ) for each = 0 , 1, , .Prove that all coefficients of are real. \n\n13 Suppose () with 0 . Let = { ()} .(a) Show that dim ()/ = deg .(b) Find a basis of ()/ .\n\n14 Suppose , () are nonconstant polynomials with no zeros in common. Let = deg and = deg . Use linear algebra as outlined below in (a)(c) to prove that there exist 1 () and 1 () such that \n\n + = 1. \n\n(a) Define 1 () 1 () + 1 () by \n\n( , ) = + . \n\nShow that the linear map is injective. (b) Show that the linear map in (a) is surjective. (c) Use (b) to conclude that there exist 1 () and 1 () \n\nsuch that + = 1 .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n# Chapter 5 \n\n# Eigenvalues and Eigenvectors \n\nLinear maps from one vector space to another vector space were the objects of study in Chapter 3. Now we begin our investigation of operators, which are linear maps from a vector space to itself. Their study constitutes the most important part of linear algebra. To learn about an operator, we might try restricting it to a smaller subspace. Asking for that restriction to be an operator will lead us to the notion of invariant subspaces. Each one-dimensional invariant subspace arises from a vector that the operator maps into a scalar multiple of the vector. This path will lead us to eigenvectors and eigenvalues. We will then prove one of the most important results in linear algebra: every operator on a finite-dimensional nonzero complex vector space has an eigenvalue. This result will allow us to show that for each operator on a finite-dimensional complex vector space, there is a basis of the vector space with respect to which the matrix of the operator has at least almost half its entries equal to 0.\n\nstanding assumptions for this chapter \n\n denotes or .\n\n denotes a vector space over .\n\n> Hans-Peter Postel CC BY\n\nStatue of Leonardo of Pisa (11701250, approximate dates ), also known as Fibonacci. Exercise 21 in Section 5D shows how linear algebra can be used to find the explicit formula for the Fibonacci sequence shown on the front cover. \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler \n\n132 Section 5A Invariant Subspaces 133 \n\n# 5A Invariant Subspaces \n\n# Eigenvalues \n\n> 5.1 definition:\n\noperator \n\nA linear map from a vector space to itself is called an operator .\n\nRecall that we defined the notation \n\n> ()\n\nto mean ( , ) .\n\nSuppose () . If 2 and \n\n = 1 ,\n\nwhere each is a nonzero subspace of , then to understand the behavior of \n\n we only need to understand the behavior of each | ; here | denotes the restriction of to the smaller domain . Dealing with | should be easier than dealing with because is a smaller vector space than .However, if we intend to apply tools useful in the study of operators (such as taking powers), then we have a problem: | may not map into itself; in other words, | may not be an operator on . Thus we are led to consider only decompositions of of the form above in which maps each into itself. Hence we now give a name to subspaces of that get mapped into themselves by . \n\n> 5.2 definition:\n\ninvariant subspace \n\nSuppose () . A subspace of is called invariant under if \n\nfor every .Thus is invariant under if | is an operator on . \n\n> 5.3 example:\n\nsubspace invariant under differentiation operator \n\nSuppose that ( ()) is defined by = . Then 4() , which is a subspace of () , is invariant under because if () has degree at most 4,then also has degree at most 4. \n\n> 5.4 example:\n\nfour invariant subspaces, not necessarily all different \n\nIf () , then the following subspaces of are all invariant under .\n\n{0} The subspace {0} is invariant under because if {0} , then = 0 \n\nand hence = 0 {0} .\n\n The subspace is invariant under because if , then .\n\nnull The subspace null is invariant under because if null , then \n\n = 0 , and hence null .\n\nrange The subspace range is invariant under because if range ,then range . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 134 Chapter 5 Eigenvalues and Eigenvectors\n\nMust an operator () have any invariant subspaces other than {0} \n\nand ? Later we will see that this question has an affirmative answer if is finite-dimensional and dim > 1 (for = ) or dim > 2 (for = ) ; see 5.19 and Exercise 29 in Section 5B. The previous example noted that null and range are invariant under .However, these subspaces do not necessarily provide easy answers to the question above about the existence of invariant subspaces other than {0} and , because \n\nnull may equal {0} and range may equal (this happens when is invertible). We will return later to a deeper study of invariant subspaces. Now we turn to an investigation of the simplest possible nontrivial invariant subspacesinvariant subspaces of dimension one. Take any with 0 and let equal the set of all scalar multiples of :\n\n = { } = span (). \n\nThen is a one-dimensional subspace of (and every one-dimensional subspace of is of this form for an appropriate choice of ). If is invariant under an operator () , then , and hence there is a scalar such that \n\n = . \n\nConversely, if = for some , then span () is a one-dimensional subspace of invariant under .The equation = , which we have just seen is intimately connected with one-dimensional invariant subspaces, is important enough that the scalars and vectors satisfying it are given special names. \n\n> 5.5 definition:\n\neigenvalue \n\nSuppose () . A number is called an eigenvalue of if there exists such that 0 and = . \n\n> The word eigenvalue is half-German, half-English. The German prefix eigen\n> means own in the sense of charac-terizing an intrinsic property.\n\nIn the definition above, we require that 0 because every scalar \n\nsatisfies 0 = 0 .The comments above show that \n\nhas a one-dimensional subspace invariant under if and only if has an eigenvalue. \n\n> 5.6 example:\n\neigenvalue \n\nDefine an operator ( 3) by \n\n( , , ) = (7 + 3 , 3 + 6 + 9 , 6) \n\nfor ( , , ) 3. Then (3 , 1, 1) = (18 , 6, 6) = 6(3 , 1, 1) . Thus 6 is an eigenvalue of . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 5A Invariant Subspaces 135\n\nThe equivalences in the next result, along with many deep results in linear algebra, are valid only in the context of finite-dimensional vector spaces. \n\n> 5.7\n\nequivalent conditions to be an eigenvalue \n\nSuppose is finite-dimensional, () , and . Then the following are equivalent. (a) is an eigenvalue of .(b) is not injective. (c) is not surjective. \n\n> Reminder: () is the identity operator. Thus = for all .\n\n(d) is not invertible. \n\n> Proof\n\nConditions (a) and (b) are equivalent because the equation = \n\nis equivalent to the equation ( ) = 0 . Conditions (b), (c), and (d) are equivalent by 3.65. \n\n> 5.8 definition:\n\neigenvector \n\nSuppose () and is an eigenvalue of . A vector is called an eigenvector of corresponding to if 0 and = .In other words, a nonzero vector is an eigenvector of an operator \n\n () if and only if is a scalar multiple of . Because = if and only if ( ) = 0 , a vector with 0 is an eigenvector of corresponding to if and only if null ( ) . \n\n> 5.9 example:\n\neigenvalues and eigenvectors \n\nSuppose ( 2) is defined by ( , ) = ( , ) .(a) First consider the case = . Then is a counterclockwise rotation by 90 \n\nabout the origin in 2. An operator has an eigenvalue if and only if there exists a nonzero vector in its domain that gets sent by the operator to a scalar multiple of itself. A 90 counterclockwise rotation of a nonzero vector in 2\n\ncannot equal a scalar multiple of itself. Conclusion: if = , then has no eigenvalues (and thus has no eigenvectors). (b) Now consider the case = . To find eigenvalues of , we must find the scalars such that ( , ) = ( , ) has some solution other than = = 0 .The equation ( , ) = ( , ) is equivalent to the simultaneous equations \n\n> 5.10\n\n = , = . \n\nSubstituting the value for given by the second equation into the first equation gives \n\n = 2. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 136 Chapter 5 Eigenvalues and Eigenvectors\n\nNow cannot equal 0 [ otherwise 5.10 implies that = 0 ; we are looking for solutions to 5.10 such that ( , ) is not the 0 vector ], so the equation above leads to the equation \n\n1 = 2.\n\nThe solutions to this equation are = and = .You can verify that and are eigenvalues of . Indeed, the eigenvectors corresponding to the eigenvalue are the vectors of the form ( , ) , with \n\n and 0 . Furthermore, the eigenvectors corresponding to the eigenvalue are the vectors of the form ( , ) , with and 0 .In the next proof, we again use the equivalence \n\n = ( ) = 0. \n\n> 5.11\n\nlinearly independent eigenvectors \n\nSuppose () . Then every list of eigenvectors of corresponding to distinct eigenvalues of is linearly independent. \n\n> Proof\n\nSuppose the desired result is false. Then there exists a smallest positive integer such that there exists a linearly dependent list 1, , of eigenvectors of corresponding to distinct eigenvalues 1, , of (note that 2 \n\nbecause an eigenvector is, by definition, nonzero). Thus there exist 1, , ,none of which are 0 (because of the minimality of ), such that \n\n 1 1 + + = 0. \n\nApply to both sides of the equation above, getting \n\n 1( 1 ) 1 + + 1 ( 1 ) 1 = 0. \n\nBecause the eigenvalues 1, , are distinct, none of the coefficients above equal 0. Thus 1, , 1 is a linearly dependent list of 1 eigenvectors of \n\ncorresponding to distinct eigenvalues, contradicting the minimality of . This contradiction completes the proof. The result above leads to a short proof of the result below, which puts an upper bound on the number of distinct eigenvalues that an operator can have. \n\n> 5.12\n\noperator cannot have more eigenvalues than dimension of vector space \n\nSuppose is finite-dimensional. Then each operator on has at most dim \n\ndistinct eigenvalues. \n\n> Proof\n\nLet () . Suppose 1, , are distinct eigenvalues of . Let \n\n 1, , be corresponding eigenvectors. Then 5.11 implies that the list 1, , \n\nis linearly independent. Thus dim (see 2.22), as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5A Invariant Subspaces 137 \n\n# Polynomials Applied to Operators \n\nThe main reason that a richer theory exists for operators (which map a vector space into itself) than for more general linear maps is that operators can be raised to powers. In this subsection we define that notion and the concept of applying a polynomial to an operator. This concept will be the key tool that we use in the next section when we prove that every operator on a nonzero finite-dimensional complex vector space has an eigenvalue. If is an operator, then makes sense (see 3.7) and is also an operator on the same vector space as . We usually write 2 instead of . More generally, we have the following definition of . \n\n> 5.13 notation:\n\n \n\nSuppose () and is a positive integer. \n\n () is defined by = \n\n> times\n\n.\n\n 0 is defined to be the identity operator on .\n\n If is invertible with inverse 1 , then () is defined by \n\n = ( 1 ) .\n\nYou should verify that if is an operator, then \n\n = + and ( ) = ,\n\nwhere and are arbitrary integers if is invertible and are nonnegative integers if is not invertible. Having defined powers of an operator, we can now define what it means to apply a polynomial to an operator. \n\n> 5.14 notation:\n\n() \n\nSuppose () and () is a polynomial given by \n\n() = 0 + 1 + 2 2 + + \n\nfor all . Then () is the operator on defined by \n\n() = 0 + 1 + 2 2 + + .\n\nThis is a new use of the symbol because we are applying to operators, not just elements of . The idea here is that to evaluate () , we simply replace with \n\n in the expression defining . Note that the constant term 0 in () becomes the operator 0 ( which is a reasonable choice because 0 = 0 0 and thus we should replace 0 with 0 0, which equals 0) .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n138 Chapter 5 Eigenvalues and Eigenvectors \n\n> 5.15 example:\n\na polynomial applied to the differentiation operator \n\nSuppose ( ()) is the differentiation operator defined by = and \n\n is the polynomial defined by () = 7 3 + 5 2. Then () = 7 3 + 5 2.Thus \n\n(()) = 7 3 + 5 \n\nfor every () .If we fix an operator () , then the function from () to () given by () is linear, as you should verify. \n\n> 5.16 definition:\n\nproduct of polynomials \n\nIf , () , then () is the polynomial defined by \n\n()() = ()() \n\nfor all .The order does not matter in taking products of polynomials of a single operator, as shown by (b) in the next result. \n\n> 5.17\n\nmultiplicative properties \n\nSuppose , () and () .Then (a) ()() = ()() ;(b) ()() = ()() .\n\nInformal proof: When a product of polynomials is expanded using the dis-tributive property, it does not matter whether the symbol is or .\n\nProof \n\n(a) Suppose () = \n\n> \n\n\n\n> = 0\n\n and () = \n\n> \n\n\n\n> = 0\n\n for all . Then \n\n()() = \n\n> \n\n\n\n> = 0\n> \n\n\n\n> = 0\n\n + .\n\nThus \n\n()() = \n\n> \n\n\n\n> = 0\n> \n\n\n\n> = 0\n\n + \n\n= ( \n\n> \n\n\n\n> = 0\n\n )( \n\n> \n\n\n\n> = 0\n\n )= ()(). \n\n(b) Using (a) twice, we have ()() = ()() = ()() = ()() . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 5A Invariant Subspaces 139\n\nWe observed earlier that if () , then the subspaces null and range \n\nare invariant under (see 5.4). Now we show that the null space and the range of every polynomial of are also invariant under . \n\n> 5.18\n\nnull space and range of () are invariant under \n\nSuppose () and () . Then null () and range () are invariant under . \n\n> Proof\n\nSuppose null () . Then () = 0 . Thus \n\n(())() = (() )() = ( ())() = (()) = (0) = 0. \n\nHence null () . Thus null () is invariant under , as desired. Suppose range () . Then there exists such that = () . Thus \n\n = (()) = ()(). \n\nHence range () . Thus range () is invariant under , as desired. \n\n# Exercises 5A \n\n1 Suppose () and is a subspace of .(a) Prove that if null , then is invariant under .(b) Prove that if range , then is invariant under .\n\n2 Suppose that () and 1, , are subspaces of invariant under .Prove that 1 + + is invariant under .\n\n3 Suppose () . Prove that the intersection of every collection of subspaces of invariant under is invariant under .\n\n4 Prove or give a counterexample: If is finite-dimensional and is a sub-space of that is invariant under every operator on , then = {0} or \n\n = .\n\n5 Suppose ( 2) is defined by ( , ) = (3 , ) . Find the eigenvalues of .\n\n6 Define ( 2) by ( , ) = ( , ) . Find all eigenvalues and eigenvec-tors of .\n\n7 Define ( 3) by ( 1, 2, 3) = (2 2, 0, 5 3). Find all eigenvalues and eigenvectors of .\n\n8 Suppose () is such that 2 = . Prove that if is an eigenvalue of ,then = 0 or = 1 .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n140 Chapter 5 Eigenvalues and Eigenvectors \n\n9 Define () () by = . Find all eigenvalues and eigenvectors of .\n\n10 Define ( 4()) by ()() = () for all . Find all eigen-values and eigenvectors of .\n\n11 Suppose is finite-dimensional, () , and . Prove that there ex-ists > 0 such that is invertible for all such that 0 < | | < .\n\n12 Suppose = , where and are nonzero subspaces of . Define \n\n () by ( + ) = for each and each . Find all eigenvalues and eigenvectors of .\n\n13 Suppose () . Suppose () is invertible. (a) Prove that and 1 have the same eigenvalues. (b) What is the relationship between the eigenvectors of and the eigenvec-tors of 1 ?\n\n14 Give an example of an operator on 4 that has no (real) eigenvalues. \n\n15 Suppose is finite-dimensional, () , and . Show that is an eigenvalue of if and only if is an eigenvalue of the dual operator \n\n ( ).\n\n16 Suppose 1, , is a basis of and () . Prove that if is an eigenvalue of , then \n\n|| max { () , 1 , } ,\n\nwhere () , denotes the entry in row , column of the matrix of with respect to the basis 1, , .\n\nSee Exercise 19 in Section 6A for a different bound on || .\n\n17 Suppose = , () , and . Prove that is an eigenvalue of \n\nif and only if is an eigenvalue of the complexification .\n\nSee Exercise 33 in Section 3B for the definition of .\n\n18 Suppose = , () , and . Prove that is an eigenvalue of the complexification if and only if is an eigenvalue of .\n\n19 Show that the forward shift operator ( ) defined by \n\n( 1, 2, ) = (0 , 1, 2, ) \n\nhas no eigenvalues. \n\n20 Define the backward shift operator ( ) by \n\n( 1, 2, 3, ) = ( 2, 3, ). \n\n(a) Show that every element of is an eigenvalue of .(b) Find all eigenvectors of . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 5A Invariant Subspaces 100 2\n\n21 Suppose () is invertible. (a) Suppose with 0 . Prove that is an eigenvalue of if and only if 1 is an eigenvalue of 1 .(b) Prove that and 1 have the same eigenvectors. \n\n22 Suppose () and there exist nonzero vectors and in such that \n\n = 3 and = 3. \n\nProve that 3 or 3 is an eigenvalue of .\n\n23 Suppose is finite-dimensional and , () . Prove that and \n\nhave the same eigenvalues. \n\n24 Suppose is an -by- matrix with entries in . Define ( ) by \n\n = , where elements of are thought of as -by-1 column vectors. (a) Suppose the sum of the entries in each row of equals 1. Prove that 1\n\nis an eigenvalue of .(b) Suppose the sum of the entries in each column of equals 1. Prove that \n\n1 is an eigenvalue of .\n\n25 Suppose () and , are eigenvectors of such that + is also an eigenvector of . Prove that and are eigenvectors of corresponding to the same eigenvalue. \n\n26 Suppose () is such that every nonzero vector in is an eigenvector of . Prove that is a scalar multiple of the identity operator. \n\n27 Suppose that is finite-dimensional and {1 , , dim 1} . Suppose \n\n () is such that every subspace of of dimension is invariant under . Prove that is a scalar multiple of the identity operator. \n\n28 Suppose is finite-dimensional and () . Prove that has at most \n\n1 + dim range distinct eigenvalues. \n\n29 Suppose ( 3) and 4 , 5, and 7 are eigenvalues of . Prove that there exists 3 such that 9 = (4 , 5, 7) .\n\n30 Suppose () and ( 2)( 3)( 4) = 0 . Suppose is an eigenvalue of . Prove that = 2 or = 3 or = 4 .\n\n31 Give an example of ( 2) such that 4 = .\n\n32 Suppose () has no eigenvalues and 4 = . Prove that 2 = .\n\n33 Suppose () and is a positive integer. (a) Prove that is injective if and only if is injective. (b) Prove that is surjective if and only if is surjective. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n142 Chapter 5 Eigenvalues and Eigenvectors \n\n34 Suppose is finite-dimensional and 1, , . Prove that the list \n\n 1, , is linearly independent if and only if there exists () such that 1, , are eigenvectors of corresponding to distinct eigenvalues. \n\n35 Suppose that 1, , is a list of distinct real numbers. Prove that the list 1 , , is linearly independent in the vector space of real-valued functions on .\n\nHint: Let = span ( 1 , , ), and define an operator () by \n\n> = \n\n. Find eigenvalues and eigenvectors of .\n\n36 Suppose that 1, , is a list of distinct positive numbers. Prove that the list cos ( 1) , , cos ( ) is linearly independent in the vector space of real-valued functions on .\n\n37 Suppose is finite-dimensional and () . Define (()) by \n\n () = \n\nfor each () . Prove that the set of eigenvalues of equals the set of eigenvalues of .\n\n38 Suppose is finite-dimensional, () , and is a subspace of \n\ninvariant under . The quotient operator / (/) is defined by \n\n(/)( + ) = + \n\nfor each .(a) Show that the definition of / makes sense (which requires using the condition that is invariant under ) and show that / is an operator on / .(b) Show that each eigenvalue of / is an eigenvalue of .\n\n39 Suppose is finite-dimensional and () . Prove that has an eigen-value if and only if there exists a subspace of of dimension dim 1 that is invariant under .\n\n40 Suppose , () and is invertible. Suppose () is a polynomial. Prove that \n\n( 1 ) = () 1 .\n\n41 Suppose () and is a subspace of invariant under . Prove that \n\n is invariant under () for every polynomial () .\n\n42 Define ( ) by ( 1, 2, 3, , ) = ( 1, 2 2, 3 3, , ).(a) Find all eigenvalues and eigenvectors of .(b) Find all subspaces of that are invariant under .\n\n43 Suppose that is finite-dimensional, dim > 1 , and () . Prove that \n\n{() ()} () . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 5B The Minimal Polynomial 143\n\n# 5B The Minimal Polynomial \n\n# Existence of Eigenvalues on Complex Vector Spaces \n\nNow we come to one of the central results about operators on finite-dimensional complex vector spaces. \n\n> 5.19\n\nexistence of eigenvalues \n\nEvery operator on a finite-dimensional nonzero complex vector space has an eigenvalue. \n\n> Proof\n\nSuppose is a finite-dimensional complex vector space of dimension \n\n > 0 and () . Choose with 0 . Then \n\n , , 2 , , \n\nis not linearly independent, because has dimension and this list has length \n\n + 1. Hence some linear combination (with not all the coefficients equal to 0)of the vectors above equals 0. Thus there exists a nonconstant polynomial of smallest degree such that \n\n() = 0. \n\nBy the first version of the fundamental theorem of algebra (see 4.12), there exists such that () = 0 . Hence there exists a polynomial () such that \n\n() = ( )() \n\nfor every (see 4.6). This implies (using 5.17) that \n\n0 = () = ( )(()). \n\nBecause has smaller degree than , we know that () 0 . Thus the equation above implies that is an eigenvalue of with eigenvector () .The proof above makes crucial use of the fundamental theorem of algebra. The comment following Exercise 16 helps explain why the fundamental theorem of algebra is so tightly connected to the result above. The hypothesis in the result above that = cannot be replaced with the hypothesis that = , as shown by Example 5.9. The next example shows that the finite-dimensional hypothesis in the result above also cannot be deleted. \n\n> 5.20 example:\n\nan operator on a complex vector space with no eigenvalues \n\nDefine ( ()) by ()() = () . If () is a nonzero poly-nomial, then the degree of is one more than the degree of , and thus cannot equal a scalar multiple of . Hence has no eigenvalues. Because () is infinite-dimensional, this example does not contradict the result above. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n144 Chapter 5 Eigenvalues and Eigenvectors \n\n# Eigenvalues and the Minimal Polynomial \n\nIn this subsection we introduce an important polynomial associated with each operator. We begin with the following definition. \n\n> 5.21 definition:\n\nmonic polynomial \n\nA monic polynomial is a polynomial whose highest-degree coefficient equals 1.For example, the polynomial 2 + 9 2 + 7 is a monic polynomial of degree 7. \n\n> 5.22\n\nexistence, uniqueness, and degree of minimal polynomial \n\nSuppose is finite-dimensional and () . Then there is a unique monic polynomial () of smallest degree such that () = 0 . Furthermore, \n\ndeg dim .\n\nProof If dim = 0 , then is the zero operator on and thus we take to be the constant polynomial 1.Now use induction on dim . Thus assume that dim > 0 and that the desired result is true for all operators on all vector spaces of smaller dimension. Let \n\n be such that 0 . The list , , , dim has length 1 + dim and thus is linearly dependent. By the linear dependence lemma (2.19), there is a smallest positive integer dim such that is a linear combination of \n\n , , , 1 . Thus there exist scalars 0, 1, 2, , 1 such that \n\n> 5.23\n\n 0 + 1 + + 1 1 + = 0. \n\nDefine a monic polynomial () by \n\n() = 0 + 1 + + 1 1 + .\n\nThen 5.23 implies that () = 0 .If is a nonnegative integer, then \n\n()( ) = (()) = (0) = 0. \n\nThe linear dependence lemma (2.19) shows that , , , 1 is linearly inde-pendent. Thus the equation above implies that dim null () . Hence \n\ndim range () = dim dim null () dim . \n\nBecause range () is invariant under (by 5.18), we can apply our induction hypothesis to the operator | range () on the vector space range () . Thus there is a monic polynomial () with \n\ndeg dim and (| range () ) = 0. \n\nHence for all we have \n\n(()())() = ()(()) = 0 \n\nbecause () range () and ()| range () = (| range () ) = 0 . Thus is a monic polynomial such that deg dim and ()() = 0 .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5B The Minimal Polynomial 145 \n\nThe paragraph above shows that there is a monic polynomial of degree at most dim that when applied to gives the 0 operator. Thus there is a monic polynomial of smallest degree with this property, completing the existence part of this result. Let () be a monic polynomial of smallest degree such that () = 0 .To prove the uniqueness part of the result, suppose () is a monic poly-nomial of the same degree as and () = 0 . Then ( )() = 0 and also \n\ndeg ( ) < deg . If were not equal to 0, then we could divide by the coefficient of the highest-order term in to get a monic polynomial (of smaller degree than ) that when applied to gives the 0 operator. Thus = 0 ,as desired. The previous result justifies the following definition. \n\n> 5.24 definition:\n\nminimal polynomial \n\nSuppose is finite-dimensional and () . Then the minimal polynomial \n\nof is the unique monic polynomial () of smallest degree such that \n\n() = 0 .To compute the minimal polynomial of an operator () , we need to find the smallest positive integer such that the equation \n\n 0 + 1 + + 1 1 = \n\nhas a solution 0, 1, , 1 . If we pick a basis of and replace in the equation above with the matrix of , then the equation above can be thought of as a system of (dim ) 2 linear equations in the unknowns 0, 1, , 1 .Gaussian elimination or another fast method of solving systems of linear equations can tell us whether a solution exists, testing successive values = 1 , 2, until a solution exists. By 5.22, a solution exists for some smallest positive integer \n\n dim . The minimal polynomial of is then 0 + 1 + + 1 1 + .Even faster (usually), pick with 0 and consider the equation \n\n> 5.25\n\n 0 + 1 + + dim 1 dim 1 = dim . \n\nUse a basis of to convert the equation above to a system of dim linear equa-tions in dim unknowns 0, 1, , dim 1 . If this system of equations has a unique solution 0, 1, , dim 1 (as happens most of the time), then the scalars \n\n 0, 1, , dim 1 , 1 are the coefficients of the minimal polynomial of (because 5.22 states that the degree of the minimal polynomial is at most dim ). \n\nThese estimates are based on testing millions of random matrices. \n\nConsider operators on 4 (thought of as 4-by-4 matrices with respect to the standard basis), and take = (1 , 0, 0, 0) \n\nin the paragraph above. The faster method described above works on over 99.8% of the 4-by-4 matrices with integer entries in the interval [10 , 10] and on over 99.999% of the 4-by-4 matrices with integer entries in [100 , 100] .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n146 Chapter 5 Eigenvalues and Eigenvectors \n\nThe next example illustrates the faster procedure discussed above. \n\n> 5.26 example:\n\nminimal polynomial of an operator on 5\n\nSuppose ( 5) and \n\n() = 0 0 0 0 3 1 0 0 0 60 1 0 0 00 0 1 0 00 0 0 1 0\n\nwith respect to the standard basis 1, 2, 3, 4, 5. Taking = 1 for 5.25, we have \n\n 1 = 2, 4 1 = ( 3 1) = 4 = 5,\n\n 2 1 = ( 1) = 2 = 3, 5 1 = ( 4 1) = 5 = 3 1 + 6 2. 3 1 = ( 2 1) = 3 = 4,\n\nThus 3 1 6 1 = 5 1. The list 1, 1, 2 1, 3 1, 4 1, which equals the list \n\n 1, 2, 3, 4, 5, is linearly independent, so no other linear combination of this list equals 5 1. Hence the minimal polynomial of is 3 6 + 5.Recall that by definition, eigenvalues of operators on and zeros of polyno-mials in () must be elements of . In particular, if = , then eigenvalues and zeros must be real numbers. \n\n> 5.27\n\neigenvalues are the zeros of the minimal polynomial \n\nSuppose is finite-dimensional and () .(a) The zeros of the minimal polynomial of are the eigenvalues of .(b) If is a complex vector space, then the minimal polynomial of has the form \n\n( 1)( ),\n\nwhere 1, , is a list of all eigenvalues of , possibly with repetitions. \n\nProof Let be the minimal polynomial of .(a) First suppose is a zero of . Then can be written in the form \n\n() = ( )() ,\n\nwhere is a monic polynomial with coefficients in (see 4.6). Because \n\n() = 0 , we have \n\n0 = ( )(()) \n\nfor all . Because deg = ( deg ) 1 and is the minimal polynomial of , there exists at least one vector such that () 0 . The equation above thus implies that is an eigenvalue of , as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 5B The Minimal Polynomial 147\n\nTo prove that every eigenvalue of is a zero of , now suppose is an eigenvalue of . Thus there exists with 0 such that = .Repeated applications of to both sides of this equation show that = \n\nfor every nonnegative integer . Thus \n\n() = (). \n\nBecause is the minimal polynomial of , we have () = 0 . Hence the equation above implies that () = 0 . Thus is a zero of , as desired. (b) To get the desired result, use (a) and the second version of the fundamental theorem of algebra (see 4.13). A nonzero polynomial has at most as many distinct zeros as its degree (see 4.8). Thus (a) of the previous result, along with the result that the minimal polynomial of an operator on has degree at most dim , gives an alternative proof of 5.12, which states that an operator on has at most dim distinct eigenvalues. Every monic polynomial is the minimal polynomial of some operator, as shown by Exercise 16, which generalizes Example 5.26. Thus 5.27(a) shows that finding exact expressions for the eigenvalues of an operator is equivalent to the problem of finding exact expressions for the zeros of a polynomial (and thus is not possible for some operators). \n\n> 5.28 example:\n\nAn operator whose eigenvalues cannot be found exactly \n\nLet ( 5) be the operator defined by \n\n( 1, 2, 3, 4, 5) = (3 5, 1 + 6 5, 2, 3, 4). \n\nThe matrix of with respect to the standard basis of 5 is the 5-by-5 matrix in Example 5.26. As we showed in that example, the minimal polynomial of is the polynomial \n\n3 6 + 5.\n\nNo zero of the polynomial above can be expressed using rational numbers, roots of rational numbers, and the usual rules of arithmetic (a proof of this would take us considerably beyond linear algebra). Because the zeros of the polynomial above are the eigenvalues of [ by 5.27(a) ], we cannot find an exact expression for any eigenvalue of in any familiar form. Numeric techniques, which we will not discuss here, show that the zeros of the polynomial above, and thus the eigenvalues of , are approximately the following five complex numbers: \n\n1.67 , 0.51 , 1.40 , 0.12 + 1.59 , 0.12 1.59. \n\nNote that the two nonreal zeros of this polynomial are complex conjugates of each other, as we expect for a polynomial with real coefficients (see 4.14). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 148 Chapter 5 Eigenvalues and Eigenvectors\n\nThe next result completely characterizes the polynomials that when applied to an operator give the 0 operator. \n\n> 5.29\n\n() = 0 is a polynomial multiple of the minimal polynomial \n\nSuppose is finite-dimensional, () , and () . Then () = 0 \n\nif and only if is a polynomial multiple of the minimal polynomial of . \n\n> Proof\n\nLet denote the minimal polynomial of .First suppose () = 0 . By the division algorithm for polynomials (4.9), there exist polynomials , () such that \n\n> 5.30\n\n = + \n\nand deg < deg . We have \n\n0 = () = ()() + () = (). \n\nThe equation above implies that = 0 (otherwise, dividing by its highest-degree coefficient would produce a monic polynomial that when applied to gives 0;this polynomial would have a smaller degree than the minimal polynomial, which would be a contradiction). Thus 5.30 becomes the equation = . Hence is a polynomial multiple of , as desired. To prove the other direction, now suppose is a polynomial multiple of .Thus there exists a polynomial () such that = . We have \n\n() = ()() = 0 () = 0 ,\n\nas desired. The next result is a nice consequence of the result above. \n\n> 5.31\n\nminimal polynomial of a restriction operator \n\nSuppose is finite-dimensional, () , and is a subspace of that is invariant under . Then the minimal polynomial of is a polynomial multiple of the minimal polynomial of | . \n\n> Proof\n\nSuppose is the minimal polynomial of . Thus () = 0 for all .In particular, \n\n() = 0 for all . \n\nThus (| ) = 0 . Now 5.29, applied to the operator | in place of , implies that is a polynomial multiple of the minimal polynomial of | .See Exercise 25 for a result about quotient operators that is analogous to the result above. The next result shows that the constant term of the minimal polynomial of an operator determines whether the operator is invertible. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 5B The Minimal Polynomial 149\n> 5.32\n\n not invertible constant term of minimal polynomial of is 0\n\nSuppose is finite-dimensional and () . Then is not invertible if and only if the constant term of the minimal polynomial of is 0. \n\n> Proof\n\nSuppose () and is the minimal polynomial of . Then \n\n is not invertible 0 is an eigenvalue of 0 is a zero of the constant term of is 0,\n\nwhere the first equivalence holds by 5.7, the second equivalence holds by 5.27(a), and the last equivalence holds because the constant term of equals (0) .\n\n# Eigenvalues on Odd-Dimensional Real Vector Spaces \n\nThe next result will be the key tool that we use to show that every operator on an odd-dimensional real vector space has an eigenvalue. \n\n> 5.33\n\neven-dimensional null space \n\nSuppose = and is finite-dimensional. Suppose also that () \n\nand , with 2 < 4 . Then dim null ( 2 + + ) is an even number. \n\n> Proof\n\nRecall that null ( 2 + + ) is invariant under (by 5.18). By replacing \n\n with null ( 2 + + ) and replacing with restricted to null ( 2 + + ) ,we can assume that 2 + + = 0 ; we now need to prove that dim is even. Suppose and are such that = . Then \n\n0 = ( 2 + + ) = ( 2 + + ) = (( + 2 )2\n\n+ 2 \n\n> 4\n\n). \n\nThe term in large parentheses above is a positive number. Thus the equation above implies that = 0 . Hence we have shown that has no eigenvectors. Let be a subspace of that is invariant under and has the largest dimension among all subspaces of that are invariant under and have even dimension. If \n\n = , then we are done; otherwise assume there exists such that .Let = span ( , ) . Then is invariant under because () = . Furthermore, dim = 2 because otherwise would be an eigen-vector of . Now \n\ndim ( + ) = dim + dim dim ( ) = dim + 2,\n\nwhere = {0} because otherwise would be a one-dimensional subspace of that is invariant under (impossible because has no eigenvectors). Because + is invariant under , the equation above shows that there exists a subspace of invariant under of even dimension larger than dim . Thus the assumption that was incorrect. Hence has even dimension. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 150 Chapter 5 Eigenvalues and Eigenvectors\n\nThe next result states that on odd-dimensional vector spaces, every operator has an eigenvalue. We already know this result for finite-dimensional complex vectors spaces (without the odd hypothesis). Thus in the proof below, we will assume that = . \n\n> 5.34\n\noperators on odd-dimensional vector spaces have eigenvalues \n\nEvery operator on an odd-dimensional vector space has an eigenvalue. \n\n> Proof\n\nSuppose = and is finite-dimensional. Let = dim , and suppose \n\n is an odd number. Let () . We will use induction on in steps of size two to show that has an eigenvalue. To get started, note that the desired result holds if dim = 1 because then every nonzero vector in is an eigenvector of .Now suppose that 3 and the desired result holds for all operators on all odd-dimensional vector spaces of dimension less than . Let denote the minimal polynomial of . If is a polynomial multiple of for some , then is an eigenvalue of [ by 5.27(a) ] and we are done. Thus we can assume that there exist , such that 2 < 4 and is a polynomial multiple of 2 + + (see 4.16). There exists a monic polynomial () such that () = ()( 2 + + ) \n\nfor all . Now \n\n0 = () = (())( 2 + + ) ,\n\nwhich means that () equals 0 on range ( 2 + + ) . Because deg < deg \n\nand is the minimal polynomial of , this implies that range ( 2 + + ) .The fundamental theorem of linear maps (3.21) tells us that \n\ndim = dim null ( 2 + + ) + dim range ( 2 + + ). \n\nBecause dim is odd (by hypothesis) and dim null ( 2 + + ) is even (by 5.33), the equation above shows that dim range ( 2 + + ) is odd. Hence range ( 2 + + ) is a subspace of that is invariant under (by 5.18) and that has odd dimension less than dim . Our induction hypothesis now implies that restricted to range ( 2 + + ) has an eigenvalue, which means that has an eigenvalue. See Exercise 23 in Section 8B and Exercise 10 in Section 9C for alternative proofs of the result above. \n\n# Exercises 5B \n\n1 Suppose () . Prove that 9 is an eigenvalue of 2 if and only if 3 or \n\n3 is an eigenvalue of .\n\n2 Suppose is a complex vector space and () has no eigenvalues. Prove that every subspace of invariant under is either {0} or infinite-dimensional. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5B The Minimal Polynomial 151 \n\n3 Suppose is an integer with > 1 and ( ) is defined by \n\n( 1, , ) = ( 1 + + , , 1 + + ). \n\n(a) Find all eigenvalues and eigenvectors of .(b) Find the minimal polynomial of .\n\nThe matrix of with respect to the standard basis of consists of all 1s. \n\n4 Suppose = , () , () is a nonconstant polynomial, and \n\n . Prove that is an eigenvalue of () if and only if = () for some eigenvalue of .\n\n5 Give an example of an operator on 2 that shows the result in Exercise 4 does not hold if is replaced with .\n\n6 Suppose ( 2) is defined by ( , ) = ( , ) . Find the minimal polynomial of .\n\n7 (a) Give an example of , ( 2) such that the minimal polynomial of \n\n does not equal the minimal polynomial of .(b) Suppose is finite-dimensional and , () . Prove that if at least one of , is invertible, then the minimal polynomial of equals the minimal polynomial of .\n\nHint: Show that if is invertible and () , then () = 1 () .\n\n8 Suppose ( 2) is the operator of counterclockwise rotation by 1.Find the minimal polynomial of .\n\nBecause dim 2 = 2 , the degree of the minimal polynomial of is at most 2.Thus the minimal polynomial of is not the tempting polynomial 180 + 1,even though 180 = .\n\n9 Suppose () is such that with respect to some basis of , all entries of the matrix of are rational numbers. Explain why all coefficients of the minimal polynomial of are rational numbers. \n\n10 Suppose is finite-dimensional, () , and . Prove that \n\nspan ( , , , ) = span ( , , , dim 1 ) \n\nfor all integers dim 1 .\n\n11 Suppose is a two-dimensional vector space, () , and the matrix of \n\n with respect to some basis of is .(a) Show that 2 ( + ) + ( ) = 0 .(b) Show that the minimal polynomial of equals \n\n{{ if = = 0 and = ,\n\n 2 ( + ) + ( ) otherwise .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n152 Chapter 5 Eigenvalues and Eigenvectors \n\n12 Define ( ) by ( 1, 2, 3, , ) = ( 1, 2 2, 3 3, , ). Find the minimal polynomial of .\n\n13 Suppose is finite-dimensional, () , and () . Prove that there exists a unique () such that () = () and deg is less than the degree of the minimal polynomial of .\n\n14 Suppose is finite-dimensional and () has minimal polynomial \n\n4 + 5 6 2 7 3 + 2 4 + 5. Find the minimal polynomial of 1 .\n\n15 Suppose is a finite-dimensional complex vector space with dim > 0 \n\nand () . Define by \n\n () = dim range ( ). \n\nProve that is not a continuous function. \n\n16 Suppose 0, , 1 . Let be the operator on whose matrix (with respect to the standard basis) is \n\n0 0\n\n1 0 1\n\n1 2\n\n 0 2 \n\n1 1 \n\n.\n\nHere all entries of the matrix are 0 except for all 1s on the line under the diagonal and the entries in the last column (some of which might also be 0). Show that the minimal polynomial of is the polynomial \n\n 0 + 1 + + 1 1 + .\n\nThe matrix above is called the companion matrix of the polynomial above. This exercise shows that every monic polynomial is the minimal polynomial of some operator. Hence a formula or an algorithm that could produce exact eigenvalues for each operator on each could then produce exact zeros for each polynomial [by 5.27 (a)]. Thus there is no such formula or algorithm. However, efficient numeric methods exist for obtaining very good approximations for the eigenvalues of an operator. \n\n17 Suppose is finite-dimensional, () , and is the minimal polynomial of . Suppose . Show that the minimal polynomial of is the polynomial defined by () = ( + ) .\n\n18 Suppose is finite-dimensional, () , and is the minimal polynomial of . Suppose \\{0} . Show that the minimal polynomial of is the polynomial defined by () = deg ( ).\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5B The Minimal Polynomial 153 \n\n19 Suppose is finite-dimensional and () . Let be the subspace of \n\n() defined by \n\n = {() ()}. \n\nProve that dim equals the degree of the minimal polynomial of .\n\n20 Suppose ( 4) is such that the eigenvalues of are 3, 5, 8. Prove that \n\n( 3) 2( 5) 2( 8) 2 = 0 .\n\n21 Suppose is finite-dimensional and () . Prove that the minimal polynomial of has degree at most 1 + dim range .\n\nIf dim range < dim 1 , then this exercise gives a better upper bound than 5.22 for the degree of the minimal polynomial of .\n\n22 Suppose is finite-dimensional and () . Prove that is invertible if and only if span ( , 2, , dim ).\n\n23 Suppose is finite-dimensional and () . Let = dim . Prove that if , then span ( , , , 1 ) is invariant under .\n\n24 Suppose is a finite-dimensional complex vector space. Suppose () \n\nis such that 5 and 6 are eigenvalues of and that has no other eigenvalues. Prove that ( 5) dim 1 ( 6) dim 1 = 0 .\n\n25 Suppose is finite-dimensional, () , and is a subspace of that is invariant under .(a) Prove that the minimal polynomial of is a polynomial multiple of the minimal polynomial of the quotient operator / .(b) Prove that (minimal polynomial of | ) (minimal polynomial of / )is a polynomial multiple of the minimal polynomial of .\n\nThe quotient operator / was defined in Exercise 38 in Section 5A. \n\n26 Suppose is finite-dimensional, () , and is a subspace of that is invariant under . Prove that the set of eigenvalues of equals the union of the set of eigenvalues of | and the set of eigenvalues of / .\n\n27 Suppose = , is finite-dimensional, and () . Prove that the minimal polynomial of equals the minimal polynomial of .\n\nThe complexification was defined in Exercise 33 of Section 3B. \n\n28 Suppose is finite-dimensional and () . Prove that the minimal polynomial of ( ) equals the minimal polynomial of .\n\nThe dual map was defined in Section 3F. \n\n29 Show that every operator on a finite-dimensional vector space of dimension at least two has an invariant subspace of dimension two. \n\nExercise 6 in Section 5C will give an improvement of this result when = .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n154 Chapter 5 Eigenvalues and Eigenvectors \n\n# 5C Upper-Triangular Matrices \n\nIn Chapter 3 we defined the matrix of a linear map from a finite-dimensional vector space to another finite-dimensional vector space. That matrix depends on a choice of basis of each of the two vector spaces. Now that we are studying operators, which map a vector space to itself, the emphasis is on using only one basis. \n\n> 5.35 definition:\n\nmatrix of an operator, () \n\nSuppose () . The matrix of with respect to a basis 1, , of is the -by- matrix \n\n() = 1,1 1, \n\n ,1 , \n\n\n\nwhose entries , are defined by \n\n = 1, 1 + + , .\n\nThe notation ( , ( 1, , )) is used if the basis is not clear from the context. Operators have square matrices (meaning that the number of rows equals the number of columns), rather than the more general rectangular matrices that we considered earlier for linear maps. \n\nThe th column of the matrix () is formed from the coefficients used to write as a linear combination of the basis 1, , .\n\nIf is an operator on and no ba-sis is specified, assume that the basis in question is the standard one (where the \n\n th basis vector is 1 in the th slot and 0\n\nin all other slots ). You can then think of the th column of () as applied to the th basis vector, where we identify \n\n -by-1 column vectors with elements of . \n\n> 5.36 example:\n\nmatrix of an operator with respect to standard basis \n\nDefine ( 3) by ( , , ) = (2 + , 5 + 3 , 8) . Then the matrix of \n\nwith respect to the standard basis of 3 is \n\n() = 2 1 00 5 30 0 8\n\n,\n\nas you should verify. A central goal of linear algebra is to show that given an operator on a finite-dimensional vector space , there exists a basis of with respect to which has a reasonably simple matrix. To make this vague formulation a bit more precise, we might try to choose a basis of such that () has many 0s. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 5C Upper-Triangular Matrices 155\n\nIf is a finite-dimensional complex vector space, then we already know enough to show that there is a basis of with respect to which the matrix of \n\nhas 0s everywhere in the first column, except possibly the first entry. In other words, there is a basis of with respect to which the matrix of looks like \n\n 0 \n\n0\n\n;\n\nhere denotes the entries in all columns other than the first column. To prove this, let be an eigenvalue of (one exists by 5.19) and let be a corresponding eigenvector. Extend to a basis of . Then the matrix of with respect to this basis has the form above. Soon we will see that we can choose a basis of with respect to which the matrix of has even more 0s. \n\n> 5.37 definition:\n\ndiagonal of a matrix \n\nThe diagonal of a square matrix consists of the entries on the line from the upper left corner to the bottom right corner. For example, the diagonal of the matrix \n\n() = 2 1 00 5 30 0 8\n\nfrom Example 5.36 consists of the entries 2, 5, 8, which are shown in red in the matrix above. \n\n> 5.38 definition:\n\nupper-triangular matrix \n\nA square matrix is called upper triangular if all entries below the diagonal are 0.For example, the 3-by-3 matrix above is upper triangular. Typically we represent an upper-triangular matrix in the form \n\n 1 \n\n0 \n\n\n\n; \n\n> We often use\n\n to denote matrix entries that we do not know or that are irrele-vant to the questions being discussed. \n\nthe 0 in the matrix above indicates that all entries below the diagonal in this \n\n -by- matrix equal 0. Upper-triangular matrices can be considered reasonably simpleif is large, then at least almost half the entries in an -by- upper-triangular matrix are 0.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n156 Chapter 5 Eigenvalues and Eigenvectors \n\nThe next result provides a useful connection between upper-triangular matrices and invariant subspaces. \n\n> 5.39\n\nconditions for upper-triangular matrix \n\nSuppose () and 1, , is a basis of . Then the following are equivalent. (a) The matrix of with respect to 1, , is upper triangular. (b) span ( 1, , ) is invariant under for each = 1 , , .(c) span ( 1, , ) for each = 1 , , .\n\nProof First suppose (a) holds. To prove that (b) holds, suppose {1 , , } .If {1 , , } , then \n\n span ( 1, , )\n\nbecause the matrix of with respect to 1, , is upper triangular. Because \n\nspan ( 1, , ) span ( 1, , ) if , we see that \n\n span ( 1, , )\n\nfor each {1 , , } . Thus span ( 1, , ) is invariant under , completing the proof that (a) implies (b). Now suppose (b) holds, so span ( 1, , ) is invariant under for each \n\n = 1 , , . In particular, span ( 1, , ) for each = 1 , , . Thus (b) implies (c). Now suppose (c) holds, so span ( 1, , ) for each = 1 , , . This means that when writing each as a linear combination of the basis vectors \n\n 1, , , we need to use only the vectors 1, , . Hence all entries under the diagonal of () are 0. Thus () is an upper-triangular matrix, completing the proof that (c) implies (a). We have shown that (a) (b) (c) (a), which shows that (a), (b), and (c) are equivalent. The next result tells us that if () and with respect to some basis of \n\nwe have \n\n() = 1 \n\n0 \n\n\n\n,\n\nthen satisfies a simple equation depending on 1, , . \n\n> 5.40\n\nequation satisfied by operator with upper-triangular matrix \n\nSuppose () and has a basis with respect to which has an upper-triangular matrix with diagonal entries 1, , . Then \n\n( 1)( ) = 0. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5C Upper-Triangular Matrices 157 \n\nProof Let 1, , denote a basis of with respect to which has an upper-triangular matrix with diagonal entries 1, , . Then 1 = 1 1, which means that ( 1) 1 = 0 , which implies that ( 1) ( ) 1 = 0 \n\nfor = 1 , , (using the commutativity of each with each ). Note that ( 2) 2 span ( 1). Thus ( 1)( 2) 2 = 0 (by the previous paragraph), which implies that ( 1) ( ) 2 = 0 for \n\n = 2 , , (using the commutativity of each with each ). Note that ( 3) 3 span ( 1, 2). Thus by the previous paragraph, \n\n( 1)( 2)( 3) 3 = 0 , which implies that ( 1) ( ) 3 =0 for = 3 , , (using the commutativity of each with each ). Continuing this pattern, we see that ( 1) ( ) = 0 for each \n\n = 1 , , . Thus ( 1) ( ) is the 0 operator because it is 0 on each vector in a basis of .Unfortunately no method exists for exactly computing the eigenvalues of an operator from its matrix. However, if we are fortunate enough to find a basis with respect to which the matrix of the operator is upper triangular, then the problem of computing the eigenvalues becomes trivial, as the next result shows. \n\n> 5.41\n\ndetermination of eigenvalues from upper-triangular matrix \n\nSuppose () has an upper-triangular matrix with respect to some basis of . Then the eigenvalues of are precisely the entries on the diagonal of that upper-triangular matrix. \n\nProof Suppose 1, , is a basis of with respect to which has an upper-triangular matrix \n\n() = 1 \n\n0 \n\n.\n\nBecause 1 = 1 1, we see that 1 is an eigenvalue of .Suppose {2 , , } . Then ( ) span ( 1, , 1 ). Thus \n\nmaps span ( 1, , ) into span ( 1, , 1 ). Because \n\ndim span ( 1, , ) = and dim span ( 1, , 1 ) = 1 ,\n\nthis implies that restricted to span ( 1, , ) is not injective (by 3.22). Thus there exists span ( 1, , ) such that 0 and ( ) = 0 . Thus \n\n is an eigenvalue of . Hence we have shown that every entry on the diagonal of () is an eigenvalue of .To prove has no other eigenvalues, let be the polynomial defined by \n\n() = ( 1) ( ). Then () = 0 (by 5.40). Hence is a polynomial multiple of the minimal polynomial of (by 5.29). Thus every zero of the minimal polynomial of is a zero of . Because the zeros of the minimal polynomial of \n\n are the eigenvalues of (by 5.27), this implies that every eigenvalue of is a zero of . Hence the eigenvalues of are all contained in the list 1, , . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 158 Chapter 5 Eigenvalues and Eigenvectors\n> 5.42 example:\n\neigenvalues via an upper-triangular matrix \n\nDefine ( 3) by ( , , ) = (2 + , 5 + 3 , 8) . The matrix of with respect to the standard basis is \n\n() = 2 1 00 5 30 0 8.\n\nNow 5.41 implies that the eigenvalues of are 2, 5, and 8.The next example illustrates 5.44: an operator has an upper-triangular matrix with respect to some basis if and only if the minimal polynomial of the operator is the product of polynomials of degree 1. \n\n> 5.43 example:\n\nwhether has an upper-triangular matrix can depend on \n\nDefine ( 4) by \n\n( 1, 2, 3, 4) = ( 2, 1, 2 1 + 3 3, 3 + 3 4). \n\nThus with respect to the standard basis of 4, the matrix of is \n\n0 1 0 01 0 0 02 0 3 00 0 1 3.\n\nYou can ask a computer to verify that the minimal polynomial of is the polyno-mial defined by \n\n() = 9 6 + 10 2 6 3 + 4.\n\nFirst consider the case = . Then the polynomial factors as \n\n() = ( 2 + 1)( 3)( 3) ,\n\nwith no further factorization of 2 + 1 as the product of two polynomials of degree \n\n1 with real coefficients. Thus 5.44 states that there does not exist a basis of 4\n\nwith respect to which has an upper-triangular matrix. Now consider the case = . Then the polynomial factors as \n\n() = ( )( + )( 3)( 3) ,\n\nwhere all factors above have the form . Thus 5.44 states that there is a basis of \n\n 4 with respect to which has an upper-triangular matrix. Indeed, you can verify that with respect to the basis (4 3 , 3 4 , 3 + , 1) , (4 + 3 , 3 + 4 , 3 , 1) ,\n\n(0 , 0, 0, 1) , (0 , 0, 1, 0) of 4, the operator has the upper-triangular matrix \n\n 0 0 00 0 00 0 3 10 0 0 3.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5C Upper-Triangular Matrices 159 \n\n> 5.44\n\nnecessary and sufficient condition to have an upper-triangular matrix \n\nSuppose is finite-dimensional and () . Then has an upper-triangular matrix with respect to some basis of if and only if the minimal polynomial of equals ( 1) ( ) for some 1, , .\n\nProof First suppose has an upper-triangular matrix with respect to some basis of . Let 1, , denote the diagonal entries of that matrix. Define a polynomial \n\n () by \n\n() = ( 1)( ). \n\nThen () = 0 , by 5.40. Hence is a polynomial multiple of the minimal polyno-mial of , by 5.29. Thus the minimal polynomial of equals ( 1) ( )\n\nfor some 1, , with { 1, , } { 1, , }.To prove the implication in the other direction, now suppose the minimal polynomial of equals ( 1) ( ) for some 1, , . We will use induction on . To get started, if = 1 then 1 is the minimal polynomial of \n\n , which implies that = 1 , which implies that the matrix of (with respect to any basis of ) is upper triangular. Now suppose > 1 and the desired result holds for all smaller positive integers. Let \n\n = range ( ). \n\nThen is invariant under [ this is a special case of 5.18 with () = ].Thus | is an operator on .If , then = ( ) for some and \n\n( 1)( 1 ) = ( 1)( ) = 0. \n\nHence ( 1) ( 1 ) is a polynomial multiple of the minimal polynomial of | , by 5.29. Thus the minimal polynomial of | is the product of at most \n\n 1 terms of the form .By our induction hypothesis, there is a basis 1, , of with respect to which | has an upper-triangular matrix. Thus for each {1 , , } , we have (using 5.39) \n\n> 5.45\n\n = (| )( ) span ( 1, , ). \n\nExtend 1, , to a basis 1, , , 1, , of . If {1 , , } , then \n\n = ( ) + .\n\nThe definition of shows that ( ) = span ( 1, , ). Thus the equation above shows that \n\n> 5.46\n\n span ( 1, , , 1, , ). \n\nFrom 5.45 and 5.46, we conclude (using 5.39) that has an upper-triangular matrix with respect to the basis 1, , , 1, , of , as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 160 Chapter 5 Eigenvalues and Eigenvectors\n\nThe set of numbers { 1, , } from the previous result equals the set of eigenvalues of (because the set of zeros of the minimal polynomial of equals the set of eigenvalues of , by 5.27), although the list 1, , in the previous result may contain repetitions. In Chapter 8 we will improve even the wonderful result below; see 8.37 and 8.46. \n\n> 5.47\n\nif = , then every operator on has an upper-triangular matrix \n\nSuppose is a finite-dimensional complex vector space and () . Then \n\n has an upper-triangular matrix with respect to some basis of . \n\n> Proof\n\nThe desired result follows immediately from 5.44 and the second version of the fundamental theorem of algebra (see 4.13). For an extension of the result above to two operators and such that \n\n = ,\n\nsee 5.80. Also, for an extension to more than two operators, see Exercise 9(b) in Section 5E. \n\nCaution: If an operator () has a upper-triangular matrix with respect to some basis 1, , of , then the eigenvalues of are exactly the entries on the diagonal of () , as shown by 5.41, and furthermore 1 is an eigenvector of \n\n . However, 2, , need not be eigenvectors of . Indeed, a basis vector is an eigenvector of if and only if all entries in the th column of the matrix of \n\nare 0, except possibly the th entry. \n\n> The row echelon form of the matrix of an operator does not give us a list of the eigenvalues of the operator. In contrast, an upper-triangular matrix with respect to some basis gives us a list of all the eigenvalues of the op-erator. However, there is no method for computing exactly such an upper-triangular matrix, even though 5.47 guarantees its existence if = .\n\nYou may recall from a previous course that every matrix of numbers can be changed to a matrix in what is called row echelon form. If one begins with a square matrix, the matrix in row echelon form will be an upper-triangular matrix. Do not confuse this upper-triangular ma-trix with the upper-triangular matrix of an operator with respect to some basis whose existence is proclaimed by 5.47 (if \n\n = )there is no connection between these upper-triangular matrices. \n\n# Exercises 5C \n\n1 Prove or give a counterexample: If () and 2 has an upper-triangular matrix with respect to some basis of , then has an upper-triangular matrix with respect to some basis of . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 5C Upper-Triangular Matrices 161\n\n2 Suppose and are upper-triangular matrices of the same size, with \n\n 1, , on the diagonal of and 1, , on the diagonal of .(a) Show that + is an upper-triangular matrix with 1 + 1, , + \n\non the diagonal. (b) Show that is an upper-triangular matrix with 1 1, , on the diagonal. \n\n> The results in this exercise are used in the proof of 5.81.\n\n3 Suppose () is invertible and 1, , is a basis of with respect to which the matrix of is upper triangular, with 1, , on the diagonal. Show that the matrix of 1 is also upper triangular with respect to the basis \n\n 1, , , with 1 1\n\n, , 1 \n\non the diagonal. \n\n4 Give an example of an operator whose matrix with respect to some basis contains only 0s on the diagonal, but the operator is invertible. \n\n> This exercise and the exercise below show that 5.41 fails without the hypoth-esis that an upper-triangular matrix is under consideration.\n\n5 Give an example of an operator whose matrix with respect to some basis contains only nonzero numbers on the diagonal, but the operator is not invertible. \n\n6 Suppose = , is finite-dimensional, and () . Prove that if \n\n {1 , , dim } , then has a -dimensional subspace invariant under .\n\n7 Suppose is finite-dimensional, () , and .(a) Prove that there exists a unique monic polynomial of smallest degree such that () = 0 .(b) Prove that the minimal polynomial of is a polynomial multiple of .\n\n8 Suppose is finite-dimensional, () , and there exists a nonzero vector such that 2 + 2 = 2 .(a) Prove that if = , then there does not exist a basis of with respect to which has an upper-triangular matrix. (b) Prove that if = and is an upper-triangular matrix that equals the matrix of with respect to some basis of , then 1 + or 1 \n\nappears on the diagonal of .\n\n9 Suppose is a square matrix with complex entries. Prove that there exists an invertible square matrix with complex entries such that 1 is an upper-triangular matrix. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 162 Chapter 5 Eigenvalues and Eigenvectors\n\n10 Suppose () and 1, , is a basis of . Show that the following are equivalent. (a) The matrix of with respect to 1, , is lower triangular. (b) span ( , , ) is invariant under for each = 1 , , .(c) span ( , , ) for each = 1 , , . \n\n> A square matrix is called lower triangular if all entries above the diagonal are 0.\n\n11 Suppose = and is finite-dimensional. Prove that if () , then there exists a basis of with respect to which has a lower-triangular matrix. \n\n12 Suppose is finite-dimensional, () has an upper-triangular matrix with respect to some basis of , and is a subspace of that is invariant under .(a) Prove that | has an upper-triangular matrix with respect to some basis of .(b) Prove that the quotient operator / has an upper-triangular matrix with respect to some basis of / . \n\n> The quotient operator / was defined in Exercise 38 in Section 5A.\n\n13 Suppose is finite-dimensional and () . Suppose there exists a subspace of that is invariant under such that | has an upper-triangular matrix with respect to some basis of and also / has an upper-triangular matrix with respect to some basis of / . Prove that has an upper-triangular matrix with respect to some basis of .\n\n14 Suppose is finite-dimensional and () . Prove that has an upper-triangular matrix with respect to some basis of if and only if the dual operator has an upper-triangular matrix with respect to some basis of the dual space .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5D Diagonalizable Operators 163 \n\n# 5D Diagonalizable Operators \n\n# Diagonal Matrices \n\n> 5.48 definition:\n\ndiagonal matrix \n\nA diagonal matrix is a square matrix that is 0 everywhere except possibly on the diagonal. \n\n> 5.49 example:\n\ndiagonal matrix \n\n8 0 00 5 00 0 5\n\nis a diagonal matrix. \n\nEvery diagonal matrix is upper tri-angular. Diagonal matrices typically have many more 0s than most upper-triangular matrices of the same size. \n\nIf an operator has a diagonal matrix with respect to some basis, then the en-tries on the diagonal are precisely the eigenvalues of the operator; this follows from 5.41 (or find an easier direct proof for diagonal matrices). \n\n> 5.50 definition:\n\ndiagonalizable \n\nAn operator on is called diagonalizable if the operator has a diagonal matrix with respect to some basis of . \n\n> 5.51 example:\n\ndiagonalization may require a different basis \n\nDefine ( 2) by \n\n( , ) = (41 + 7 , 20 + 74). \n\nThe matrix of with respect to the standard basis of 2 is \n\n41 720 74 ,\n\nwhich is not a diagonal matrix. However, is diagonalizable. Specifically, the matrix of with respect to the basis (1 , 4) , (7 , 5) is \n\n69 00 46 \n\nbecause (1 , 4) = (69 , 276) = 69(1 , 4) and (7 , 5) = (322 , 230) = 46(7 , 5) .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n164 Chapter 5 Eigenvalues and Eigenvectors \n\nFor , we will find it convenient to have a name and a notation for the set of vectors that an operator maps to times the vector. \n\n> 5.52 definition:\n\neigenspace, ( , ) \n\nSuppose () and . The eigenspace of corresponding to is the subspace ( , ) of defined by \n\n( , ) = null ( ) = { = }. \n\nHence ( , ) is the set of all eigenvectors of corresponding to , along with the 0 vector. For () and , the set ( , ) is a subspace of because the null space of each linear map on is a subspace of . The definitions imply that is an eigenvalue of if and only if ( , ) {0} . \n\n> 5.53 example:\n\neigenspaces of an operator \n\nSuppose the matrix of an operator () with respect to a basis 1, 2, 3\n\nof is the matrix in Example 5.49. Then \n\n(8 , ) = span ( 1), (5 , ) = span ( 2, 3). \n\nIf is an eigenvalue of an operator () , then restricted to ( , ) is just the operator of multiplication by . \n\n> 5.54\n\nsum of eigenspaces is a direct sum \n\nSuppose () and 1, , are distinct eigenvalues of . Then \n\n( 1, ) + + ( , ) \n\nis a direct sum. Furthermore, if is finite-dimensional, then \n\ndim ( 1, ) + + dim ( , ) dim . \n\nProof To show that ( 1, ) + + ( , ) is a direct sum, suppose \n\n 1 + + = 0 ,\n\nwhere each is in ( , ) . Because eigenvectors corresponding to distinct eigenvalues are linearly independent (by 5.11), this implies that each equals 0.Thus ( 1, ) + + ( , ) is a direct sum (by 1.45), as desired. Now suppose is finite-dimensional. Then \n\ndim ( 1, ) + + dim ( , ) = dim (( 1, ) ( , )) dim ,\n\nwhere the first line follows from 3.94 and the second line follows from 2.37. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5D Diagonalizable Operators 165 \n\n# Conditions for Diagonalizability \n\nThe following characterizations of diagonalizable operators will be useful. \n\n> 5.55\n\nconditions equivalent to diagonalizability \n\nSuppose is finite-dimensional and () . Let 1, , denote the distinct eigenvalues of . Then the following are equivalent. (a) is diagonalizable. (b) has a basis consisting of eigenvectors of .(c) = ( 1, ) ( , ) .(d) dim = dim ( 1, ) + + dim ( , ) .\n\nProof An operator () has a diagonal matrix \n\n 1 00 \n\n\n\nwith respect to a basis 1, , of if and only if = for each . Thus (a) and (b) are equivalent. Suppose (b) holds; thus has a basis consisting of eigenvectors of . Hence every vector in is a linear combination of eigenvectors of , which implies that \n\n = ( 1, ) + + ( , ). \n\nNow 5.54 shows that (c) holds, proving that (b) implies (c). That (c) implies (d) follows immediately from 3.94. Finally, suppose (d) holds; thus \n\n> 5.56\n\ndim = dim ( 1, ) + + dim ( , ). \n\nChoose a basis of each ( , ) ; put all these bases together to form a list 1, , \n\nof eigenvectors of , where = dim (by 5.56). To show that this list is linearly independent, suppose \n\n 1 1 + + = 0 ,\n\nwhere 1, , . For each = 1 , , , let denote the sum of all the terms \n\n such that ( , ) . Thus each is in ( , ) , and \n\n 1 + + = 0. \n\nBecause eigenvectors corresponding to distinct eigenvalues are linearly indepen-dent (see 5.11), this implies that each equals 0. Because each is a sum of terms , where the s were chosen to be a basis of ( , ) , this implies that all s equal 0. Thus 1, , is linearly independent and hence is a basis of \n\n(by 2.38). Thus (d) implies (b), completing the proof. For additional conditions equivalent to diagonalizability, see 5.62, Exercises 5 and 15 in this section, Exercise 24 in Section 7B, and Exercise 15 in Section 8A. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 166 Chapter 5 Eigenvalues and Eigenvectors\n\nAs we know, every operator on a nonzero finite-dimensional complex vector space has an eigenvalue. However, not every operator on a nonzero finite-dimensional complex vector space has enough eigenvectors to be diagonalizable, as shown by the next example. \n\n> 5.57 example:\n\nan operator that is not diagonalizable \n\nDefine an operator ( 3) by ( , , ) = ( , , 0) . The matrix of with respect to the standard basis of 3 is \n\n0 1 00 0 10 0 0\n\n,\n\nwhich is an upper-triangular matrix but is not a diagonal matrix. As you should verify, 0 is the only eigenvalue of and furthermore \n\n(0 , ) = {( , 0, 0) 3 }. \n\nHence conditions (b), (c), and (d) of 5.55 fail (of course, because these conditions are equivalent, it is sufficient to check that only one of them fails). Thus condition (a) of 5.55 also fails. Hence is not diagonalizable, regardless of whether = \n\nor = .The next result shows that if an operator has as many distinct eigenvalues as the dimension of its domain, then the operator is diagonalizable. \n\n> 5.58\n\nenough eigenvalues implies diagonalizability \n\nSuppose is finite-dimensional and () has dim distinct eigenvalues. Then is diagonalizable. \n\n> Proof\n\nSuppose has distinct eigenvalues 1, , dim . For each , let \n\nbe an eigenvector corresponding to the eigenvalue . Because eigenvectors corre-sponding to distinct eigenvalues are linearly independent (see 5.11), 1, , dim \n\nis linearly independent. A linearly independent list of dim vectors in is a basis of (see 2.38); thus \n\n 1, , dim is a basis of . With respect to this basis consisting of eigenvectors, \n\n has a diagonal matrix. In later chapters we will find additional conditions that imply that certain operators are diagonalizable. For example, see the real spectral theorem (7.29) and the complex spectral theorem (7.31). The result above gives a sufficient condition for an operator to be diagonal-izable. However, this condition is not necessary. For example, the operator \n\non 3 defined by ( , , ) = (6 , 6 , 7) has only two eigenvalues ( 6 and 7) and \n\ndim 3 = 3 , but is diagonalizable (by the standard basis of 3).\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5D Diagonalizable Operators 167 \n\nFor a spectacular application of these techniques, see Exercise 21, which shows how to use diagonalization to find an exact formula for the th term of the Fibonacci sequence. \n\nThe next example illustrates the im-portance of diagonalization, which can be used to compute high powers of an operator, taking advantage of the equa-tion = if is an eigenvector of \n\n with eigenvalue . \n\n> 5.59 example:\n\nusing diagonalization to compute 100 \n\nDefine ( 3) by ( , , ) = (2 + , 5 + 3 , 8) . With respect to the standard basis, the matrix of is \n\n2 1 00 5 30 0 8.\n\nThe matrix above is an upper-triangular matrix but it is not a diagonal matrix. By 5.41, the eigenvalues of are 2, 5, and 8. Because is an operator on a vector space of dimension three and has three distinct eigenvalues, 5.58 assures us that there exists a basis of 3 with respect to which has a diagonal matrix. To find this basis, we only have to find an eigenvector for each eigenvalue. In other words, we have to find a nonzero solution to the equation \n\n( , , ) = ( , , ) \n\nfor = 2 , then for = 5 , and then for = 8 . Solving these simple equations shows that for = 2 we have an eigenvector (1 , 0, 0) , for = 5 we have an eigenvector (1 , 3, 0) , and for = 8 we have an eigenvector (1 , 6, 6) .Thus (1 , 0, 0) , (1 , 3, 0) , (1 , 6, 6) is a basis of 3 consisting of eigenvectors of ,and with respect to this basis the matrix of is the diagonal matrix \n\n2 0 00 5 00 0 8.\n\nTo compute 100 (0 , 0, 1) , for example, write (0 , 0, 1) as a linear combination of our basis of eigenvectors: \n\n(0 , 0, 1) = 16 (1 , 0, 0) 13 (1 , 3, 0) + 16 (1 , 6, 6). \n\nNow apply 100 to both sides of the equation above, getting \n\n 100 (0 , 0, 1) = 16 ( 100 (1 , 0, 0)) 13 ( 100 (1 , 3, 0)) + 16 ( 100 (1 , 6, 6)) = 16 (2 100 (1 , 0, 0) 2 5 100 (1 , 3, 0) + 8100 (1 , 6, 6)) = 16 (2 100 2 5 100 + 8100 , 6 8 100 6 5 100 , 6 8 100 ). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 168 Chapter 5 Eigenvalues and Eigenvectors\n\nWe saw earlier that an operator on a finite-dimensional vector space has an upper-triangular matrix with respect to some basis of if and only if the minimal polynomial of equals ( 1) ( ) for some 1, , (see 5.44). As we previously noted (see 5.47), this condition is always satisfied if = .Our next result 5.62 states that an operator () has a diagonal matrix with respect to some basis of if and only if the minimal polynomial of equals \n\n( 1) ( ) for some distinct 1, , . Before formally stating this result, we give two examples of using it. \n\n> 5.60 example:\n\ndiagonalizable, but with no known exact eigenvalues \n\nDefine ( 5) by \n\n( 1, 2, 3, 4, 5) = (3 5, 1 + 6 5, 2, 3, 4). \n\nThe matrix of is shown in Example 5.26, where we showed that the minimal polynomial of is 3 6 + 5.As mentioned in Example 5.28, no exact expression is known for any of the zeros of this polynomial, but numeric techniques show that the zeros of this polynomial are approximately 1.67 , 0.51 , 1.40 , 0.12 + 1.59 , 0.12 1.59 .The software that produces these approximations is accurate to more than three digits. Thus these approximations are good enough to show that the five numbers above are distinct. The minimal polynomial of equals the fifth degree monic polynomial with these zeros. Now 5.62 shows that is diagonalizable. \n\n> 5.61 example:\n\nshowing that an operator is not diagonalizable \n\nDefine ( 3) by \n\n( 1, 2, 3) = (6 1 + 3 2 + 4 3, 6 2 + 2 3, 7 3). \n\nThe matrix of with respect to the standard basis of 3 is \n\n6 3 40 6 20 0 7.\n\nThe matrix above is an upper-triangular matrix but is not a diagonal matrix. Might \n\n have a diagonal matrix with respect to some other basis of 3?To answer this question, we will find the minimal polynomial of . First note that the eigenvalues of are the diagonal entries of the matrix above (by 5.41). Thus the zeros of the minimal polynomial of are 6, 7 [ by 5.27(a) ]. The diagonal of the matrix above tells us that ( 6) 2( 7) = 0 (by 5.40). The minimal polynomial of has degree at most 3 (by 5.22). Putting all this together, we see that the minimal polynomial of is either ( 6)( 7) or ( 6) 2( 7) .A simple computation shows that ( 6)( 7) 0 . Thus the minimal polynomial of is ( 6) 2( 7) .Now 5.62 shows that is not diagonalizable. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5D Diagonalizable Operators 169 \n\n> 5.62\n\nnecessary and sufficient condition for diagonalizability \n\nSuppose is finite-dimensional and () . Then is diagonalizable if and only if the minimal polynomial of equals ( 1) ( ) for some list of distinct numbers 1, , .\n\nProof First suppose is diagonalizable. Thus there is a basis 1, , of \n\nconsisting of eigenvectors of . Let 1, , be the distinct eigenvalues of .Then for each , there exists with ( ) = 0 . Thus \n\n( 1)( ) = 0 ,\n\nwhich implies that the minimal polynomial of equals ( 1) ( ).To prove the implication in the other direction, now suppose the minimal polynomial of equals ( 1) ( ) for some list of distinct numbers \n\n 1, , . Thus \n\n> 5.63\n\n( 1)( ) = 0. \n\nWe will prove that is diagonalizable by induction on . To get started, suppose = 1 . Then 1 = 0 , which means that is a scalar multiple of the identity operator, which implies that is diagonalizable. Now suppose that > 1 and the desired result holds for all smaller values of \n\n . The subspace range ( ) is invariant under [ this is a special case of 5.18 with () = ]. Thus restricted to range ( ) is an operator on \n\nrange ( ) .If range ( ) , then = ( ) for some , and 5.63 implies \n\n> 5.64\n\n( 1)( 1 ) = ( 1)( ) = 0. \n\nHence ( 1) ( 1 ) is a polynomial multiple of the minimal polynomial of restricted to range ( ) [by 5.29]. Thus by our induction hypothesis, there is a basis of range ( ) consisting of eigenvectors of .Suppose that range ( ) null ( ) . Then = . Now 5.64 implies that \n\n0 = ( 1)( 1 ) = ( 1)( 1 ). \n\nBecause 1, , are distinct, the equation above implies that = 0 . Hence \n\nrange ( ) null ( ) = {0} .Thus range ( ) +null ( ) is a direct sum (by 1.46) whose dimension is dim (by 3.94 and 3.21). Hence range ( ) null ( ) = . Every nonzero vector in null ( ) is an eigenvector of with eigenvalue .Earlier in this proof we saw that there is a basis of range ( ) consisting of eigenvectors of . Adjoining to that basis a basis of null ( ) gives a basis of consisting of eigenvectors of . The matrix of with respect to this basis is a diagonal matrix, as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 170 Chapter 5 Eigenvalues and Eigenvectors\n\nNo formula exists for the zeros of polynomials of degree 5 or greater. However, the previous result can be used to determine whether an operator on a complex vector space is diagonalizable without even finding approximations of the zeros of the minimal polynomialsee Exercise 15. The next result will be a key tool when we prove a result about the simultaneous diagonalization of two operators; see 5.76. Note how the use of a characterization of diagonalizable operators in terms of the minimal polynomial (see 5.62) leads to a short proof of the next result. \n\n> 5.65\n\nrestriction of diagonalizable operator to invariant subspace \n\nSuppose () is diagonalizable and is a subspace of that is invariant under . Then | is a diagonalizable operator on . \n\n> Proof\n\nBecause the operator is diagonalizable, the minimal polynomial of \n\nequals ( 1) ( ) for some list of distinct numbers 1, , (by 5.62). The minimal polynomial of is a polynomial multiple of the minimal polynomial of | (by 5.31). Hence the minimal polynomial of | has the form required by 5.62, which shows that | is diagonalizable. \n\n# Gershgorin Disk Theorem \n\n> 5.66 definition:\n\nGershgorin disks \n\nSuppose () and 1, , is a basis of . Let denote the matrix of \n\n with respect to this basis. A Gershgorin disk of with respect to the basis \n\n 1, , is a set of the form \n\n{{ | , | \n\n> \n> \n> = 1 \n\n| , |}},\n\nwhere {1 , , } .Because there are choices for in the definition above, has Gershgorin disks. If = , then for each {1 , , } , the corresponding Gershgorin disk is a closed disk in centered at , , which is the th entry on the diagonal of .The radius of this closed disk is the sum of the absolute values of the entries in row of , excluding the diagonal entry. If = , then the Gershgorin disks are closed intervals in .In the special case that the square matrix above is a diagonal matrix, each Gershgorin disk consists of a single point that is a diagonal entry of (and each eigenvalue of is one of those points, as required by the next result). One consequence of our next result is that if the nondiagonal entries of are small, then each eigenvalue of is near a diagonal entry of .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5D Diagonalizable Operators 171 \n\n5.67 Gershgorin disk theorem \n\nSuppose () and 1, , is a basis of . Then each eigenvalue of \n\nis contained in some Gershgorin disk of with respect to the basis 1, , .\n\nProof Suppose is an eigenvalue of . Let be a corresponding eigenvector. There exist 1, , such that \n\n5.68 = 1 1 + + .\n\nLet denote the matrix of with respect to the basis 1, , . Applying \n\nto both sides of the equation above gives \n\n = \n\n> \n\n\n\n> = 1\n\n 5.69 \n\n=\n\n> \n\n\n\n> = 1\n\n \n\n\n\n> = 1\n\n , \n\n=\n\n> \n\n\n\n> = 1\n\n(\n\n> \n\n\n\n> = 1\n\n , ) .5.70 \n\nLet {1 , , } be such that \n\n| | = max {| 1|, , | |}. \n\nUsing 5.68, we see that the coefficient of on the left side of 5.69 equals ,which must equal the coefficient of on the right side of 5.70. In other words, \n\n =\n\n> \n\n\n\n> = 1\n\n , .\n\nSubtract , from each side of the equation above and then divide both sides by to get \n\n| , | = \n\n> \n\n\n\n> = 1 \n\n , \n\n \n\n \n\n\n\n> \n\n\n\n> = 1 \n\n| , |. \n\nThus is in the th Gershgorin disk with respect to the basis 1, , .\n\nThe Gershgorin disk theorem is named for Semyon Aronovich Gershgorin, who published this result in 1931. \n\nExercise 22 gives a nice application of the Gershgorin disk theorem. Exercise 23 states that the radius of each Gershgorin disk could be changed to the sum of the absolute values of corresponding column entries (instead of row entries), excluding the diagonal entry, and the theorem above would still hold. \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler 172 Chapter 5 Eigenvalues and Eigenvectors \n\n# Exercises 5D \n\n1 Suppose is a finite-dimensional complex vector space and () .(a) Prove that if 4 = , then is diagonalizable. (b) Prove that if 4 = , then is diagonalizable. (c) Give an example of an operator ( 2) such that 4 = 2 and is not diagonalizable. \n\n2 Suppose () has a diagonal matrix with respect to some basis of . Prove that if , then appears on the diagonal of precisely \n\ndim ( , ) times. \n\n3 Suppose is finite-dimensional and () . Prove that if the operator \n\nis diagonalizable, then = null range .\n\n4 Suppose is finite-dimensional and () . Prove that the following are equivalent. (a) = null range .(b) = null + range .(c) null range = {0} .\n\n5 Suppose is a finite-dimensional complex vector space and () .Prove that is diagonalizable if and only if \n\n = null ( ) range ( ) \n\nfor every .\n\n6 Suppose ( 5) and dim (8 , ) = 4 . Prove that 2 or 6 is invertible. \n\n7 Suppose () is invertible. Prove that \n\n( , ) = ( 1 , 1 )\n\nfor every with 0 .\n\n8 Suppose is finite-dimensional and () . Let 1, , denote the distinct nonzero eigenvalues of . Prove that \n\ndim ( 1, ) + + dim ( , ) dim range . \n\n9 Suppose , ( 3) each have 2, 6, 7 as eigenvalues. Prove that there exists an invertible operator ( 3) such that = 1 .\n\n10 Find , ( 4) such that and each have 2, 6, 7 as eigenvalues, and \n\n have no other eigenvalues, and there does not exist an invertible operator \n\n ( 4) such that = 1 .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5D Diagonalizable Operators 173 \n\n11 Find ( 3) such that 6 and 7 are eigenvalues of and such that does not have a diagonal matrix with respect to any basis of 3.\n\n12 Suppose ( 3) is such that 6 and 7 are eigenvalues of . Furthermore, suppose does not have a diagonal matrix with respect to any basis of 3.Prove that there exists ( 1, 2, 3) 3 such that \n\n( 1, 2, 3) = (6 + 8 1, 7 + 8 2, 13 + 8 3). \n\n13 Suppose is a diagonal matrix with distinct entries on the diagonal and \n\nis a matrix of the same size as . Show that = if and only if is a diagonal matrix. \n\n14 (a) Give an example of a finite-dimensional complex vector space and an operator on that vector space such that 2 is diagonalizable but is not diagonalizable. (b) Suppose = , is a positive integer, and () is invertible. Prove that is diagonalizable if and only if is diagonalizable. \n\n15 Suppose is a finite-dimensional complex vector space, () , and \n\nis the minimal polynomial of . Prove that the following are equivalent. (a) is diagonalizable. (b) There does not exist such that is a polynomial multiple of \n\n( ) 2.(c) and its derivative have no zeros in common. (d) The greatest common divisor of and is the constant polynomial 1.\n\nThe greatest common divisor of and is the monic polynomial of largest degree such that and are both polynomial multiples of . The Euclidean algorithm for polynomials (look it up ) can quickly determine the greatest common divisor of two polynomials, without requiring any information about the zeros of the polynomials. Thus the equivalence of (a)\n\nand (d) above shows that we can determine whether is diagonalizable without knowing anything about the zeros of .\n\n16 Suppose that () is diagonalizable. Let 1, , denote the distinct eigenvalues of . Prove that a subspace of is invariant under if and only if there exist subspaces 1, , of such that ( , ) for each and = 1 .\n\n17 Suppose is finite-dimensional. Prove that () has a basis consisting of diagonalizable operators. \n\n18 Suppose that () is diagonalizable and is a subspace of that is invariant under . Prove that the quotient operator / is a diagonalizable operator on / .\n\nThe quotient operator / was defined in Exercise 38 in Section 5A. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n174 Chapter 5 Eigenvalues and Eigenvectors \n\n19 Prove or give a counterexample: If () and there exists a subspace \n\nof that is invariant under such that | and / are both diagonalizable, then is diagonalizable. \n\nSee Exercise 13 in Section 5C for an analogous statement about upper-triangular matrices. \n\n20 Suppose is finite-dimensional and () . Prove that is diagonaliz-able if and only if the dual operator is diagonalizable. \n\n21 The Fibonacci sequence 0, 1, 2, is defined by \n\n 0 = 0 , 1 = 1 , and = 2 + 1 for 2. \n\nDefine ( 2) by ( , ) = ( , + ) .(a) Show that (0 , 1) = ( , + 1) for each nonnegative integer .(b) Find the eigenvalues of .(c) Find a basis of 2 consisting of eigenvectors of .(d) Use the solution to (c) to compute (0 , 1) . Conclude that \n\n = 1\n\n> \n\n51 + 52 \n\n> \n\n 1 52 \n\n> \n\n\n\nfor each nonnegative integer .(e) Use (d) to conclude that if is a nonnegative integer, then the Fibonacci number is the integer that is closest to \n\n1\n\n> \n\n51 + 52 \n\n> \n\n.\n\nEach is a nonnegative integer, even though the right side of the formula in (d) does not look like an integer. The number \n\n> 1\n\n+ 52\n\nis called the golden ratio .\n\n22 Suppose () and is an -by- matrix that is the matrix of with respect to some basis of . Prove that if \n\n| , | > \n\n> \n> \n> = 1 \n\n| , |\n\nfor each {1 , , } , then is invertible. \n\nThis exercise states that if the diagonal entries of the matrix of are large compared to the nondiagonal entries, then is invertible. \n\n23 Suppose the definition of the Gershgorin disks is changed so that the radius of the th disk is the sum of the absolute values of the entries in column (instead of row) of , excluding the diagonal entry. Show that the Gershgorin disk theorem (5.67) still holds with this changed definition. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5E Commuting Operators 175 \n\n# 5E Commuting Operators \n\n> 5.71 definition:\n\ncommute \n\n Two operators and on the same vector space commute if = .\n\n Two square matrices and of the same size commute if = .For example, if is the identity operator on and , then commutes with every operator on .As another example, if is an operator then 2 and 3 commute. More generally, if , () , then () and () commute [see 5.17(b) ]. \n\n> 5.72 example:\n\npartial differentiation operators commute \n\nSuppose is a nonnegative integer. Let ( 2) denote the real vector space of polynomials (with real coefficients) in two real variables and of degree at most ,with the usual operations of addition and scalar multiplication of real-valued functions. Thus the elements of ( 2) are functions on 2 of the form \n\n> 5.73\n\n = \n\n> + \n\n , ,\n\nwhere the indices and take on all nonnegative integer values such that + ,each , is in , and denotes the function on 2 defined by ( , ) .Define operators , ( ( 2)) by \n\n = = \n\n> + \n\n , 1 and = = \n\n> + \n\n , 1 ,\n\nwhere is as in 5.73. The operators and are called partial differentiation operators because each of these operators differentiates with respect to one of the variables while pretending that the other variable is a constant. The operators and commute because if is as in 5.73, then \n\n( ) = \n\n> + \n\n , 1 1 = ( ). \n\nThe equation = on ( 2) illustrates a more general result that the order of partial differentiation does not matter for nice functions. \n\nAll 214,358,881 (which equals 11 8)\n\nordered pairs of the 2-by-2 matrices under consideration were checked by a computer to discover that only 674,609 of these ordered pairs of matrices commute. \n\nCommuting matrices are unusual. For example, there are 214,358,881 or-dered pairs of 2-by-2 matrices all of whose entries are integers in the inter-val [5 , 5] . Only about 0.3% of these ordered pairs of matrices commute. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 176 Chapter 5 Eigenvalues and Eigenvectors\n\nThe next result shows that two operators commute if and only if their matrices (with respect to the same basis) commute. \n\n> 5.74\n\ncommuting operators correspond to commuting matrices \n\nSuppose , () and 1, , is a basis of . Then and commute if and only if ( , ( 1, , )) and ( , ( 1, , )) commute. \n\n> Proof\n\nWe have \n\n and commute = () = () () () = () () () and () commute ,\n\nas desired. The next result shows that if two operators commute, then every eigenspace for one operator is invariant under the other operator. This result, which we will use several times, is one of the main reasons why a pair of commuting operators behaves better than a pair of operators that does not commute. \n\n> 5.75\n\neigenspace is invariant under commuting operator \n\nSuppose , () commute and . Then ( , ) is invariant under . \n\n> Proof\n\nSuppose ( , ) . Then \n\n() = () = () = () = () = . \n\nThe equation above shows that ( , ) . Thus ( , ) is invariant under .Suppose we have two operators, each of which is diagonalizable. If we want to do computations involving both operators (for example, involving their sum), then we want the two operators to be diagonalizable by the same basis, which according to the next result is possible when the two operators commute. \n\n> 5.76\n\nsimultaneous diagonalizability commutativity \n\nTwo diagonalizable operators on the same vector space have diagonal matrices with respect to the same basis if and only if the two operators commute. \n\n> Proof\n\nFirst suppose , () have diagonal matrices with respect to the same basis. The product of two diagonal matrices of the same size is the diagonal matrix obtained by multiplying the corresponding elements of the two diagonals. Thus any two diagonal matrices of the same size commute. Thus and commute, by 5.74. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 5E Commuting Operators 177 \n\nTo prove the implication in the other direction, now suppose that , () \n\nare diagonalizable operators that commute. Let 1, , denote the distinct eigenvalues of . Because is diagonalizable, 5.55(c) shows that \n\n> 5.77\n\n = ( 1, ) ( , ). \n\nFor each = 1 , , , the subspace ( , ) is invariant under (by 5.75). Because is diagonalizable, 5.65 implies that | ( ,) is diagonalizable for each . Hence for each = 1 , , , there is a basis of ( , ) consisting of eigenvectors of . Putting these bases together gives a basis of (because of 5.77), with each vector in this basis being an eigenvector of both and . Thus \n\nand both have diagonal matrices with respect to this basis, as desired. See Exercise 2 for an extension of the result above to more than two operators. Suppose is a finite-dimensional nonzero complex vector space. Then every operator on has an eigenvector (see 5.19). The next result shows that if two operators on commute, then there is a vector in that is an eigenvector for both operators (but the two commuting operators might not have a common eigenvalue). For an extension of the next result to more than two operators, see Exercise 9(a). \n\n> 5.78\n\ncommon eigenvector for commuting operators \n\nEvery pair of commuting operators on a finite-dimensional nonzero complex vector space has a common eigenvector. \n\nProof Suppose is a finite-dimensional nonzero complex vector space and \n\n , () commute. Let be an eigenvalue of (5.19 tells us that does indeed have an eigenvalue). Thus ( , ) {0} . Also, ( , ) is invariant under (by 5.75). Thus | ( ,) has an eigenvector (again using 5.19), which is an eigenvector for both and , completing the proof. \n\n> 5.79 example:\n\ncommon eigenvector for partial differentiation operators \n\nLet ( 2) be as in Example 5.72 and let , ( ( 2)) be the commuting partial differentiation operators in that example. As you can verify, 0\n\nis the only eigenvalue of each of these operators. Also \n\n(0 , ) = { \n\n> \n> \n> = 0\n\n 0, , } ,\n\n(0 , ) = { \n\n> \n> \n> = 0\n\n 0, , }. \n\nThe intersection of these two eigenspaces is the set of common eigenvectors of the two operators. Because (0 , ) (0 , ) is the set of constant functions, we see that and indeed have a common eigenvector, as promised by 5.78. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n178 Chapter 5 Eigenvalues and Eigenvectors \n\nThe next result extends 5.47 (the existence of a basis that gives an upper-triangular matrix) to two commuting operators. \n\n> 5.80\n\ncommuting operators are simultaneously upper triangularizable \n\nSuppose is a finite-dimensional complex vector space and , are commuting operators on . Then there is a basis of with respect to which both and have upper-triangular matrices. \n\nProof Let = dim . We will use induction on . The desired result holds if \n\n = 1 because all 1-by-1 matrices are upper triangular. Now suppose > 1 and the desired result holds for all complex vector spaces whose dimension is 1 .Let 1 be any common eigenvector of and (using 5.78). Hence 1 \n\nspan ( 1) and 1 span ( 1). Let be a subspace of such that \n\n = span ( 1) ;\n\nsee 2.33 for the existence of . Define a linear map by \n\n( 1 + ) = \n\nfor each and each . Define , () by = () and = () \n\nfor each . To apply our induction hypothesis to and , we must first show that these two operators on commute. To do this, suppose . Then there exists such that \n\n() = (()) = ( 1) = (( 1)) = (()) ,\n\nwhere the last equality holds because 1 is an eigenvector of and 1 = 0 .Similarly, \n\n() = (()). \n\nBecause the operators and commute, the last two displayed equations show that () = () . Hence and commute. Thus we can use our induction hypothesis to state that there exists a basis \n\n 2, , of such that and both have upper-triangular matrices with respect to this basis. The list 1, , is a basis of .If {2 , , } , then there exist , such that \n\n = 1 + and = 1 + .\n\nBecause and have upper-triangular matrices with respect to 2, , , we know that span ( 2, , ) and span ( 2, , ). Hence the equations above imply that \n\n span ( 1, , ) and span ( 1, , ). \n\nThus and have upper-triangular matrices with respect to 1, , , as desired. Exercise 9(b) extends the result above to more than two operators. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 5E Commuting Operators 179\n\nIn general, it is not possible to determine the eigenvalues of the sum or product of two operators from the eigenvalues of the two operators. However, the next result shows that something nice happens when the two operators commute. \n\n> 5.81\n\neigenvalues of sum and product of commuting operators \n\nSuppose is a finite-dimensional complex vector space and , are commut-ing operators on . Then \n\n every eigenvalue of + is an eigenvalue of plus an eigenvalue of ,\n\n every eigenvalue of is an eigenvalue of times an eigenvalue of . \n\n> Proof\n\nThere is a basis of with respect to which both and have upper-triangular matrices (by 5.80). With respect to that basis, \n\n( + ) = () + () and () = () () ,\n\nas stated in 3.35 and 3.43. The definition of matrix addition shows that each entry on the diagonal of \n\n( + ) equals the sum of the corresponding entries on the diagonals of () \n\nand () . Similarly, because () and () are upper-triangular matrices, the definition of matrix multiplication shows that each entry on the diagonal of \n\n() equals the product of the corresponding entries on the diagonals of () \n\nand () . Furthermore, ( + ) and () are upper-triangular matrices (see Exercise 2 in Section 5C). Every entry on the diagonal of () is an eigenvalue of , and every entry on the diagonal of () is an eigenvalue of (by 5.41). Every eigenvalue of + is on the diagonal of ( + ) , and every eigenvalue of is on the diagonal of () (these assertions follow from 5.41). Putting all this together, we conclude that every eigenvalue of + is an eigenvalue of plus an eigenvalue of , and every eigenvalue of is an eigenvalue of times an eigenvalue of .\n\n# Exercises 5E \n\n1 Give an example of two commuting operators , on 4 such that there is a subspace of 4 that is invariant under but not under and there is a subspace of 4 that is invariant under but not under .\n\n2 Suppose is a subset of () and every element of is diagonalizable. Prove that there exists a basis of with respect to which every element of \n\nhas a diagonal matrix if and only if every pair of elements of commutes. \n\n> This exercise extends 5.76, which considers the case in which contains only two elements. For this exercise, may contain any number of elements, and may even be an infinite set.\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 180 Chapter 5 Eigenvalues and Eigenvectors\n\n3 Suppose , () are such that = . Suppose () .(a) Prove that null () is invariant under .(b) Prove that range () is invariant under . \n\n> See 5.18 for the special case = .\n\n4 Prove or give a counterexample: If is a diagonal matrix and is an upper-triangular matrix of the same size as , then and commute. \n\n5 Prove that a pair of operators on a finite-dimensional vector space commute if and only if their dual operators commute. \n\n> See 3.118 for the definition of the dual of an operator.\n\n6 Suppose that is a nonzero finite-dimensional complex vector space and \n\n , () commute. Prove that there exist , such that \n\nrange ( ) + range ( ) . \n\n7 Suppose is a complex vector space, () is diagonalizable, and \n\n () commutes with . Prove that there is a basis of such that has a diagonal matrix with respect to this basis and has an upper-triangular matrix with respect to this basis. \n\n8 Suppose = 3 in Example 5.72 and , are the commuting partial differentiation operators on 3( 2) from that example. Find a basis of \n\n 3( 2) with respect to which and each have an upper-triangular matrix. \n\n9 Suppose is a finite-dimensional nonzero complex vector space. Suppose that () is such that and commute for all , .(a) Prove that there is a vector in that is an eigenvector for every element of .(b) Prove that there is a basis of with respect to which every element of \n\n has an upper-triangular matrix. \n\n> This exercise extends 5.78 and 5.80, which consider the case in which \n> contains only two elements. For this exercise, may contain any number of elements, and may even be an infinite set.\n\n10 Give an example of two commuting operators , on a finite-dimensional real vector space such that + has an eigenvalue that does not equal an eigenvalue of plus an eigenvalue of and has an eigenvalue that does not equal an eigenvalue of times an eigenvalue of .\n\n> This exercise shows that 5.81 does not hold on real vector spaces.\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n# Chapter 6 \n\n# Inner Product Spaces \n\nIn making the definition of a vector space, we generalized the linear structure (addition and scalar multiplication) of 2 and 3. We ignored geometric features such as the notions of length and angle. These ideas are embedded in the concept of inner products, which we will investigate in this chapter. Every inner product induces a norm, which you can think of as a length. This norm satisfies key properties such as the Pythagorean theorem, the triangle inequality, the parallelogram equality, and the CauchySchwarz inequality. The notion of perpendicular vectors in Euclidean geometry gets renamed to orthogonal vectors in the context of an inner product space. We will see that orthonormal bases are tremendously useful in inner product spaces. The Gram Schmidt procedure constructs such bases. This chapter will conclude by putting together these tools to solve minimization problems. \n\nstanding assumptions for this chapter \n\n denotes or .\n\n and denote vector spaces over .\n\n> Matthew Petroff CC BY-SA\n\nThe George Peabody Library, now part of Johns Hopkins University, opened while James Sylvester (18141897 ) was the universitys first mathematics professor. Sylvesters publications include the first use of the word matrix in mathematics. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n> 181\n\n182 Chapter 6 Inner Product Spaces \n\n# 6A Inner Products and Norms \n\n# Inner Products \n\nThis vector has norm 2 + 2.\n\nTo motivate the concept of inner product, think of vectors in 2 and 3 as arrows with initial point at the origin. The length of a vector in 2 or 3 is called the \n\nnorm of and is denoted by . Thus for = ( , ) 2, we have \n\n = 2 + 2.\n\nSimilarly, if = ( , , ) 3, then = 2 + 2 + 2.Even though we cannot draw pictures in higher dimensions, the generalization to is easy: we define the norm of = ( 1, , ) by \n\n = 12 + + 2.\n\nThe norm is not linear on . To inject linearity into the discussion, we introduce the dot product. \n\n> 6.1 definition:\n\ndot product \n\nFor , , the dot product of and , denoted by , is defined by \n\n = 1 1 + + ,\n\nwhere = ( 1, , ) and = ( 1, , ).\n\nIf we think of a vector as a point instead of as an arrow, then should be interpreted to mean the distance from the origin to the point .\n\nThe dot product of two vectors in \n\nis a number, not a vector. Notice that \n\n = 2 for all . Furthermore, the dot product on has the following properties. \n\n 0 for all .\n\n = 0 if and only if = 0 .\n\n For fixed, the map from to that sends to is linear. \n\n = for all , .An inner product is a generalization of the dot product. At this point you may be tempted to guess that an inner product is defined by abstracting the properties of the dot product discussed in the last paragraph. For real vector spaces, that guess is correct. However, so that we can make a definition that will be useful for both real and complex vector spaces, we need to examine the complex case before making the definition. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 6A Inner Products and Norms 183\n\nRecall that if = + , where , , then \n\n the absolute value of , denoted by || , is defined by || = 2 + 2;\n\n the complex conjugate of , denoted by , is defined by = ;\n\n || 2 = .See Chapter 4 for the definitions and the basic properties of the absolute value and complex conjugate. For = ( 1, , ) , we define the norm of by \n\n = | 1|2 + + | |2.\n\nThe absolute values are needed because we want to be a nonnegative number. Note that \n\n 2 = 1 1 + + .\n\nWe want to think of 2 as the inner product of with itself, as we did in . The equation above thus suggests that the inner product of the vector \n\n = ( 1, , ) with should equal \n\n 1 1 + + .\n\nIf the roles of the and were interchanged, the expression above would be replaced with its complex conjugate. Thus we should expect that the inner product of with equals the complex conjugate of the inner product of with . With that motivation, we are now ready to define an inner product on , which may be a real or a complex vector space. One comment about the notation used in the next definition: \n\n For , the notation 0 means is real and nonnegative. \n\n> 6.2 definition:\n\ninner product \n\nAn inner product on is a function that takes each ordered pair ( , ) of elements of to a number , and has the following properties. \n\npositivity \n\n , 0 for all .\n\ndefiniteness \n\n , = 0 if and only if = 0 .\n\nadditivity in first slot \n\n + , = , + , for all , , .\n\nhomogeneity in first slot \n\n , = , for all and all , .\n\nconjugate symmetry \n\n , = , for all , .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n184 Chapter 6 Inner Product Spaces \n\nMost mathematicians define inner products as above, but many physicists use a definition that requires homo-geneity in the second slot instead of the first slot. \n\nEvery real number equals its complex conjugate. Thus if we are dealing with a real vector space, then in the last con-dition above we can dispense with the complex conjugate and simply state that \n\n , = , for all , . \n\n> 6.3 example:\n\ninner products \n\n(a) The Euclidean inner product on is defined by \n\n( 1, , ), ( 1, , ) = 1 1 + + \n\nfor all ( 1, , ), ( 1, , ) .(b) If 1, , are positive numbers, then an inner product can be defined on \n\nby \n\n( 1, , ), ( 1, , ) = 1 1 1 + + \n\nfor all ( 1, , ), ( 1, , ) .(c) An inner product can be defined on the vector space of continuous real-valued functions on the interval [1 , 1] by \n\n , = 11 \n\n \n\nfor all , continuous real-valued functions on [1 , 1] .(d) An inner product can be defined on () by \n\n , = (0)(0) + 11 \n\n \n\nfor all , () .(e) An inner product can be defined on () by \n\n , = 0\n\n()() \n\nfor all , () . \n\n> 6.4 definition:\n\ninner product space \n\nAn inner product space is a vector space along with an inner product on .The most important example of an inner product space is with the Euclidean inner product given by (a) in the example above. When is referred to as an inner product space, you should assume that the inner product is the Euclidean inner product unless explicitly told otherwise. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 6A Inner Products and Norms 185\n\nSo that we do not have to keep repeating the hypothesis that and are inner product spaces, we make the following assumption. \n\n> 6.5 notation:\n\n , \n\nFor the rest of this chapter and the next chapter, and denote inner product spaces over .Note the slight abuse of language here. An inner product space is a vector space along with an inner product on that vector space. When we say that a vector space is an inner product space, we are also thinking that an inner product on \n\n is lurking nearby or is clear from the context (or is the Euclidean inner product if the vector space is ). \n\n> 6.6\n\nbasic properties of an inner product \n\n(a) For each fixed , the function that takes to , is a linear map from to .(b) 0 , = 0 for every .(c) , 0 = 0 for every .(d) , + = , + , for all , , .(e) , = , for all and all , .\n\n> Proof\n\n(a) For , the linearity of , follows from the conditions of additivity and homogeneity in the first slot in the definition of an inner product. (b) Every linear map takes 0 to 0. Thus (b) follows from (a). (c) If , then the conjugate symmetry property in the definition of an inner product and (b) show that , 0 = 0 , = 0 = 0 .(d) Suppose , , . Then \n\n , + = + , = , + , = , + , = , + , . \n\n(e) Suppose and , . Then \n\n , = , = , = , = , . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n186 Chapter 6 Inner Product Spaces \n\n# Norms \n\nOur motivation for defining inner products came initially from the norms of vectors on 2 and 3. Now we see that each inner product determines a norm. \n\n> 6.7 definition:\n\nnorm, \n\nFor , the norm of , denoted by , is defined by \n\n = , . \n\n> 6.8 example:\n\nnorms \n\n(a) If ( 1, , ) (with the Euclidean inner product), then \n\n( 1, , ) = | 1|2 + + | |2.\n\n(b) For in the vector space of continuous real-valued functions on [1 , 1] and with inner product given as in 6.3(c), we have \n\n = 11 \n\n 2. \n\n> 6.9\n\nbasic properties of the norm \n\nSuppose .(a) = 0 if and only if = 0 .(b) = || for all .\n\nProof \n\n(a) The desired result holds because , = 0 if and only if = 0 .(b) Suppose . Then \n\n 2 = , = , = , = || 2 2.\n\nTaking square roots now gives the desired equality. The proof of (b) in the result above illustrates a general principle: working with norms squared is usually easier than working directly with norms. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 6A Inner Products and Norms 187\n\nNow we come to a crucial definition. \n\n> 6.10 definition:\n\northogonal \n\nTwo vectors , are called orthogonal if , = 0 . \n\n> The word orthogonal comes from the Greek word orthogonios , which means right-angled.\n\nIn the definition above, the order of the two vectors does not matter, because \n\n , = 0 if and only if , = 0 . In-stead of saying and are orthogonal, sometimes we say is orthogonal to .Exercise 15 asks you to prove that if , are nonzero vectors in 2, then \n\n , = cos ,\n\nwhere is the angle between and (thinking of and as arrows with initial point at the origin). Thus two nonzero vectors in 2 are orthogonal (with respect to the Euclidean inner product) if and only if the cosine of the angle between them is 0, which happens if and only if the vectors are perpendicular in the usual sense of plane geometry. Thus you can think of the word orthogonal as a fancy word meaning perpendicular .We begin our study of orthogonality with an easy result. \n\n> 6.11\n\northogonality and 0\n\n(a) 0 is orthogonal to every vector in .(b) 0 is the only vector in that is orthogonal to itself. \n\n> Proof\n\n(a) Recall that 6.6(b) states that 0 , = 0 for every .(b) If and , = 0 , then = 0 (by definition of inner product). For the special case = 2, the next theorem was known over 3,500 years ago in Babylonia and then rediscovered and proved over 2,500 years ago in Greece. Of course, the proof below is not the original proof. \n\n> 6.12\n\nPythagorean theorem \n\nSuppose , . If and are orthogonal, then \n\n + 2 = 2 + 2. \n\n> Proof\n\nSuppose , = 0 . Then \n\n + 2 = + , + = , + , + , + , = 2 + 2. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 188 Chapter 6 Inner Product Spaces\n\nSuppose , , with 0 . We would like to write as a scalar multiple of plus a vector orthogonal to , as suggested in the picture here. \n\n> An orthogonal decomposition:\n> expressed as a scalar multiple of plus a vector orthogonal to .\n\nTo discover how to write as a scalar multiple of plus a vector orthogonal to , let denote a scalar. Then \n\n = + ( ). \n\nThus we need to choose so that is orthogonal to ( ) . Hence we want \n\n0 = , = , 2.\n\nThe equation above shows that we should choose to be , / 2. Making this choice of , we can write \n\n = , 2 + ( , 2 ). \n\nAs you should verify, the equation displayed above explicitly writes as a scalar multiple of plus a vector orthogonal to . Thus we have proved the following key result. \n\n> 6.13\n\nan orthogonal decomposition \n\nSuppose , , with 0 . Set = , 2 and = , 2 . Then \n\n = + and , = 0. \n\nThe orthogonal decomposition 6.13 will be used in the proof of the Cauchy Schwarz inequality, which is our next result and is one of the most important inequalities in mathematics. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6A Inner Products and Norms 189 \n\n> 6.14\n\nCauchySchwarz inequality \n\nSuppose , . Then \n\n| , | . \n\nThis inequality is an equality if and only if one of , is a scalar multiple of the other. \n\nProof If = 0 , then both sides of the desired inequality equal 0. Thus we can assume that 0 . Consider the orthogonal decomposition \n\n = , 2 + \n\ngiven by 6.13, where is orthogonal to . By the Pythagorean theorem, \n\n 2 = , 2 \n\n> 2\n\n+ 2\n\n= , 2\n\n 2 + 2\n\n , 2\n\n 2 .6.15 \n\nMultiplying both sides of this inequality by 2 and then taking square roots gives the desired inequality. \n\nAugustin-Louis Cauchy (17891857 )\n\nproved 6.16 (a) in 1821. In 1859, Cauchys student Viktor Bunyakovsky \n\n(18041889 ) proved integral inequal-ities like the one in 6.16 (b). A few decades later, similar discoveries by Hermann Schwarz (18431921 ) at-tracted more attention and led to the name of this inequality. \n\nThe proof in the paragraph above shows that the CauchySchwarz inequal-ity is an equality if and only if 6.15 is an equality. This happens if and only if = 0 . But = 0 if and only if \n\nis a multiple of (see 6.13). Thus the CauchySchwarz inequality is an equal-ity if and only if is a scalar multiple of \n\nor is a scalar multiple of (or both; the phrasing has been chosen to cover cases in which either or equals 0). \n\n> 6.16 example:\n\nCauchySchwarz inequality \n\n(a) If 1, , , 1, , , then \n\n( 1 1 + + )2 ( 12 + + 2)( 12 + + 2),\n\nas follows from applying the CauchySchwarz inequality to the vectors \n\n( 1, , ), ( 1, , ) , using the usual Euclidean inner product. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 190 Chapter 6 Inner Product Spaces\n\n(b) If , are continuous real-valued functions on [1 , 1] , then \n\n11 \n\n 2\n\n ( 11 \n\n 2)( 11 \n\n 2),\n\nas follows from applying the CauchySchwarz inequality to Example 6.3(c). \n\n> In this triangle, the length of\n> + is less than the length of plus the length of .\n\nThe next result, called the triangle inequality, has the geometric interpretation that the length of each side of a triangle is less than the sum of the lengths of the other two sides. Note that the triangle inequality implies that the shortest polygonal path between two points is a single line segment (a polygonal path consists of line segments). \n\n> 6.17\n\ntriangle inequality \n\nSuppose , . Then \n\n + + . \n\nThis inequality is an equality if and only if one of , is a nonnegative real multiple of the other. \n\n> Proof\n\nWe have \n\n + 2 = + , + = , + , + , + , = , + , + , + , = 2 + 2 + 2 Re , 2 + 2 + 2 , 6.18 \n\n 2 + 2 + 2 6.19 \n\n= ( + ) 2,\n\nwhere 6.19 follows from the CauchySchwarz inequality (6.14). Taking square roots of both sides of the inequality above gives the desired inequality. The proof above shows that the triangle inequality is an equality if and only if we have equality in 6.18 and 6.19. Thus we have equality in the triangle inequality if and only if \n\n> 6.20\n\n , = . \n\nIf one of , is a nonnegative real multiple of the other, then 6.20 holds. Con-versely, suppose 6.20 holds. Then the condition for equality in the Cauchy Schwarz inequality (6.14) implies that one of , is a scalar multiple of the other. This scalar must be a nonnegative real number, by 6.20, completing the proof. For the reverse triangle inequality, see Exercise 20. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6A Inner Products and Norms 191 \n\nThe diagonals of this parallelogram are + and .\n\nThe next result is called the parallel-ogram equality because of its geometric interpretation: in every parallelogram, the sum of the squares of the lengths of the diagonals equals the sum of the squares of the lengths of the four sides. Note that the proof here is more straightforward than the usual proof in Euclidean geometry. \n\n> 6.21\n\nparallelogram equality \n\nSuppose , . Then \n\n + 2 + 2 = 2( 2 + 2). \n\nProof We have \n\n + 2 + 2 = + , + + , = 2 + 2 + , + , \n\n+ 2 + 2 , , = 2( 2 + 2),\n\nas desired. \n\n# Exercises 6A \n\n1 Prove or give a counterexample: If 1, , , then \n\n> \n> \n> = 1\n> \n> \n> = 1\n\n , 0. \n\n2 Suppose () . Define , 1 by \n\n , 1 = , \n\nfor all , . Show that , 1 is an inner product on if and only if is injective. \n\n3 (a) Show that the function taking an ordered pair (( 1, 2), ( 1, 2)) of elements of 2 to | 1 1| + | 2 2| is not an inner product on 2.(b) Show that the function taking an ordered pair (( 1, 2, 3), ( 1, 2, 3)) \n\nof elements of 3 to 1 1 + 3 3 is not an inner product on 3.\n\n4 Suppose () is such that for every . Prove that \n\n 2 is injective. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 192 Chapter 6 Inner Product Spaces\n\n5 Suppose is a real inner product space. (a) Show that + , = 2 2 for every , .(b) Show that if , have the same norm, then + is orthogonal to \n\n .(c) Use (b) to show that the diagonals of a rhombus are perpendicular to each other. \n\n6 Suppose , . Prove that , = 0 + for all .\n\n7 Suppose , . Prove that + = + for all , if and only if = .\n\n8 Suppose , , , , and 2 + 2 + 2 + 2 + 2 1 . Prove that \n\n + + + 4 + 9 10 .\n\n9 Suppose , and = = 1 and , = 1 . Prove that = .\n\n10 Suppose , and 1 and 1 . Prove that \n\n> \n\n1 21 2 1 , . \n\n11 Find vectors , 2 such that is a scalar multiple of (1 , 3) , is orthog-onal to (1 , 3) , and (1 , 2) = + .\n\n12 Suppose , , , are positive numbers. (a) Prove that ( + + + )( 1 + 1 + 1 + 1 ) 16 .(b) For which positive numbers , , , is the inequality above an equality? \n\n13 Show that the square of an average is less than or equal to the average of the squares. More precisely, show that if 1, , , then the square of the average of 1, , is less than or equal to the average of 12, , 2.\n\n14 Suppose and 0 . Prove that / is the unique closest element on the unit sphere of to . More precisely, prove that if and = 1 ,then \n\n ,\n\nwith equality only if = / .\n\n15 Suppose , are nonzero vectors in 2. Prove that \n\n , = cos ,\n\nwhere is the angle between and (thinking of and as arrows with initial point at the origin). \n\n> Hint: Use the law of cosines on the triangle formed by , , and .\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6A Inner Products and Norms 193 \n\n16 The angle between two vectors (thought of as arrows with initial point at the origin) in 2 or 3 can be defined geometrically. However, geometry is not as clear in for > 3 . Thus the angle between two nonzero vectors \n\n , is defined to be \n\narccos , ,\n\nwhere the motivation for this definition comes from Exercise 15. Explain why the CauchySchwarz inequality is needed to show that this definition makes sense. \n\n17 Prove that \n\n(\n\n> \n> \n> = 1\n\n )\n\n> 2\n\n ( \n\n> \n> \n> = 1\n\n 2)( \n\n> \n> \n> = 1\n\n 2\n\n )\n\nfor all real numbers 1, , and 1, , .\n\n18 (a) Suppose [1 , ) [0 , ) is continuous. Show that \n\n(1\n\n ) 2\n\n 1\n\n 2( ()) 2 . \n\n(b) For which continuous functions [1 , ) [0 , ) is the inequality in (a) an equality with both sides finite? \n\n19 Suppose 1, , is a basis of and () . Prove that if is an eigenvalue of , then \n\n|| 2 \n\n> \n> \n> = 1\n> \n> \n> = 1\n\n|() , |2,\n\nwhere () , denotes the entry in row , column of the matrix of with respect to the basis 1, , .\n\n20 Prove that if , , then .\n\nThe inequality above is called the reverse triangle inequality . For the reverse triangle inequality when = , see Exercise 2 in Chapter 4. \n\n21 Suppose , are such that \n\n = 3 , + = 4 , = 6. \n\nWhat number does equal? \n\n22 Show that if , , then \n\n + 2 + 2.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n194 Chapter 6 Inner Product Spaces \n\n23 Suppose 1, , are such that 1 for each = 1 , , . Show that there exist 1, , {1 , 1} such that \n\n 1 1 + + . \n\n24 Prove or give a counterexample: If is the norm associated with an inner product on 2, then there exists ( , ) 2 such that ( , ) max {|| , ||} .\n\n25 Suppose > 0 . Prove that there is an inner product on 2 such that the associated norm is given by \n\n( , ) = (|| + || )1/ \n\nfor all ( , ) 2 if and only if = 2 .\n\n26 Suppose is a real inner product space. Prove that \n\n , = + 2 2\n\n4\n\nfor all , .\n\n27 Suppose is a complex inner product space. Prove that \n\n , = + 2 2 + + 2 2 4\n\nfor all , .\n\n28 A norm on a vector space is a function \n\n [0 , ) \n\nsuch that = 0 if and only if = 0 , = || for all and all \n\n , and + + for all , . Prove that a norm satisfying the parallelogram equality comes from an inner product (in other words, show that if is a norm on satisfying the parallelogram equality, then there is an inner product , on such that = , 1/2 for all ) .\n\n29 Suppose 1, , are inner product spaces. Show that the equation \n\n( 1, , ), ( 1, , ) = 1, 1 + + , \n\ndefines an inner product on 1 .\n\nIn the expression above on the right, for each = 1 , , , the inner product \n\n> \n\n, denotes the inner product on . Each of the spaces 1, , may have a different inner product, even though the same notation is used here. \n\n30 Suppose is a real inner product space. For , , , , define \n\n + , + = , + , + ( , , ). \n\n(a) Show that , makes into a complex inner product space. (b) Show that if , , then \n\n , = , and + 2 = 2 + 2.\n\nSee Exercise 8 in Section 1B for the definition of the complexification .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6A Inner Products and Norms 195 \n\n31 Suppose , , . Prove that \n\n 12 ( + ) 2 = 2 + 2\n\n2 2\n\n4 .\n\n32 Suppose that is a subset of with the property that , implies \n\n> 12\n\n( + ) . Let . Show that there is at most one point in that is closest to . In other words, show that there is at most one such that \n\n \n\nfor all .\n\n33 Suppose , are differentiable functions from to .(a) Show that \n\n () , () = () , () + () , (). \n\n(b) Suppose is a positive number and () = for every . Show that () , () = 0 for every .(c) Interpret the result in (b) geometrically in terms of the tangent vector to a curve lying on a sphere in centered at the origin. \n\nA function is called differentiable if there exist differentiable functions 1, , from to such that () = ( 1() , , ()) for each \n\n> \n\n. Furthermore, for each , the derivative () is defined by \n\n> () = ( 1()\n\n, , ()) .\n\n34 Use inner products to prove Apolloniuss identity: In a triangle with sides of length , , and , let be the length of the line segment from the midpoint of the side of length to the opposite vertex. Then \n\n 2 + 2 = 12 2 + 2 2.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n196 Chapter 6 Inner Product Spaces \n\n35 Fix a positive integer . The Laplacian of a twice differentiable real-valued function on is the function on defined by \n\n = 2 12 + + 2 2 .\n\nThe function is called harmonic if = 0 .A polynomial on is a linear combination (with coefficients in ) of functions of the form 1 1 , where 1, , are nonnegative integers. Suppose is a polynomial on . Prove that there exists a harmonic polynomial on such that () = () for every with = 1 .\n\nThe only fact about harmonic functions that you need for this exercise is that if is a harmonic function on and () = 0 for all with \n\n> = 1\n\n, then = 0 .Hint: A reasonable guess is that the desired harmonic polynomial is of the form + (1 2) for some polynomial . Prove that there is a polynomial \n\n> \n\non such that + (1 2) is harmonic by defining an operator on a suitable vector space by \n\n> = ((1 2))\n\nand then showing that is injective and hence surjective. \n\nIn realms of numbers, where the secrets lie, A noble truth emerges from the deep, Cauchy and Schwarz, their wisdom they apply, An inequality for all to keep. Two vectors, by this bond, are intertwined, As inner products weave a gilded thread, Their magnitude, by providence, confined, A bound to which their destiny is wed. Though shadows fall, and twilight dims the day, This inequality will stand the test, To guide us in our quest, to light the way, And in its truth, our understanding rest. So sing, ye muses, of this noble feat, CauchySchwarz, the bound that none can beat. \n\nwritten by ChatGPT with input Shakespearean sonnet on CauchySchwarz inequality \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6B Orthonormal Bases 197 \n\n# 6B Orthonormal Bases \n\n# Orthonormal Lists and the GramSchmidt Procedure \n\n> 6.22 definition:\n\northonormal \n\n A list of vectors is called orthonormal if each vector in the list has norm 1\n\nand is orthogonal to all the other vectors in the list. \n\n In other words, a list 1, , of vectors in is orthonormal if \n\n , = {{1 if = ,\n\n0 if \n\nfor all , {1 , , } . \n\n> 6.23 example:\n\northonormal lists \n\n(a) The standard basis of is an orthonormal list. (b) ( 1 \n\n> 3\n\n, 1 \n\n> 3\n\n, 1 \n\n> 3\n\n), ( 1 \n\n> 2\n\n, 1 \n\n> 2\n\n, 0) is an orthonormal list in 3.(c) ( 1 \n\n> 3\n\n, 1 \n\n> 3\n\n, 1 \n\n> 3\n\n), ( 1 \n\n> 2\n\n, 1 \n\n> 2\n\n, 0) , ( 1 \n\n> 6\n\n, 1 \n\n> 6\n\n, 2 \n\n> 6\n\n) is an orthonormal list in 3.(d) Suppose is a positive integer. Then, as Exercise 4 asks you to verify, \n\n1\n\n> \n\n2 , cos \n\n> \n\n , cos 2 \n\n> \n\n , , cos \n\n> \n\n , sin \n\n> \n\n , sin 2 \n\n> \n\n , , sin \n\n> \n\n \n\nis an orthonormal list of vectors in [ , ] , the vector space of continuous real-valued functions on [ , ] with inner product \n\n , = \n\n . \n\nThe orthonormal list above is often used for modeling periodic phenomena, such as tides. (e) Suppose we make 2() into an inner product space using the inner product given by \n\n , = 11 \n\n \n\nfor all , 2() . The standard basis 1, , 2 of 2() is not an orthonor-mal list because the vectors in that list do not have norm 1. Dividing each vector by its norm gives the list 1/ 2, 3/2 , 5/2 2, in which each vector has norm 1, and the second vector is orthogonal to the first and third vectors. However, the first and third vectors are not orthogonal. Thus this is not an orthonormal list. Soon we will see how to construct an orthonormal list from the standard basis 1, , 2 (see Example 6.34). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n198 Chapter 6 Inner Product Spaces \n\nOrthonormal lists are particularly easy to work with, as illustrated by the next result. \n\n> 6.24\n\nnorm of an orthonormal linear combination \n\nSuppose 1, , is an orthonormal list of vectors in . Then \n\n 1 1 + + 2 = | 1|2 + + | |2\n\nfor all 1, , .\n\nProof Because each has norm 1, this follows from repeated applications of the Pythagorean theorem (6.12). The result above has the following important corollary. \n\n> 6.25\n\northonormal lists are linearly independent \n\nEvery orthonormal list of vectors is linearly independent. \n\nProof Suppose 1, , is an orthonormal list of vectors in and 1, , \n\nare such that \n\n 1 1 + + = 0. \n\nThen | 1|2 + + | |2 = 0 (by 6.24), which means that all the s are 0. Thus \n\n 1, , is linearly independent. Now we come to an important inequality. \n\n> 6.26\n\nBessels inequality \n\nSuppose 1, , is an orthonormal list of vectors in . If then \n\n , 1 2 + + , 2 2.\n\nProof Suppose . Then \n\n = , 1 1 + + , \n\n> \n\n+ , 1 1 , \n\n> \n\n.\n\nLet and be defined as in the equation above. If {1 , , } , then \n\n , = , , , = 0. This implies that , = 0. The Pythagorean theorem now implies that \n\n 2 = 2 + 2\n\n 2\n\n= , 1 2 + + , 2,\n\nwhere the last line comes from 6.24. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6B Orthonormal Bases 199 \n\nThe next definition introduces one of the most useful concepts in the study of inner product spaces. \n\n> 6.27 definition:\n\northonormal basis \n\nAn orthonormal basis of is an orthonormal list of vectors in that is also a basis of .For example, the standard basis is an orthonormal basis of . \n\n> 6.28\n\northonormal lists of the right length are orthonormal bases \n\nSuppose is finite-dimensional. Then every orthonormal list of vectors in \n\nof length dim is an orthonormal basis of .\n\nProof By 6.25, every orthonormal list of vectors in is linearly independent. Thus every such list of the right length is a basissee 2.38. \n\n> 6.29 example:\n\nan orthonormal basis of 4\n\nAs mentioned above, the standard basis is an orthonormal basis of 4. We now show that \n\n( 12 , 12 , 12 , 12 ), ( 12 , 12 , 12 , 12 ), ( 12 , 12 , 12 , 12 ), ( 12 , 12 , 12 , 12 )\n\nis also an orthonormal basis of 4.We have \n\n( 12 , 12 , 12 , 12 ) = 122 + 122 + 122 + 122 = 1. \n\nSimilarly, the other three vectors in the list above also have norm 1.Note that \n\n( 12 , 12 , 12 , 12 ), ( 12 , 12 , 12 , 12 ) = 12 12 + 12 12 + 12 ( 12 ) + 12 ( 12 ) = 0. \n\nSimilarly, the inner product of any two distinct vectors in the list above also equals 0.Thus the list above is orthonormal. Because we have an orthonormal list of length four in the four-dimensional vector space 4, this list is an orthonormal basis of 4 (by 6.28). In general, given a basis 1, , of and a vector , we know that there is some choice of scalars 1, , such that \n\n = 1 1 + + .\n\nComputing the numbers 1, , that satisfy the equation above can be a long computation for an arbitrary basis of . The next result shows, however, that this is easy for an orthonormal basisjust take = , .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n200 Chapter 6 Inner Product Spaces \n\nThe formula below for is called Parsevals identity. It was published in 1799 in the context of Fourier series. \n\nNotice how the next result makes each inner product space of dimension \n\n behave like , with the role of the coordinates of a vector in played by \n\n , 1, , , . \n\n> 6.30\n\nwriting a vector as a linear combination of an orthonormal basis \n\nSuppose 1, , is an orthonormal basis of and , . Then (a) = , 1 1 + + , ;(b) 2 = , 1 2 + + , 2;(c) , = , 1 , 1 + + , , .\n\nProof Because 1, , is a basis of , there exist scalars 1, , such that \n\n = 1 1 + + .\n\nBecause 1, , is orthonormal, taking the inner product of both sides of this equation with gives , = . Thus (a) holds. Now (b) follows immediately from (a) and 6.24. Take the inner product of with each side of (a) and then get (c) by using conjugate linearity [6.6(d) and 6.6(e)] in the second slot of the inner product. \n\n> 6.31 example:\n\nfinding coefficients for a linear combination \n\nSuppose we want to write the vector (1 , 2, 4, 7) 4 as a linear combination of the orthonormal basis \n\n( 12 , 12 , 12 , 12 ), ( 12 , 12 , 12 , 12 ), ( 12 , 12 , 12 , 12 ), ( 12 , 12 , 12 , 12 )\n\nof 4 from Example 6.29. Instead of solving a system of four linear equations in four unknowns, as typically would be required if we were working with a nonorthonormal basis, we simply evaluate four inner products and use 6.30(a), getting that (1 , 2, 4, 7) equals \n\n7( 12 , 12 , 12 , 12 ) 4( 12 , 12 , 12 , 12 ) + ( 12 , 12 , 12 , 12 ) + 2( 12 , 12 , 12 , 12 ). \n\nNow that we understand the usefulness of orthonormal bases, how do we go about finding them? For example, does () with inner product as in 6.3(c) have an orthonormal basis? The next result will lead to answers to these questions. \n\nJrgen Gram (18501916 ) and Erhard Schmidt (18761959 ) popularized this algorithm that constructs orthonormal lists. \n\nThe algorithm used in the next proof is called the GramSchmidt procedure .It gives a method for turning a linearly independent list into an orthonormal list with the same span as the original list. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6B Orthonormal Bases 201 \n\n> 6.32\n\nGramSchmidt procedure \n\nSuppose 1, , is a linearly independent list of vectors in . Let 1 = 1.For = 2 , , , define inductively by \n\n = , 1 12 1 , 1 1 2 1 .\n\nFor each = 1 , , , let = \n\n . Then 1, , is an orthonormal list of vectors in such that \n\nspan ( 1, , ) = span ( 1, , )\n\nfor each = 1 , , .\n\nProof We will show by induction on that the desired conclusion holds. To get started with = 1 , note that because 1 = 1/ 1, we have 1 = 1 ; also, \n\nspan ( 1) = span ( 1) because 1 is a nonzero multiple of 1.Suppose 1 < and the list 1, , 1 generated by 6.32 is an orthonor-mal list such that \n\n> 6.33\n\nspan ( 1, , 1 ) = span ( 1, , 1 ). \n\nBecause 1, , is linearly independent, we have span ( 1, , 1 ). Thus \n\n span ( 1, , 1 ) = span ( 1, , 1 ), which implies that 0 . Hence we are not dividing by 0 in the definition of given in 6.32. Dividing a vector by its norm produces a new vector with norm 1; thus = 1 .Let {1 , , 1} . Then \n\n , = 1 , = 1 , 1 12 1 , 1 1 2 1 , = 1 ( , , ) = 0. \n\nThus 1, , is an orthonormal list. From the definition of given in 6.32, we see that span ( 1, , ).Combining this information with 6.33 shows that \n\nspan ( 1, , ) span ( 1, , ). \n\nBoth lists above are linearly independent (the s by hypothesis, and the s by orthonormality and 6.25). Thus both subspaces above have dimension , and hence they are equal, completing the induction step and thus completing the proof. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n202 Chapter 6 Inner Product Spaces \n\n> 6.34 example:\n\nan orthonormal basis of 2() \n\nSuppose we make 2() into an inner product space using the inner product given by \n\n , = 11 \n\n \n\nfor all , 2() . We know that 1, , 2 is a basis of 2() , but it is not an orthonormal basis. We will find an orthonormal basis of 2() by applying the GramSchmidt procedure with 1 = 1 , 2 = , and 3 = 2.To get started, take 1 = 1 = 1 . Thus 12 = 11 1 = 2 . Hence the formula in 6.32 tells us that \n\n 2 = 2 2, 1 12 1 = , 1 12 = ,\n\nwhere the last equality holds because , 1 = 11 = 0 .The formula above for 2 implies that 22 = 11 2 = 23 . Now the formula in 6.32 tells us that \n\n 3 = 3 3, 1 12 1 3, 2 22 2 = 2 12 2, 1 32 2, = 2 13 .\n\nThe formula above for 3 implies that \n\n 32 = 11 \n\n( 2 13 )2\n\n = 11 \n\n( 4 23 2 + 19 ) = 845 .\n\nNow dividing each of 1, 2, 3 by its norm gives us the orthonormal list \n\n> 12\n\n, 32 , 45 8 ( 2 13 ). \n\nThe orthonormal list above has length three, which is the dimension of 2() .Hence this orthonormal list is an orthonormal basis of 2() [by 6.28]. Now we can answer the question about the existence of orthonormal bases. \n\n> 6.35\n\nexistence of orthonormal basis \n\nEvery finite-dimensional inner product space has an orthonormal basis. \n\nProof Suppose is finite-dimensional. Choose a basis of . Apply the Gram Schmidt procedure (6.32) to it, producing an orthonormal list of length dim .By 6.28, this orthonormal list is an orthonormal basis of .Sometimes we need to know not only that an orthonormal basis exists, but also that every orthonormal list can be extended to an orthonormal basis. In the next corollary, the GramSchmidt procedure shows that such an extension is always possible. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 6B Orthonormal Bases 203\n> 6.36\n\nevery orthonormal list extends to an orthonormal basis \n\nSuppose is finite-dimensional. Then every orthonormal list of vectors in \n\ncan be extended to an orthonormal basis of . \n\n> Proof\n\nSuppose 1, , is an orthonormal list of vectors in . Then 1, , \n\nis linearly independent (by 6.25). Hence this list can be extended to a basis \n\n 1, , , 1, , of (see 2.32). Now apply the GramSchmidt procedure (6.32) to 1, , , 1, , , producing an orthonormal list \n\n 1, , , 1, , ;\n\nhere the formula given by the GramSchmidt procedure leaves the first vectors unchanged because they are already orthonormal. The list above is an orthonormal basis of by 6.28. Recall that a matrix is called upper triangular if it looks like this: \n\n\n\n# \n\n0 \n\n\n\n,\n\nwhere the 0 in the matrix above indicates that all entries below the diagonal equal 0, and asterisks are used to denote entries on and above the diagonal. In the last chapter, we gave a necessary and sufficient condition for an operator to have an upper-triangular matrix with respect to some basis (see 5.44). Now that we are dealing with inner product spaces, we would like to know whether there exists an orthonormal basis with respect to which we have an upper-triangular matrix. The next result shows that the condition for an operator to have an upper-triangular matrix with respect to some orthonormal basis is the same as the condition to have an upper-triangular matrix with respect to an arbitrary basis. \n\n> 6.37\n\nupper-triangular matrix with respect to some orthonormal basis \n\nSuppose is finite-dimensional and () . Then has an upper-triangular matrix with respect to some orthonormal basis of if and only if the minimal polynomial of equals ( 1) ( ) for some 1, , . \n\n> Proof\n\nSuppose has an upper-triangular matrix with respect to some basis \n\n 1, , of . Thus span ( 1, , ) is invariant under for each = 1 , , \n\n(see 5.39). Apply the GramSchmidt procedure to 1, , , producing an orthonormal basis 1, , of . Because \n\nspan ( 1, , ) = span ( 1, , )\n\nfor each (see 6.32), we conclude that span ( 1, , ) is invariant under for each = 1 , , . Thus, by 5.39, has an upper-triangular matrix with respect to the orthonormal basis 1, , . Now use 5.44 to complete the proof. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n204 Chapter 6 Inner Product Spaces \n\nIssai Schur (18751941 ) published a proof of the next result in 1909. \n\nFor complex vector spaces, the next result is an important application of the result above. See Exercise 20 for a ver-sion of Schurs theorem that applies simultaneously to more than one operator. \n\n> 6.38\n\nSchurs theorem \n\nEvery operator on a finite-dimensional complex inner product space has an upper-triangular matrix with respect to some orthonormal basis. \n\nProof The desired result follows from the second version of the fundamental theorem of algebra (4.13) and 6.37. \n\n# Linear Functionals on Inner Product Spaces \n\nBecause linear maps into the scalar field play a special role, we defined a special name for them and their vector space in Section 3F. Those definitions are repeated below in case you skipped Section 3F. \n\n> 6.39 definition:\n\nlinear functional, dual space, \n\n A linear functional on is a linear map from to .\n\n The dual space of , denoted by , is the vector space of all linear functionals on . In other words, = ( , ) . \n\n> 6.40 example:\n\nlinear functional on 3\n\nThe function 3 defined by \n\n( 1, 2, 3) = 2 1 5 2 + 3\n\nis a linear functional on 3. We could write this linear functional in the form \n\n() = , \n\nfor every 3, where = (2 , 5 , 1) . \n\n> 6.41 example:\n\nlinear functional on 5() \n\nThe function 5() defined by \n\n() = 11 \n\n()( cos ()) \n\nis a linear functional on 5() .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6B Orthonormal Bases 205 \n\nThe next result is named in honor of Frigyes Riesz (18801956 ), who proved several theorems early in the twentieth century that look very much like the result below. \n\nIf , then the map that sends \n\nto , is a linear functional on . The next result states that every linear func-tional on is of this form. For example, we can take = (2 , 5 , 1) in Example 6.40. Suppose we make the vector space 5() into an inner product space by defining , = 11 . Let be as in Example 6.41. It is not obvious that there exists 5() such that \n\n> 11\n\n()( cos ()) = , \n\nfor every 5() [ we cannot take () = cos () because that choice of is not an element of 5()] . The next result tells us the somewhat surprising result that there indeed exists a polynomial 5() such that the equation above holds for all 5() . \n\n> 6.42\n\nRiesz representation theorem \n\nSuppose is finite-dimensional and is a linear functional on . Then there is a unique vector such that \n\n() = , \n\nfor every .\n\nProof First we show that there exists a vector such that () = , for every . Let 1, , be an orthonormal basis of . Then \n\n() = ( , 1 1 + + , )= , 1( 1) + + , ( )= , ( 1) 1 + + ( ) \n\nfor every , where the first equality comes from 6.30(a). Thus setting \n\n> 6.43\n\n = ( 1) 1 + + ( ) ,\n\nwe have () = , for every , as desired. Now we prove that only one vector has the desired behavior. Suppose \n\n 1, 2 are such that \n\n() = , 1 = , 2\n\nfor every . Then \n\n0 = , 1 , 2 = , 1 2\n\nfor every . Taking = 1 2 shows that 1 2 = 0 . Thus 1 = 2,completing the proof of the uniqueness part of the result. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n206 Chapter 6 Inner Product Spaces \n\n> 6.44 example:\n\ncomputation illustrating Riesz representation theorem \n\nSuppose we want to find a polynomial 2() such that \n\n> 6.45 11\n\n()( cos ()) = 11 \n\n \n\nfor every polynomial 2() . To do this, we make 2() into an inner product space by defining , to be the right side of the equation above for , 2() .Note that the left side of the equation above does not equal the inner product in 2() of and the function cos () because this last function is not a polynomial. Define a linear functional on 2() by letting \n\n() = 11 \n\n()( cos ()) \n\nfor each 2() . Now use the orthonormal basis from Example 6.34 and apply formula 6.43 from the proof of the Riesz representation theorem to see that if 2() , then () = , , where \n\n() = ( 11 \n\n> 12\n\ncos () ) 12 + (11 \n\n> 32\n\n cos () ) 32 \n\n+ (11 \n\n> 45 8\n\n( 2 13 ) cos () ) 45 8 ( 2 13 ). \n\nA bit of calculus applied to the equation above shows that \n\n() = 15 2 2 (1 3 2). \n\nThe same procedure shows that if we want to find 5() such that 6.45 holds for all 5() , then we should take \n\n() = 105 8 4 ((27 2 2) + (24 2 270) 2 + (315 30 2) 4). \n\nSuppose is finite-dimensional and a linear functional on . Then 6.43 gives a formula for the vector that satisfies \n\n() = , \n\nfor all . Specifically, we have \n\n = ( 1) 1 + + ( ) .\n\nThe right side of the equation above seems to depend on the orthonormal basis \n\n 1, , as well as on . However, 6.42 tells us that is uniquely determined by . Thus the right side of the equation above is the same regardless of which orthonormal basis 1, , of is chosen. For two additional different proofs of the Riesz representation theorem, see 6.58 and also Exercise 13 in Section 6C. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6B Orthonormal Bases 207 \n\n# Exercises 6B \n\n1 Suppose 1, , is a list of vectors in such that \n\n 1 1 + + 2 = | 1|2 + + | |2\n\nfor all 1, , . Show that 1, , is an orthonormal list. \n\nThis exercise provides a converse to 6.24. \n\n2 (a) Suppose . Show that both \n\n(cos , sin ) , ( sin , cos ) and (cos , sin ) , (sin , cos ) \n\nare orthonormal bases of 2.(b) Show that each orthonormal basis of 2 is of the form given by one of the two possibilities in (a). \n\n3 Suppose 1, , is an orthonormal list in and . Prove that \n\n 2 = , 1 2 + + , 2 span ( 1, , ). \n\n4 Suppose is a positive integer. Prove that \n\n1\n\n> \n\n2 , cos \n\n> \n\n , cos 2 \n\n> \n\n , , cos \n\n> \n\n , sin \n\n> \n\n , sin 2 \n\n> \n\n , , sin \n\n> \n\n \n\nis an orthonormal list of vectors in [ , ] , the vector space of continuous real-valued functions on [ , ] with inner product \n\n , = \n\n . \n\nHint: The following formulas should help. \n\n(sin )( cos ) = sin ( ) + sin ( + ) 2(sin )( sin ) = cos ( ) cos ( + ) 2(cos )( cos ) = cos ( ) + cos ( + ) 2\n\n5 Suppose [ , ] is continuous. For each nonnegative integer ,define \n\n = 1 \n\n> \n\n () cos () and = 1 \n\n> \n\n () sin () . \n\nProve that 02\n\n2 + \n\n> \n> = 1\n\n( 2 + 2) \n\n 2.\n\nThe inequality above is actually an equality for all continuous functions \n\n [ , ] . However, proving that this inequality is an equality involves Fourier series techniques beyond the scope of this book. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n208 Chapter 6 Inner Product Spaces \n\n6 Suppose 1, , is an orthonormal basis of .(a) Prove that if 1, , are vectors in such that \n\n < 1\n\n> \n\n \n\nfor each , then 1, , is a basis of .(b) Show that there exist 1, , such that \n\n 1\n\n> \n\n \n\nfor each , but 1, , is not linearly independent. \n\nThis exercise states in (a) that an appropriately small perturbation of an orthonormal basis is a basis. Then (b) shows that the number 1/ on the right side of the inequality in (a) cannot be higher. \n\n7 Suppose ( 3) has an upper-triangular matrix with respect to the basis \n\n(1 , 0, 0) , (1, 1, 1), (1 , 1, 2) . Find an orthonormal basis of 3 with respect to which has an upper-triangular matrix. \n\n8 Make 2() into an inner product space by defining , = 10 for all \n\n , 2() .(a) Apply the GramSchmidt procedure to the basis 1, , 2 to produce an orthonormal basis of 2() .(b) The differentiation operator (the operator that takes to ) on 2() \n\nhas an upper-triangular matrix with respect to the basis 1, , 2, which is not an orthonormal basis. Find the matrix of the differentiation operator on 2() with respect to the orthonormal basis produced in (a) and verify that this matrix is upper triangular, as expected from the proof of 6.37. \n\n9 Suppose 1, , is the result of applying the GramSchmidt procedure to a linearly independent list 1, , in . Prove that , > 0 for each \n\n = 1 , , .\n\n10 Suppose 1, , is a linearly independent list in . Explain why the orthonormal list produced by the formulas of the GramSchmidt procedure (6.32) is the only orthonormal list 1, , in such that , > 0 and \n\nspan ( 1, , ) = span ( 1, , ) for each = 1 , , .\n\nThe result in this exercise is used in the proof of 7.58. \n\n11 Find a polynomial 2() such that ( 12 ) = 10 for every 2() .\n\n12 Find a polynomial 2() such that \n\n> 10\n\n() cos () = 10\n\n \n\nfor every 2() .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6B Orthonormal Bases 209 \n\n13 Show that a list 1, , of vectors in is linearly dependent if and only if the GramSchmidt formula in 6.32 produces = 0 for some {1 , , } .\n\nThis exercise gives an alternative to Gaussian elimination techniques for determining whether a list of vectors in an inner product space is linearly dependent. \n\n14 Suppose is a real inner product space and 1, , is a linearly indepen-dent list of vectors in . Prove that there exist exactly 2 orthonormal lists \n\n 1, , of vectors in such that \n\nspan ( 1, , ) = span ( 1, , )\n\nfor all {1 , , } .\n\n15 Suppose , 1 and , 2 are inner products on such that , 1 = 0 if and only if , 2 = 0 . Prove that there is a positive number such that \n\n , 1 = , 2 for every , .\n\nThis exercise shows that if two inner products have the same pairs of orthogonal vectors, then each of the inner products is a scalar multiple of the other inner product. \n\n16 Suppose is finite-dimensional. Suppose , 1, , 2 are inner products on \n\n with corresponding norms 1 and 2. Prove that there exists a positive number such that 1 2 for every .\n\n17 Suppose = and is finite-dimensional. Prove that if is an operator on such that 1 is the only eigenvalue of and for all ,then is the identity operator. \n\n18 Suppose 1, , is a linearly independent list in . Show that there exists \n\n such that , = 1 for all {1 , , } .\n\n19 Suppose 1, , is a basis of . Prove that there exists a basis 1, , \n\nof such that \n\n , = {{0 if ,\n\n1 if = . \n\n20 Suppose = , is finite-dimensional, and () is such that \n\n = \n\nfor all , . Prove that there is an orthonormal basis of with respect to which every element of has an upper-triangular matrix. \n\nThis exercise strengthens Exercise 9 (b) in Section 5E (in the context of inner product spaces ) by asserting that the basis in that exercise can be chosen to be orthonormal. \n\n21 Suppose = , is finite-dimensional, () , and all eigenvalues of have absolute value less than 1. Let > 0 . Prove that there exists a positive integer such that for every . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 210 Chapter 6 Inner Product Spaces\n\n22 Suppose [1 , 1] is the vector space of continuous real-valued functions on the interval [1 , 1] with inner product given by \n\n , = 11 \n\n \n\nfor all , [1 , 1] . Let be the linear functional on [1 , 1] defined by ( ) = (0) . Show that there does not exist [1 , 1] such that \n\n( ) = , \n\nfor every [1 , 1] . \n\n> This exercise shows that the Riesz representation theorem (6.42 )does not hold on infinite-dimensional vector spaces without additional hypotheses on and .\n\n23 For all , , define ( , ) = .(a) Show that is a metric on .(b) Show that if is finite-dimensional, then is a complete metric on \n\n(meaning that every Cauchy sequence converges). (c) Show that every finite-dimensional subspace of is a closed subset of (with respect to the metric ). \n\n> This exercise requires familiarity with metric spaces.\n\northogonality at the Supreme Court \n\nLaw professor Richard Friedman presenting a case before the U.S. Supreme Court in 2010: \n\nMr. Friedman : I think that issue is entirely orthogonal to the issue here because the Commonwealth is acknowledging \n\nChief Justice Roberts : Im sorry. Entirely what? \n\nMr. Friedman : Orthogonal. Right angle. Unrelated. Irrelevant. \n\nChief Justice Roberts : Oh. \n\nJustice Scalia : What was that adjective? I liked that. \n\nMr. Friedman : Orthogonal. \n\nChief Justice Roberts : Orthogonal. \n\nMr. Friedman : Right, right. \n\nJustice Scalia : Orthogonal, ooh. (Laughter.) \n\nJustice Kennedy : I knew this case presented us a problem. (Laughter.) \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6C Orthogonal Complements and Minimization Problems 211 \n\n# 6C Orthogonal Complements and Minimization Problems \n\n# Orthogonal Complements \n\n> 6.46 definition:\n\northogonal complement, \n\nIf is a subset of , then the orthogonal complement of , denoted by , is the set of all vectors in that are orthogonal to every vector in :\n\n = { , = 0 for every }. \n\nThe orthogonal complement depends on as well as on . However, the inner product space should always be clear from the context and thus it can be omitted from the notation. \n\n> 6.47 example:\n\northogonal complements \n\n If = 3 and is the subset of consisting of the single point (2 , 3, 5) , then \n\n is the plane {( , , ) 3 2 + 3 + 5 = 0} .\n\n If = 3 and is the plane {( , , ) 3 2 + 3 + 5 = 0} , then is the line {(2 , 3 , 5) } .\n\n More generally, if is a plane in 3 containing the origin, then is the line containing the origin that is perpendicular to .\n\n If is a line in 3 containing the origin, then is the plane containing the origin that is perpendicular to .\n\n If = 5 and = {( , , 0, 0, 0) 5 , } , then \n\n = {(0 , 0, , , ) 5 , , }. If 1, , , 1, , is an orthonormal basis of , then \n\n(span ( 1, , )) = span ( 1, , ). \n\nWe begin with some straightforward consequences of the definition. \n\n> 6.48\n\nproperties of orthogonal complement \n\n(a) If is a subset of , then is a subspace of .(b) {0} = .(c) = {0} .(d) If is a subset of , then {0} .(e) If and are subsets of and , then . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 212 Chapter 6 Inner Product Spaces Proof\n\n(a) Suppose is a subset of . Then , 0 = 0 for every ; thus 0 .Suppose , . If , then \n\n , + = , + , = 0 + 0 = 0. \n\nThus + , which shows that is closed under addition. Similarly, suppose and . If , then \n\n , = , = 0 = 0. \n\nThus , which shows that is closed under scalar multiplication. Thus is a subspace of .(b) Suppose that . Then 0 , = 0 , which implies that {0} . Thus \n\n{0} = .(c) Suppose that . Then , = 0 , which implies that = 0 . Thus \n\n = {0} .(d) Suppose is a subset of and . Then , = 0 , which implies that = 0 . Thus {0} .(e) Suppose and are subsets of and . Suppose . Then \n\n , = 0 for every , which implies that , = 0 for every .Hence . Thus .Recall that if and are subspaces of , then is the direct sum of and \n\n (written = ) if each element of can be written in exactly one way as a vector in plus a vector in (see 1.41). Furthermore, this happens if and only if = + and = {0} (see 1.46). The next result shows that every finite-dimensional subspace of leads to a natural direct sum decomposition of . See Exercise 16 for an example showing that the result below can fail without the hypothesis that the subspace is finite-dimensional. \n\n> 6.49\n\ndirect sum of a subspace and its orthogonal complement \n\nSuppose is a finite-dimensional subspace of . Then \n\n = . \n\n> Proof\n\nFirst we will show that \n\n = + .\n\nTo do this, suppose that . Let 1, , be an orthonormal basis of . We want to write as the sum of a vector in and a vector orthogonal to .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6C Orthogonal Complements and Minimization Problems 213 \n\nWe have \n\n> 6.50\n\n = , 1 1 + + , \n\n> \n\n+ , 1 1 , \n\n> \n\n.\n\nLet and be defined as in the equation above (as was done in the proof of 6.26). Because each , we see that . Because 1, , is an orthonormal list, for each = 1 , , we have \n\n , = , , = 0. \n\nThus is orthogonal to every vector in span ( 1, , ), which shows that .Hence we have written = + , where and , completing the proof that = + .From 6.48(d), we know that = {0} . Now equation = + \n\nimplies that = (see 1.46). Now we can see how to compute dim from dim . \n\n> 6.51\n\ndimension of orthogonal complement \n\nSuppose is finite-dimensional and is a subspace of . Then \n\ndim = dim dim . \n\nProof The formula for dim follows immediately from 6.49 and 3.94. The next result is an important consequence of 6.49. \n\n> 6.52\n\northogonal complement of the orthogonal complement \n\nSuppose is a finite-dimensional subspace of . Then \n\n = ( ).\n\nProof First we will show that \n\n> 6.53\n\n ( ).\n\nTo do this, suppose . Then , = 0 for every (by the definition of ). Because is orthogonal to every vector in , we have ( ),completing the proof of 6.53. To prove the inclusion in the other direction, suppose ( ). By 6.49, we can write = + , where and . We have = .Because ( ) and ( ) (from 6.53), we have ( ). Thus \n\n ( ), which implies that = 0 [by 6.48(d)], which implies that = , which implies that . Thus ( ) , which along with 6.53 completes the proof. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n214 Chapter 6 Inner Product Spaces \n\nExercise 16 (a) shows that the result below is not true without the hypothesis that is finite-dimensional. \n\nSuppose is a subspace of and we want to show that = . In some situations, the easiest way to do this is to show that the only vector orthogonal to \n\n is 0, and then use the result below. For example, the result below is useful for Exercise 4. \n\n> 6.54\n\n = {0} = (for a finite-dimensional subspace of )Suppose is a finite-dimensional subspace of . Then \n\n = {0} = . \n\nProof First suppose = {0} . Then by 6.52, = ( ) = {0} = , as desired. Conversely, if = , then = = {0} by 6.48(c). We now define an operator for each finite-dimensional subspace of . \n\n> 6.55 definition:\n\northogonal projection, \n\nSuppose is a finite-dimensional subspace of . The orthogonal projection \n\nof onto is the operator () defined as follows: For each ,write = + , where and . Then let = .The direct sum decomposition = given by 6.49 shows that each \n\n can be uniquely written in the form = + with and .Thus is well defined. See the figure that accompanies the proof of 6.61 for the picture describing that you should keep in mind. \n\n> 6.56 example:\n\northogonal projection onto one-dimensional subspace \n\nSuppose with 0 and is the one-dimensional subspace of \n\ndefined by = span () .If , then \n\n = , 2 + ( , 2 ) ,\n\nwhere the first term on the right is in span () (and thus is in ) and the second term on the right is orthogonal to (and thus is in ). Thus equals the first term on the right. In other words, we have the formula \n\n = , 2 \n\nfor every .The formula above becomes = if = and becomes = 0 if \n\n {} . These equations are special cases of (b) and (c) in the next result. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6C Orthogonal Complements and Minimization Problems 215 \n\n> 6.57\n\nproperties of orthogonal projection \n\nSuppose is a finite-dimensional subspace of . Then (a) () ;(b) = for every ;(c) = 0 for every ;(d) range = ;(e) null = ;(f) for every ;(g) 2 = ;(h) for every ;(i) if 1, , is an orthonormal basis of and , then \n\n = , 1 1 + + , .\n\nProof \n\n(a) To show that is a linear map on , suppose 1, 2 . Write \n\n 1 = 1 + 1 and 2 = 2 + 2\n\nwith 1, 2 and 1, 2 . Thus 1 = 1 and 2 = 2. Now \n\n 1 + 2 = ( 1 + 2) + ( 1 + 2),\n\nwhere 1 + 2 and 1 + 2 . Thus \n\n ( 1 + 2) = 1 + 2 = 1 + 2.\n\nSimilarly, suppose and . Write = + , where \n\nand . Then = + with and . Thus \n\n () = = .Hence is a linear map from to .(b) Suppose . We can write = + 0, where and 0 . Thus \n\n = .(c) Suppose . We can write = 0 + , where 0 and . Thus \n\n = 0 .(d) The definition of implies that range . Furthermore, (b) implies that range . Thus range = .(e) The inclusion null follows from (c). To prove the inclusion in the other direction, note that if null then the decomposition given by 6.49 must be = 0 + , where 0 and . Thus null .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n216 Chapter 6 Inner Product Spaces \n\n(f) If and = + with and , then \n\n = = .\n\n(g) If and = + with and , then \n\n( 2) = ( ) = = = . \n\n(h) If and = + with and , then \n\n 2 = 2 2 + 2 = 2,\n\nwhere the last equality comes from the Pythagorean theorem. (i) The formula for follows from equation 6.50 in the proof of 6.49. In the previous section we proved the Riesz representation theorem (6.42), whose key part states that every linear functional on a finite-dimensional inner product space is given by taking the inner product with some fixed vector. Seeing a different proof often provides new insight. Thus we now give a new proof of the key part of the Riesz representation theorem using orthogonal complements instead of orthonormal bases as in our previous proof. The restatement below of the Riesz representation theorem provides an iden-tification of with . We will prove only the onto part of the result below because the more routine one-to-one part of the result can be proved as in 6.42. Intuition behind this new proof: If , , and () = , for all \n\n , then ( null ) . However, (null ) is a one-dimensional subspace of (except for the trivial case in which = 0 ), as follows from 6.51 and 3.21. Thus we can obtain by choosing any nonzero element of (null ) and then multiplying by an appropriate scalar, as is done in the proof below. \n\n> 6.58\n\nRiesz representation theorem, revisited \n\nSuppose is finite-dimensional. For each , define by \n\n () = , \n\nfor each . Then is a one-to-one function from onto .\n\nCaution: The function is a linear mapping from to if = .However, this function is not linear if \n\n> = \n\nbecause = if .\n\nProof To show that is surjective, suppose . If = 0 , then = 0.Thus assume 0 . Hence null ,which implies that (null ) {0} (by 6.49 with = null ). Let ( null ) be such that 0 . Let \n\n> 6.59\n\n = () 2 . \n\nThen ( null ) . Also, 0 (because null ). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6C Orthogonal Complements and Minimization Problems 217 \n\nTaking the norm of both sides of 6.59 gives \n\n> 6.60\n\n = |()| .\n\nApplying to both sides of 6.59 and then using 6.60, we have \n\n() = |()| 2\n\n 2 = 2.\n\nNow suppose . Using the equation above, we have \n\n = ( () () ) + () 2 . \n\nThe first term in parentheses above is in null and hence is orthogonal to . Thus taking the inner product of both sides of the equation above with shows that \n\n , = () 2 , = (). \n\nThus = , showing that is surjective, as desired. See Exercise 13 for yet another proof of the Riesz representation theorem. \n\n# Minimization Problems \n\nThe remarkable simplicity of the solu-tion to this minimization problem has led to many important applications of inner product spaces outside of pure mathematics. \n\nThe following problem often arises: Given a subspace of and a point \n\n , find a point such that \n\n is as small as possible. The next result shows that = is the unique solution of this minimization problem. \n\n> 6.61\n\nminimizing distance to a subspace \n\nSuppose is a finite-dimensional subspace of , , and . Then \n\n . \n\nFurthermore, the inequality above is an equality if and only if = .\n\nProof We have \n\n 2 2 + 2\n\n> 6.62\n\n= ( ) + ( ) 2\n\n= 2, \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 218 Chapter 6 Inner Product Spaces\n> is the closest point in to .\n\nwhere the first line above holds because 0 2,the second line above comes from the Pythagorean the-orem [which applies because by 6.57(f), and ] , and the third line above holds by simple algebra. Taking square roots gives the desired inequality. The inequality proved above is an equality if and only if 6.62 is an equality, which happens if and only if \n\n = 0 , which happens if and only if = .The last result is often combined with the formula 6.57(i) to compute explicit solutions to minimization problems, as in the following example. \n\n> 6.63 example:\n\nusing linear algebra to approximate the sine function \n\nSuppose we want to find a polynomial with real coefficients and of degree at most 5 that approximates the sine function as well as possible on the interval \n\n[ , ] , in the sense that \n\n> \n\nsin () 2 \n\nis as small as possible. Let [ , ] denote the real inner product space of continuous real-valued functions on [ , ] with inner product \n\n> 6.64\n\n , = \n\n . \n\nLet [ , ] be the function defined by () = sin . Let denote the subspace of [ , ] consisting of the polynomials with real coefficients and of degree at most 5. Our problem can now be reformulated as follows: Find such that is as small as possible. \n\n> A computer that can integrate is useful here.\n\nTo compute the solution to our ap-proximation problem, first apply the GramSchmidt procedure (using the in-ner product given by 6.64) to the basis 1, , 2, 3, 4, 5 of , producing an ortho-normal basis 1, 2, 3, 4, 5, 6 of . Then, again using the inner product given by 6.64, compute using 6.57(i) (with = 6 ). Doing this computation shows that is the function defined by \n\n> 6.65\n\n() = 0.987862 0.155271 3 + 0.00564312 5,\n\nwhere the s that appear in the exact answer have been replaced with a good decimal approximation. By 6.61, the polynomial above is the best approximation to the sine function on [ , ] using polynomials of degree at most 5 ( here best approximation means in the sense of minimizing | sin ()| 2 ) . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 6C Orthogonal Complements and Minimization Problems 219\n\nTo see how good this approximation is, the next figure shows the graphs of both the sine function and our approximation given by 6.65 over the interval \n\n[ , ] . \n\n> Graphs on [ ,] of the sine function (red )and its best fifth degree polynomial approximation (blue )from 6.65.\n\nOur approximation 6.65 is so accurate that the two graphs are almost identical our eyes may see only one graph! Here the red graph is placed almost exactly over the blue graph. If you are viewing this on an electronic device, enlarge the picture above by 400% near or to see a small gap between the two graphs. Another well-known approximation to the sine function by a polynomial of degree 5 is given by the Taylor polynomial defined by \n\n> 6.66\n\n() = 3\n\n3! + 5\n\n5! .\n\nTo see how good this approximation is, the next picture shows the graphs of both the sine function and the Taylor polynomial over the interval [ , ] . \n\n> Graphs on [ ,] of the sine function (red )\n> and the Taylor polynomial (blue )from 6.66.\n\nThe Taylor polynomial of degree 5 is an excellent approximation to sin for \n\n near 0. But the picture above shows that for || > 2 , the Taylor polynomial is not so accurate, especially compared to 6.65. For example, taking = 3 , our approximation 6.65 estimates sin 3 with an error of approximately 0.001 , but the Taylor polynomial 6.66 estimates sin 3 with an error of approximately 0.4 . Thus at = 3 , the error in the Taylor polynomial is hundreds of times larger than the error given by 6.65. Linear algebra has helped us discover an approximation to the sine function that improves upon what we learned in calculus! \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n220 Chapter 6 Inner Product Spaces \n\n# Pseudoinverse \n\nSuppose ( , ) and . Consider the problem of finding such that \n\n = . \n\nFor example, if = and = , then the equation above could represent a system of linear equations in unknowns 1, , , where = ( 1, , ).If is invertible, then the unique solution to the equation above is = 1 .However, if is not invertible, then for some there may not exist any solutions of the equation above, and for some there may exist infinitely many solutions of the equation above. If is not invertible, then we can still try to do as well as possible with the equation above. For example, if the equation above has no solutions, then instead of solving the equation = 0 , we can try to find such that \n\nis as small as possible. As another example, if the equation above has infinitely many solutions , then among all those solutions we can try to find one such that is as small as possible. The pseudoinverse will provide the tool to solve the equation above as well as possible, even when is not invertible. We need the next result to define the pseudoinverse. In the next two proofs, we will use without further comment the result that if \n\n is finite-dimensional and ( , ) , then null , (null ) , and range are all finite-dimensional. \n\n> 6.67\n\nrestriction of a linear map to obtain a one-to-one and onto map \n\nSuppose is finite-dimensional and ( , ) . Then | (null ) is an injective map of (null ) onto range .\n\nProof Suppose that ( null ) and | (null ) = 0 . Hence = 0 and thus ( null ) ( null ) , which implies that = 0 [ by 6.48(d) ]. Hence \n\nnull | (null ) = {0} , which implies that | (null ) is injective, as desired. Clearly range | (null ) range . To prove the inclusion in the other direction, suppose range . Hence there exists such that = . There exist \n\n null and ( null ) such that = + (by 6.49). Now \n\n| (null ) = = = 0 = ,\n\nwhich shows that range | (null ) . Hence range range | (null ) , complet-ing the proof that range | (null ) = range .\n\nTo produce the pseudoinverse notation \n\n> \n\nin TEX, type T \\dagger .\n\nNow we can define the pseudoinverse \n\n (pronounced dagger) of a linear map . In the next definition (and from now on), think of | (null ) as an invertible linear map from (null ) onto range ,as is justified by the result above. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6C Orthogonal Complements and Minimization Problems 221 \n\n> 6.68 definition:\n\npseudoinverse, \n\nSuppose that is finite-dimensional and ( , ) . The pseudoinverse \n\n ( , ) of is the linear map from to defined by \n\n = (| (null ) )1 range \n\nfor each .Recall that range = 0 if ( range ) and range = if range .Thus if ( range ) , then = 0 , and if range , then is the unique element of (null ) such that ( ) = .The pseudoinverse behaves much like an inverse, as we will see. \n\n> 6.69\n\nalgebraic properties of the pseudoinverse \n\nSuppose is finite-dimensional and ( , ) .(a) If is invertible, then = 1 .(b) = range = the orthogonal projection of onto range .(c) = (null ) = the orthogonal projection of onto (null ) .\n\nProof \n\n(a) Suppose is invertible. Then (null ) = and range = . Thus \n\n| (null ) = and range is the identity operator on . Hence = 1 .(b) Suppose range . Thus \n\n = (| (null ) )1 = = range . \n\nIf ( range ) , then = 0 . Hence = 0 = range . Thus \n\nand range agree on range and on (range ) . Hence these two linear maps are equal (by 6.49). (c) Suppose ( null ) . Because range , the definition of shows that \n\n () = (| (null ) )1 () = = (null ) . \n\nIf null , then = 0 = (null ) . Thus and (null ) agree on \n\n(null ) and on null . Hence these two linear maps are equal (by 6.49). \n\nThe pseudoinverse is also called the MoorePenrose inverse. \n\nSuppose that ( , ) . If is surjective, then is the identity opera-tor on , as follows from (b) in the result above. If is injective, then is the identity operator on , as follows from (c) in the result above. For additional algebraic properties of the pseudoinverse, see Exercises 1923. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 222 Chapter 6 Inner Product Spaces\n\nFor ( , ) and , we now return to the problem of finding \n\nthat solves the equation \n\n = . \n\nAs we noted earlier, if is invertible, then = 1 is the unique solution, but if is not invertible, then 1 is not defined. However, the pseudoinverse is defined. Taking = makes as close to as possible, as shown by (a) of the next result. Thus the pseudoinverse provides what is called a best fit to the equation above. Among all vectors that make as close as possible to , the vector \n\n has the smallest norm, as shown by combining (b) in the next result with the condition for equality in (a). \n\n> 6.70\n\npseudoinverse provides best approximate solution or best solution \n\nSuppose is finite-dimensional, ( , ) , and .(a) If , then \n\n( ) ,\n\nwith equality if and only if + null .(b) If + null , then \n\n ,\n\nwith equality if and only if = .\n\n> Proof\n\n(a) Suppose . Then \n\n = ( ) + ( ). \n\nThe first term in parentheses above is in range . Because the operator \n\nis the orthogonal projection of onto range [ by 6.69(b) ], the second term in parentheses above is in (range ) [see 6.57(f) ].Thus the Pythagorean theorem implies the desired inequality that the norm of the second term in parentheses above is less than or equal to , with equality if and only if the first term in parentheses above equals 0. Hence we have equality if and only if null , which is equivalent to the statement that + null , completing the proof of (a). (b) Suppose + null . Hence null . Now \n\n = ( ) + . \n\nThe definition of implies that ( null ) . Thus the Pythagorean theorem implies that , with equality if and only if = .A formula for will be given in the next chapter (see 7.78). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6C Orthogonal Complements and Minimization Problems 223 \n\n> 6.71 example:\n\npseudoinverse of a linear map from 4 to 3\n\nSuppose ( 4, 3) is defined by \n\n( , , , ) = ( + + , 2 + , 0). \n\nThis linear map is neither injective nor surjective, but we can compute its pseudo-inverse. To do this, first note that range = {( , , 0) , } . Thus \n\n range ( , , ) = ( , , 0) \n\nfor each ( , , ) 3. Also, \n\nnull = {( , , , ) 4 + + = 0 and 2 + = 0}. \n\nThe list (1 , 1, 0, 0) , (1 , 0, 1, 2) of two vectors in null spans null because if ( , , , ) null then \n\n( , , , ) = (1 , 1, 0, 0) + (1 , 0, 1, 2). \n\nBecause the list (1 , 1, 0, 0) , (1 , 0, 1, 2) is linearly independent, this list is a basis of null .Now suppose ( , , ) 3. Then \n\n> 6.72\n\n ( , , ) = (| (null ) )1 range ( , , ) = (| (null ) )1 ( , , 0). \n\nThe right side of the equation above is the vector ( , , , ) 4 such that \n\n( , , , ) = ( , , 0) and ( , , , ) ( null ) . In other words, , , , must satisfy the following equations: \n\n + + = 2 + = + = 0 + 2 = 0 ,\n\nwhere the first two equations are equivalent to the equation ( , , , ) = ( , , 0) \n\nand the last two equations come from the condition for ( , , , ) to be orthogo-nal to each of the basis vectors (1 , 1, 0, 0) , (1 , 0, 1, 2) in this basis of null .Thinking of and as constants and , , , as unknowns, we can solve the system above of four equations in four unknowns, getting \n\n = 111 (5 2) , = 111 (5 2) , = 111 ( + 4) , = 111 (2 + 3). \n\nHence 6.72 tells us that \n\n ( , , ) = 111 (5 2 , 5 2 , + 4 , 2 + 3). \n\nThe formula above for shows that ( , , ) = ( , , 0) for all ( , , ) 3,which illustrates the equation = range from 6.69(b). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n224 Chapter 6 Inner Product Spaces \n\n# Exercises 6C \n\n1 Suppose 1, , . Prove that \n\n{ 1, , } = ( span ( 1, , )) .\n\n2 Suppose is a subspace of with basis 1, , and \n\n 1, , , 1, , \n\nis a basis of . Prove that if the GramSchmidt procedure is applied to the basis of above, producing a list 1, , , 1, , , then 1, , is an orthonormal basis of and 1, , is an orthonormal basis of .\n\n3 Suppose is the subspace of 4 defined by \n\n = span ((1 , 2, 3, 4) , (5 , 4, 3, 2)). \n\nFind an orthonormal basis of and an orthonormal basis of .\n\n4 Suppose 1, , is a list of vectors in with = 1 for each = 1 , , \n\nand \n\n 2 = , 1 2 + + , 2\n\nfor all . Prove that 1, , is an orthonormal basis of .\n\nThis exercise provides a converse to 6.30 (b).\n\n5 Suppose that is finite-dimensional and is a subspace of . Show that \n\n = , where is the identity operator on .\n\n6 Suppose is finite-dimensional and ( , ) . Show that \n\n = (null ) = range . \n\n7 Suppose that and are finite-dimensional subspaces of . Prove that \n\n = 0 if and only if , = 0 for all and all .\n\n8 Suppose is a finite-dimensional subspace of and . Define a linear functional by \n\n() = , \n\nfor all . By the Riesz representation theorem (6.42) as applied to the inner product space , there exists a unique vector such that \n\n() = , \n\nfor all . Show that = .\n\n9 Suppose is finite-dimensional. Suppose () is such that 2 = \n\nand every vector in null is orthogonal to every vector in range . Prove that there exists a subspace of such that = .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 6C Orthogonal Complements and Minimization Problems 225 \n\n10 Suppose is finite-dimensional and () is such that 2 = and \n\n \n\nfor every . Prove that there exists a subspace of such that = .\n\n11 Suppose () and is a finite-dimensional subspace of . Prove that \n\n is invariant under = .\n\n12 Suppose is finite-dimensional, () , and is a subspace of . Prove that \n\n and are both invariant under = .\n\n13 Suppose = and is finite-dimensional. For each , let denote the linear functional on defined by \n\n () = , \n\nfor all .(a) Show that is an injective linear map from to .(b) Use (a) and a dimension-counting argument to show that is an isomorphism from onto .\n\nThe purpose of this exercise is to give an alternative proof of the Riesz representation theorem (6.42 and 6.58 ) when = . Thus you should not use the Riesz representation theorem as a tool in your solution. \n\n14 Suppose that 1, , is an orthonormal basis of . Explain why the dual basis (see 3.112) of 1, , is 1, , under the identification of with \n\n provided by the Riesz representation theorem (6.58). \n\n15 In 4, let \n\n = span ((1 , 1, 0, 0) , (1 , 1, 1, 2)). \n\nFind such that (1 , 2, 3, 4) is as small as possible. \n\n16 Suppose [1 , 1] is the vector space of continuous real-valued functions on the interval [1 , 1] with inner product given by \n\n , = 11 \n\n \n\nfor all , [1 , 1] . Let be the subspace of [1 , 1] defined by \n\n = { [1 , 1] (0) = 0}. \n\n(a) Show that = {0} .(b) Show that 6.49 and 6.52 do not hold without the finite-dimensional hypothesis. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n226 Chapter 6 Inner Product Spaces \n\n17 Find 3() such that (0) = 0 , (0) = 0 , and 10\n\n2 + 3 () 2 is as small as possible. \n\n18 Find 5() that makes \n\nsin () 2 as small as possible. \n\nThe polynomial 6.65 is an excellent approximation to the answer to this exercise, but here you are asked to find the exact solution, which involves powers of . A computer that can perform symbolic integration should help. \n\n19 Suppose is finite-dimensional and () is an orthogonal projection of onto some subspace of . Prove that = .\n\n20 Suppose is finite-dimensional and ( , ) . Show that \n\nnull = ( range ) and range = ( null ) .\n\n21 Suppose ( 3, 2) is defined by \n\n( , , ) = ( + + , 2 + 3). \n\n(a) For ( , ) 2, find a formula for ( , ) .(b) Verify that the equation = range from 6.69(b) holds with the formula for obtained in (a). (c) Verify that the equation = (null ) from 6.69(c) holds with the formula for obtained in (a). \n\n22 Suppose is finite-dimensional and ( , ) . Prove that \n\n = and = .\n\nBoth formulas above clearly hold if is invertible because in that case we can replace with 1 .\n\n23 Suppose and are finite-dimensional and ( , ) . Prove that \n\n( ) = . \n\nThe equation above is analogous to the equation ( 1 )1 = that holds if \n\n> \n\nis invertible. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n# Chapter 7 \n\n# Operators on Inner Product Spaces \n\nThe deepest results related to inner product spaces deal with the subject to which we now turnlinear maps and operators on inner product spaces. As we will see, good theorems can be proved by exploiting properties of the adjoint. The hugely important spectral theorem will provide a complete description of self-adjoint operators on real inner product spaces and of normal operators on complex inner product spaces. We will then use the spectral theorem to help understand positive operators and unitary operators, which will lead to unitary matrices and matrix factorizations. The spectral theorem will also lead to the popular singular value decomposition, which will lead to the polar decomposition. The most important results in the rest of this book are valid only in finite dimensions. Thus from now on we assume that and are finite-dimensional. \n\nstanding assumptions for this chapter \n\n denotes or .\n\n and are nonzero finite-dimensional inner product spaces over . Petar Miloevi CC BY-SA \n\nMarket square in Lviv, a city that has had several names and has been in several countries because of changing international borders. From 1772 until 1918, the city was in Austria and was called Lemberg. Between World War I and World War II, the city was in Poland and was called Lww. During this time, mathematicians in Lww, particularly Stefan Banach (18921945 ) and his colleagues, developed the basic results of modern functional analysis, using tools of analysis to study infinite-dimensional vector spaces. Since the end of World War II, Lviv has been in Ukraine, which was part of the Soviet Union until Ukraine became an independent country in 1991. \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler \n\n227 228 Chapter 7 Operators on Inner Product Spaces \n\n# 7A Self-Adjoint and Normal Operators \n\n# Adjoints \n\n> 7.1 definition:\n\nadjoint, \n\nSuppose ( , ) . The adjoint of is the function such that \n\n , = , \n\nfor every and every .\n\nThe word adjoint has another meaning in linear algebra. In case you en-counter the second meaning elsewhere, be warned that the two meanings for adjoint are unrelated to each other. \n\nTo see why the definition above makes sense, suppose ( , ) . Fix \n\n . Consider the linear functional \n\n , \n\non that maps to , ; this linear functional depends on and . By the Riesz representation theorem (6.42), there exists a unique vector in such that this linear functional is given by taking the inner product with it. We call this unique vector . In other words, is the unique vector in such that \n\n , = , \n\nfor every .In the equation above, the inner product on the left takes place in and the inner product on the right takes place in . However, we use the same notation \n\n , for both inner products. \n\n> 7.2 example:\n\nadjoint of a linear map from 3 to 2\n\nDefine 3 2 by \n\n( 1, 2, 3) = ( 2 + 3 3, 2 1). \n\nTo compute , suppose ( 1, 2, 3) 3 and ( 1, 2) 2. Then \n\n( 1, 2, 3), ( 1, 2) = ( 2 + 3 3, 2 1), ( 1, 2) = 2 1 + 3 3 1 + 2 1 2\n\n= ( 1, 2, 3), (2 2, 1, 3 1). \n\nThe equation above and the definition of the adjoint imply that \n\n ( 1, 2) = (2 2, 1, 3 1). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7A Self-Adjoint and Normal Operators 229 \n\n> 7.3 example:\n\nadjoint of a linear map with range of dimension at most 1\n\nFix and . Define ( , ) by \n\n = , \n\nfor each . To compute , suppose and . Then \n\n , = , , = , , = , , . \n\nThus \n\n = , . \n\nThe two examples above and the proof below use a common technique for computing : start with a formula for , then manipulate it to get just in the first slot; the entry in the second slot will then be .\n\nIn the two examples above, turned out to be not just a function from to \n\n but a linear map from to . This behavior is true in general, as shown by the next result. \n\n> 7.4\n\nadjoint of a linear map is a linear map \n\nIf ( , ) , then ( , ) .\n\nProof Suppose ( , ) . If and 1, 2 , then \n\n , 1 + 2 = , 1 + , 2= , 1 + , 2= , 1 + 2. \n\nThe equation above shows that \n\n ( 1 + 2) = 1 + 2.\n\nIf , , and , then \n\n , = , = , = , . \n\nThe equation above shows that \n\n () = . \n\nThus is a linear map, as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n230 Chapter 7 Operators on Inner Product Spaces \n\n> 7.5\n\nproperties of the adjoint \n\nSuppose ( , ) . Then (a) ( + ) = + for all ( , ) ;(b) () = for all ;(c) ( ) = ;(d) () = for all ( , ) (here is a finite-dimensional inner product space over ); (e) = , where is the identity operator on ;(f) if is invertible, then is invertible and ( )1 = ( 1 ).\n\nProof Suppose and .(a) If ( , ) , then \n\n( + ) , = , + , = , + , = , + . \n\nThus ( + ) = + , as desired. (b) If , then \n\n() , = , = , = , . \n\nThus () = , as desired. (c) We have \n\n , = , = , = , . \n\nThus ( ) = , as desired. (d) Suppose ( , ) and . Then \n\n() , = () , = , = , ( ). \n\nThus () = ( ) , as desired. (e) Suppose . Then \n\n , = , . \n\nThus = , as desired. (f) Suppose is invertible. Take adjoints of both sides of the equation 1 = ,then use (d) and (e) to show that ( 1 ) = . Similarly, the equation \n\n 1 = implies ( 1 ) = . Thus ( 1 ) is the inverse of , as desired. If = , then the map is a linear map from ( , ) to ( , ) ,as follows from (a) and (b) of the result above. However, if = , then this map is not linear because of the complex conjugate that appears in (b). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7A Self-Adjoint and Normal Operators 231 \n\nThe next result shows the relationship between the null space and the range of a linear map and its adjoint. \n\n> 7.6\n\nnull space and range of \n\nSuppose ( , ) . Then (a) null = ( range ) ;(b) range = ( null ) ;(c) null = ( range );(d) range = ( null ).\n\nProof We begin by proving (a). Let . Then \n\n null = 0 , = 0 for all , = 0 for all ( range ) .\n\nThus null = ( range ) , proving (a). If we take the orthogonal complement of both sides of (a), we get (d), where we have used 6.52. Replacing with in (a) gives (c), where we have used 7.5(c). Finally, replacing with in (d) gives (b). As we will soon see, the next definition is intimately connected to the matrix of the adjoint of a linear map. \n\n> 7.7 definition:\n\nconjugate transpose, \n\nThe conjugate transpose of an -by- matrix is the -by- matrix \n\nobtained by interchanging the rows and columns and then taking the complex conjugate of each entry. In other words, if {1 , , } and {1 , , } ,then \n\n( ) , = , . \n\n> 7.8 example:\n\nconjugate transpose of a 2-by-3 matrix \n\nIf a matrix has only real entries, then = t, where t denotes the transpose of (the matrix obtained by interchanging the rows and the columns ).\n\nThe conjugate transpose of the 2-by-3\n\nmatrix 2 3 + 4 76 5 8 is the 3-by-2\n\nmatrix \n\n2 63 4 57 8 .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n232 Chapter 7 Operators on Inner Product Spaces \n\nThe adjoint of a linear map does not depend on a choice of basis. Thus we frequently emphasize adjoints of linear maps instead of transposes or conjugate transposes of matrices. \n\nThe next result shows how to compute the matrix of from the matrix of .\n\nCaution: With respect to nonorthonor-mal bases, the matrix of does not nec-essarily equal the conjugate transpose of the matrix of . \n\n> 7.9\n\nmatrix of equals conjugate transpose of matrix of \n\nLet ( , ) . Suppose 1, , is an orthonormal basis of and \n\n 1, , is an orthonormal basis of . Then ( , ( 1, , ), ( 1, , )) \n\nis the conjugate transpose of ( , ( 1, , ), ( 1, , )) . In other words, \n\n( ) = ( ()) .\n\nProof In this proof, we will write () and ( ) instead of the longer expressions ( , ( 1, , ), ( 1, , )) and ( , ( 1, , ), ( 1, , )) .Recall that we obtain the th column of () by writing as a linear combination of the s; the scalars used in this linear combination then become the th column of () . Because 1, , is an orthonormal basis of , we know how to write as a linear combination of the s [see 6.30(a) ]:\n\n = , 1 1 + + , .\n\nThus the entry in row , column , of () is , . \n\nIn the statement above, replace with and interchange 1, , and \n\n 1, , . This shows that the entry in row , column , of ( ) is , ,which equals , , which equals , , which equals the complex conjugate of the entry in row , column , of () . Thus ( ) = ( ()) .The Riesz representation theorem as stated in 6.58 provides an identification of \n\n with its dual space defined in 3.110. Under this identification, the orthogonal complement of a subset corresponds to the annihilator 0 of . If \n\nis a subspace of , then the formulas for the dimensions of and 0 become identical under this identificationsee 3.125 and 6.51. \n\nBecause orthogonal complements and adjoints are easier to deal with than annihilators and dual maps, there is no need to work with annihilators and dual maps in the context of inner product spaces. \n\nSuppose is a linear map. Under the identification of with and the identification of with , the ad-joint map corresponds to the dual map defined in 3.118, as Exercise 32 asks you to verify. Under this identification, the formulas for \n\nnull and range [7.6(a) and (b) ] then become identical to the formulas for \n\nnull and range [3.128(a) and 3.130(b) ]. Furthermore, the theorem about the matrix of (7.9) is analogous to the theorem about the matrix of (3.132). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7A Self-Adjoint and Normal Operators 233\n\n# Self-Adjoint Operators \n\nNow we switch our attention to operators on inner product spaces. Instead of considering linear maps from to , we will focus on linear maps from to ;recall that such linear maps are called operators. \n\n> 7.10 definition:\n\nself-adjoint \n\nAn operator () is called self-adjoint if = .If () and 1, , is an orthonormal basis of , then is self-adjoint if and only if ( , ( 1, , )) = ( , ( 1, , )) , as follows from 7.9. \n\n> 7.11 example:\n\ndetermining whether is self-adjoint from its matrix \n\nSuppose and is the operator on 2 whose matrix (with respect to the standard basis) is \n\n() = 2 3 7 .\n\nThe matrix of (with respect to the standard basis) is \n\n( ) = 2 3 7 .\n\nThus () = ( ) if and only if = 3 . Hence the operator is self-adjoint if and only if = 3 .A good analogy to keep in mind is that the adjoint on () plays a role similar to that of the complex conjugate on . A complex number is real if and only if \n\n = ; thus a self-adjoint operator ( = ) is analogous to a real number. \n\n> An operator () is self-adjoint if and only if\n> , = ,\n> for all , .\n\nWe will see that the analogy discussed above is reflected in some important prop-erties of self-adjoint operators, beginning with eigenvalues in the next result. If = , then by definition every eigenvalue is real, so the next result is interesting only when = . \n\n> 7.12\n\neigenvalues of self-adjoint operators \n\nEvery eigenvalue of a self-adjoint operator is real. \n\n> Proof\n\nSuppose is a self-adjoint operator on . Let be an eigenvalue of ,and let be a nonzero vector in such that = . Then \n\n 2 = , = , = , = , = 2.\n\nThus = , which means that is real, as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 234 Chapter 7 Operators on Inner Product Spaces\n\nThe next result is false for real inner product spaces. As an example, consider the operator ( 2) that is a counterclockwise rotation of 90 around the origin; thus ( , ) = ( , ) . Notice that is orthogonal to for every 2,even though 0 . \n\n> 7.13\n\n is orthogonal to for all = 0 (assuming = )Suppose is a complex inner product space and () . Then \n\n , = 0 for every = 0. \n\n> Proof\n\nIf , , then \n\n , = ( + ) , + ( ) , 4\n\n+ ( + ) , + ( ) , 4 ,\n\nas can be verified by computing the right side. Note that each term on the right side is of the form , for appropriate .Now suppose , = 0 for every . Then the equation above implies that , = 0 for all , , which then implies that = 0 for every \n\n(take = ). Hence = 0 , as desired. \n\n> The next result provides another good example of how self-adjoint operators behave like real numbers.\n\nThe next result is false for real inner product spaces, as shown by considering any operator on a real inner product space that is not self-adjoint. \n\n> 7.14\n\n , is real for all is self-adjoint (assuming = )Suppose is a complex inner product space and () . Then \n\n is self-adjoint , for every . \n\n> Proof\n\nIf , then \n\n> 7.15\n\n , = , = , . \n\nNow \n\n is self-adjoint = 0 ( ) , = 0 for every , , = 0 for every , for every ,\n\nwhere the second equivalence follows from 7.13 as applied to and the third equivalence follows from 7.15. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7A Self-Adjoint and Normal Operators 235\n\nOn a real inner product space , a nonzero operator might satisfy , = 0 \n\nfor all . However, the next result shows that this cannot happen for a self-adjoint operator. \n\n> 7.16\n\n self-adjoint and , = 0 for all = 0 \n\nSuppose is a self-adjoint operator on . Then \n\n , = 0 for every = 0. \n\n> Proof\n\nWe have already proved this (without the hypothesis that is self-adjoint) when is a complex inner product space (see 7.13). Thus we can assume that \n\nis a real inner product space. If , , then \n\n> 7.17\n\n , = ( + ) , + ( ) , 4 ,\n\nas can be proved by computing the right side using the equation \n\n , = , = , ,\n\nwhere the first equality holds because is self-adjoint and the second equality holds because we are working in a real inner product space. Now suppose , = 0 for every . Because each term on the right side of 7.17 is of the form , for appropriate , this implies that , = 0 \n\nfor all , . This implies that = 0 for every (take = ). Hence \n\n = 0 , as desired. \n\n# Normal Operators \n\n> 7.18 definition:\n\nnormal \n\n An operator on an inner product space is called normal if it commutes with its adjoint. \n\n In other words, () is normal if = .Every self-adjoint operator is normal, because if is self-adjoint then = \n\nand hence commutes with . \n\n> 7.19 example:\n\nan operator that is normal but not self-adjoint \n\nLet be the operator on 2 whose matrix (with respect to the standard basis) is \n\n2 3 3 2 .\n\nThus ( , ) = (2 3 , 3 + 2) . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 236 Chapter 7 Operators on Inner Product Spaces\n\nThis operator is not self-adjoint because the entry in row 2, column 1 (which equals 3) does not equal the complex conjugate of the entry in row 1, column 2 (which equals 3 ). The matrix of equals \n\n2 3 3 2 2 33 2 , which equals 13 00 13 .\n\nSimilarly, the matrix of equals \n\n2 33 2 2 3 3 2 , which equals 13 00 13 .\n\nBecause and have the same matrix, we see that = . Thus is normal. In the next section we will see why normal operators are worthy of special attention. The next result provides a useful characterization of normal operators. \n\n> 7.20\n\n is normal if and only if and have the same norm \n\nSuppose () . Then \n\n is normal = for every . \n\n> Proof\n\nWe have \n\n is normal = 0 ( ) , = 0 for every , = , for every , = , for every 2 = 2 for every = for every ,\n\nwhere we used 7.16 to establish the second equivalence (note that the operator \n\n is self-adjoint). The next result presents several consequences of the result above. Compare (e) of the next result to Exercise 3. That exercise states that the eigenvalues of the adjoint of each operator are equal (as a set) to the complex conjugates of the eigenvalues of the operator. The exercise says nothing about eigenvectors, because an operator and its adjoint may have different eigenvectors. However, (e) of the next result implies that a normal operator and its adjoint have the same eigenvectors. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7A Self-Adjoint and Normal Operators 237\n> 7.21\n\nrange, null space, and eigenvectors of a normal operator \n\nSuppose () is normal. Then (a) null = null ;(b) range = range ;(c) = null range ;(d) is normal for every ;(e) if and , then = if and only if = .\n\n> Proof\n\n(a) Suppose . Then \n\n null = 0 = 0 null ,\n\nwhere the middle equivalence above follows from 7.20. Thus null = null .(b) We have \n\nrange = ( null ) = ( null ) = range ,\n\nwhere the first equality comes from 7.6(d), the second equality comes from (a) in this result, and the third equality comes from 7.6(b). (c) We have \n\n = ( null ) ( null ) = null range = null range ,\n\nwhere the first equality comes from 6.49, the second equality comes from 7.6(b), and the third equality comes from (b) in this result. (d) Suppose . Then \n\n( )( ) = ( )( ) = + || 2 = + || 2 = ( )( ) = ( ) ( ). \n\nThus commutes with its adjoint. Hence is normal. (e) Suppose and . Then (d) and 7.20 imply that \n\n( ) = ( ) = ( ). \n\nThus ( ) = 0 if and only if ( ) = 0 . Hence = if and only if = . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 238 Chapter 7 Operators on Inner Product Spaces\n\nBecause every self-adjoint operator is normal, the next result applies in partic-ular to self-adjoint operators. \n\n> 7.22\n\northogonal eigenvectors for normal operators \n\nSuppose () is normal. Then eigenvectors of corresponding to distinct eigenvalues are orthogonal. \n\n> Proof\n\nSuppose , are distinct eigenvalues of , with corresponding eigen-vectors , . Thus = and = . From 7.21(e) we have = .Thus \n\n( ) , = , , = , , = 0. \n\nBecause , the equation above implies that , = 0 . Thus and are orthogonal, as desired. As stated here, the next result makes sense only when = . However, see Exercise 12 for a version that makes sense when = and when = .Suppose = and () . Under the analogy between () and ,with the adjoint on () playing a similar role to that of the complex conjugate on \n\n , the operators and as defined by 7.24 correspond to the real and imaginary parts of . Thus the informal title of the result below should make sense. \n\n> 7.23\n\n is normal the real and imaginary parts of commute \n\nSuppose = and () . Then is normal if and only if there exist commuting self-adjoint operators and such that = + . \n\n> Proof\n\nFirst suppose is normal. Let \n\n> 7.24\n\n = + \n\n2 and = \n\n2 .\n\nThen and are self-adjoint and = + . A quick computation shows that \n\n> 7.25\n\n = \n\n2 .\n\nBecause is normal, the right side of the equation above equals 0. Thus the operators and commute, as desired. To prove the implication in the other direction, now suppose there exist com-muting self-adjoint operators and such that = + . Then = .Adding the last two equations and then dividing by 2 produces the equation for \n\nin 7.24. Subtracting the last two equations and then dividing by 2 produces the equation for in 7.24. Now 7.24 implies 7.25. Because and commute, 7.25 implies that is normal, as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7A Self-Adjoint and Normal Operators 239 \n\n# Exercises 7A \n\n1 Suppose is a positive integer. Define ( ) by \n\n( 1, , ) = (0 , 1, , 1 ). \n\nFind a formula for ( 1, , ).\n\n2 Suppose ( , ) . Prove that \n\n = 0 = 0 = 0 = 0. \n\n3 Suppose () and . Prove that \n\n is an eigenvalue of is an eigenvalue of .\n\n4 Suppose () and is a subspace of . Prove that \n\n is invariant under is invariant under .\n\n5 Suppose ( , ) . Suppose 1, , is an orthonormal basis of and \n\n 1, , is an orthonormal basis of . Prove that \n\n 12 + + 2 = 12 + + 2.\n\nThe numbers 12, , 2 in the equation above depend on the ortho-normal basis 1, , , but the right side of the equation does not depend on 1, , . Thus the equation above shows that the sum on the left side does not depend on which orthonormal basis 1, , is used. \n\n6 Suppose ( , ) . Prove that (a) is injective is surjective; (b) is surjective is injective. \n\n7 Prove that if ( , ) , then (a) dim null = dim null + dim dim ;(b) dim range = dim range .\n\n8 Suppose is an -by- matrix with entries in . Use (b) in Exercise 7 to prove that the row rank of equals the column rank of .\n\nThis exercise asks for yet another alternative proof of a result that was previously proved in 3.57 and 3.133. \n\n9 Prove that the product of two self-adjoint operators on is self-adjoint if and only if the two operators commute. \n\n10 Suppose = and () . Prove that is self-adjoint if and only if \n\n , = , \n\nfor all . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 240 Chapter 7 Operators on Inner Product Spaces\n\n11 Define an operator 2 2 by ( , ) = ( , ) .(a) Find a formula for .(b) Show that is normal but not self-adjoint. (c) Find all eigenvalues of . \n\n> If = , then is the operator on 2of counterclockwise rotation by 90 .\n\n12 An operator () is called skew if \n\n = . \n\nSuppose that () . Prove that is normal if and only if there exist commuting operators and such that is self-adjoint, is a skew operator, and = + .\n\n13 Suppose = . Define (()) by = for all () .(a) Find all eigenvalues of .(b) Find the minimal polynomial of .\n\n14 Define an inner product on 2() by , = 10 . Define an operator \n\n ( 2()) by \n\n( 2 + + ) = . \n\n(a) Show that with this inner product, the operator is not self-adjoint. (b) The matrix of with respect to the basis 1, , 2 is \n\n0 0 00 1 00 0 0.\n\nThis matrix equals its conjugate transpose, even though is not self-adjoint. Explain why this is not a contradiction. \n\n15 Suppose () is invertible. Prove that (a) is self-adjoint 1 is self-adjoint; (b) is normal 1 is normal. \n\n16 Suppose = .(a) Show that the set of self-adjoint operators on is a subspace of () .(b) What is the dimension of the subspace of () in (a) [in terms of \n\ndim ]? \n\n17 Suppose = . Show that the set of self-adjoint operators on is not a subspace of () .\n\n18 Suppose dim 2 . Show that the set of normal operators on is not a subspace of () .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7A Self-Adjoint and Normal Operators 241 \n\n19 Suppose () and for every . Prove that is normal. \n\nThis exercise fails on infinite-dimensional inner product spaces, leading to what are called hyponormal operators, which have a well-developed theory. \n\n20 Suppose () is such that 2 = . Prove that the following are equivalent. (a) is self-adjoint. (b) is normal. (c) There is a subspace of such that = .\n\n21 Suppose 8() 8() is the differentiation operator defined by \n\n = . Prove that there does not exist an inner product on 8() that makes a normal operator. \n\n22 Give an example of an operator ( 3) such that is normal but not self-adjoint. \n\n23 Suppose is a normal operator on . Suppose also that , satisfy the equations \n\n = = 2 , = 3 , = 4. \n\nShow that ( + ) = 10 .\n\n24 Suppose () and \n\n 0 + 1 + 2 2 + + 1 1 + \n\nis the minimal polynomial of . Prove that the minimal polynomial of is \n\n 0 + 1 + 2 2 + + 1 1 + .\n\nThis exercise shows that the minimal polynomial of equals the minimal polynomial of if = .\n\n25 Suppose () . Prove that is diagonalizable if and only if is diagonalizable. \n\n26 Fix , . Define () by = , for every .(a) Prove that if is a real vector space, then is self-adjoint if and only if the list , is linearly dependent. (b) Prove that is normal if and only if the list , is linearly dependent. \n\n27 Suppose () is normal. Prove that \n\nnull = null and range = range \n\nfor every positive integer .\n\n28 Suppose () is normal. Prove that if , then the minimal polynomial of is not a polynomial multiple of ( ) 2.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n242 Chapter 7 Operators on Inner Product Spaces \n\n29 Prove or give a counterexample: If () and there is an orthonormal basis 1, , of such that = for each = 1 , , , then is normal. \n\n30 Suppose that ( 3) is normal and (1 , 1, 1) = (2 , 2, 2) . Suppose \n\n( 1, 2, 3) null . Prove that 1 + 2 + 3 = 0 .\n\n31 Fix a positive integer . In the inner product space of continuous real-valued functions on [ , ] with inner product , = , let \n\n = span (1 , cos , cos 2 , , cos , sin , sin 2 , , sin ). \n\n(a) Define () by = . Show that = . Conclude that \n\nis normal but not self-adjoint. (b) Define () by = . Show that is self-adjoint. \n\n32 Suppose is a linear map. Show that under the standard identifica-tion of with (see 6.58) and the corresponding identification of with \n\n , the adjoint map corresponds to the dual map .More precisely, show that \n\n ( ) = \n\nfor all , where and are defined as in 6.58. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7B Spectral Theorem 243\n\n# 7B Spectral Theorem \n\nRecall that a diagonal matrix is a square matrix that is 0 everywhere except possibly on the diagonal. Recall that an operator on is called diagonalizable if the operator has a diagonal matrix with respect to some basis of . Recall also that this happens if and only if there is a basis of consisting of eigenvectors of the operator (see 5.55). The nicest operators on are those for which there is an orthonormal basis of with respect to which the operator has a diagonal matrix. These are precisely the operators () such that there is an orthonormal basis of consisting of eigenvectors of . Our goal in this section is to prove the spectral theorem, which characterizes these operators as the self-adjoint operators when = and as the normal operators when = .The spectral theorem is probably the most useful tool in the study of operators on inner product spaces. Its extension to certain infinite-dimensional inner product spaces (see, for example, Section 10D of the authors book Measure, Integration & Real Analysis ) plays a key role in functional analysis. Because the conclusion of the spectral theorem depends on , we will break the spectral theorem into two pieces, called the real spectral theorem and the complex spectral theorem. \n\n# Real Spectral Theorem \n\nTo prove the real spectral theorem, we will need two preliminary results. These preliminary results hold on both real and complex inner product spaces, but they are not needed for the proof of the complex spectral theorem. \n\n> This completing-the-square technique can be used to derive the quadratic formula.\n\nYou could guess that the next result is true and even discover its proof by think-ing about quadratic polynomials with real coefficients. Specifically, suppose \n\n , and 2 < 4 . Let be a real number. Then \n\n 2 + + = ( + 2 )\n\n> 2\n\n+ ( 2\n\n4 ) > 0. \n\nIn particular, 2 + + is an invertible real number (a convoluted way of saying that it is not 0). Replacing the real number with a self-adjoint operator (recall the analogy between real numbers and self-adjoint operators) leads to the next result. \n\n> 7.26\n\ninvertible quadratic expressions \n\nSuppose () is self-adjoint and , are such that 2 < 4 . Then \n\n 2 + + \n\nis an invertible operator. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n244 Chapter 7 Operators on Inner Product Spaces Proof Let be a nonzero vector in . Then \n\n( 2 + + ) , = 2 , + , + , = , + , + 2\n\n 2 || + 2\n\n= ( || 2 )\n\n> 2\n\n+ ( 2\n\n4 ) 2\n\n> 0 ,\n\nwhere the third line above holds by the CauchySchwarz inequality (6.14). The last inequality implies that ( 2 + + ) 0 . Thus 2 + + is injective, which implies that it is invertible (see 3.65). The next result will be a key tool in our proof of the real spectral theorem. \n\n> 7.27\n\nminimal polynomial of self-adjoint operator \n\nSuppose () is self-adjoint. Then the minimal polynomial of equals \n\n( 1) ( ) for some 1, , .\n\nProof First suppose = . The zeros of the minimal polynomial of are the eigenvalues of [ by 5.27(a) ]. All eigenvalues of are real (by 7.12). Thus the second version of the fundamental theorem of algebra (see 4.13) tells us that the minimal polynomial of has the desired form. Now suppose = . By the factorization of a polynomial over (see 4.16) there exist 1, , and 1, , , 1, , with 2 < 4 for each \n\nsuch that the minimal polynomial of equals \n\n> 7.28\n\n( 1)( )( 2 + 1 + 1)( 2 + + );\n\nhere either or might equal 0, meaning that there are no terms of the corre-sponding form. Now \n\n( 1)( )( 2 + 1 + 1)( 2 + + ) = 0. \n\nIf > 0 , then we could multiply both sides of the equation above on the right by the inverse of 2 + + (which is an invertible operator by 7.26) to obtain a polynomial expression of that equals 0. The corresponding polynomial would have degree two less than the degree of 7.28, violating the minimality of the degree of the polynomial with this property. Thus we must have = 0 , which means that the minimal polynomial in 7.28 has the form ( 1) ( ), as desired. The result above along with 5.27(a) implies that every self-adjoint operator has an eigenvalue. In fact, as we will see in the next result, self-adjoint operators have enough eigenvectors to form a basis. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7B Spectral Theorem 245\n\nThe next result, which gives a complete description of the self-adjoint operators on a real inner product space, is one of the major theorems in linear algebra. \n\n> 7.29\n\nreal spectral theorem \n\nSuppose = and () . Then the following are equivalent. (a) is self-adjoint. (b) has a diagonal matrix with respect to some orthonormal basis of .(c) has an orthonormal basis consisting of eigenvectors of . \n\n> Proof\n\nFirst suppose (a) holds, so is self-adjoint. Our results on minimal poly-nomials, specifically 6.37 and 7.27, imply that has an upper-triangular matrix with respect to some orthonormal basis of . With respect to this orthonormal basis, the matrix of is the transpose of the matrix of . However, = .Thus the transpose of the matrix of equals the matrix of . Because the matrix of is upper-triangular, this means that all entries of the matrix above and below the diagonal are 0. Hence the matrix of is a diagonal matrix with respect to the orthonormal basis. Thus (a) implies (b). Conversely, now suppose (b) holds, so has a diagonal matrix with respect to some orthonormal basis of . That diagonal matrix equals its transpose. Thus with respect to that basis, the matrix of equals the matrix of . Hence = ,proving that (b) implies (a). The equivalence of (b) and (c) follows from the definitions [or see the proof that (a) and (b) are equivalent in 5.55 ]. \n\n> 7.30 example:\n\nan orthonormal basis of eigenvectors for an operator \n\nConsider the operator on 3 whose matrix (with respect to the standard basis) is 14 13 813 14 88 8 7 .\n\nThis matrix with real entries equals its transpose; thus is self-adjoint. As you can verify, (1 , 1 , 0) \n\n> \n\n2 , (1 , 1, 1) \n\n> \n\n3 , (1 , 1, 2) \n\n> \n\n6\n\nis an orthonormal basis of 3 consisting of eigenvectors of . With respect to this basis, the matrix of is the diagonal matrix \n\n27 0 00 9 00 0 15 .\n\nSee Exercise 17 for a version of the real spectral theorem that applies simulta-neously to more than one operator. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n246 Chapter 7 Operators on Inner Product Spaces \n\n# Complex Spectral Theorem \n\nThe next result gives a complete description of the normal operators on a complex inner product space. \n\n> 7.31\n\ncomplex spectral theorem \n\nSuppose = and () . Then the following are equivalent. (a) is normal. (b) has a diagonal matrix with respect to some orthonormal basis of .(c) has an orthonormal basis consisting of eigenvectors of .\n\nProof First suppose (a) holds, so is normal. By Schurs theorem (6.38), there is an orthonormal basis 1, , of with respect to which has an upper-triangular matrix. Thus we can write \n\n> 7.32\n\n( , ( 1, , )) = 1,1 1, \n\n 0 , \n\n.\n\nWe will show that this matrix is actually a diagonal matrix. We see from the matrix above that \n\n 12 = | 1,1|2,\n\n 12 = | 1,1|2 + | 1,2|2 + + | 1, |2.\n\nBecause is normal, 1 = 1 (see 7.20). Thus the two equations above imply that all entries in the first row of the matrix in 7.32, except possibly the first entry 1,1, equal 0.Now 7.32 implies \n\n 22 = | 2,2|2\n\n(because 1,2 = 0 , as we showed in the paragraph above) and \n\n 22 = | 2,2|2 + | 2,3|2 + + | 2, |2.\n\nBecause is normal, 2 = 2. Thus the two equations above imply that all entries in the second row of the matrix in 7.32, except possibly the diagonal entry 2,2, equal 0.Continuing in this fashion, we see that all nondiagonal entries in the matrix 7.32 equal 0. Thus (b) holds, completing the proof that (a) implies (b). Now suppose (b) holds, so has a diagonal matrix with respect to some orthonormal basis of . The matrix of (with respect to the same basis) is obtained by taking the conjugate transpose of the matrix of ; hence also has a diagonal matrix. Any two diagonal matrices commute; thus commutes with ,which means that is normal. In other words, (a) holds, completing the proof that (b) implies (a). The equivalence of (b) and (c) follows from the definitions (also see 5.55). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7B Spectral Theorem 247\n\nSee Exercises 13 and 20 for alternative proofs that (a) implies (b) in the previous result. Exercises 14 and 15 interpret the real spectral theorem and the complex spectral theorem by expressing the domain space as an orthogonal direct sum of eigenspaces. See Exercise 16 for a version of the complex spectral theorem that applies simultaneously to more than one operator. The main conclusion of the complex spectral theorem is that every normal operator on a complex finite-dimensional inner product space is diagonalizable by an orthonormal basis, as illustrated by the next example. \n\n> 7.33 example:\n\nan orthonormal basis of eigenvectors for an operator \n\nConsider the operator ( 2) defined by ( , ) = (2 3 , 3 + 2) .The matrix of (with respect to the standard basis) is \n\n2 3 3 2 .\n\nAs we saw in Example 7.19, is a normal operator. As you can verify, 1 \n\n> 2\n\n( , 1) , 1 \n\n> 2\n\n( , 1) \n\nis an orthonormal basis of 2 consisting of eigenvectors of , and with respect to this basis the matrix of is the diagonal matrix \n\n2 + 3 00 2 3 .\n\n# Exercises 7B \n\n1 Prove that a normal operator on a complex inner product space is self-adjoint if and only if all its eigenvalues are real. \n\n> This exercise strengthens the analogy (for normal operators )between self-adjoint operators and real numbers.\n\n2 Suppose = . Suppose () is normal and has only one eigenvalue. Prove that is a scalar multiple of the identity operator. \n\n3 Suppose = and () is normal. Prove that the set of eigenvalues of is contained in {0 , 1} if and only if there is a subspace of such that \n\n = .\n\n4 Prove that a normal operator on a complex inner product space is skew (meaning it equals the negative of its adjoint) if and only if all its eigenvalues are purely imaginary (meaning that they have real part equal to 0). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 248 Chapter 7 Operators on Inner Product Spaces\n\n5 Prove or give a counterexample: If ( 3) is a diagonalizable operator, then is normal (with respect to the usual inner product). \n\n6 Suppose is a complex inner product space and () is a normal operator such that 9 = 8. Prove that is self-adjoint and 2 = .\n\n7 Give an example of an operator on a complex vector space such that \n\n 9 = 8 but 2 .\n\n8 Suppose = and () . Prove that is normal if and only if every eigenvector of is also an eigenvector of .\n\n9 Suppose = and () . Prove that is normal if and only if there exists a polynomial () such that = () .\n\n10 Suppose is a complex inner product space. Prove that every normal operator on has a square root. \n\n> An operator () is called a square root of () if 2= . We will discuss more about square roots of operators in Sections 7C and 8C.\n\n11 Prove that every self-adjoint operator on has a cube root. \n\n> An operator () is called a cube root of () if 3= .\n\n12 Suppose is a complex vector space and () is normal. Prove that if is an operator on that commutes with , then commutes with .\n\n> The result in this exercise is called Fugledes theorem.\n\n13 Without using the complex spectral theorem, use the version of Schurs theorem that applies to two commuting operators (take = { , } in Exercise 20 in Section 6B ) to give a different proof that if = and \n\n () is normal, then has a diagonal matrix with respect to some orthonormal basis of .\n\n14 Suppose = and () . Prove that is self-adjoint if and only if all pairs of eigenvectors corresponding to distinct eigenvalues of are orthogonal and = ( 1, ) ( , ) , where 1, , denote the distinct eigenvalues of .\n\n15 Suppose = and () . Prove that is normal if and only if all pairs of eigenvectors corresponding to distinct eigenvalues of are orthogonal and = ( 1, ) ( , ) , where 1, , denote the distinct eigenvalues of .\n\n16 Suppose = and () . Prove that there is an orthonormal basis of with respect to which every element of has a diagonal matrix if and only if and are commuting normal operators for all , .\n\n> This exercise extends the complex spectral theorem to the context of a collection of commuting normal operators.\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7B Spectral Theorem 249 \n\n17 Suppose = and () . Prove that there is an orthonormal basis of with respect to which every element of has a diagonal matrix if and only if and are commuting self-adjoint operators for all , .\n\nThis exercise extends the real spectral theorem to the context of a collection of commuting self-adjoint operators. \n\n18 Give an example of a real inner product space , an operator () ,and real numbers , with 2 < 4 such that \n\n 2 + + \n\nis not invertible. \n\nThis exercise shows that the hypothesis that is self-adjoint cannot be deleted in 7.26, even for real vector spaces. \n\n19 Suppose () is self-adjoint and is a subspace of that is invariant under .(a) Prove that is invariant under .(b) Prove that | () is self-adjoint. (c) Prove that | ( ) is self-adjoint. \n\n20 Suppose () is normal and is a subspace of that is invariant under .(a) Prove that is invariant under .(b) Prove that is invariant under .(c) Prove that (| ) = ( )| .(d) Prove that | () and | ( ) are normal operators. \n\nThis exercise can be used to give yet another proof of the complex spectral theorem (use induction on dim and the result that has an eigenvector ).\n\n21 Suppose that is a self-adjoint operator on a finite-dimensional inner product space and that 2 and 3 are the only eigenvalues of . Prove that \n\n 2 5 + 6 = 0. \n\n22 Give an example of an operator ( 3) such that 2 and 3 are the only eigenvalues of and 2 5 + 6 0 .\n\n23 Suppose () is self-adjoint, , and > 0 . Suppose there exists \n\n such that = 1 and \n\n < . \n\nProve that has an eigenvalue such that < .\n\nThis exercise shows that for a self-adjoint operator, a number that is close to satisfying an equation that would make it an eigenvalue is close to an eigenvalue. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n250 Chapter 7 Operators on Inner Product Spaces \n\n24 Suppose is a finite-dimensional vector space and () .(a) Suppose = . Prove that is diagonalizable if and only if there is a basis of such that the matrix of with respect to this basis equals its transpose. (b) Suppose = . Prove that is diagonalizable if and only if there is a basis of such that the matrix of with respect to this basis commutes with its conjugate transpose. \n\nThis exercise adds another equivalence to the list of conditions equivalent to diagonalizability in 5.55. \n\n25 Suppose that () and there is an orthonormal basis 1, , of \n\nconsisting of eigenvectors of , with corresponding eigenvalues 1, , .Show that if {1 , , } , then the pseudoinverse satisfies the equation \n\n ={{{{\n\n> 1 \n\n if 0 ,\n\n0 if = 0. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7C Positive Operators 251 \n\n# 7C Positive Operators \n\n> 7.34 definition:\n\npositive operator \n\nAn operator () is called positive if is self-adjoint and \n\n , 0 \n\nfor all .If is a complex vector space, then the requirement that be self-adjoint can be dropped from the definition above (by 7.14). \n\n> 7.35 example:\n\npositive operators \n\n(a) Let ( 2) be the operator whose matrix (using the standard basis) is \n\n( 2 1 1 1 ). Then is self-adjoint and ( , ) , ( , ) = 2|| 22 Re () +|| 2\n\n= | | 2 + || 2 0 for all ( , ) 2. Thus is a positive operator. (b) If is a subspace of , then the orthogonal projection is a positive operator, as you should verify. (c) If () is self-adjoint and , are such that 2 < 4 , then 2+ + \n\nis a positive operator, as shown by the proof of 7.26. \n\n> 7.36 definition:\n\nsquare root \n\nAn operator is called a square root of an operator if 2 = . \n\n> 7.37 example:\n\nsquare root of an operator \n\nIf ( 3) is defined by ( 1, 2, 3) = ( 3, 0, 0) , then the operator \n\n ( 3) defined by ( 1, 2, 3) = ( 2, 3, 0) is a square root of because \n\n 2 = , as you can verify. \n\nBecause positive operators correspond to nonnegative numbers, better termi-nology would use the term nonnegative operators. However, operator theorists consistently call these positive opera-tors, so we follow that custom. Some mathematicians use the term positive semidefinite operator , which means the same as positive operator. \n\nThe characterizations of the positive operators in the next result correspond to characterizations of the nonnegative numbers among . Specifically, a num-ber is nonnegative if and only if it has a nonnegative square root, cor-responding to condition (d). Also, is nonnegative if and only if it has a real square root, corresponding to condition (e). Finally, is nonnegative if and only if there exists such that = , corresponding to condition (f). See Exercise 20 for another condition that is equivalent to being a positive operator. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 252 Chapter 7 Operators on Inner Product Spaces\n> 7.38\n\ncharacterizations of positive operators \n\nLet () . Then the following are equivalent. (a) is a positive operator. (b) is self-adjoint and all eigenvalues of are nonnegative. (c) With respect to some orthonormal basis of , the matrix of is a diagonal matrix with only nonnegative numbers on the diagonal. (d) has a positive square root. (e) has a self-adjoint square root. (f) = for some () . \n\n> Proof\n\nWe will prove that (a) (b) (c) (d) (e) (f) (a). First suppose (a) holds, so that is positive, which implies that is self-adjoint (by definition of positive operator). To prove the other condition in (b), suppose \n\n is an eigenvalue of . Let be an eigenvector of corresponding to . Then \n\n0 , = , = , . \n\nThus is a nonnegative number. Hence (b) holds, showing that (a) implies (b). Now suppose (b) holds, so that is self-adjoint and all eigenvalues of are nonnegative. By the spectral theorem (7.29 and 7.31), there is an orthonormal basis 1, , of consisting of eigenvectors of . Let 1, , be the eigenval-ues of corresponding to 1, , ; thus each is a nonnegative number. The matrix of with respect to 1, , is the diagonal matrix with 1, , on the diagonal, which shows that (b) implies (c). Now suppose (c) holds. Suppose 1, , is an orthonormal basis of such that the matrix of with respect to this basis is a diagonal matrix with nonnegative numbers 1, , on the diagonal. The linear map lemma (3.4) implies that there exists () such that \n\n = \n\nfor each = 1 , , . As you should verify, is a positive operator. Furthermore, \n\n 2 = = for each , which implies that 2 = . Thus is a positive square root of . Hence (d) holds, which shows that (c) implies (d). Every positive operator is self-adjoint (by definition of positive operator). Thus (d) implies (e). Now suppose (e) holds, meaning that there exists a self-adjoint operator on \n\n such that = 2. Then = ( because = ) . Hence (e) implies (f). Finally, suppose (f) holds. Let () be such that = . Then \n\n = ( ) = ( ) = = . Hence is self-adjoint. To complete the proof that (a) holds, note that \n\n , = , = , 0 \n\nfor every . Thus is positive, showing that (f) implies (a). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7C Positive Operators 253 \n\nEvery nonnegative number has a unique nonnegative square root. The next result shows that positive operators enjoy a similar property. \n\n> 7.39\n\neach positive operator has only one positive square root \n\nEvery positive operator on has a unique positive square root. \n\nA positive operator can have infinitely many square roots (although only one of them can be positive ). For example, the identity operator on has infinitely many square roots if dim > 1 .\n\nProof Suppose () is positive. Suppose is an eigenvector of .Hence there exists a real number 0 \n\nsuch that = .Let be a positive square root of .We will prove that = . This will imply that the behavior of on the eigenvectors of is uniquely determined. Because there is a basis of consisting of eigenvectors of (by the spectral theorem), this will imply that is uniquely determined. To prove that = , note that the spectral theorem asserts that there is an orthonormal basis 1, , of consisting of eigenvectors of . Because is a positive operator, all its eigenvalues are nonnegative. Thus there exist nonnegative numbers 1, , such that = for each = 1 , , .Because 1, , is a basis of , we can write \n\n = 1 1 + + \n\nfor some numbers 1, , . Thus \n\n = 1 1 1 + + .\n\nHence \n\n = = 2 = 1 1 1 + + .\n\nThe equation above implies that \n\n 1 1 + + = 1 1 1 + + .\n\nThus ( ) = 0 for each = 1 , , . Hence \n\n = \n\n> { = }\n\n .\n\nThus \n\n = \n\n> { = }\n\n = ,\n\nas desired. The notation defined below makes sense thanks to the result above. \n\n> 7.40 notation: \n\n \n\nFor a positive operator, denotes the unique positive square root of .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n254 Chapter 7 Operators on Inner Product Spaces \n\n> 7.41 example:\n\nsquare root of positive operators \n\nDefine operators , on 2 (with the usual Euclidean inner product) by \n\n( , ) = ( , 2) and ( , ) = ( + , + ). \n\nThen with respect to the standard basis of 2 we have \n\n> 7.42\n\n() = 1 00 2 and () = 1 11 1 .\n\nEach of these matrices equals its transpose; thus and are self-adjoint. If ( , ) 2, then \n\n( , ) , ( , ) = 2 + 2 2 0 \n\nand \n\n( , ) , ( , ) = 2 + 2 + 2 = ( + ) 2 0. \n\nThus and are positive operators. The standard basis of 2 is an orthonormal basis consisting of eigenvectors of \n\n . Note that \n\n( 1 \n\n> 2\n\n, 1 \n\n> 2\n\n), ( 1 \n\n> 2\n\n, 1 \n\n> 2\n\n)\n\nis an orthonormal basis of eigenvectors of , with eigenvalue 2 for the first eigenvector and eigenvalue 0 for the second eigenvector. Thus has the same eigenvectors, with eigenvalues 2 and 0.You can verify that \n\n( ) = 1 00 2\n\nand ( ) = \n\n> 1\n> 21\n> 21\n> 21\n> 2\n\n\n\nwith respect to the standard basis by showing that the squares of the matrices above are the matrices in 7.42 and that each matrix above is the matrix of a positive operator. The statement of the next result does not involve a square root, but the clean proof makes nice use of the square root of a positive operator. \n\n> 7.43\n\n positive and , = 0 = 0 \n\nSuppose is a positive operator on and is such that , = 0 .Then = 0 .\n\nProof We have \n\n0 = , = , = , = 2\n\n.\n\nHence = 0 . Thus = ( ) = 0 , as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7C Positive Operators 255\n\n# Exercises 7C \n\n1 Suppose () . Prove that if both and are positive operators, then \n\n = 0 .\n\n2 Suppose ( 4) is the operator whose matrix (with respect to the standard basis) is \n\n2 1 0 01 2 1 00 1 2 1 0 0 1 2.\n\nShow that is an invertible positive operator. \n\n3 Suppose is a positive integer and ( ) is the operator whose matrix (with respect to the standard basis) consists of all 1s. Show that is a positive operator. \n\n4 Suppose is an integer with > 1 . Show that there exists an -by- matrix \n\n such that all of the entries of are positive numbers and = , but the operator on whose matrix (with respect to the standard basis) equals is not a positive operator. \n\n5 Suppose () is self-adjoint. Prove that is a positive operator if and only if for every orthonormal basis 1, , of , all entries on the diagonal of ( , ( 1, , )) are nonnegative numbers. \n\n6 Prove that the sum of two positive operators on is a positive operator. \n\n7 Suppose () is an invertible positive operator and () is a positive operator. Prove that + is invertible. \n\n8 Suppose () . Prove that is a positive operator if and only if the pseudoinverse is a positive operator. \n\n9 Suppose () is a positive operator and ( , ) . Prove that \n\n is a positive operator on .\n\n10 Suppose is a positive operator on . Suppose , are such that \n\n = and = . \n\nProve that = .\n\n11 Suppose is a positive operator on and is a subspace of invariant under . Prove that | () is a positive operator on .\n\n12 Suppose () is a positive operator. Prove that is a positive operator for every positive integer .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n256 Chapter 7 Operators on Inner Product Spaces \n\n13 Suppose () is self-adjoint and .(a) Prove that is a positive operator if and only if is less than or equal to every eigenvalue of .(b) Prove that is a positive operator if and only if is greater than or equal to every eigenvalue of .\n\n14 Suppose is a positive operator on and 1, , . Prove that \n\n> \n> \n> = 1\n> \n> \n> = 1\n\n , 0. \n\n15 Suppose () is self-adjoint. Prove that there exist positive operators \n\n , () such that \n\n = and = + and = = 0. \n\n16 Suppose is a positive operator on . Prove that \n\nnull = null and range = range . \n\n17 Suppose that () is a positive operator. Prove that there exists a polynomial with real coefficients such that = () .\n\n18 Suppose and are positive operators on . Prove that is a positive operator if and only if and commute. \n\n19 Show that the identity operator on 2 has infinitely many self-adjoint square roots. \n\n20 Suppose () and 1, , is an orthonormal basis of . Prove that \n\nis a positive operator if and only if there exist 1, , such that \n\n , = , \n\nfor all , = 1 , , .\n\nThe numbers { , } , = 1 , , are the entries in the matrix of with respect to the orthonormal basis 1, , .\n\n21 Suppose is a positive integer. The -by- Hilbert matrix is the -by- \n\nmatrix whose entry in row , column is 1 + 1 . Suppose () is an operator whose matrix with respect to some orthonormal basis of is the \n\n -by- Hilbert matrix. Prove that is a positive invertible operator. \n\nExample: The 4-by-4 Hilbert matrix is \n\n1 121314121314151314151614151617\n\n. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7C Positive Operators 257\n\n22 Suppose () is a positive operator and is such that = 1 \n\nand for all with = 1 . Show that is an eigenvector of corresponding to the largest eigenvalue of .\n\n23 For () and , , define , by , = , .(a) Suppose () . Prove that , is an inner product on if and only if is an invertible positive operator (with respect to the original inner product , ). (b) Prove that every inner product on is of the form , for some positive invertible operator () .\n\n24 Suppose and are positive operators on . Prove that \n\nnull ( + ) = null null . \n\n25 Let be the second derivative operator in Exercise 31(b) in Section 7A. Show that is a positive operator. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n258 Chapter 7 Operators on Inner Product Spaces \n\n# 7D Isometries, Unitary Operators, and Matrix Factorization \n\n# Isometries \n\nLinear maps that preserve norms are sufficiently important to deserve a name. \n\n> 7.44 definition:\n\nisometry \n\nA linear map ( , ) is called an isometry if \n\n = \n\nfor every . In other words, a linear map is an isometry if it preserves norms. \n\nThe Greek word isos means equal; the Greek word metron means measure. Thus isometry literally means equal measure. \n\nIf ( , ) is an isometry and \n\n is such that = 0 , then \n\n = = 0 = 0 ,\n\nwhich implies that = 0 . Thus every isometry is injective. \n\n> 7.45 example:\n\northonormal basis maps to orthonormal list isometry \n\nSuppose 1, , is an orthonormal basis of and 1, , is an orthonormal list in . Let ( , ) be the linear map such that = for each \n\n = 1 , , . To show that is an isometry, suppose . Then \n\n> 7.46\n\n = , 1 1 + + , \n\nand \n\n> 7.47\n\n 2 = , 1 2 + + , 2,\n\nwhere we have used 6.30(b). Applying to both sides of 7.46 gives \n\n = , 1 1 + + , = , 1 1 + + , .\n\nThus \n\n> 7.48\n\n 2 = , 1 2 + + | , | 2.\n\nComparing 7.47 and 7.48 shows that = . Thus is an isometry. The next result gives conditions equivalent to being an isometry. The equiv-alence of (a) and (c) shows that a linear map is an isometry if and only if it preserves inner products. The equivalence of (a) and (d) shows that a linear map is an isometry if and only if it maps some orthonormal basis to an orthonormal list. Thus the isometries given by Example 7.45 include all isometries. Furthermore, a linear map is an isometry if and only if it maps every orthonormal basis to an orthonormal list [because whether or not (a) holds does not depend on the basis \n\n 1, , ].\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7D Isometries, Unitary Operators, and Matrix Factorization 259 \n\nThe equivalence of (a) and (e) in the next result shows that a linear map is an isometry if and only if the columns of its matrix (with respect to any orthonormal bases) form an orthonormal list. Here we are identifying the columns of an -by- \n\nmatrix with elements of and then using the Euclidean inner product on . \n\n> 7.49\n\ncharacterizations of isometries \n\nSuppose ( , ) . Suppose 1, , is an orthonormal basis of and \n\n 1, , is an orthonormal basis of . Then the following are equivalent. (a) is an isometry. (b) = .(c) , = , for all , .(d) 1, , is an orthonormal list in .(e) The columns of ( , ( 1, , ), ( 1, , )) form an orthonormal list in with respect to the Euclidean inner product. \n\nProof First suppose (a) holds, so is an isometry. If then \n\n( ) , = , , = 2 , = 2 2 = 0. \n\nHence the self-adjoint operator equals 0 (by 7.16). Thus = , proving that (a) implies (b). Now suppose (b) holds, so = . If , then \n\n , = , = , = , ,\n\nproving that (b) implies (c). Now suppose that (c) holds, so , = , for all , . Thus if \n\n , {1 , , } , then \n\n , = , . \n\nHence 1, , is an orthonormal list in , proving that (c) implies (d). Now suppose that (d) holds, so 1, , is an orthonormal list in . Let \n\n = ( , ( 1, , ), ( 1, , )) . If , {1 , , } , then \n\n> 7.50\n> \n> \n> = 1\n\n , , = \n\n> \n> \n> = 1\n\n , , \n\n> \n> = 1\n\n , = , = {{1 if = ,\n\n0 if . \n\nThe left side of 7.50 is the inner product in of columns and of . Thus the columns of form an orthonormal list in , proving that (d) implies (e). Now suppose (e) holds, so the columns of the matrix defined in the paragraph above form an orthonormal list in . Then 7.50 shows that 1, , is an orthonormal list in . Thus Example 7.45, with 1, , playing the role of \n\n 1, , , shows that is an isometry, proving that (e) implies (a). See Exercises 1 and 11 for additional conditions that are equivalent to being an isometry. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 260 Chapter 7 Operators on Inner Product Spaces\n\n# Unitary Operators \n\nIn this subsection, we confine our attention to linear maps from a vector space to itself. In other words, we will be working with operators. \n\n> 7.51 definition:\n\nunitary operator \n\nAn operator () is called unitary if is an invertible isometry. \n\n> Although the words unitary and isometry mean the same thing for operators on finite-dimensional inner product spaces, remember that a uni-tary operator maps a vector space to itself, while an isometry maps a vector space to another (possibly different )\n> vector space.\n\nAs previously noted, every isometry is injective. Every injective operator on a finite-dimensional vector space is in-vertible (see 3.65). A standing assump-tion for this chapter is that is a finite-dimensional inner product space. Thus we could delete the word invertible from the definition above without chang-ing the meaning. The unnecessary word invertible has been retained in the definition above for consistency with the definition readers may encounter when learning about inner product spaces that are not necessarily finite-dimensional. \n\n> 7.52 example:\n\nrotation of 2\n\nSuppose and is the operator on 2 whose matrix with respect to the standard basis of 2 is \n\n\n\ncos sin \n\nsin cos .\n\nThe two columns of this matrix form an orthonormal list in 2; hence is an isometry [by the equivalence of (a) and (e) in 7.49 ]. Thus is a unitary operator. If = , then is the operator of counterclockwise rotation by radians around the origin of 2. This observation gives us another way to think about why \n\n is an isometry, because each rotation around the origin of 2 preserves norms. The next result (7.53) lists several conditions that are equivalent to being a unitary operator. All the conditions equivalent to being an isometry in 7.49 should be added to this list. The extra conditions in 7.53 arise because of limiting the context to linear maps from a vector space to itself. For example, 7.49 shows that a linear map ( , ) is an isometry if and only if = , while 7.53 shows that an operator () is a unitary operator if and only if = = .Another difference is that 7.49(d) mentions an orthonormal list, while 7.53(d) mentions an orthonormal basis. Also, 7.49(e) mentions the columns of () ,while 7.53(e) mentions the rows of () . Furthermore, () in 7.49(e) is with respect to an orthonormal basis of and an orthonormal basis of , while () \n\nin 7.53(e) is with respect to a single basis of doing double duty. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7D Isometries, Unitary Operators, and Matrix Factorization 261\n> 7.53\n\ncharacterizations of unitary operators \n\nSuppose () . Suppose 1, , is an orthonormal basis of . Then the following are equivalent. (a) is a unitary operator. (b) = = .(c) is invertible and 1 = .(d) 1, , is an orthonormal basis of .(e) The rows of ( , ( 1, , )) form an orthonormal basis of with respect to the Euclidean inner product. (f) is a unitary operator. \n\n> Proof\n\nFirst suppose (a) holds, so is a unitary operator. Hence \n\n = \n\nby the equivalence of (a) and (b) in 7.49. Multiply both sides of this equation by \n\n 1 on the right, getting = 1 . Thus = 1 = , as desired, proving that (a) implies (b). The definitions of invertible and inverse show that (b) implies (c). Now suppose (c) holds, so is invertible and 1 = . Thus = . Hence \n\n 1, , is an orthonormal list in , by the equivalence of (b) and (d) in 7.49. The length of this list equals dim . Thus 1, , is an orthonormal basis of , proving that (c) implies (d). Now suppose (d) holds, so 1, , is an orthonormal basis of . The equivalence of (a) and (d) in 7.49 shows that is a unitary operator. Thus \n\n( ) = = ,\n\nwhere the last equation holds because we already showed that (a) implies (b) in this result. The equation above and the equivalence of (a) and (b) in 7.49 show that \n\nis an isometry. Thus the columns of ( , ( 1, , )) form an orthonormal basis of [by the equivalence of (a) and (e) of 7.49 ]. The rows of ( , ( 1, , )) \n\nare the complex conjugates of the columns of ( , ( 1, , )) . Thus the rows of ( , ( 1, , )) form an orthonormal basis of , proving that (d) implies (e). Now suppose (e) holds. Thus the columns of ( , ( 1, , )) form an orthonormal basis of . The equivalence of (a) and (e) in 7.49 shows that is an isometry, proving that (e) implies (f). Now suppose (f) holds, so is a unitary operator. The chain of implications we have already proved in this result shows that (a) implies (f). Applying this result to shows that ( ) is a unitary operator, proving that (f) implies (a). We have shown that (a) (b) (c) (d) (e) (f) (a), completing the proof. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 262 Chapter 7 Operators on Inner Product Spaces\n\nRecall our analogy between and () . Under this analogy, a complex number corresponds to an operator () , and corresponds to . The real numbers ( = ) correspond to the self-adjoint operators ( = ), and the nonnegative numbers correspond to the (badly named) positive operators. Another distinguished subset of is the unit circle, which consists of the complex numbers such that || = 1 . The condition || = 1 is equivalent to the condition = 1 . Under our analogy, this corresponds to the condition = ,which is equivalent to being a unitary operator. Hence the analogy shows that the unit circle in corresponds to the set of unitary operators. In the next two results, this analogy appears in the eigenvalues of unitary operators. Also see Exercise 15 for another example of this analogy. \n\n> 7.54\n\neigenvalues of unitary operators have absolute value 1\n\nSuppose is an eigenvalue of a unitary operator. Then || = 1 . \n\n> Proof\n\nSuppose () is a unitary operator and is an eigenvalue of . Let \n\n be such that 0 and = . Then \n\n|| = = = . \n\nThus || = 1 , as desired. The next result characterizes unitary operators on finite-dimensional complex inner product spaces, using the complex spectral theorem as the main tool. \n\n> 7.55\n\ndescription of unitary operators on complex inner product spaces \n\nSuppose = and () . Then the following are equivalent. (a) is a unitary operator. (b) There is an orthonormal basis of consisting of eigenvectors of whose corresponding eigenvalues all have absolute value 1. \n\n> Proof\n\nSuppose (a) holds, so is a unitary operator. The equivalence of (a) and (b) in 7.53 shows that is normal. Thus the complex spectral theorem (7.31) shows that there is an orthonormal basis 1, , of consisting of eigenvectors of . Every eigenvalue of has absolute value 1 (by 7.54), completing the proof that (a) implies (b). Now suppose (b) holds. Let 1, , be an orthonormal basis of consisting of eigenvectors of whose corresponding eigenvalues 1, , all have absolute value 1. Then 1, , is also an orthonormal basis of because \n\n , = , = , = {{0 if ,\n\n1 if = \n\nfor all , = 1 , , . Thus the equivalence of (a) and (d) in 7.53 shows that is unitary, proving that (b) implies (a). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7D Isometries, Unitary Operators, and Matrix Factorization 263\n\n# QR Factorization \n\nIn this subsection, we shift our attention from operators to matrices. This switch should give you good practice in identifying an operator with a square matrix (after picking a basis of the vector space on which the operator is defined). You should also become more comfortable with translating concepts and results back and forth between the context of operators and the context of square matrices. When starting with -by- matrices instead of operators, unless otherwise specified assume that the associated operators live on (with the Euclidean inner product) and that their matrices are computed with respect to the standard basis of .We begin by making the following definition, transferring the notion of a unitary operator to a unitary matrix. \n\n> 7.56 definition:\n\nunitary matrix \n\nAn -by- matrix is called unitary if its columns form an orthonormal list in .In the definition above, we could have replaced orthonormal list in with orthonormal basis of because every orthonormal list of length in an -dimensional inner product space is an orthonormal basis. If () and \n\n 1, , and 1, , are orthonormal bases of , then is a unitary operator if and only if ( , ( 1, , ), ( 1, , )) is a unitary matrix, as shown by the equivalence of (a) and (e) in 7.49. Also note that we could also have replaced columns in the definition above with rows by using the equivalence between conditions (a) and (e) in 7.53. The next result, whose proof will be left as an exercise for the reader, gives some equivalent conditions for a square matrix to be unitary. In (c), denotes the matrix product of and , identifying elements of with -by-1 matrices (sometimes called column vectors). The norm in (c) below is the usual Euclidean norm on that comes from the Euclidean inner product. In (d), denotes the conjugate transpose of the matrix , which corresponds to the adjoint of the associated operator. \n\n> 7.57\n\ncharacterizations of unitary matrices \n\nSuppose is an -by- matrix. Then the following are equivalent. (a) is a unitary matrix. (b) The rows of form an orthonormal list in .(c) = for every .(d) = = , the -by- matrix with 1s on the diagonal and 0s elsewhere. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n264 Chapter 7 Operators on Inner Product Spaces \n\nThe QR factorization stated and proved below is the main tool in the widely used QR algorithm (not discussed here) for finding good approximations to eigenvalues and eigenvectors of square matrices. In the result below, if the matrix \n\n is in , , then the matrices and are also in , . \n\n> 7.58\n\nQR factorization \n\nSuppose is a square matrix with linearly independent columns. Then there exist unique matrices and such that is unitary, is upper triangular with only positive numbers on its diagonal, and \n\n = . \n\nProof Let 1, , denote the columns of , thought of as elements of . Apply the GramSchmidt procedure (6.32) to the list 1, , , getting an orthonormal basis 1, , of such that \n\n> 7.59\n\nspan ( 1, , ) = span ( 1, , )\n\nfor each = 1 , , . Let be the -by- matrix defined by \n\n , = , ,\n\nwhere , denotes the entry in row , column of . If > , then is orthogonal to span ( 1, , ) and hence is orthogonal to (by 7.59). In other words, if \n\n > then , = 0 . Thus is an upper-triangular matrix. Let be the unitary matrix whose columns are 1, , . If {1 , , } ,then the th column of equals a linear combination of the columns of , with the coefficients for the linear combination coming from the th column of see 3.51(a). Hence the th column of equals \n\n , 1 1 + + , ,\n\nwhich equals [by 6.30(a) ], the th column of . Thus = , as desired. The equations defining the GramSchmidt procedure (see 6.32) show that each equals a positive multiple of plus a linear combination of 1, , 1 .Thus each , is a positive number. Hence all entries on the diagonal of are positive numbers, as desired. Finally, to show that and are unique, suppose we also have = , where is unitary and is upper triangular with only positive numbers on its diagonal. Let 1, , denote the columns of . Thinking of matrix multiplication as above, we see that each is a linear combination of 1, , , with the coefficients com-ing from the th column of . This implies that span ( 1, , ) = span ( 1, , )\n\nand , > 0 . The uniqueness of the orthonormal lists satisfying these condi-tions (see Exercise 10 in Section 6B) now shows that = for each = 1 , , .Hence = , which then implies that = , completing the proof of unique-ness. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7D Isometries, Unitary Operators, and Matrix Factorization 265 \n\nThe proof of the QR factorization shows that the columns of the unitary matrix can be computed by applying the GramSchmidt procedure to the columns of the matrix to be factored. The next example illustrates the computation of the QR factorization based on the proof that we just completed. \n\n> 7.60 example:\n\nQR factorization of a 3-by-3 matrix \n\nTo find the QR factorization of the matrix \n\n = 1 2 10 1 4 0 3 2\n\n,\n\nfollow the proof of 7.58. Thus set 1, 2, 3 equal to the columns of :\n\n 1 = (1 , 0, 0) , 2 = (2 , 1, 3) , 3 = (1 , 4 , 2). \n\nApply the GramSchmidt procedure to 1, 2, 3, producing the orthonormal list \n\n 1 = (1 , 0, 0) , 2 = (0 , 1 \n\n> 10\n\n, 3 \n\n> 10\n\n), 3 = (0 , 3 \n\n> 10\n\n, 1 \n\n> 10\n\n). \n\nStill following the proof of 7.58, let be the unitary matrix whose columns are \n\n 1, 2, 3:\n\n = 1 0 00 1 \n\n> 10\n\n 3\n\n> 10\n\n0 3\n\n> 10 1\n> 10\n\n.\n\nAs in the proof of 7.58, let be the 3-by-3 matrix whose entry in row , column \n\nis , , which gives \n\n = 1 2 10 10 10 5\n\n0 0 710 5\n\n.\n\nNote that is indeed an upper-triangular matrix with only positive numbers on the diagonal, as required by the QR factorization. Now matrix multiplication can verify that = is the desired factorization of :\n\n = 1 0 00 1 \n\n> 10\n\n 3\n\n> 10\n\n0 3\n\n> 10 1\n> 10\n\n1 2 10 10 10 5\n\n0 0 710 5\n\n= 1 2 10 1 4 0 3 2= . \n\nThus = , as expected. The QR factorization will be the major tool used in the proof of the Cholesky factorization (7.63) in the next subsection. For another nice application of the QR factorization, see the proof of Hadamards inequality (9.66). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 266 Chapter 7 Operators on Inner Product Spaces\n\nIf a QR factorization is available, then it can be used to solve a corresponding system of linear equations without using Gaussian elimination. Specifically, suppose is an -by- square matrix with linearly independent columns. Suppose that and we want to solve the equation = for = ( 1, , ) \n\n(as usual, we are identifying elements of with -by-1 column vectors). Suppose = , where is unitary and is upper triangular with only positive numbers on its diagonal ( and are computable from using just the GramSchmidt procedure, as shown in the proof of 7.58). The equation = is equivalent to the equation = . Multiplying both sides of this last equation by on the left and using 7.57(d) gives the equation \n\n = . \n\nThe matrix is the conjugate transpose of the matrix . Thus computing \n\n is straightforward. Because is an upper-triangular matrix with positive numbers on its diagonal, the system of linear equations represented by the equation above can quickly be solved by first solving for , then for 1 , and so on. \n\n# Cholesky Factorization \n\nWe begin this subsection with a characterization of positive invertible operators in terms of inner products. \n\n> 7.61\n\npositive invertible operator \n\nA self-adjoint operator () is a positive invertible operator if and only if , > 0 for every nonzero . \n\n> Proof\n\nFirst suppose is a positive invertible operator. If and 0 , then because is invertible we have 0 . This implies that , 0 (by 7.43). Hence , > 0 .To prove the implication in the other direction, suppose now that , > 0 \n\nfor every nonzero . Thus 0 for every nonzero . Hence is injective. Thus is invertible, as desired. The next definition transfers the result above to the language of matrices. Here we are using the usual Euclidean inner product on and identifying elements of \n\n with -by-1 column vectors. \n\n> 7.62 definition:\n\npositive definite \n\nA matrix , is called positive definite if = and \n\n , > 0 \n\nfor every nonzero . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7D Isometries, Unitary Operators, and Matrix Factorization 267\n\nA matrix is upper triangular if and only if its conjugate transpose is lower triangular (meaning that all entries above the diagonal are 0). The factorization below, which has important consequences in computational linear algebra, writes a positive definite matrix as the product of a lower triangular matrix and its conjugate transpose. Our next result is solely about matrices, although the proof makes use of the identification of results about operators with results about square matrices. In the result below, if the matrix is in , , then the matrix is also in , . \n\n> 7.63\n\nCholesky factorization \n\nSuppose is a positive definite matrix. Then there exists a unique upper-triangular matrix with only positive numbers on its diagonal such that \n\n = . \n\n> Proof\n\nBecause is positive definite, there exists an invertible square matrix \n\nof the same size as such that = [ by the equivalence of (a) and (f) in 7.38 ].Let = be the QR factorization of (see 7.58), where is unitary and \n\nis upper triangular with only positive numbers on its diagonal. Then = . \n\n> Andr-Louis Cholesky (18751918 )\n> discovered this factorization, which was published posthumously in 1924.\n\nThus \n\n = = = ,\n\nas desired. To prove the uniqueness part of this result, suppose is an upper-triangular matrix with only positive numbers on its diagonal and = . The matrix is invertible because is invertible (see Exercise 11 in Section 3D). Multiplying both sides of the equation = by 1 on the right gives the equation 1 = .Let be the matrix from the first paragraph of this proof. Then \n\n( 1 )( 1 ) = ( )1 1 \n\n= ( )1 1 \n\n= ( )1 \n\n= . \n\nThus 1 is unitary. Hence = ( 1 ) is a factorization of as the product of a unitary matrix and an upper-triangular matrix with only positive numbers on its diagonal. The uniqueness of the QR factorization, as stated in 7.58, now implies that = .In the first paragraph of the proof above, we could have chosen to be the unique positive definite matrix that is a square root of (see 7.39). However, the proof was presented with the more general choice of because for specific positive definite matrices , it may be easier to find a different choice of . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 268 Chapter 7 Operators on Inner Product Spaces\n\n# Exercises 7D \n\n1 Suppose dim 2 and ( , ) . Prove that is an isometry if and only if 1, 2 is an orthonormal list in for every orthonormal list 1, 2\n\nof length two in .\n\n2 Suppose ( , ) . Prove that is a scalar multiple of an isometry if and only if preserves orthogonality. \n\n> The phrase preserves orthogonality means that , = 0 for all\n> , such that , = 0 .\n\n3 (a) Show that the product of two unitary operators on is a unitary operator. (b) Show that the inverse of a unitary operator on is a unitary operator. \n\n> This exercise shows that the set of unitary operators on is a group, where the group operation is the usual product of two operators.\n\n4 Suppose = and , () are self-adjoint. Show that + is unitary if and only if = and 2 + 2 = .\n\n5 Suppose () . Prove that the following are equivalent. (a) is a self-adjoint unitary operator. (b) = 2 for some orthogonal projection on .(c) There exists a subspace of such that = for every and \n\n = for every .\n\n6 Suppose 1, 2 are both normal operators on 3 with 2, 5, 7 as eigenvalues. Prove that there exists a unitary operator ( 3) such that 1 = 2 .\n\n7 Give an example of two self-adjoint operators 1, 2 ( 4) such that the eigenvalues of both operators are 2, 5, 7 but there does not exist a unitary operator ( 4) such that 1 = 2 . Be sure to explain why there is no unitary operator with the required property. \n\n8 Prove or give a counterexample: If () and there exists an orthonormal basis 1, , of such that = 1 for each , then is a unitary operator. \n\n9 Suppose = and () . Suppose every eigenvalue of has absolute value 1 and for every . Prove that is a unitary operator. \n\n10 Suppose = and () is a self-adjoint operator such that \n\nfor all .(a) Show that 2 is a positive operator. (b) Show that + 2 is a unitary operator. \n\n11 Suppose () . Prove that is a unitary operator if and only if \n\n{ and 1} = { 1}. \n\n12 Prove or give a counterexample: If () is invertible and 1 = \n\nfor every , then is unitary. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7D Isometries, Unitary Operators, and Matrix Factorization 269 \n\n13 Explain why the columns of a square matrix of complex numbers form an orthonormal list in if and only if the rows of the matrix form an orthonormal list in .\n\n14 Suppose with = 1 and . Also suppose dim 2 . Prove that there exists a unitary operator () such that , = if and only if || 1 .\n\n15 Suppose is a unitary operator on such that is invertible. (a) Prove that ( + )( ) 1 is a skew operator (meaning that it equals the negative of its adjoint). (b) Prove that if = , then ( + )( ) 1 is a self-adjoint operator. \n\nThe function ( + 1)( 1) 1 maps the unit circle in (except for the point 1) to . Thus (b) illustrates the analogy between the unitary operators and the unit circle in , along with the analogy between the self-adjoint operators and .\n\n16 Suppose = and () is self-adjoint. Prove that ( + )( ) 1 \n\nis a unitary operator and 1 is not an eigenvalue of this operator. \n\n17 Explain why the characterizations of unitary matrices given by 7.57 hold. \n\n18 A square matrix is called symmetric if it equals its transpose. Prove that if \n\n is a symmetric matrix with real entries, then there exists a unitary matrix \n\n with real entries such that is a diagonal matrix. \n\n19 Suppose is a positive integer. For this exercise, we adopt the notation that a typical element of is denoted by = ( 0, 1, , 1 ). Define linear functionals 0, 1, , 1 on by \n\n ( 0, 1, , 1 ) = 1\n\n> \n\n \n\n> 1\n> \n> = 0\n\n 2/ .\n\nThe discrete Fourier transform is the operator defined by \n\n = ( 0() , 1() , , 1 ()). \n\n(a) Show that is a unitary operator on .(b) Show that if ( 0, , 1 ) and is defined to equal 0, then \n\n1 ( 0, 1, , 1 ) = ( , 1 , , 1). \n\n(c) Show that 4 = .\n\nThe discrete Fourier transform has many important applications in data analysis. The usual Fourier transform involves expressions of the form \n\n> () 2 \n\nfor complex-valued integrable functions defined on .\n\n20 Suppose is a square matrix with linearly independent columns. Prove that there exist unique matrices and such that is lower triangular with only positive numbers on its diagonal, is unitary, and = .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n270 Chapter 7 Operators on Inner Product Spaces \n\n# 7E Singular Value Decomposition \n\n# Singular Values \n\nWe will need the following result in this section. \n\n> 7.64\n\nproperties of \n\nSuppose ( , ) . Then (a) is a positive operator on ;(b) null = null ;(c) range = range ;(d) dim range = dim range = dim range .\n\nProof \n\n(a) We have \n\n( ) = ( ) = . \n\nThus is self-adjoint. If , then \n\n( ) , = () , = , = 2 0. \n\nThus is a positive operator. (b) First suppose null . Then \n\n 2 = , = , = 0 , = 0. \n\nThus = 0 , proving that null null .The inclusion in the other direction is clear, because if and = 0 ,then = 0 .Thus null = null , completing the proof of (b). (c) We already know from (a) that is self-adjoint. Thus \n\nrange = ( null ) = ( null ) = range ,\n\nwhere the first and last equalities come from 7.6 and the second equality comes from (b). (d) To verify the first equation in (d), note that \n\ndim range = dim (null ) = dim dim null = dim range ,\n\nwhere the first equality comes from 7.6(d), the second equality comes from 6.51, and the last equality comes from the fundamental theorem of linear maps (3.21). The equality dim range = dim range follows from (c). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7E Singular Value Decomposition 271\n\nThe eigenvalues of an operator tell us something about the behavior of the operator. Another collection of numbers, called the singular values, is also useful. Eigenspaces and the notation (used in the examples) were defined in 5.52. \n\n> 7.65 definition:\n\nsingular values \n\nSuppose ( , ) . The singular values of are the nonnegative square roots of the eigenvalues of , listed in decreasing order, each included as many times as the dimension of the corresponding eigenspace of . \n\n> 7.66 example:\n\nsingular values of an operator on 4\n\nDefine ( 4) by ( 1, 2, 3, 4) = (0 , 3 1, 2 2, 3 4). A calculation shows that \n\n ( 1, 2, 3, 4) = (9 1, 4 2, 0, 9 4),\n\nas you should verify. Thus the standard basis of 4 diagonalizes , and we see that the eigenvalues of are 9, 4, and 0. Also, the dimensions of the eigenspaces corresponding to the eigenvalues are \n\ndim (9 , ) = 2 and dim (4 , ) = 1 and dim (0 , ) = 1. \n\nTaking nonnegative square roots of these eigenvalues of and using dimension information from above, we conclude that the singular values of are 3, 3, 2, 0.The only eigenvalues of are 3 and 0. Thus in this case, the collection of eigenvalues did not pick up the number 2 that appears in the definition (and hence the behavior) of , but the list of singular values does include 2. \n\n> 7.67 example:\n\nsingular values of a linear map from 4 to 3\n\nSuppose ( 4, 3) has matrix (with respect to the standard bases) \n\n0 0 0 5 0 0 0 01 1 0 0.\n\nYou can verify that the matrix of is \n\n1 1 0 01 1 0 00 0 0 00 0 0 25 \n\nand that the eigenvalues of the operator are 25 , 2, 0, with dim (25 , ) = 1 ,\n\ndim (2 , ) = 1 , and dim (0 , ) = 2 . Thus the singular values of are \n\n5, 2, 0, 0.See Exercise 2 for a characterization of the positive singular values. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 100 Chapter 7 Operators on Inner Product Spaces\n> 7.68\n\nrole of positive singular values \n\nSuppose that ( , ) . Then (a) is injective 0 is not a singular value of ;(b) the number of positive singular values of equals dim range ;(c) is surjective number of positive singular values of equals dim . \n\n> Proof\n\nThe linear map is injective if and only if null = {0} , which happens if and only if null = {0} [ by 7.64(b) ], which happens if and only if 0 is not an eigenvalue of , which happens if and only if 0 is not a singular value of ,completing the proof of (a). The spectral theorem applied to shows that dim range equals the num-ber of positive eigenvalues of (counting repetitions). Thus 7.64(d) implies that dim range equals the number of positive singular values of , proving (b). Use (b) and 2.39 to show that (c) holds. The table below compares eigenvalues with singular values. \n\n> list of eigenvalues list of singular values\n> context: vector spaces context: inner product spaces defined only for linear maps from a vector space to itself defined for linear maps from an inner product space to a possibly different inner product space can be arbitrary real numbers (if = )or complex numbers (if = )are nonnegative numbers can be the empty list if = length of list equals dimension of domain includes 0 operator is not invertible includes 0 linear map is not injective no standard order, especially if = always listed in decreasing order\n\nThe next result nicely characterizes isometries in terms of singular values. \n\n> 7.69\n\nisometries characterized by having all singular values equal 1\n\nSuppose that ( , ) . Then \n\n is an isometry all singular values of equal 1. \n\n> Proof\n\nWe have \n\n is an isometry = all eigenvalues of equal 1 all singular values of equal 1,where the first equivalence comes from 7.49 and the second equivalence comes from the spectral theorem (7.29 or 7.31) applied to the self-adjoint operator .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7E Singular Value Decomposition 273 \n\n# SVD for Linear Maps and for Matrices \n\nThe singular value decomposition is useful in computational linear alge-bra because good techniques exist for approximating eigenvalues and eigen-vectors of positive operators such as \n\n , whose eigenvalues and eigenvec-tors lead to the singular value decom-position. \n\nThe next result shows that every linear map from to has a remarkably clean description in terms of its singular val-ues and orthonormal lists in and .In the next section we will see several important applications of the singular value decomposition (often called the SVD). \n\n> 7.70\n\nsingular value decomposition \n\nSuppose ( , ) and the positive singular values of are 1, , .Then there exist orthonormal lists 1, , in and 1, , in such that \n\n> 7.71\n\n = 1 , 1 1 + + , \n\nfor every .\n\nProof Let 1, , denote the singular values of (thus = dim ). Because \n\n is a positive operator [see 7.64(a) ], the spectral theorem implies that there exists an orthonormal basis 1, , of with \n\n> 7.72\n\n = 2 \n\nfor each = 1 , , .For each = 1 , , , let \n\n> 7.73\n\n = \n\n \n\n.\n\nIf , {1 , , } , then \n\n , = 1 \n\n , = 1 \n\n , = \n\n \n\n , = {{0 if ,\n\n1 if = . \n\nThus 1, , is an orthonormal list in .If {1 , , } and > , then = 0 and hence = 0 (by 7.72), which implies that = 0 [ by 7.64(b) ].Suppose . Then \n\n = ( , 1 1 + + , )= , 1 1 + + , \n\n= 1 , 1 1 + + , ,\n\nwhere the last index in the first line switched from to in the second line because = 0 if > (as noted in the paragraph above) and the third line follows from 7.73. The equation above is our desired result. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n274 Chapter 7 Operators on Inner Product Spaces \n\nSuppose ( , ) , the positive singular values of are 1, , , and \n\n 1, , and 1, , are as in the singular value decomposition 7.70. The orthonormal list 1, , can be extended to an orthonormal basis 1, , dim \n\nof and the orthonormal list 1, , can be extended to an orthonormal basis \n\n 1, , dim of . The formula 7.71 shows that \n\n ={{ if 1 ,\n\n0 if < dim . \n\nThus the matrix of with respect to the orthonormal bases ( 1, , dim ) and \n\n( 1, , dim ) has the simple form \n\n( , ( 1, , dim ), ( 1, , dim )) , ={{ if 1 = ,\n\n0 otherwise .\n\nIf dim = dim (as happens, for example, if = ), then the matrix described in the paragraph above is a diagonal matrix. If we extend the definition of diagonal matrix as follows to apply to matrices that are not necessarily square, then we have proved the wonderful result that every linear map from to has a diagonal matrix with respect to appropriate orthonormal bases. \n\n> 7.74 definition:\n\ndiagonal matrix \n\nAn -by- matrix is called a diagonal matrix if all entries of the matrix are 0 except possibly , for = 1 , , min { , } .The table below compares the spectral theorem (7.29 and 7.31) with the singular value decomposition (7.70). \n\nspectral theorem singular value decomposition \n\ndescribes only self-adjoint operators (when = ) or normal operators (when \n\n> = \n\n)describes arbitrary linear maps from an inner product space to a possibly different inner product space produces a single orthonormal basis produces two orthonormal lists, one for domain space and one for range space, that are not necessarily the same even when range space equals domain space different proofs depending on whether \n\n> = \n\nor = \n\nsame proof works regardless of whether \n\n> = \n\nor = \n\nThe singular value decomposition gives us a new way to understand the adjoint and the inverse of a linear map. Specifically, the next result shows that given a singular value decomposition of a linear map ( , ) , we can obtain the adjoint of simply by interchanging the roles of the s and the s (see 7.77). Similarly, we can obtain the pseudoinverse (see 6.68) of by interchanging the roles of the s and the s and replacing each positive singular value of \n\nwith 1/ (see 7.78). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7E Singular Value Decomposition 275 \n\nRecall that the pseudoinverse in 7.78 below equals the inverse 1 if is invertible [see 6.69(a) ]. \n\n> 7.75\n\nsingular value decomposition of adjoint and pseudoinverse \n\nSuppose ( , ) and the positive singular values of are 1, , .Suppose 1, , and 1, , are orthonormal lists in and such that \n\n> 7.76\n\n = 1 , 1 1 + + , \n\nfor every . Then \n\n> 7.77\n\n = 1 , 1 1 + + , \n\nand \n\n> 7.78\n\n = , 1 1\n\n 1 + + , \n\n \n\nfor every .\n\nProof If and then \n\n , = 1 , 1 1 + + , , = 1 , 1 1, + + , , = , 1 , 1 1 + + , . \n\nThis implies that \n\n = 1 , 1 1 + + , ,\n\nproving 7.77. To prove 7.78, suppose . Let \n\n = , 1 1\n\n 1 + + , \n\n .\n\nApply to both sides of the equation above, getting \n\n = , 1 1\n\n 1 + + , \n\n \n\n= , 1 1 + + , \n\n= range ,\n\nwhere the second line holds because 7.76 implies that = if = 1 , , ,and the last line above holds because 7.76 implies that 1, , spans range \n\nand thus is an orthonormal basis of range [ and hence 6.57(i) applies ]. The equation above, the observation that ( null ) [see Exercise 8(b) ], and the definition of (see 6.68) show that = , proving 7.78. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n276 Chapter 7 Operators on Inner Product Spaces \n\n> 7.79 example:\n\nfinding a singular value decomposition \n\nDefine ( 4, 3) by ( 1, 2, 3, 4) = (5 4, 0, 1 + 2). We want to find a singular value decomposition of . The matrix of (with respect to the standard bases) is \n\n0 0 0 5 0 0 0 01 1 0 0.\n\nThus, as discussed in Example 7.67, the matrix of is \n\n1 1 0 01 1 0 00 0 0 00 0 0 25 \n\n,\n\nand the positive eigenvalues of are 25 , 2, with dim (25 , ) = 1 and \n\ndim (2 , ) = 1 . Hence the positive singular values of are 5, 2.Thus to find a singular value decomposition of , we must find an orthonormal list 1, 2 in 4 and an orthonormal list 1, 2 in 3 such that \n\n = 5 , 1 1 + 2 , 2 2\n\nfor all 4.An orthonormal basis of (25 , ) is the vector (0 , 0, 0, 1) ; an orthonormal basis of (2 , ) is the vector ( 1 \n\n> 2\n\n, 1 \n\n> 2\n\n, 0, 0) . Thus, following the proof of 7.70, we take \n\n 1 = (0 , 0, 0, 1) and 2 = ( 1\n\n> \n\n2 , 1\n\n> \n\n2 , 0, 0) \n\nand \n\n 1 = 1\n\n5 = (1 , 0, 0) and 2 = 2\n\n> \n\n2 = (0 , 0, 1). \n\nThen, as expected, we see that 1, 2 is an orthonormal list in 4 and 1, 2 is an orthonormal list in 3 and \n\n = 5 , 1 1 + 2 , 2 2\n\nfor all 4. Thus we have found a singular value decomposition of .The next result translates the singular value decomposition from the context of linear maps to the context of matrices. Specifically, the following result gives a factorization of an arbitrary matrix as the product of three nice matrices. The proof gives an explicit construction of these three matrices in terms of the singular value decomposition. In the next result, the phrase orthonormal columns should be interpreted to mean that the columns are orthonormal with respect to the standard Euclidean inner product. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7E Singular Value Decomposition 277 \n\n> 7.80\n\nmatrix version of SVD \n\nSuppose is a -by- matrix of rank 1 . Then there exist a -by- matrix \n\n with orthonormal columns, an -by- diagonal matrix with positive numbers on the diagonal, and an -by- matrix with orthonormal columns such that \n\n = .\n\nProof Let be the linear map whose matrix with respect to the standard bases equals . Then dim range = (by 3.78). Let \n\n> 7.81\n\n = 1 , 1 1 + + , \n\nbe a singular value decomposition of . Let \n\n = the -by- matrix whose columns are 1, , ,\n\n = the -by- diagonal matrix whose diagonal entries are 1, , ,\n\n = the -by- matrix whose columns are 1, , .\n\nLet 1, , denote the standard basis of . If {1 , , } then \n\n( ) = ( ) = = 0. \n\nThus = .Multiply both sides of this last equation by (the conjugate transpose of )on the right to get \n\n = .\n\nNote that the rows of are the complex conjugates of 1, , . Thus if \n\n {1 , , } , then the definition of matrix multiplication shows that = ;hence = . Thus = for all span ( 1, , ).If ( span ( 1, , )) , then = 0 (as follows from 7.81) and = 0 \n\n(as follows from the definition of matrix multiplication). Hence = for all ( span ( 1, , )) .Because and agree on span ( 1, , ) and on (span ( 1, , )) ,we conclude that = . Thus the displayed equation above becomes \n\n = ,\n\nas desired. Note that the matrix in the result above has entries. In comparison, the matrices , , and above have a total of \n\n( + + ) \n\nentries. Thus if and are large numbers and the rank is considerably less than and , then the number of entries that must be stored on a computer to represent is considerably less than .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n278 Chapter 7 Operators on Inner Product Spaces \n\n# Exercises 7E \n\n1 Suppose ( , ) . Show that = 0 if and only if all singular values of \n\n are 0.\n\n2 Suppose ( , ) and > 0 . Prove that is a singular value of if and only if there exist nonzero vectors and such that \n\n = and = . \n\nThe vectors , satisfying both equations above are called a Schmidt pair .Erhard Schmidt introduced the concept of singular values in 1907. \n\n3 Give an example of ( 2) such that 0 is the only eigenvalue of and the singular values of are 5, 0.\n\n4 Suppose that ( , ) , 1 is the largest singular value of , and is the smallest singular value of . Prove that \n\n{ and = 1} = [ , 1]. \n\n5 Suppose ( 2) is defined by ( , ) = (4 , ) . Find the singular values of .\n\n6 Find the singular values of the differentiation operator ( 2()) \n\ndefined by = , where the inner product on 2() is as in Example 6.34. \n\n7 Suppose that () is self-adjoint or that = and () is normal. Let 1, , be the eigenvalues of , each included in this list as many times as the dimension of the corresponding eigenspace. Show that the singular values of are | 1|, , | |, after these numbers have been sorted into decreasing order. \n\n8 Suppose ( , ) . Suppose 1 2 > 0 and 1, , is an orthonormal list in and 1, , is an orthonormal list in such that \n\n = 1 , 1 1 + + , \n\nfor every .(a) Prove that 1, , is an orthonormal basis of range .(b) Prove that 1, , is an orthonormal basis of (null ) .(c) Prove that 1, , are the positive singular values of .(d) Prove that if {1 , , } , then is an eigenvector of with corresponding eigenvalue 2.(e) Prove that \n\n = 12 , 1 1 + + 2 , \n\nfor all .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7E Singular Value Decomposition 279 \n\n9 Suppose ( , ) . Show that and have the same positive singular values. \n\n10 Suppose ( , ) has singular values 1, , . Prove that if is an invertible linear map, then 1 has singular values \n\n1 \n\n, , 1 1\n\n.\n\n11 Suppose that ( , ) and 1, , is an orthonormal basis of . Let \n\n 1, , denote the singular values of .(a) Prove that 12 + + 2 = 12 + + 2.(b) Prove that if = and is a positive operator, then \n\n 1, 1 + + , = 1 + + .\n\nSee the comment after Exercise 5 in Section 7A. \n\n12 (a) Give an example of a finite-dimensional vector space and an operator \n\non it such that the singular values of 2 do not equal the squares of the singular values of .(b) Suppose () is normal. Prove that the singular values of 2\n\nequal the squares of the singular values of .\n\n13 Suppose 1, 2 () . Prove that 1 and 2 have the same singular values if and only if there exist unitary operators 1, 2 () such that \n\n 1 = 1 2 2.\n\n14 Suppose ( , ) . Let denote the smallest singular value of . Prove that for every .\n\n15 Suppose () and 1 are the singular values of . Prove that if is an eigenvalue of , then 1 || .\n\n16 Suppose ( , ) . Prove that ( ) = ( ).\n\nCompare the result in this exercise to the analogous result for invertible linear maps [see 7.5 ( f )].\n\n17 Suppose () . Prove that is self-adjoint if and only if is self-adjoint. Matrices unfold Singular values gleam like stars Order in chaos shines \n\nwritten by ChatGPT with input haiku about SVD \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n280 Chapter 7 Operators on Inner Product Spaces \n\n# 7F Consequences of Singular Value Decomposition \n\n# Norms of Linear Maps \n\nThe singular value decomposition leads to the following upper bound for . \n\n> 7.82\n\nupper bound for \n\nSuppose ( , ) . Let 1 be the largest singular value of . Then \n\n 1 \n\nfor all .\n\nFor a lower bound on , look at Exercise 14 in Section 7E. \n\nProof Let 1, , denote the positive singular values of , and let 1, , be an orthonormal list in and 1, , be an orthonormal list in that provide a singular value decomposition of . Thus \n\n> 7.83\n\n = 1 , 1 1 + + , \n\nfor all . Hence if then \n\n 2 = 12 , 1 2 + + 2 , 2\n\n 12( , 1 2 + + , 2) 12 2,\n\nwhere the last inequality follows from Bessels inequality (6.26). Taking square roots of both sides of the inequality above shows that 1 , as desired. Suppose ( , ) and 1 is the largest singular value of . The result above shows that \n\n> 7.84\n\n 1 for all with 1. \n\nTaking = 1 in 7.83 shows that 1 = 1 1. Because 1 = 1 , this implies that \n\n 1 = 1. Thus because 1 = 1 , the inequality in 7.84 leads to the equation \n\n> 7.85\n\nmax { and 1} = 1.\n\nThe equation above is the motivation for the following definition, which defines the norm of to be the left side of the equation above without needing to refer to singular values or the singular value decomposition. \n\n> 7.86 definition:\n\nnorm of a linear map, \n\nSuppose ( , ) . Then the norm of , denoted by , is defined by \n\n = max { and 1}. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7F Consequences of Singular Value Decomposition 281\n\nIn general, the maximum of an infinite set of nonnegative numbers need not exist. However, the discussion before 7.86 shows that the maximum in the definition of the norm of a linear map from to does indeed exist (and equals the largest singular value of ). We now have two different uses of the word norm and the notation . Our first use of this notation was in connection with an inner product on , when we defined = , for each . Our second use of the norm notation and terminology is with the definition we just made of for ( , ) . The norm for ( , ) does not usually come from taking an inner product of with itself (see Exercise 21). You should be able to tell from the context and from the symbols used which meaning of the norm is intended. The properties of the norm on ( , ) listed below look identical to properties of the norm on an inner product space (see 6.9 and 6.17). The inequality in (d) is called the triangle inequality , thus using the same terminology that we used for the norm on . For the reverse triangle inequality, see Exercise 1. \n\n> 7.87\n\nbasic properties of norms of linear maps \n\nSuppose ( , ) . Then (a) 0 ;(b) = 0 = 0 ;(c) = || for all ;(d) + + for all ( , ) .\n\n> Proof\n\n(a) Because 0 for every , the definition of implies that 0 .(b) Suppose = 0 . Thus = 0 for all with 1 . If with \n\n 0 , then = ( ) = 0 ,\n\nwhere the last equality holds because / has norm 1. Because = 0 for all , we have = 0 .Conversely, if = 0 then = 0 for all and hence = 0 .(c) Suppose . Then \n\n = max { and 1} = || max { and 1} = || . \n\n(d) Suppose ( , ) . The definition of + implies that there exists \n\n such that 1 and + = ( + ) . Now \n\n + = ( + ) = + + + ,\n\ncompleting the proof of (d). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 282 Chapter 7 Operators on Inner Product Spaces\n\nFor , ( , ) , the quantity is often called the distance between \n\n and . Informally, think of the condition that is a small number as meaning that and are close together. For example, Exercise 9 asserts that for every () , there is an invertible operator as close to as we wish. \n\n> 7.88\n\nalternative formulas for \n\nSuppose ( , ) . Then (a) = the largest singular value of ;(b) = max { and = 1} ;(c) = the smallest number such that for all .\n\n> Proof\n\n(a) See 7.85. (b) Let be such that 0 < 1 . Let = / . Then \n\n = = 1 and = ( ) = . \n\nThus when finding the maximum of with 1 , we can restrict attention to vectors in with norm 1, proving (b). (c) Suppose and 0 . Then the definition of implies that \n\n( ) ,\n\nwhich implies that \n\n> 7.89\n\n . \n\nNow suppose 0 and for all . This implies that \n\n \n\nfor all with 1 . Taking the maximum of the left side of the inequality above over all with 1 shows that . Thus is the smallest number such that for all .When working with norms of linear maps, you will probably frequently use the inequality 7.89. For computing an approximation of the norm of a linear map given the matrix of with respect to some orthonormal bases, 7.88(a) is likely to be most useful. The matrix of is quickly computable from matrix multiplication. Then a computer can be asked to find an approximation for the largest eigenvalue of (excellent numeric algorithms exist for this purpose). Then taking the square root and using 7.88(a) gives an approximation for the norm of (which usually cannot be computed exactly). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7F Consequences of Singular Value Decomposition 283\n\nYou should verify all assertions in the example below. \n\n> 7.90 example:\n\nnorms \n\n If denotes the usual identity operator on , then = 1 .\n\n If ( ) and the matrix of with respect to the standard basis of \n\nconsists of all 1s, then = .\n\n If () and has an orthonormal basis consisting of eigenvectors of \n\n with corresponding eigenvalues 1, , , then is the maximum of the numbers | 1|, , | |.\n\n Suppose ( 5) is the operator whose matrix (with respect to the stan-dard basis) is the 5-by-5 matrix whose entry in row , column is 1/( 2 + ) .Standard mathematical software shows that the largest singular value of is approximately 0.8 and the smallest singular value of is approximately 10 6 .Thus 0.8 and (using Exercise 10 in Section 7E) 1 10 6. It is not possible to find exact formulas for these norms. A linear map and its adjoint have the same norm, as shown by the next result. \n\n> 7.91\n\nnorm of the adjoint \n\nSuppose ( , ) . Then = . \n\n> Proof\n\nSuppose . Then \n\n 2 = , = , . \n\nThe inequality above implies that \n\n ,\n\nwhich along with 7.88(c) implies that .Replacing with in the inequality and then using the equation \n\n( ) = shows that . Thus = , as desired. You may want to construct an alternative proof of the result above using Exercise 9 in Section 7E, which asserts that a linear map and its adjoint have the same positive singular values. \n\n# Approximation by Linear Maps with Lower-Dimensional Range \n\nThe next result is a spectacular application of the singular value decomposition. It says that to best approximate a linear map by a linear map whose range has dimension at most , chop off the singular value decomposition after the first \n\n terms. Specifically, the linear map in the next result has the property that \n\ndim range = and minimizes the distance to among all linear maps with range of dimension at most . This result leads to algorithms for compressing huge matrices while preserving their most important information. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n284 Chapter 7 Operators on Inner Product Spaces \n\n7.92 best approximation by linear map whose range has dimension \n\nSuppose ( , ) and 1 are the positive singular values of .Suppose 1 < . Then \n\nmin { ( , ) and dim range } = + 1.\n\nFurthermore, if \n\n = 1 , 1 1 + + , \n\nis a singular value decomposition of and ( , ) is defined by \n\n = 1 , 1 1 + + , \n\nfor each , then dim range = and = + 1.\n\nProof If then \n\n( ) 2 = + 1 , + 1 + 1 + + , 2\n\n= + 12 , + 1 2 + + 2 , 2\n\n + 12( , + 1 2 + + , 2) + 12 2.\n\nThus + 1. The equation ( ) + 1 = + 1 + 1 now shows that \n\n = + 1.Suppose ( , ) and dim range . Thus 1, , + 1, which is a list of length + 1, is linearly dependent. Hence there exist 1, , + 1 , not all 0, such that \n\n 1 1 + + + 1 + 1 = 0. \n\nNow 1 1 + + + 1 + 1 0 because 1, , + 1 are not all 0. We have \n\n( )( 1 1 + + + 1 + 1) 2 = ( 1 1 + + + 1 + 1) 2\n\n= 1 1 1 + + + 1 + 1 + 12\n\n= 12 | 1|2 + + + 12 | + 1|2\n\n + 12(| 1|2 + + | + 1|2)= + 12 1 1 + + + 1 + 12.\n\nBecause 1 1 + + + 1 + 1 0 , the inequality above implies that \n\n + 1.\n\nThus = minimizes among ( , ) with dim range .For other examples of the use of the singular value decomposition in best approximation, see Exercise 22, which finds a subspace of given dimension on which the restriction of a linear map is as small as possible, and Exercise 27, which finds a unitary operator that is as close as possible to a given operator. \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler Section 7F Consequences of Singular Value Decomposition 285 \n\n# Polar Decomposition \n\nRecall our discussion before 7.54 of the analogy between complex numbers \n\nwith || = 1 and unitary operators. Continuing with this analogy, note that every complex number except 0 can be written in the form \n\n> = ( || )|| = ( || )\n\n,\n\nwhere the first factor, namely, /|| , has absolute value 1.Our analogy leads us to guess that every operator () can be written as a unitary operator times . That guess is indeed correct. The corresponding result is called the polar decomposition, which gives a beautiful description of an arbitrary operator on .Note that if () , then is a positive operator [as was shown in 7.64(a) ]. Thus the operator makes sense and is well defined as a positive operator on .The polar decomposition that we are about to state and prove says that every operator on is the product of a unitary operator and a positive operator. Thus we can write an arbitrary operator on as the product of two nice operators, each of which comes from a class that we can completely describe and that we understand reasonably well. The unitary operators are described by 7.55 if = ;the positive operators are described by the real and complex spectral theorems (7.29 and 7.31). Specifically, consider the case = , and suppose \n\n> = \n\nis a polar decomposition of an operator () , where is a unitary operator. Then there is an orthonormal basis of with respect to which has a diagonal matrix, and there is an orthonormal basis of with respect to which has a diagonal matrix. Warning: There may not exist an orthonormal basis that simultaneously puts the matrices of both and into these nice diagonal forms may require one orthonormal basis and may require a different orthonormal basis. However (still assuming that = ), if is normal, then an orthonormal basis of can be chosen such that both and have diagonal matrices with respect to this basissee Exercise 31. The converse is also true: If () \n\nand = for some unitary operator () such that and both have diagonal matrices with respect to the same orthonormal basis of , then \n\nis normal. This holds because then has a diagonal matrix with respect to this same orthonormal basis, which implies that is normal [by the equivalence of (c) and (a) in 7.31 ].\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n286 Chapter 7 Operators on Inner Product Spaces \n\nThe polar decomposition below is valid on both real and complex inner product spaces and for all operators on those spaces. \n\n> 7.93\n\npolar decomposition \n\nSuppose () . Then there exists a unitary operator () such that \n\n = . \n\nProof Let 1, , be the positive singular values of , and let 1, , and \n\n 1, , be orthonormal lists in such that \n\n> 7.94\n\n = 1 , 1 1 + + , \n\nfor every . Extend 1, , and 1, , to orthonormal bases 1, , \n\nand 1, , of .Define () by \n\n = , 1 1 + + , \n\nfor each . Then \n\n 2 = , 1 1 + + , 2\n\n= , 1 2 + + , 2\n\n= 2.\n\nThus is a unitary operator. Applying to both sides of 7.94 and then using the formula for given by 7.77 shows that \n\n = 12 , 1 1 + + 2 , \n\nfor every . Thus if , then \n\n> \n\n = 1 , 1 1 + + , \n\nbecause the operator that sends to the right side of the equation above is a positive operator whose square equals . Now \n\n = ( 1 , 1 1 + + , )= 1 , 1 1 + + , \n\n= ,\n\nwhere the last equation follows from 7.94. Exercise 27 shows that the unitary operator produced in the proof above is as close as a unitary operator can be to .Alternative proofs of the polar decomposition directly use the spectral theorem, avoiding the singular value decomposition. However, the proof above seems cleaner than those alternative proofs. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7F Consequences of Singular Value Decomposition 287 \n\n# Operators Applied to Ellipsoids and Parallelepipeds \n\n> 7.95 definition:\n\nball, \n\nThe ball in of radius 1 centered at 0, denoted by , is defined by \n\n = { < 1}. \n\nThe ball in 2.\n\nIf dim = 2 , the word disk is sometimes used instead of \n\nball . However, using ball in all dimensions is less confusing. Similarly, if dim = 2 , then the word ellipse is sometimes used instead of the word ellipsoid that we are about to define. Again, using ellipsoid in all dimensions is less confusing. You can think of the ellipsoid defined below as obtained by starting with the ball and then stretching by a factor of \n\n along each -axis. \n\n> 7.96 definition:\n\nellipsoid, ( 1 1, , ), principal axes \n\nSuppose that 1, , is an orthonormal basis of and 1, , are positive numbers. The ellipsoid ( 1 1, , ) with principal axes 1 1, , is defined by \n\n( 1 1, , ) = { | , 1| 2\n\n 12 + + | , | 2\n\n 2 < 1}. \n\nThe ellipsoid notation ( 1 1, , ) does not explicitly include the inner product space , even though the definition above depends on . However, the in-ner product space should be clear from the context and also from the requirement that 1, , be an orthonormal basis of . \n\n> 7.97 example:\n\nellipsoids \n\nThe ellipsoid (2 1, 2) in 2, where \n\n 1, 2 is the standard basis of 2.\n\nThe ellipsoid (2 1, 2) in 2, where \n\n 1 = ( 1 \n\n> 2\n\n, 1 \n\n> 2\n\n) and 2 = ( 1 \n\n> 2\n\n, 1 \n\n> 2\n\n).\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n288 Chapter 7 Operators on Inner Product Spaces \n\nThe ellipsoid \n\n(4 1, 3 2, 2 3) in 3,where 1, 2, 3 is the standard basis of 3.\n\nThe ellipsoid ( 1, , ) equals the ball in for every orthonormal basis \n\n 1, , of [ by Parsevals identity 6.30(b) ]. \n\n> 7.98 notation:\n\n() \n\nFor a function defined on and , define () by \n\n() = { }. \n\nThus if is a function defined on , then () = range .The next result states that every invertible operator () maps the ball \n\n in onto an ellipsoid in . The proof shows that the principal axes of this ellipsoid come from the singular value decomposition of . \n\n> 7.99\n\ninvertible operator takes ball to ellipsoid \n\nSuppose () is invertible. Then maps the ball in onto an ellipsoid in .\n\nProof Suppose has singular value decomposition \n\n> 7.100\n\n = 1 , 1 1 + + , \n\nfor all ; here 1, , are the singular values of and 1, , and 1, , \n\nare both orthonormal bases of . We will show that () = ( 1 1, , ).First suppose . Because is invertible, none of the singular values \n\n 1, , equals 0 (see 7.68). Thus 7.100 implies that \n\n , 1 2\n\n 12 + + , 2\n\n 2 = | , 1| 2 + + | , | 2 < 1. \n\nThus ( 1 1, , ). Hence () ( 1 1, , ).To prove inclusion in the other direction, now suppose ( 1 1, , ).Let \n\n = , 1 1\n\n 1 + + , \n\n .\n\nThen < 1 and 7.100 implies that = , 1 1 + + , = . Thus \n\n() ( 1 1, , ).\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7F Consequences of Singular Value Decomposition 289 \n\nWe now use the previous result to show that invertible operators take all ellipsoids, not just the ball of radius 1, to ellipsoids. \n\n> 7.101\n\ninvertible operator takes ellipsoids to ellipsoids \n\nSuppose () is invertible and is an ellipsoid in . Then () is an ellipsoid in .\n\nProof There exist an orthonormal basis 1, , of and positive numbers \n\n 1, , such that = ( 1 1, , ). Define () by \n\n( 1 1 + + ) = 1 1 1 + + .\n\nThen maps the ball of onto , as you can verify. Thus \n\n() = (()) = ()(). \n\nThe equation above and 7.99, applied to , show that () is an ellipsoid in .Recall (see 3.95) that if and then + is defined by \n\n + = { + }. \n\nGeometrically, the sets and + look the same, but they are in different locations. In the following definition, if dim = 2 then the word parallelogram is often used instead of parallelepiped . \n\n> 7.102 definition:\n\n( 1, , ), parallelepiped \n\nSuppose 1, , is a basis of . Let \n\n( 1, , ) = { 1 1 + + 1, , (0 , 1)}. \n\nA parallelepiped is a set of the form + ( 1, , ) for some . The vectors 1, , are called the edges of this parallelepiped. \n\n> 7.103 example:\n\nparallelepipeds \n\nThe parallelepiped \n\n> (0.3\n\n, 0.5) + ((1 , 0) , (1 , 1)) in 2. A parallelepiped in 3.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n290 Chapter 7 Operators on Inner Product Spaces \n\n> 7.104\n\ninvertible operator takes parallelepipeds to parallelepipeds \n\nSuppose , 1, , is a basis of , and () is invertible. Then \n\n( + ( 1, , )) = + ( 1, , ). \n\nProof Because is invertible, the list 1, , is a basis of . The linearity of implies that \n\n( + 1 1 + + ) = + 1 1 + + \n\nfor all 1, , (0 , 1) . Thus ( + ( 1, , )) = + ( 1, , ).Just as the rectangles are distinguished among the parallelograms in 2, we give a special name to the parallelepipeds in whose defining edges are orthogonal to each other. \n\n> 7.105 definition:\n\nbox \n\nA box in is a set of the form \n\n + ( 1 1, , ),\n\nwhere and 1, , are positive numbers and 1, , is an ortho-normal basis of .Note that in the special case of 2 each box is a rectangle, but the terminology \n\nbox can be used in all dimensions. \n\n> 7.106 example:\n\nboxes \n\nThe box (1 , 0) + ( 2 1, 2 2), where \n\n 1 = ( 1 \n\n> 2\n\n, 1 \n\n> 2\n\n) and 2 = ( 1 \n\n> 2\n\n, 1 \n\n> 2\n\n).\n\nThe box ( 1, 2 2, 3), where 1, 2, 3\n\nis the standard basis of 3.\n\nSuppose () is invertible. Then maps every parallelepiped in \n\nto a parallelepiped in (by 7.104). In particular, maps every box in to a parallelepiped in . This raises the question of whether maps some boxes in \n\n to boxes in . The following result answers this question, with the help of the singular value decomposition. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7F Consequences of Singular Value Decomposition 291 \n\n> 7.107\n\nevery invertible operator takes some boxes to boxes \n\nSuppose () is invertible. Suppose has singular value decomposition \n\n = 1 , 1 1 + + , ,\n\nwhere 1, , are the singular values of and 1, , and 1, , are orthonormal bases of and the equation above holds for all . Then \n\nmaps the box + ( 1 1, , ) onto the box + ( 1 1 1, , ) for all positive numbers 1, , and all .\n\nProof If 1, , (0 , 1) and 1, , are positive numbers and , then \n\n( + 1 1 1 + + ) = + 1 1 1 1 + + .\n\nThus ( + ( 1 1, , )) = + ( 1 1 1, , ).\n\n# Volume via Singular Values \n\nOur goal in this subsection is to understand how an operator changes the volume of subsets of its domain. Because notions of volume belong to analysis rather than to linear algebra, we will work only with an intuitive notion of volume. Our intuitive approach to volume can be converted into appropriate correct definitions, correct statements, and correct proofs using the machinery of analysis. Our intuition about volume works best in real inner product spaces. Thus the assumption that = will appear frequently in the rest of this subsection. If dim = , then by volume we will mean -dimensional volume. You should be familiar with this concept in 3. When = 2 , this is usually called area instead of volume, but for consistency we use the word volume in all dimensions. The most fundamental intuition about volume is that the volume of a box (whose defining edges are by definition orthogonal to each other) is the product of the lengths of the defining edges. Thus we make the following definition. \n\n> 7.108 definition:\n\nvolume of a box \n\nSuppose = . If and 1, , are positive numbers and 1, , is an orthonormal basis of , then \n\nvolume ( + ( 1 1, , )) = 1 .\n\nThe definition above agrees with the familiar formulas for the area (which we are calling the volume) of a rectangle in 2 and for the volume of a box in 3. For example, the first box in Example 7.106 has two-dimensional volume (or area) 2\n\nbecause the defining edges of that box have length 2 and 2. The second box in Example 7.106 has three-dimensional volume 2 because the defining edges of that box have length 1, 2, and 1. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 292 Chapter 7 Operators on Inner Product Spaces\n> Volume of this ball sum of the volumes of the five boxes.\n\nTo define the volume of a subset of , approximate the subset by a finite collection of disjoint boxes, and then add up the volumes of the approximating collection of boxes. As we approximate a subset of more accurately by disjoint unions of more boxes, we get a better approximation to the volume. These ideas should remind you of how the Riemann integral is defined by approximating the area under a curve by a disjoint collection of rectangles. This discussion leads to the following nonrigorous but intuitive definition. \n\n> 7.109 definition:\n\nvolume \n\nSuppose = and . Then the volume of , denoted by volume , is approximately the sum of the volumes of a collection of disjoint boxes that approximate .We are ignoring many reasonable questions by taking an intuitive approach to volume. For example, if we approximate by boxes with respect to one basis, do we get the same volume if we approximate by boxes with respect to a different basis? If 1 and 2 are disjoint subsets of , is volume ( 1 2) = \n\nvolume 1 + volume 2? Provided that we consider only reasonably nice subsets of , techniques of analysis show that both these questions have affirmative answers that agree with our intuition about volume. \n\n> 7.110 example:\n\nvolume change by a linear map \n\n> Each box here has twice the width and the same height as the boxes in the previous figure.\n\nSuppose that ( 2) is defined by \n\n = 2 , 1 1 + , 2 2, where 1, 2 is the standard basis of 2. This linear map stretches vectors along the 1-axis by a factor of 2 and leaves vectors along the 2-axis unchanged. The ball approximated by five boxes above gets mapped by to the ellipsoid shown here. Each of the five boxes in the original figure gets mapped to a box of twice the width and the same height as in the original figure. Hence each box gets mapped to a box of twice the volume (area) as in the original figure. The sum of the volumes of the five new boxes approximates the volume of the ellipsoid. Thus changes the volume of the ball by a factor of 2.In the example above, maps boxes with respect to the basis 1, 2 to boxes with respect to the same basis; thus we can see how changes volume. In general, an operator maps boxes to parallelepipeds that are not boxes. However, if we choose the right basis (coming from the singular value decomposition!), then boxes with respect to that basis get mapped to boxes with respect to a possibly different basis, as shown in 7.107. This observation leads to a natural proof of the following result. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 7F Consequences of Singular Value Decomposition 293\n> 7.111\n\nvolume changes by a factor of the product of the singular values \n\nSuppose = , () is invertible, and . Then \n\nvolume () = ( product of singular values of )( volume ). \n\n> Proof\n\nSuppose has singular value decomposition \n\n = 1 , 1 1 + + , \n\nfor all , where 1, , and 1, , are orthonormal bases of .Approximate by boxes of the form + ( 1 1, , ), which have volume \n\n 1 . The operator maps each box + ( 1 1, , ) onto the box \n\n + ( 1 1 1, , ), which has volume ( 1 )( 1 ).The operator maps a collection of boxes that approximate onto a collection of boxes that approximate () . Because changes the volume of each box in a collection that approximates by a factor of 1 , the linear map changes the volume of by the same factor. Suppose () . As we will see when we get to determinants, the product of the singular values of equals |det | ; see 9.60 and 9.61. \n\n# Properties of an Operator as Determined by Its Eigenvalues \n\nWe conclude this chapter by presenting the table below. The context of this table is a finite-dimensional complex inner product space. The first column of the table shows a property that a normal operator on such a space might have. The second column of the table shows a subset of such that the operator has the corresponding property if and only if all eigenvalues of the operator lie in the specified subset. For example, the first row of the table states that a normal operator is invertible if and only if all its eigenvalues are nonzero (this first row is the only one in the table that does not need the hypothesis that the operator is normal). Make sure you can explain why all results in the table hold. For example, the last row of the table holds because the norm of an operator equals its largest singular value (by 7.85) and the singular values of a normal operator, assuming \n\n = , equal the absolute values of the eigenvalues (by Exercise 7 in Section 7E). \n\nproperties of a normal operator eigenvalues are contained in \n\ninvertible \\{0} \n\nself-adjoint \n\nskew { Re = 0} \n\northogonal projection {0 , 1} \n\npositive [0 , ) \n\nunitary { || = 1} \n\nnorm is less than 1 { || < 1} \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 294 Chapter 7 Operators on Inner Product Spaces\n\n# Exercises 7F \n\n1 Prove that if , ( , ) , then . \n\n> The inequality above is called the reverse triangle inequality .\n\n2 Suppose that () is self-adjoint or that = and () is normal. Prove that \n\n = max {|| is an eigenvalue of }. \n\n3 Suppose ( , ) and . Prove that \n\n = = 2 . \n\n4 Suppose ( , ) , , and = . Prove that if and \n\n , = 0 , then , = 0 .\n\n5 Suppose is a finite-dimensional inner product space, ( , ) , and \n\n ( , ) . Prove that \n\n . \n\n6 Prove or give a counterexample: If , () , then = .\n\n7 Show that defining ( , ) = for , ( , ) makes a metric on ( , ) .\n\n> This exercise is intended for readers who are familiar with metric spaces.\n\n8 (a) Prove that if () and < 1 , then is invertible. (b) Suppose that () is invertible. Prove that if () and \n\n < 1/ 1 , then is invertible. \n\n> This exercise shows that the set of invertible operators in () is an open subset of () , using the metric defined in Exercise 7.\n\n9 Suppose () . Prove that for every > 0 , there exists an invertible operator () such that 0 < < .\n\n10 Suppose dim > 1 and () is not invertible. Prove that for every \n\n > 0 , there exists () such that 0 < < and is not invertible. \n\n11 Suppose = and () . Prove that for every > 0 there exists a diagonalizable operator () such that 0 < < .\n\n12 Suppose () is a positive operator. Show that = .\n\n13 Suppose , () are positive operators. Show that \n\n max { , } + . \n\n14 Suppose and are subspaces of such that < 1 . Prove that \n\ndim = dim .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 7F Consequences of Singular Value Decomposition 295 \n\n15 Define ( 3) by \n\n( 1, 2, 3) = ( 3, 2 1, 3 2). \n\nFind (explicitly) a unitary operator ( 3) such that = .\n\n16 Suppose () is a positive invertible operator. Prove that there exists \n\n > 0 such that is a positive operator for every self-adjoint operator \n\n () with < .\n\n17 Prove that if and is the linear functional on defined by the equation () = , , then = .\n\nHere we are thinking of the scalar field as an inner product space with \n\n> \n\n, = for all , . Thus means the norm of as a linear map from to .\n\n18 Suppose 1, , is an orthonormal basis of and ( , ) .(a) Prove that max { 1, , } ( 12 + + 2)1/2 .(b) Prove that = ( 12 ++ 2)1/2 if and only if dim range 1 .\n\nHere 1, , is an arbitrary orthonormal basis of , not necessarily connected with a singular value decomposition of . If 1, , is the list of singular values of , then the right side of the inequality above equals \n\n> ( 12\n\n+ + 2)1/2 , as was shown in Exercise 11 (a) in Section 7E. \n\n19 Prove that if ( , ) , then = 2.\n\nThis formula for leads to the important subject of -algebras. \n\n20 Suppose () is normal. Prove that = for every positive integer .\n\n21 Suppose dim > 1 and dim > 1 . Prove that the norm on ( , ) does not come from an inner product. In other words, prove that there does not exist an inner product on ( , ) such that \n\nmax { and 1} = , \n\nfor all ( , ) .\n\n22 Suppose ( , ) . Let = dim and let 1 denote the singular values of . Prove that if 1 , then \n\nmin {| is a subspace of with dim = } = + 1.\n\n23 Suppose ( , ) . Show that is uniformly continuous with respect to the metrics on and that arise from the norms on those spaces (see Exercise 23 in Section 6B). \n\n24 Suppose () is invertible. Prove that \n\n 1 = 1 is a unitary operator .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n296 Chapter 7 Operators on Inner Product Spaces \n\n25 Fix , with 0 . Define () by = , for every .Prove that \n\n> \n\n = , \n\nfor every .\n\n26 Suppose () . Prove that is invertible if and only if there exists a unique unitary operator () such that = .\n\n27 Suppose () and 1, , are the singular values of . Let 1, , \n\nand 1, , be orthonormal bases of such that \n\n = 1 , 1 1 + + , \n\nfor all . Define () by \n\n = , 1 1 + + , .\n\n(a) Show that is unitary and = max {| 1 1| , , | 1|} .(b) Show that if () is unitary, then .\n\nThis exercise finds a unitary operator that is as close as possible (among the unitary operators ) to a given operator .\n\n28 Suppose () . Prove that there exists a unitary operator () \n\nsuch that = .\n\n29 Suppose () .(a) Use the polar decomposition to show that there exists a unitary operator \n\n () such that = .(b) Show how (a) implies that and have the same singular values. \n\n30 Suppose () , () is a unitary operator, and () is a positive operator such that = . Prove that = .\n\nThis exercise shows that if we write as the product of a unitary operator and a positive operator (as in the polar decomposition 7.93 ), then the positive operator equals .\n\n31 Suppose = and () is normal. Prove that there exists a unitary operator () such that = and such that and both have diagonal matrices with respect to the same orthonormal basis of .\n\n32 Suppose that ( , ) and 0 . Let 1, , denote the positive singular values of . Show that there exists an orthonormal basis 1, , \n\nof (null ) such that \n\n(( 1\n\n 1\n\n, , \n\n \n\n)) \n\nequals the ball in range of radius 1 centered at 0.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n# Chapter 8 \n\n# Operators on Complex Vector Spaces \n\nIn this chapter we delve deeper into the structure of operators, with most of the attention on complex vector spaces. Some of the results in this chapter apply to both real and complex vector spaces; thus we do not make a standing assumption that = . Also, an inner product does not help with this material, so we return to the general setting of a finite-dimensional vector space. Even on a finite-dimensional complex vector space, an operator may not have enough eigenvectors to form a basis of the vector space. Thus we will consider the closely related objects called generalized eigenvectors. We will see that for each operator on a finite-dimensional complex vector space, there is a basis of the vector space consisting of generalized eigenvectors of the operator. The generalized eigenspace decomposition then provides a good description of arbitrary operators on a finite-dimensional complex vector space. Nilpotent operators, which are operators that when raised to some power equal 0, have an important role in these investigations. Nilpotent operators provide a key tool in our proof that every invertible operator on a finite-dimensional complex vector space has a square root and in our approach to Jordan form. This chapter concludes by defining the trace and proving its key properties. \n\nstanding assumptions for this chapter \n\n denotes or .\n\n denotes a finite-dimensional nonzero vector space over .\n\n> David Iliff CC BY-SA\n\nThe Long Room of the Old Library at the University of Dublin, where William Hamilton \n\n(18051865 ) was a student and then a faculty member. Hamilton proved a special case of what we now call the CayleyHamilton theorem in 1853. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n> 297\n\n298 Chapter 8 Operators on Complex Vector Spaces \n\n# 8A Generalized Eigenvectors and Nilpotent Operators \n\n# Null Spaces of Powers of an Operator \n\nWe begin this chapter with a study of null spaces of powers of an operator. \n\n> 8.1\n\nsequence of increasing null spaces \n\nSuppose () . Then \n\n{0} = null 0 null 1 null null + 1 . \n\nProof Suppose is a nonnegative integer and null . Then = 0 ,which implies that + 1 = ( ) = (0) = 0 . Thus null + 1. Hence \n\nnull null + 1, as desired. \n\nFor similar results about decreasing sequences of ranges, see Exercises 6, 7, and 8. \n\nThe following result states that if two consecutive terms in the sequence of sub-spaces above are equal, then all later terms in the sequence are equal. \n\n> 8.2\n\nequality in the sequence of null spaces \n\nSuppose () and is a nonnegative integer such that \n\nnull = null + 1.\n\nThen \n\nnull = null + 1 = null + 2 = null + 3 = . \n\nProof Let be a positive integer. We want to prove that \n\nnull + = null + + 1.\n\nWe already know from 8.1 that null + null + + 1.To prove the inclusion in the other direction, suppose null + + 1. Then \n\n + 1( ) = + + 1 = 0. \n\nHence \n\n null + 1 = null .\n\nThus + = ( ) = 0 , which means that null + . This implies that \n\nnull + + 1 null + , completing the proof. The result above raises the question of whether there exists a nonnegative integer such that null = null + 1. The next result shows that this equality holds at least when equals the dimension of the vector space on which \n\noperates. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8A Generalized Eigenvectors and Nilpotent Operators 299 \n\n> 8.3\n\nnull spaces stop growing \n\nSuppose () . Then \n\nnull dim = null dim + 1 = null dim + 2 = . \n\nProof We only need to prove that null dim = null dim + 1 (by 8.2). Suppose this is not true. Then, by 8.1 and 8.2, we have \n\n{0} = null 0 null 1 null dim null dim + 1,\n\nwhere the symbol means contained in but not equal to. At each of the strict inclusions in the chain above, the dimension increases by at least 1. Thus \n\ndim null dim + 1 dim + 1, a contradiction because a subspace of cannot have a larger dimension than dim .It is not true that = null range for every () . However, the next result can be a useful substitute. \n\n> 8.4\n\n is the direct sum of null dim and range dim \n\nSuppose () . Then \n\n = null dim range dim .\n\nProof Let = dim . First we show that \n\n> 8.5\n\n(null ) ( range ) = {0}. \n\nSuppose ( null ) ( range ). Then = 0 , and there exists \n\nsuch that = . Applying to both sides of the last equation shows that \n\n = 2 . Hence 2 = 0 , which implies that = 0 (by 8.3). Thus \n\n = = 0 , completing the proof of 8.5. Now 8.5 implies that null + range is a direct sum (by 1.46). Also, \n\ndim (null range ) = dim null + dim range = dim ,\n\nwhere the first equality above comes from 3.94 and the second equality comes from the fundamental theorem of linear maps (3.21). The equation above implies that null range = (see 2.39), as desired. For an improvement of the result above, see Exercise 19. \n\n> 8.6 example:\n\n 3 = null 3 range 3 for ( 3)\n\nSuppose ( 3) is defined by \n\n( 1, 2, 3) = (4 2, 0, 5 3). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 300 Chapter 8 Operators on Complex Vector Spaces\n\nThen null = {( 1, 0, 0) 1 } and range = {( 1, 0, 3) 1, 3 } . Thus \n\nnull range {0} . Hence null + range is not a direct sum. Also note that \n\nnull + range 3. However, we have 3( 1, 2, 3) = (0 , 0, 125 3). Thus we see that \n\nnull 3 = {( 1, 2, 0) 1, 2 } and range 3 = {(0 , 0, 3) 3 }. \n\nHence 3 = null 3 range 3, as expected by 8.4. \n\n# Generalized Eigenvectors \n\nSome operators do not have enough eigenvectors to lead to good descriptions of their behavior. Thus in this subsection we introduce the concept of generalized eigenvectors, which will play a major role in our description of the structure of an operator. To understand why we need more than eigenvectors, lets examine the question of describing an operator by decomposing its domain into invariant subspaces. Fix \n\n () . We seek to describe by finding a nice direct sum decomposition \n\n = 1 ,\n\nwhere each is a subspace of invariant under . The simplest possible nonzero invariant subspaces are one-dimensional. A decomposition as above in which each is a one-dimensional subspace of invariant under is possible if and only if has a basis consisting of eigenvectors of (see 5.55). This happens if and only if has an eigenspace decomposition \n\n> 8.7\n\n = ( 1, ) ( , ) ,\n\nwhere 1, , are the distinct eigenvalues of (see 5.55). The spectral theorem in the previous chapter shows that if is an inner product space, then a decomposition of the form 8.7 holds for every self-adjoint operator if = and for every normal operator if = because operators of those types have enough eigenvectors to form a basis of (see 7.29 and 7.31). However, a decomposition of the form 8.7 may not hold for more general operators, even on a complex vector space. An example was given by the operator in 5.57, which does not have enough eigenvectors for 8.7 to hold. Generalized eigenvectors and generalized eigenspaces, which we now introduce, will remedy this situation. \n\n> 8.8 definition:\n\ngeneralized eigenvector \n\nSuppose () and is an eigenvalue of . A vector is called a \n\ngeneralized eigenvector of corresponding to if 0 and \n\n( ) = 0 \n\nfor some positive integer . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 8A Generalized Eigenvectors and Nilpotent Operators 301\n> Generalized eigenvalues are not de-fined because doing so would not lead to anything new. Reason: if ( ) \n> is not injective for some positive inte-ger , then is not injective, and hence is an eigenvalue of .\n\nA nonzero vector is a general-ized eigenvector of corresponding to \n\nif and only if \n\n( ) dim = 0 ,\n\nas follows from applying 8.1 and 8.3 to the operator .As we know, an operator on a complex vector space may not have enough eigenvectors to form a basis of the domain. The next result shows that on a complex vector space there are enough generalized eigenvectors to do this. \n\n> 8.9\n\na basis of generalized eigenvectors \n\nSuppose = and () . Then there is a basis of consisting of generalized eigenvectors of . \n\n> Proof\n\nLet = dim . We will use induction on . To get started, note that the desired result holds if = 1 because then every nonzero vector in is an eigenvector of . \n\n> This step is where we use the hypothesis that = , because if = then \n> may not have any eigenvalues.\n\nNow suppose > 1 and the de-sired result holds for all smaller values of dim . Let be an eigenvalue of .Applying 8.4 to shows that \n\n = null ( ) range ( ) .\n\nIf null ( ) = , then every nonzero vector in is a generalized eigen-vector of , and thus in this case there is a basis of consisting of generalized eigenvectors of . Hence we can assume that null ( ) , which implies that range ( ) {0} .Also, null ( ) {0} , because is an eigenvalue of . Thus we have \n\n0 < dim range ( ) < . \n\nFurthermore, range ( ) is invariant under [ by 5.18 with () = ( ) ].Let (range ( ) ) equal restricted to range ( ) . Our induction hypothesis applied to the operator implies that there is a basis of range ( ) \n\nconsisting of generalized eigenvectors of , which of course are generalized eigenvectors of . Adjoining that basis of range () to a basis of null () \n\ngives a basis of consisting of generalized eigenvectors of .If = and dim > 1 , then some operators on have the property that there exists a basis of consisting of generalized eigenvectors of the operator, and (unlike what happens when = ) other operators do not have this property. See Exercise 11 for a necessary and sufficient condition that determines whether an operator has this property. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n302 Chapter 8 Operators on Complex Vector Spaces \n\n> 8.10 example:\n\ngeneralized eigenvectors of an operator on 3\n\nDefine ( 3) by \n\n( 1, 2, 3) = (4 2, 0, 5 3)\n\nfor each ( 1, 2, 3) 3. A routine use of the definition of eigenvalue shows that the eigenvalues of are 0 and 5. Furthermore, the eigenvectors corresponding to the eigenvalue 0 are the nonzero vectors of the form ( 1, 0, 0) , and the eigenvectors corresponding to the eigenvalue 5 are the nonzero vectors of the form (0 , 0, 3).Hence this operator does not have enough eigenvectors to span its domain 3.We compute that 3( 1, 2, 3) = (0 , 0, 125 3). Thus 8.1 and 8.3 imply that the generalized eigenvectors of corresponding to the eigenvalue 0 are the nonzero vectors of the form ( 1, 2, 0) .We also have ( 5) 3( 1, 2, 3) = (125 1 + 300 2, 125 2, 0) . Thus the generalized eigenvectors of corresponding to the eigenvalue 5 are the nonzero vectors of the form (0 , 0, 3).The paragraphs above show that each of the standard basis vectors of 3 is a generalized eigenvector of . Thus 3 indeed has a basis consisting of generalized eigenvectors of , as promised by 8.9. If is an eigenvector of () , then the corresponding eigenvalue is uniquely determined by the equation = , which can be satisfied by only one \n\n (because 0 ). However, if is a generalized eigenvector of , then it is not obvious that the equation ( ) dim = 0 can be satisfied by only one \n\n . Fortunately, the next result tells us that all is well on this issue. \n\n> 8.11\n\ngeneralized eigenvector corresponds to a unique eigenvalue \n\nSuppose () . Then each generalized eigenvector of corresponds to only one eigenvalue of .\n\nProof Suppose is a generalized eigenvector of corresponding to eigen-values and of . Let be the smallest positive integer such that () = 0 .Let = dim . Then \n\n0 = ( ) = (( ) + ( )) =\n\n> \n> \n> = 0\n\n ( ) ( ) ,\n\nwhere 0 = 1 and the values of the other binomial coefficients do not matter. Apply the operator ( ) 1 to both sides of the equation above, getting \n\n0 = ( ) ( ) 1 . \n\nBecause ( ) 1 0 , the equation above implies that = , as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8A Generalized Eigenvectors and Nilpotent Operators 303 \n\nWe saw earlier (5.11) that eigenvectors corresponding to distinct eigenvalues are linearly independent. Now we prove a similar result for generalized eigen-vectors, with a proof that roughly follows the pattern of the proof of that earlier result. \n\n> 8.12\n\nlinearly independent generalized eigenvectors \n\nSuppose that () . Then every list of generalized eigenvectors of \n\ncorresponding to distinct eigenvalues of is linearly independent. \n\nProof Suppose the desired result is false. Then there exists a smallest positive integer such that there exists a linearly dependent list 1, , of generalized eigenvectors of corresponding to distinct eigenvalues 1, , of (note that \n\n 2 because a generalized eigenvector is, by definition, nonzero). Thus there exist 1, , , none of which are 0 (because of the minimality of ), such that \n\n 1 1 + + = 0. \n\nLet = dim . Apply ( ) to both sides of the equation above, getting \n\n> 8.13\n\n 1( ) 1 + + 1 ( ) 1 = 0. \n\nSuppose {1 , , 1} . Then \n\n( ) 0 \n\nbecause otherwise would be a generalized eigenvector of corresponding to the distinct eigenvalues and , which would contradict 8.11. However, \n\n( ) (( ) ) = ( ) (( ) ) = 0. \n\nThus the last two displayed equations show that ( ) is a generalized eigenvector of corresponding to the eigenvalue . Hence \n\n( ) 1, , ( ) 1 \n\nis a linearly dependent list (by 8.13) of 1 generalized eigenvectors correspond-ing to distinct eigenvalues, contradicting the minimality of . This contradiction completes the proof. \n\n# Nilpotent Operators \n\n> 8.14 definition:\n\nnilpotent \n\nAn operator is called nilpotent if some power of it equals 0.Thus an operator () is nilpotent if and only if every nonzero vector in \n\n is a generalized eigenvector of corresponding to the eigenvalue 0.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n304 Chapter 8 Operators on Complex Vector Spaces \n\n> 8.15 example:\n\nnilpotent operators \n\n(a) The operator ( 4) defined by \n\n( 1, 2, 3, 4) = (0 , 0, 1, 2)\n\nis nilpotent because 2 = 0 .(b) The operator on 3 whose matrix (with respect to the standard basis) is \n\n3 9 07 9 64 0 6 \n\nis nilpotent, as can be shown by cubing the matrix above to get the zero matrix. (c) The operator of differentiation on () is nilpotent because the ( + 1) th \n\nderivative of every polynomial of degree at most equals 0. Note that on this space of dimension + 1, we need to raise the nilpotent operator to the power + 1 to get the 0 operator. \n\nThe Latin word nil means nothing or zero; the Latin word potens means having power. Thus nilpotent literally means having a power that is zero. \n\nThe next result shows that when rais-ing a nilpotent operator to a power, we never need to use a power higher than the dimension of the space. For a slightly stronger result, see Exercise 18. \n\n> 8.16\n\nnilpotent operator raised to dimension of domain is 0\n\nSuppose () is nilpotent. Then dim = 0 .\n\nProof Because is nilpotent, there exists a positive integer such that = 0 .Thus null = . Now 8.1 and 8.3 imply that null dim = . Thus dim = 0 . \n\n> 8.17\n\neigenvalues of nilpotent operator \n\nSuppose () .(a) If is nilpotent, then 0 is an eigenvalue of and has no other eigenvalues. (b) If = and 0 is the only eigenvalue of , then is nilpotent. \n\nProof \n\n(a) To prove (a), suppose is nilpotent. Hence there is a positive integer such that = 0 . This implies that is not injective. Thus 0 is an eigenvalue of . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 8A Generalized Eigenvectors and Nilpotent Operators 305\n\nTo show that has no other eigenvalues, suppose is an eigenvalue of .Then there exists a nonzero vector such that \n\n = . \n\nRepeatedly applying to both sides of this equation shows that \n\n = = 0. \n\nThus = 0 , as desired. (b) Suppose = and 0 is the only eigenvalue of . By 5.27(b), the minimal polynomial of equals for some positive integer . Thus = 0 . Hence \n\n is nilpotent. Exercise 23 shows that the hypothesis that = cannot be deleted in (b) of the result above. Given an operator on , we want to find a basis of such that the matrix of the operator with respect to this basis is as simple as possible, meaning that the matrix contains many 0s. The next result shows that if is nilpotent, then we can choose a basis of such that the matrix of with respect to this basis has more than half of its entries equal to 0. Later in this chapter we will do even better. \n\n> 8.18\n\nminimal polynomial and upper-triangular matrix of nilpotent operator \n\nSuppose () . Then the following are equivalent. (a) is nilpotent. (b) The minimal polynomial of is for some positive integer .(c) There is a basis of with respect to which the matrix of has the form \n\n0 \n\n0 0\n\n,\n\nwhere all entries on and below the diagonal equal 0. \n\n> Proof\n\nSuppose (a) holds, so is nilpotent. Thus there exists a positive integer \n\n such that = 0 . Now 5.29 implies that is a polynomial multiple of the minimal polynomial of . Thus the minimal polynomial of is for some positive integer , proving that (a) implies (b). Now suppose (b) holds, so the minimal polynomial of is for some positive integer . This implies, by 5.27(a), that 0 ( which is the only zero of ) is the only eigenvalue of . This further implies, by 5.44, that there is a basis of with respect to which the matrix of is upper triangular. This also implies, by 5.41, that all entries on the diagonal of this matrix are 0, proving that (b) implies (c). Now suppose (c) holds. Then 5.40 implies that dim = 0 . Thus is nilpotent, proving that (c) implies (a). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n306 Chapter 8 Operators on Complex Vector Spaces \n\n# Exercises 8A \n\n1 Suppose () . Prove that if dim null 4 = 8 and dim null 6 = 9 , then \n\ndim null = 9 for all integers 5 .\n\n2 Suppose () , is a positive integer, , and 1 0 but \n\n = 0 . Prove that , , 2 , , 1 is linearly independent. \n\nThe result in this exercise is used in the proof of 8.45. \n\n3 Suppose () . Prove that \n\n = null range null 2 = null . \n\n4 Suppose () , , and is a positive integer such that the minimal polynomial of is a polynomial multiple of ( ) . Prove that \n\ndim null ( ) . \n\n5 Suppose () and is a positive integer. Prove that \n\ndim null dim null . \n\nHint: Exercise 21 in Section 3B may be useful. \n\n6 Suppose () . Show that \n\n = range 0 range 1 range range + 1 . \n\n7 Suppose () and is a nonnegative integer such that \n\nrange = range + 1.\n\nProve that range = range for all > .\n\n8 Suppose () . Prove that \n\nrange dim = range dim + 1 = range dim + 2 = . \n\n9 Suppose () and is a nonnegative integer. Prove that \n\nnull = null + 1 range = range + 1.\n\n10 Define ( 2) by ( , ) = ( , 0) . Find all generalized eigenvectors of .\n\n11 Suppose that () . Prove that there is a basis of consisting of generalized eigenvectors of if and only if the minimal polynomial of \n\nequals ( 1) ( ) for some 1, , .\n\nAssume = because the case = follows from 5.27 (b) and 8.9. This exercise states that the condition for there to be a basis of consisting of generalized eigenvectors of is the same as the condition for there to be a basis with respect to which has an upper-triangular matrix (see 5.44 ).\n\nCaution: If has an upper-triangular matrix with respect to a basis \n\n 1, , of , then 1 is an eigenvector of but it is not necessarily true that 2, , are generalized eigenvectors of .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8A Generalized Eigenvectors and Nilpotent Operators 307 \n\n12 Suppose () is such that every vector in is a generalized eigenvector of . Prove that there exists such that is nilpotent. \n\n13 Suppose , () and is nilpotent. Prove that is nilpotent. \n\n14 Suppose () is nilpotent and 0 . Prove is not diagonalizable. \n\n15 Suppose = and () . Prove that is diagonalizable if and only if every generalized eigenvector of is an eigenvector of .\n\nFor = , this exercise adds another equivalence to the list of conditions for diagonalizability in 5.55. \n\n16 (a) Give an example of nilpotent operators , on the same vector space such that neither + nor is nilpotent. (b) Suppose , () are nilpotent and = . Prove that + and \n\n are nilpotent. \n\n17 Suppose () is nilpotent and is a positive integer such that = 0 .(a) Prove that is invertible and that ( ) 1 = + + + 1 .(b) Explain how you would guess the formula above. \n\n18 Suppose () is nilpotent. Prove that 1 + dimrange = 0 .\n\nIf dim range < dim 1 , then this exercise improves 8.16. \n\n19 Suppose () is not nilpotent. Show that \n\n = null dim 1 range dim 1 .\n\nFor operators that are not nilpotent, this exercise improves 8.4. \n\n20 Suppose is an inner product space and () is normal and nilpotent. Prove that = 0 .\n\n21 Suppose () is such that null dim 1 null dim . Prove that is nilpotent and that dim null = for every integer with 0 dim .\n\n22 Suppose ( 5) is such that range 4 range 5. Prove that is nilpotent. \n\n23 Give an example of an operator on a finite-dimensional real vector space such that 0 is the only eigenvalue of but is not nilpotent. \n\nThis exercise shows that the implication (b) (a) in 8.17 does not hold without the hypothesis that = .\n\n24 For each item in Example 8.15, find a basis of the domain vector space such that the matrix of the nilpotent operator with respect to that basis has the upper-triangular form promised by 8.18(c). \n\n25 Suppose that is an inner product space and () is nilpotent. Show that there is an orthonormal basis of with respect to which the matrix of \n\nhas the upper-triangular form promised by 8.18(c). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n308 Chapter 8 Operators on Complex Vector Spaces \n\n# 8B Generalized Eigenspace Decomposition \n\n# Generalized Eigenspaces \n\n> 8.19 definition:\n\ngeneralized eigenspace, ( , ) \n\nSuppose () and . The generalized eigenspace of correspond-ing to , denoted by ( , ) , is defined by \n\n( , ) = { ( ) = 0 for some positive integer }. \n\nThus ( , ) is the set of generalized eigenvectors of corresponding to ,along with the 0 vector. Because every eigenvector of is a generalized eigenvector of (take = 1 \n\nin the definition of generalized eigenvector), each eigenspace is contained in the corresponding generalized eigenspace. In other words, if () and ,then ( , ) ( , ) .The next result implies that if () and , then the generalized eigenspace ( , ) is a subspace of (because the null space of each linear map on is a subspace of ). \n\n> 8.20\n\ndescription of generalized eigenspaces \n\nSuppose () and . Then ( , ) = null ( ) dim .\n\nProof Suppose null ( ) dim . The definitions imply ( , ) . Thus \n\n( , ) null ( ) dim .Conversely, suppose ( , ) . Thus there is a positive integer such that null ( ) . From 8.1 and 8.3 (with replacing ), we get \n\n null ( ) dim . Thus ( , ) null ( ) dim , completing the proof. \n\n> 8.21 example:\n\ngeneralized eigenspaces of an operator on 3\n\nDefine ( 3) by \n\n( 1, 2, 3) = (4 2, 0, 5 3). \n\nIn Example 8.10, we saw that the eigenvalues of are 0 and 5, and we found the corresponding sets of generalized eigenvectors. Taking the union of those sets with {0} , we have \n\n(0 , ) = {( 1, 2, 0) 1, 2 } and (5 , ) = {(0 , 0, 3) 3 }. \n\nNote that 3 = (0 , ) (5 , ) .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8B Generalized Eigenspace Decomposition 309 \n\nIn Example 8.21, the domain space 3 is the direct sum of the generalized eigenspaces of the operator in that example. Our next result shows that this behavior holds in general. Specifically, the following major result shows that if \n\n = and () , then is the direct sum of the generalized eigenspaces of , each of which is invariant under and on which is a nilpotent operator plus a scalar multiple of the identity. Thus the next result achieves our goal of decomposing into invariant subspaces on which has a known behavior. As we will see, the proof follows from putting together what we have learned about generalized eigenspaces and then using our result that for each operator \n\n () , there exists a basis of consisting of generalized eigenvectors of . \n\n> 8.22\n\ngeneralized eigenspace decomposition \n\nSuppose = and () . Let 1, , be the distinct eigenvalues of . Then (a) ( , ) is invariant under for each = 1 , , ;(b) ( )| ( ,) is nilpotent for each = 1 , , ;(c) = ( 1, ) ( , ) .\n\nProof \n\n(a) Suppose {1 , , } . Then 8.20 shows that \n\n( , ) = null ( ) dim .\n\nThus 5.18, with () = ( )dim , implies that ( , ) is invariant under ,proving (a). (b) Suppose {1 , , } . If ( , ) , then ( ) dim = 0 (by 8.20). Thus (( )| ( ,) )dim = 0 . Hence ( )| ( ,) is nilpotent, proving (b). (c) To show that ( 1, ) + + ( , ) is a direct sum, suppose \n\n 1 + + = 0 ,\n\nwhere each is in ( , ) . Because generalized eigenvectors of cor-responding to distinct eigenvalues are linearly independent (by 8.12), this implies that each equals 0. Thus ( 1, ) + + ( , ) is a direct sum (by 1.45). Finally, each vector in can be written as a finite sum of generalized eigen-vectors of (by 8.9). Thus \n\n = ( 1, ) ( , ) ,\n\nproving (c). For the analogous result when = , see Exercise 8. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 310 Chapter 8 Operators on Complex Vector Spaces\n\n# Multiplicity of an Eigenvalue \n\nIf is a complex vector space and () , then the decomposition of pro-vided by the generalized eigenspace decomposition (8.22) can be a powerful tool. The dimensions of the subspaces involved in this decomposition are sufficiently important to get a name, which is given in the next definition. \n\n> 8.23 definition:\n\nmultiplicity \n\n Suppose () . The multiplicity of an eigenvalue of is defined to be the dimension of the corresponding generalized eigenspace ( , ) .\n\n In other words, the multiplicity of an eigenvalue of equals \n\ndim null ( ) dim .\n\nThe second bullet point above holds because ( , ) = null ( ) dim (see 8.20). \n\n> 8.24 example:\n\nmultiplicity of each eigenvalue of an operator \n\nSuppose ( 3) is defined by \n\n( 1, 2, 3) = (6 1 + 3 2 + 4 3, 6 2 + 2 3, 7 3). \n\nThe matrix of (with respect to the standard basis) is \n\n6 3 40 6 20 0 7.\n\nThe eigenvalues of are the diagonal entries 6 and 7, as follows from 5.41. You can verify that the generalized eigenspaces of are as follows: \n\n(6 , ) = span ((1 , 0, 0) , (0 , 1, 0)) and (7 , ) = span ((10 , 2, 1)). \n\n> In this example, the multiplicity of each eigenvalue equals the number of times that eigenvalue appears on the diago-nal of an upper-triangular matrix rep-resenting the operator. This behavior always happens, as we will see in 8.31.\n\nThus the eigenvalue 6 has multiplicity 2\n\nand the eigenvalue 7 has multiplicity 1.The direct sum 3 = (6 , ) (7 , ) \n\nis the generalized eigenspace decom-position promised by 8.22. A basis of 3 consisting of generalized eigen-vectors of , as promised by 8.9, is \n\n(1 , 0, 0) , (0 , 1, 0) , (10 , 2, 1) . There does not exist a basis of 3 consisting of eigen-vectors of this operator. In the example above, the sum of the multiplicities of the eigenvalues of \n\nequals 3, which is the dimension of the domain of . The next result shows that this holds for all operators on finite-dimensional complex vector spaces. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 8B Generalized Eigenspace Decomposition 311\n> 8.25\n\nsum of the multiplicities equals dim \n\nSuppose = and () . Then the sum of the multiplicities of all eigenvalues of equals dim . \n\n> Proof\n\nThe desired result follows from the generalized eigenspace decomposition (8.22) and the formula for the dimension of a direct sum (see 3.94). The terms algebraic multiplicity and geometric multiplicity are used in some books. In case you encounter this terminology, be aware that the algebraic multi-plicity is the same as the multiplicity defined here and the geometric multiplicity is the dimension of the corresponding eigenspace. In other words, if () \n\nand is an eigenvalue of , then algebraic multiplicity of = dim null ( ) dim = dim ( , ) ,\n\ngeometric multiplicity of = dim null ( ) = dim ( , ). \n\nNote that as defined above, the algebraic multiplicity also has a geometric meaning as the dimension of a certain null space. The definition of multiplicity given here is cleaner than the traditional definition that involves determinants; 9.62 implies that these definitions are equivalent. If is an inner product space, () is normal, and is an eigenvalue of , then the algebraic multiplicity of equals the geometric multiplicity of ,as can be seen from applying Exercise 27 in Section 7A to the normal operator \n\n . As a special case, the singular values of ( , ) (here and are both finite-dimensional inner product spaces) depend on the multiplicities (either algebraic or geometric) of the eigenvalues of the self-adjoint operator .The next definition associates a monic polynomial with each operator on a finite-dimensional complex vector space. \n\n> 8.26 definition:\n\ncharacteristic polynomial \n\nSuppose = and () . Let 1, , denote the distinct eigenvalues of , with multiplicities 1, , . The polynomial \n\n( 1) 1 ( ) \n\nis called the characteristic polynomial of . \n\n> 8.27 example:\n\nthe characteristic polynomial of an operator \n\nSuppose ( 3) is defined as in Example 8.24. Because the eigenvalues of \n\n are 6, with multiplicity 2, and 7, with multiplicity 1, we see that the characteristic polynomial of is ( 6) 2( 7) .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n312 Chapter 8 Operators on Complex Vector Spaces \n\n> 8.28\n\ndegree and zeros of characteristic polynomial \n\nSuppose = and () . Then (a) the characteristic polynomial of has degree dim ;(b) the zeros of the characteristic polynomial of are the eigenvalues of .\n\nProof Our result about the sum of the multiplicities (8.25) implies (a). The definition of the characteristic polynomial implies (b). Most texts define the characteristic polynomial using determinants (the two definitions are equivalent by 9.62). The approach taken here, which is considerably simpler, leads to the following nice proof of the CayleyHamilton theorem. \n\n> 8.29\n\nCayleyHamilton theorem \n\nSuppose = , () , and is the characteristic polynomial of . Then \n\n() = 0 .\n\nProof Let 1, , be the distinct eigenvalues of , and let = dim ( , ) .For each {1 , , } , we know that ( )| ( ,) is nilpotent. Thus we have \n\nArthur Cayley (18211895 ) published three mathematics papers before com-pleting his undergraduate degree. \n\n( ) |( ,) = 0 \n\n(by 8.16) for each {1 , , } .The generalized eigenspace decom-position (8.22) states that every vector in is a sum of vectors in \n\n( 1, ) , , ( , ) . Thus to prove that () = 0 , we only need to show that ()| ( ,) = 0 for each .Fix {1 , , } . We have \n\n() = ( 1) 1 ( ) .\n\nThe operators on the right side of the equation above all commute, so we can move the factor ( ) to be the last term in the expression on the right. Because ( ) |( ,) = 0 , we have ()| ( ,) = 0 , as desired. The next result implies that if the minimal polynomial of an operator () \n\nhas degree dim (as happens almost alwayssee the paragraphs following 5.24), then the characteristic polynomial of equals the minimal polynomial of . \n\n> 8.30\n\ncharacteristic polynomial is a multiple of minimal polynomial \n\nSuppose = and () . Then the characteristic polynomial of is a polynomial multiple of the minimal polynomial of .\n\nProof The desired result follows immediately from the CayleyHamilton theo-rem (8.29) and 5.29. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8B Generalized Eigenspace Decomposition 313 \n\nNow we can prove that the result suggested by Example 8.24 holds for all operators on finite-dimensional complex vector spaces. \n\n> 8.31\n\nmultiplicity of an eigenvalue equals number of times on diagonal \n\nSuppose = and () . Suppose 1, , is a basis of such that \n\n( , ( 1, , )) is upper triangular. Then the number of times that each eigenvalue of appears on the diagonal of ( , ( 1, , )) equals the multiplicity of as an eigenvalue of .\n\nProof Let = ( , ( 1, , )) . Thus is an upper-triangular matrix. Let \n\n 1, , denote the entries on the diagonal of . Thus for each {1 , , } ,we have \n\n = + \n\nfor some span ( 1, , 1 ). Hence if {1 , , } and 0 , then is not a linear combination of 1, , 1 . The linear dependence lemma (2.19) now implies that the list of those such that 0 is linearly independent. Let denote the number of indices {1 , , } such that = 0 . The conclusion of the previous paragraph implies that \n\ndim range . \n\nBecause = dim = dim null + dim range , the inequality above implies that \n\n> 8.32\n\ndim null . \n\nThe matrix of the operator with respect to the basis 1, , is the upper-triangular matrix , which has diagonal entries 1 , , [see Exercise 2(b) in Section 5C ]. Because = 0 if and only if = 0 , the number of times that 0\n\nappears on the diagonal of equals . Thus applying 8.32 with replaced with \n\n , we have \n\n> 8.33\n\ndim null . \n\nFor an eigenvalue of , let denote the multiplicity of as an eigenvalue of and let denote the number of times that appears on the diagonal of .Replacing in 8.33 with , we see that \n\n> 8.34\n\n \n\nfor each eigenvalue of . The sum of the multiplicities over all eigenvalues \n\n of equals , the dimension of (by 8.25). The sum of the numbers over all eigenvalues of also equals , because the diagonal of has length .Thus summing both sides of 8.34 over all eigenvalues of produces an equality. Hence 8.34 must actually be an equality for each eigenvalue of .Thus the multiplicity of as an eigenvalue of equals the number of times that \n\n appears on the diagonal of , as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 100\n\nChapter 8 Operators on Complex Vector Spaces \n\n# Block Diagonal Matrices \n\nOften we can understand a matrix better by thinking of it as composed of smaller matrices. \n\nTo interpret our results in matrix form, we make the following definition, gener-alizing the notion of a diagonal matrix. If each matrix in the definition below is a 1-by-1 matrix, then we actually have a diagonal matrix. \n\n> 8.35 definition:\n\nblock diagonal matrix \n\nA block diagonal matrix is a square matrix of the form \n\n 1 00 \n\n\n\n,\n\nwhere 1, , are square matrices lying along the diagonal and all other entries of the matrix equal 0. \n\n> 8.36 example:\n\na block diagonal matrix \n\nThe 5-by-5 matrix \n\n = ( 4 ) 0 0 0 0002 3 0 20 00 0000 00 01 70 1\n\nis a block diagonal matrix with \n\n = 1 0 2\n\n0 3\n\n\n\n,\n\nwhere \n\n 1 = ( 4 ) , 2 = 2 3 0 2\n\n, 3 = 1 70 1.\n\nHere the inner matrices in the 5-by-5 matrix above are blocked off to show how we can think of it as a block diagonal matrix. Note that in the example above, each of 1, 2, 3 is an upper-triangular matrix whose diagonal entries are all equal. The next result shows that with respect to an appropriate basis, every operator on a finite-dimensional complex vector space has a matrix of this form. Note that this result gives us many more zeros in the matrix than are needed to make it upper triangular. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8B Generalized Eigenspace Decomposition 315 \n\n> 8.37\n\nblock diagonal matrix with upper-triangular blocks \n\nSuppose = and () . Let 1, , be the distinct eigenvalues of , with multiplicities 1, , . Then there is a basis of with respect to which has a block diagonal matrix of the form \n\n 1 00 \n\n\n\n,\n\nwhere each is a -by- upper-triangular matrix of the form \n\n = \n\n0 \n\n.\n\nProof Each ( )| ( ,) is nilpotent (see 8.22). For each , choose a basis of ( , ) , which is a vector space of dimension , such that the matrix of \n\n( )| ( ,) with respect to this basis is as in 8.18(c). Thus with respect to this basis, the matrix of | ( ,) , which equals ( )| ( ,) + | ( ,) ,looks like the desired form shown above for .The generalized eigenspace decomposition (8.22) shows that putting together the bases of the ( , ) s chosen above gives a basis of . The matrix of with respect to this basis has the desired form. \n\n> 8.38 example:\n\nblock diagonal matrix via generalized eigenvectors \n\nLet ( 3) be defined by ( 1, 2, 3) = (6 1 + 3 2 + 4 3, 6 2 + 2 3, 7 3).The matrix of (with respect to the standard basis) is \n\n6 3 40 6 20 0 7\n\n,\n\nwhich is an upper-triangular matrix but is not of the form promised by 8.37. As we saw in Example 8.24, the eigenvalues of are 6 and 7; also, \n\n(6 , ) = span ((1 , 0, 0) , (0 , 1, 0)) and (7 , ) = span ((10 , 2, 1)). \n\nWe also saw that a basis of 3 consisting of generalized eigenvectors of is \n\n(1 , 0, 0) , (0 , 1, 0) , (10 , 2, 1). \n\nThe matrix of with respect to this basis is \n\n6 30 6 000 0 ( 7 ) \n\n,\n\nwhich is a matrix of the block diagonal form promised by 8.37. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n316 Chapter 8 Operators on Complex Vector Spaces \n\n# Exercises 8B \n\n1 Define ( 2) by ( , ) = ( , ) . Find the generalized eigenspaces corresponding to the distinct eigenvalues of .\n\n2 Suppose () is invertible. Prove that ( , ) = ( 1 , 1 ) for every \n\n with 0 .\n\n3 Suppose () . Suppose () is invertible. Prove that and \n\n 1 have the same eigenvalues with the same multiplicities. \n\n4 Suppose dim 2 and () is such that null dim 2 null dim 1 .Prove that has at most two distinct eigenvalues. \n\n5 Suppose () and 3 and 8 are eigenvalues of . Let = dim . Prove that = ( null 2 ) ( range 2 ).\n\n6 Suppose () and is an eigenvalue of . Explain why the exponent of in the factorization of the minimal polynomial of is the smallest positive integer such that ( ) |( ,) = 0 .\n\n7 Suppose () and is an eigenvalue of with multiplicity . Prove that ( , ) = null ( ) .\n\nIf < dim , then this exercise improves 8.20. \n\n8 Suppose () and 1, , are the distinct eigenvalues of . Prove that \n\n = ( 1, ) ( , ) \n\nif and only if the minimal polynomial of equals ( 1) 1 ( ) \n\nfor some positive integers 1, , .\n\nThe case = follows immediately from 5.27 (b) and the generalized eigenspace decomposition (8.22 ); thus this exercise is interesting only when \n\n> = \n\n.\n\n9 Suppose = and () . Prove that there exist , () \n\nsuch that = + , the operator is diagonalizable, is nilpotent, and \n\n = .\n\n10 Suppose is a complex inner product space, 1, , is an orthonormal basis of , and () . Let 1, , be the eigenvalues of , each included as many times as its multiplicity. Prove that \n\n| 1|2 + + | |2 12 + + 2.\n\nSee the comment after Exercise 5 in Section 7A. \n\n11 Give an example of an operator on 4 whose characteristic polynomial equals ( 7) 2( 8) 2.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8B Generalized Eigenspace Decomposition 317 \n\n12 Give an example of an operator on 4 whose characteristic polynomial equals ( 1)( 5) 3 and whose minimal polynomial equals ( 1)( 5) 2.\n\n13 Give an example of an operator on 4 whose characteristic and minimal polynomials both equal ( 1) 2( 3) .\n\n14 Give an example of an operator on 4 whose characteristic polynomial equals \n\n( 1) 2( 3) and whose minimal polynomial equals ( 1)( 3) .\n\n15 Let be the operator on 4 defined by ( 1, 2, 3, 4) = (0 , 1, 2, 3). Find the characteristic polynomial and the minimal polynomial of .\n\n16 Let be the operator on 6 defined by \n\n( 1, 2, 3, 4, 5, 6) = (0 , 1, 2, 0, 4, 0). \n\nFind the characteristic polynomial and the minimal polynomial of .\n\n17 Suppose = and () is such that 2 = . Prove that the characteris-tic polynomial of is (1) , where = dim null and = dim range .\n\n18 Suppose () and is an eigenvalue of . Explain why the following four numbers equal each other. (a) The exponent of in the factorization of the minimal polynomial of .(b) The smallest positive integer such that ( ) |( ,) = 0 .(c) The smallest positive integer such that \n\nnull ( ) = null ( ) + 1.\n\n(d) The smallest positive integer such that \n\nrange ( ) = range ( ) + 1.\n\n19 Suppose = and () is a unitary operator. Prove that the constant term in the characteristic polynomial of has absolute value 1.\n\n20 Suppose that = and 1, , are nonzero subspaces of such that \n\n = 1 .\n\nSuppose () and each is invariant under . For each , let \n\ndenote the characteristic polynomial of | . Prove that the characteristic polynomial of equals 1 .\n\n21 Suppose , () are monic polynomials with the same zeros and is a polynomial multiple of . Prove that there exists ( deg ) such that the characteristic polynomial of is and the minimal polynomial of is .\n\nThis exercise implies that every monic polynomial is the characteristic polynomial of some operator. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n318 Chapter 8 Operators on Complex Vector Spaces \n\n22 Suppose and are block diagonal matrices of the form \n\n = 1 00 \n\n\n\n, = 1 00 \n\n\n\n,\n\nwhere and are square matrices of the same size for each = 1 , , .Show that is a block diagonal matrix of the form \n\n = 1 1 00 \n\n.\n\n23 Suppose = , () , and .(a) Show that + ( , ) if and only if ( , ).(b) Show that the multiplicity of as an eigenvalue of equals the multiplicity of as an eigenvalue of .(c) Use (b) and the result about the sum of the multiplicities (8.25) to show that if dim is an odd number, then has a real eigenvalue. (d) Use (c) and the result about real eigenvalues of (Exercise 17 in Section 5A) to show that if dim is an odd number, then has an eigenvalue (thus giving an alternative proof of 5.34). \n\nSee Exercise 33 in Section 3B for the definition of the complexification .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8C Consequences of Generalized Eigenspace Decomposition 319 \n\n# 8C Consequences of Generalized Eigenspace Decomposition \n\n# Square Roots of Operators \n\nRecall that a square root of an operator () is an operator () such that 2 = (see 7.36). Every complex number has a square root, but not every operator on a complex vector space has a square root. For example, the operator on 3 defined by ( 1, 2, 3) = ( 2, 3, 0) does not have a square root, as you are asked to show in Exercise 1. The noninvertibility of that operator is no accident, as we will soon see. We begin by showing that the identity plus any nilpotent operator has a square root. \n\n> 8.39\n\nidentity plus nilpotent has a square root \n\nSuppose () is nilpotent. Then + has a square root. \n\nProof Consider the Taylor series for the function 1 + : \n\n> 8.40 \n\n1 + = 1 + 1 + 2 2 + . \n\nBecause 1 = 12 , the formula above implies that 1 + 2 is a good estimate for 1 + when is small. \n\nWe do not find an explicit formula for the coefficients or worry about whether the infinite sum converges because we use this equation only as motivation. Because is nilpotent, = 0 for some positive integer . In 8.40, suppose we replace with and 1 with . Then the infinite sum on the right side becomes a finite sum (because = 0 for all \n\n ). Thus we guess that there is a square root of + of the form \n\n + 1 + 2 2 + + 1 1 .\n\nHaving made this guess, we can try to choose 1, 2, , 1 such that the operator above has its square equal to + . Now \n\n( + 1 + 2 2 + 3 3 + + 1 1 )2\n\n= + 2 1 + (2 2 + 12) 2 + (2 3 + 2 1 2) 3 + \n\n+ (2 1 + terms involving 1, , 2 ) 1 .\n\nWe want the right side of the equation above to equal + . Hence choose 1\n\nsuch that 2 1 = 1 (thus 1 = 1/2 ). Next, choose 2 such that 2 2 + 12 = 0 (thus \n\n 2 = 1/8 ). Then choose 3 such that the coefficient of 3 on the right side of the equation above equals 0 (thus 3 = 1/16 ). Continue in this fashion for each \n\n = 4 , , 1 , at each step solving for so that the coefficient of on the right side of the equation above equals 0. Actually we do not care about the explicit formula for the s. We only need to know that some choice of the s gives a square root of + .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n320 Chapter 8 Operators on Complex Vector Spaces \n\nThe previous lemma is valid on real and complex vector spaces. However, the result below holds only on complex vector spaces. For example, the operator of multiplication by 1 on the one-dimensional real vector space has no square root. \n\nRepresentation of a complex number with polar coordinates. \n\nFor the proof below, we need to know that every has a square root in . To show this, write \n\n = ( cos + sin ) ,\n\nwhere is the length of the line segment in the complex plane from the origin to and is the angle of that line segment with the positive horizontal axis. Then \n\n> \n\n( cos 2 + sin 2 )\n\nis a square root of , as you can verify by showing that the square of the complex number above equals . \n\n> 8.41\n\nover , invertible operators have square roots \n\nSuppose is a complex vector space and () is invertible. Then has a square root. \n\nProof Let 1, , be the distinct eigenvalues of . For each , there exists a nilpotent operator (( , )) such that | ( ,) = + [see 8.22(b) ].Because is invertible, none of the s equals 0, so we can write \n\n| ( ,) = + \n\n \n\n\n\nfor each . Because / is nilpotent, + / has a square root (by 8.39). Multiplying a square root of the complex number by a square root of + / ,we obtain a square root of | ( ,) .By the generalized eigenspace decomposition (8.22), a typical vector \n\ncan be written uniquely in the form \n\n = 1 + + ,\n\nwhere each is in ( , ) . Using this decomposition, define an operator \n\n () by \n\n = 1 1 + + .\n\nYou should verify that this operator is a square root of , completing the proof. By imitating the techniques in this subsection, you should be able to prove that if is a complex vector space and () is invertible, then has a th root for every positive integer .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8C Consequences of Generalized Eigenspace Decomposition 321 \n\n# Jordan Form \n\nWe know that if is a complex vector space, then for every () there is a basis of with respect to which has a nice upper-triangular matrix (see 8.37). In this subsection we will see that we can do even betterthere is a basis of \n\nwith respect to which the matrix of contains 0s everywhere except possibly on the diagonal and the line directly above the diagonal. We begin by looking at two examples of nilpotent operators. \n\n> 8.42 example:\n\nnilpotent operator with nice matrix \n\nLet be the operator on 4 defined by \n\n( 1, 2, 3, 4) = (0 , 1, 2, 3). \n\nThen 4 = 0 ; thus is nilpotent. If = (1 , 0, 0, 0) , then 3 , 2 , , is a basis of 4. The matrix of with respect to this basis is \n\n0 1 0 00 0 1 00 0 0 10 0 0 0.\n\nThe next example of a nilpotent operator has more complicated behavior than the example above. \n\n> 8.43 example:\n\nnilpotent operator with slightly more complicated matrix \n\nLet be the operator on 6 defined by \n\n( 1, 2, 3, 4, 5, 6) = (0 , 1, 2, 0, 4, 0). \n\nThen 3 = 0 ; thus is nilpotent. In contrast to the nice behavior of the nilpotent operator of the previous example, for this nilpotent operator there does not exist a vector 6 such that 5 , 4 , 3 , 2 , , is a basis of 6. However, if we take 1 = (1 , 0, 0, 0, 0, 0) , 2 = (0 , 0, 0, 1, 0, 0) , and 3 = (0 , 0, 0, 0, 0, 1) , then \n\n 2 1, 1, 1, 2, 2, 3 is a basis of 6. The matrix of with respect to this basis is \n\n0 1 00 0 10 0 00 00 00 00000 0 00 0 0 0 10 0 000 0 0 0 0 ( 0 ) .\n\nHere the inner matrices are blocked off to show that we can think of the 6-by-6\n\nmatrix above as a block diagonal matrix consisting of a 3-by-3 block with 1s on the line above the diagonal and 0s elsewhere, a 2-by-2 block with 1 above the diagonal and 0s elsewhere, and a 1-by-1 block containing 0. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 322 Chapter 8 Operators on Complex Vector Spaces\n\nOur next goal is to show that every nilpotent operator () behaves similarly to the operator in the previous example. Specifically, there is a finite collection of vectors 1, , such that there is a basis of consisting of the vectors of the form , as varies from 1 to and varies (in reverse order) from 0 to the largest nonnegative integer such that 0 . With respect to this basis, the matrix of looks like the matrix in the previous example. More specifically, has a block diagonal matrix with respect to this basis, with each block a square matrix that is 0 everywhere except on the line above the diagonal. In the next definition, the diagonal of each is filled with some eigenvalue \n\n of , the line directly above the diagonal of is filled with 1s, and all other entries in are 0 (to understand why each is an eigenvalue of , see 5.41). The s need not be distinct. Also, may be a 1-by-1 matrix ( ) containing just an eigenvalue of . If each is 0, then the next definition captures the behavior described in the paragraph above (recall that if is nilpotent, then 0 is the only eigenvalue of ). \n\n> 8.44 definition:\n\nJordan basis \n\nSuppose () . A basis of is called a Jordan basis for if with respect to this basis has a block diagonal matrix \n\n 1 00 \n\n\n\nin which each is an upper-triangular matrix of the form \n\n = 1 0 10 \n\n.\n\nMost of the work in proving that every operator on a finite-dimensional com-plex vector space has a Jordan basis occurs in proving the special case below of nilpotent operators. This special case holds on real vector spaces as well as complex vector spaces. \n\n> 8.45\n\nevery nilpotent operator has a Jordan basis \n\nSuppose () is nilpotent. Then there is a basis of that is a Jordan basis for . \n\n> Proof\n\nWe will prove this result by induction on dim . To get started, note that the desired result holds if dim = 1 (because in that case, the only nilpotent operator is the 0 operator). Now assume that dim > 1 and that the desired result holds on all vector spaces of smaller dimension. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8C Consequences of Generalized Eigenspace Decomposition 323 \n\nLet be the smallest positive integer such that = 0 . Thus there exists \n\n such that 1 0 . Let \n\n = span ( , , , 1 ). \n\nThe list , , , 1 is linearly independent (see Exercise 2 in Section 8A). If = , then writing this list in reverse order gives a Jordan basis for and we are done. Thus we can assume that .Note that is invariant under . By our induction hypothesis, there is a basis of that is a Jordan basis for | . The strategy of our proof is that we will find a subspace of such that is also invariant under and = . Again by our induction hypothesis, there will be a basis of that is a Jordan basis for \n\n| . Putting together the Jordan bases for | and | , we will have a Jordan basis for .Let be such that ( 1 ) 0 . Let \n\n = { ( ) = 0 for each = 0 , , 1}. \n\nThen is a subspace of that is invariant under ( the invariance holds because if then ( ()) = 0 for = 0 , , 1 , where the case = 1 \n\nholds because = 0) . We will show that = , which by the previous paragraph will complete the proof. To show that + is a direct sum, suppose with 0 . Because \n\n , there exist 0, , 1 such that \n\n = 0 + 1 + + 1 1 . \n\nLet be the smallest index such that 0 . Apply 1 to both sides of the equation above, getting \n\n 1 = 1 ,\n\nwhere we have used the equation = 0 . Now apply to both sides of the equation above, getting \n\n( 1 ) = ( 1 ) 0. \n\nThe equation above shows that . Hence we have proved that = {0} ,which implies that + is a direct sum (see 1.46). To show that = , define by \n\n = (() , () , , ( 1 )). \n\nThus null = . Hence \n\ndim = dim null = dim dim range dim ,\n\nwhere the second equality comes from the fundamental theorem of linear maps (3.21). Using the inequality above, we have \n\ndim ( ) = dim + dim + (dim ) = dim . \n\nThus = (by 2.39), completing the proof. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n324 Chapter 8 Operators on Complex Vector Spaces \n\nCamille Jordan (18381922 ) pub-lished a proof of 8.46 in 1870. \n\nNow the generalized eigenspace de-composition allows us to extend the pre-vious result to operators that may not be nilpotent. Doing this requires that we deal with complex vector spaces. \n\n> 8.46\n\nJordan form \n\nSuppose = and () . Then there is a basis of that is a Jordan basis for .\n\nProof Let 1, , be the distinct eigenvalues of . The generalized eigenspace decomposition states that \n\n = ( 1, ) ( , ) ,\n\nwhere each ( )| ( ,) is nilpotent (see 8.22). Thus 8.45 implies that some basis of each ( , ) is a Jordan basis for ( )| ( ,) . Put these bases together to get a basis of that is a Jordan basis for .\n\n# Exercises 8C \n\n1 Suppose ( 3) is the operator defined by ( 1, 2, 3) = ( 2, 3, 0) .Prove that does not have a square root. \n\n2 Define ( 5) by ( 1, 2, 3, 4, 5) = (2 2, 3 3, 4, 4 5, 0) .(a) Show that is nilpotent. (b) Find a square root of + .\n\n3 Suppose is a complex vector space. Prove that every invertible operator on has a cube root. \n\n4 Suppose is a real vector space. Prove that the operator on has a square root if and only if dim is an even number. \n\n5 Suppose ( 2) is the operator defined by ( , ) = ( , 9 + 5) .Find a Jordan basis for .\n\n6 Find a basis of 4() that is a Jordan basis for the differentiation operator \n\n on 4() defined by = .\n\n7 Suppose () is nilpotent and 1, , is a Jordan basis for . Prove that the minimal polynomial of is + 1, where is the length of the longest consecutive string of 1s that appears on the line directly above the diagonal in the matrix of with respect to 1, , .\n\n8 Suppose () and 1, , is a basis of that is a Jordan basis for .Describe the matrix of 2 with respect to this basis. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8C Consequences of Generalized Eigenspace Decomposition 325 \n\n9 Suppose () is nilpotent. Explain why there exist 1, , \n\nand nonnegative integers 1, , such that (a) and (b) below both hold. (a) 1 1, , 1, 1, , , , , is a basis of .(b) 1 + 1 1 = = + 1 = 0 .\n\n10 Suppose () and 1, , is a basis of that is a Jordan basis for .Describe the matrix of with respect to the basis , , 1 obtained by reversing the order of the s. \n\n11 Suppose () . Explain why every vector in each Jordan basis for is a generalized eigenvector of .\n\n12 Suppose () is diagonalizable. Show that () is a diagonal matrix with respect to every Jordan basis for .\n\n13 Suppose () is nilpotent. Prove that if 1, , are vectors in and \n\n 1, , are nonnegative integers such that \n\n 1 1, , 1, 1, , , , , is a basis of \n\nand \n\n 1 + 1 1 = = + 1 = 0 ,\n\nthen 1 1, , is a basis of null .\n\nThis exercise shows that = dim null . Thus the positive integer that appears above depends only on and not on the specific Jordan basis chosen for .\n\n14 Suppose = and () . Prove that there does not exist a direct sum decomposition of into two nonzero subspaces invariant under if and only if the minimal polynomial of is of the form ( ) dim for some \n\n .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n326 Chapter 8 Operators on Complex Vector Spaces \n\n# 8D Trace: A Connection Between Matrices and Operators \n\nWe begin this section by defining the trace of a square matrix. After developing some properties of the trace of a square matrix, we will use this concept to define the trace of an operator. \n\n> 8.47 definition:\n\ntrace of a matrix \n\nSuppose is a square matrix with entries in . The trace of , denoted tr ,is defined to be the sum of the diagonal entries of . \n\n> 8.48 example:\n\ntrace of a 3-by-3 matrix \n\nSuppose \n\n = 3 1 2 3 2 3 1 2 0.\n\nThe diagonal entries of , which are shown in red above, are 3, 2, and 0. Thus \n\ntr = 3 + 2 + 0 = 5 .Matrix multiplication is not commutative, but the next result shows that the order of matrix multiplication does not matter to the trace. \n\n> 8.49\n\ntrace of equals trace of \n\nSuppose is an -by- matrix and is an -by- matrix. Then \n\ntr () = tr (). \n\nProof Suppose \n\n = 1,1 1, \n\n ,1 , \n\n\n\n, = 1,1 1, \n\n ,1 , \n\n.\n\nThe th term on the diagonal of the -by- matrix equals = 1 , , . Thus \n\ntr () = \n\n> \n\n\n\n> = 1\n> \n\n\n\n> = 1\n\n , , \n\n=\n\n> \n\n\n\n> = 1 \n\n\n\n> = 1\n\n , , \n\n=\n\n> \n\n\n\n> = 1\n\n( th term on diagonal of the -by- matrix ) = tr () ,\n\nas desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8D Trace: A Connection Between Matrices and Operators 327 \n\nWe want to define the trace of an operator () to be the trace of the matrix of with respect to some basis of . However, this definition should not depend on the choice of basis. The following result will make this possible. \n\n> 8.50\n\ntrace of matrix of operator does not depend on basis \n\nSuppose () . Suppose 1, , and 1, , are bases of . Then \n\ntr ( , ( 1, , )) = tr ( , ( 1, , )). \n\nProof Let = ( , ( 1, , )) and = ( , ( 1, , )) . The change-of-basis formula tells us that there exists an invertible -by- matrix such that \n\n = 1 (see 3.84). Thus \n\ntr = tr (( 1 )) = tr (( 1 )) = tr (( 1 )) = tr ,\n\nwhere the second line comes from 8.49. Because of 8.50, the following definition now makes sense. \n\n> 8.51 definition:\n\ntrace of an operator \n\nSuppose () . The trace of , denoted tr , is defined by \n\ntr = tr ( , ( 1, , )) ,\n\nwhere 1, , is any basis of .Suppose () and is an eigenvalue of . Recall that we defined the multiplicity of to be the dimension of the generalized eigenspace ( , ) (see 8.23); we proved that this multiplicity equals dim null ( ) dim (see 8.20). Recall also that if is a complex vector space, then the sum of the multiplicities of all eigenvalues of equals dim (see 8.25). In the following result, the sum of the eigenvalues with each eigenvalue included as many times as its multiplicity means that if 1, , are the distinct eigenvalues of with multiplicities 1, , , then the sum is \n\n 1 1 + + .\n\nOr if you prefer to work with a list of not-necessarily-distinct eigenvalues, with each eigenvalue included as many times as its multiplicity, then the eigenvalues could be denoted by 1, , (where equals dim ) and the sum is \n\n 1 + + . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 328 Chapter 8 Operators on Complex Vector Spaces\n> 8.52\n\non complex vector spaces, trace equals sum of eigenvalues \n\nSuppose = and () . Then tr equals the sum of the eigenvalues of , with each eigenvalue included as many times as its multiplicity. \n\n> Proof\n\nThere is a basis of with respect to which has an upper-triangular matrix with the diagonal entries of the matrix consisting of the eigenvalues of ,with each eigenvalue included as many times as its multiplicitysee 8.37. Thus the definition of the trace of an operator along with 8.50, which allows us to use a basis of our choice, implies that tr equals the sum of the eigenvalues of , with each eigenvalue included as many times as its multiplicity. \n\n> 8.53 example:\n\ntrace of an operator on 3\n\nSuppose ( 3) is defined by \n\n( 1, 2, 3) = (3 1 2 2 3, 3 1 + 2 2 3 3, 1 + 2 2). \n\nThen the matrix of with respect to the standard basis of 3 is \n\n3 1 2 3 2 3 1 2 0.\n\nAdding up the diagonal entries of this matrix, we see that tr = 5 .The eigenvalues of are 1, 2 + 3 , and 2 3 , each with multiplicity 1, as you can verify. The sum of these eigenvalues, each included as many times as its multiplicity, is 1 + (2 + 3) + (2 3) , which equals 5, as expected by 8.52. The trace has a close connection with the characteristic polynomial. Suppose \n\n = , () , and 1, , are the eigenvalues of , with each eigenvalue included as many times as its multiplicity. Then by definition (see 8.26), the characteristic polynomial of equals \n\n( 1)( ). \n\nExpanding the polynomial above, we can write the characteristic polynomial of \n\nin the form \n\n ( 1 + + ) 1 + + (1) ( 1 ). \n\nThe expression above immediately leads to the next result. Also see 9.65, which does not require the hypothesis that = . \n\n> 8.54\n\ntrace and characteristic polynomial \n\nSuppose = and () . Let = dim . Then tr equals the negative of the coefficient of 1 in the characteristic polynomial of . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 8D Trace: A Connection Between Matrices and Operators 329\n\nThe next result gives a nice formula for the trace of an operator on an inner product space. \n\n> 8.55\n\ntrace on an inner product space \n\nSuppose is an inner product space, () , and 1, , is an orthonor-mal basis of . Then \n\ntr = 1, 1 + + , . \n\n> Proof\n\nThe desired formula follows from the observation that the entry in row ,column of ( , ( 1, , )) equals , [use 6.30(a) with = ]. The algebraic properties of the trace as defined on square matrices translate to algebraic properties of the trace as defined on operators, as shown in the next result. \n\n> 8.56\n\ntrace is linear \n\nThe function tr () is a linear functional on () such that \n\ntr () = tr () \n\nfor all , () . \n\n> Proof\n\nChoose a basis of . All matrices of operators in this proof will be with respect to that basis. Suppose , () .If , then \n\ntr () = tr () = tr ( ()) = tr () = tr ,\n\nwhere the first and last equalities come from the definition of the trace of an operator, the second equality comes from 3.38, and the third equality follows from the definition of the trace of a square matrix. Also, \n\ntr ( +) = tr ( +) = tr (() +()) = tr () +tr () = tr +tr ,\n\nwhere the first and last equalities come from the definition of the trace of an operator, the second equality comes from 3.35, and the third equality follows from the definition of the trace of a square matrix. The two paragraphs above show that tr () is a linear functional on () .Furthermore, \n\ntr () = tr () = tr (() ()) = tr (() ()) = tr () = tr () ,\n\nwhere the second and fourth equalities come from 3.43 and the crucial third equality comes from 8.49. The equations tr () = tr () and tr = dim uniquely characterize the trace among the linear functionals on () see Exercise 10. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 330 Chapter 8 Operators on Complex Vector Spaces\n> The statement of the next result does not involve traces, but the short proof uses traces. When something like this happens in mathematics, then usually a good definition lurks in the back-ground.\n\nThe equation tr () = tr () leads to our next result, which does not hold on infinite-dimensional vector spaces (see Exercise 13). However, additional hy-potheses on , , and lead to an infinite-dimensional generalization of the result below, with important applications to quantum theory. \n\n> 8.57\n\nidentity operator is not the difference of and \n\nThere do not exist operators , () such that = . \n\n> Proof\n\nSuppose , () . Then \n\ntr ( ) = tr () tr () = 0 ,\n\nwhere both equalities come from 8.56. The trace of equals dim , which is not 0.Because and have different traces, they cannot be equal. \n\n# Exercises 8D \n\n1 Suppose is an inner product space and , . Define an operator \n\n () by = , . Find a formula for tr .\n\n2 Suppose () satisfies 2 = . Prove that \n\ntr = dim range . \n\n3 Suppose () and 5 = . Prove that the real and imaginary parts of \n\ntr are both integers. \n\n4 Suppose is an inner product space and () . Prove that \n\ntr = tr . \n\n5 Suppose is an inner product space. Suppose () is a positive operator and tr = 0 . Prove that = 0 .\n\n6 Suppose is an inner product space and , () are orthogonal projections. Prove that tr () 0 .\n\n7 Suppose ( 3) is the operator whose matrix is \n\n51 12 21 60 40 28 57 68 1.\n\nSomeone tells you (accurately) that 48 and 24 are eigenvalues of . Without using a computer or writing anything down, find the third eigenvalue of .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 8D Trace: A Connection Between Matrices and Operators 331 \n\n8 Prove or give a counterexample: If , () , then tr () = ( tr )( tr ) .\n\n9 Suppose () is such that tr () = 0 for all () . Prove that \n\n = 0 .\n\n10 Prove that the trace is the only linear functional () such that \n\n() = () \n\nfor all , () and () = dim .\n\nHint: Suppose that 1, , is a basis of . For , {1 , , } , define \n\n , () by , ( 1 1 + + ) = . Prove that \n\n( , ) = {{1 if = ,\n\n0 if . \n\nThen for () , use the equation = = 1 \n\n> = 1\n\n() , , to show that () = tr .\n\n11 Suppose and are inner product spaces and ( , ) . Prove that if \n\n 1, , is an orthonormal basis of and 1, , is an orthonormal basis of , then \n\ntr ( ) = \n\n> \n\n\n\n> = 1 \n\n\n\n> = 1\n\n| , | 2.\n\nThe numbers , are the entries of the matrix of with respect to the orthonormal bases 1, , and 1, , . These numbers depend on the bases, but tr ( ) does not depend on a choice of bases. Thus this exercise shows that the sum of the squares of the absolute values of the matrix entries does not depend on which orthonormal bases are used. \n\n12 Suppose and are finite-dimensional inner product spaces. (a) Prove that , = tr ( ) defines an inner product on ( , ) .(b) Suppose 1, , is an orthonormal basis of and 1, , is an orthonormal basis of . Show that the inner product on ( , ) from (a) is the same as the standard inner product on , where we identify each element of ( , ) with its matrix (with respect to the bases just mentioned) and then with an element of .\n\nCaution: The norm of a linear map ( , ) as defined by 7.86 is not the same as the norm that comes from the inner product in (a) above. Unless explicitly stated otherwise, always assume that refers to the norm as defined by 7.86. The norm that comes from the inner product in (a) is called the Frobenius norm or the HilbertSchmidt norm .\n\n13 Find , ( ()) such that = .\n\nHint: Make an appropriate modification of the operators in Example 3.9. This exercise shows that additional hypotheses are needed on and to extend 8.57 to the setting of infinite-dimensional vector spaces. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n# Chapter 9 \n\n# Multilinear Algebra and Determinants \n\nWe begin this chapter by investigating bilinear forms and quadratic forms on a vector space. Then we will move on to multilinear forms. We will show that the vector space of alternating -linear forms has dimension one on a vector space of dimension . This result will allow us to give a clean basis-free definition of the determinant of an operator. This approach to the determinant via alternating multilinear forms leads to straightforward proofs of key properties of determinants. For example, we will see that the determinant is multiplicative, meaning that det () = ( det )( det ) for all operators and on the same vector space. We will also see that is invertible if and only if det 0 . Another important result states that the determinant of an operator on a complex vector space equals the product of the eigenvalues of the operator, with each eigenvalue included as many times as its multiplicity. The chapter concludes with an introduction to tensor products. \n\nstanding assumptions for this chapter \n\n denotes or .\n\n and denote finite-dimensional nonzero vector spaces over .\n\n> Daniel Schwen CC BY-SA\n\nThe Mathematical Institute at the University of Gttingen. This building opened in 1930, when Emmy Noether (18821935 ) had already been a research mathematician and faculty member at the university for 15 years (the first eight years without salary ).Noether was fired by the Nazi government in 1933. By then Noether and her collaborators had created many of the foundations of modern algebra, including an abstract algebra viewpoint that contributed to the development of linear algebra. \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler \n\n332 Section 9A Bilinear Forms and Quadratic Forms 333 \n\n# 9A Bilinear Forms and Quadratic Forms \n\n# Bilinear Forms \n\nA bilinear form on is a function from to that is linear in each slot separately, meaning that if we hold either slot fixed then we have a linear function in the other slot. Here is the formal definition. \n\n> 9.1 definition:\n\nbilinear form \n\nA bilinear form on is a function such that \n\n ( , ) and ( , ) \n\nare both linear functionals on for every .\n\nRecall that the term linear functional ,used in the definition above, means a linear function that maps into the scalar field . Thus the term bilinear functional would be more consistent terminology than bilinear form , which unfortunately has become standard. \n\nFor example, if is a real inner prod-uct space, then the function that takes an ordered pair ( , ) to , is a bilinear form on . If is a nonzero complex inner product space, then this function is not a bilinear form because the inner product is not linear in the sec-ond slot (complex scalars come out of the second slot as their complex conjugates). If = , then a bilinear form differs from an inner product in that an inner product requires symmetry [meaning that ( , ) = ( , ) for all , ] \n\nand positive definiteness [meaning that ( , ) > 0 for all \\{0}] , but these properties are not required for a bilinear form. \n\n> 9.2 example:\n\nbilinear forms \n\n The function 3 3 defined by \n\n(( 1, 2, 3), ( 1, 2, 3)) = 1 2 5 2 3 + 2 3 1\n\nis a bilinear form on 3.\n\n Suppose is an -by- matrix with , in row , column . Define a bilinear form on by \n\n (( 1, , ), ( 1, , )) = \n\n> \n> \n> = 1 \n> \n> = 1\n\n , .\n\nThe first bullet point is a special case of this bullet point with = 3 and \n\n = 0 1 00 0 5 2 0 0.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n334 Chapter 9 Multilinear Algebra and Determinants \n\n Suppose is a real inner product space and () . Then the function \n\n defined by \n\n( , ) = , \n\nis a bilinear form on .\n\n If is a positive integer, then the function () () defined by \n\n( , ) = (2) (3) \n\nis a bilinear form on () .\n\n Suppose , . Then the function defined by \n\n( , ) = () () \n\nis a bilinear form on .\n\n More generally, suppose that 1, , , 1, , . Then the function \n\n defined by \n\n( , ) = 1() 1() + + () () \n\nis a bilinear form on .A bilinear form on is a function from to . Because is a vector space, this raises the question of whether a bilinear form can also be a linear map from to . Note that none of the bilinear forms in 9.2 are linear maps except in some special cases in which the bilinear form is the zero map. Exercise 3 shows that a bilinear form on is a linear map on only if = 0 . \n\n> 9.3 definition:\n\n (2) \n\nThe set of bilinear forms on is denoted by (2) .With the usual operations of addition and scalar multiplication of functions, \n\n (2) is a vector space. For an operator on an -dimensional vector space and a basis 1, , \n\nof , we used an -by- matrix to provide information about . We now do the same thing for bilinear forms on . \n\n> 9.4 definition:\n\nmatrix of a bilinear form, () \n\nSuppose is a bilinear form on and 1, , is a basis of . The matrix of \n\n with respect to this basis is the -by- matrix () whose entry () , \n\nin row , column is given by \n\n() , = ( , ). \n\nIf the basis 1, , is not clear from the context, then the notation \n\n( , ( 1, , )) is used. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9A Bilinear Forms and Quadratic Forms 335 \n\nRecall that , denotes the vector space of -by- matrices with entries in \n\nand that dim , = 2 (see 3.39 and 3.40). \n\n9.5 dim (2) = ( dim ) 2\n\nSuppose 1, , is a basis of . Then the map () is an isomorphism of (2) onto , . Furthermore, dim (2) = ( dim ) 2.\n\nProof The map () is clearly a linear map of (2) into , . For , ,define a bilinear form on by \n\n ( 1 1 + + , 1 1 + + ) = \n\n> \n\n\n\n> = 1 \n\n\n\n> = 1\n\n , \n\nfor 1, , , 1, , ( if = and 1, , is the standard basis of ,this is the same as the bilinear form in the second bullet point of Example 9.2 ).The linear map () from (2) to , and the linear map from \n\n , to (2) are inverses of each other because () = for all (2) and \n\n( ) = for all , , as you should verify. Thus both maps are isomorphisms and the two spaces that they connect have the same dimension. Hence dim (2) = dim , = 2 = ( dim ) 2.Recall that t denotes the transpose of a matrix . The matrix t is obtained by interchanging the rows and the columns of .\n\n9.6 composition of a bilinear form and an operator \n\nSuppose is a bilinear form on and () . Define bilinear forms \n\nand on by \n\n( , ) = ( , ) and ( , ) = ( , ). \n\nLet 1, , be a basis of . Then \n\n() = () () and () = () t (). \n\nProof If , {1 , , } , then \n\n() , = ( , )= ( , )= ( , \n\n\n\n> = 1\n\n() , )=\n\n> \n\n\n\n> = 1\n\n( , )() , \n\n= ( () ()) , .\n\nThus () = () () . The proof that () = () t () is similar. \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler 336 Chapter 9 Multilinear Algebra and Determinants \n\nThe result below shows how the matrix of a bilinear form changes if we change the basis. The formula in the result below should be compared to the change-of-basis formula for the matrix of an operator (see 3.84). The two formulas are similar, except that the transpose t appears in the formula below and the inverse \n\n 1 appears in the change-of-basis formula for the matrix of an operator. \n\n> 9.7\n\nchange-of-basis formula \n\nSuppose (2) . Suppose 1, , and 1, , are bases of . Let \n\n = ( , ( 1, , )) and = ( , ( 1, , )) \n\nand = ( , ( 1, , ), ( 1, , )) . Then \n\n = t. \n\nProof The linear map lemma (3.4) tells us that there exists an operator () \n\nsuch that = for each = 1 , , . The definition of the matrix of an operator with respect to a basis implies that \n\n( , ( 1, , )) = . \n\nDefine bilinear forms , on by \n\n( , ) = ( , ) and ( , ) = ( , ) = ( , ). \n\nThen ( , ) = ( , ) = ( , ) for all , {1 , , } . Thus \n\n = ( , ( 1, , )) = t( , ( 1, , )) = t ,\n\nwhere the second and third lines each follow from 9.6. \n\n> 9.8 example:\n\nthe matrix of a bilinear form on 2() \n\nDefine a bilinear form on 2() by ( , ) = (2) (3) . Let \n\n = ( , (1 , 2 , ( 3) 2)) and = ( , (1 , , 2)) \n\nand = ( , (1 , 2 , ( 3) 2), (1 , , 2)) . Then \n\n = 0 1 00 0 00 1 0\n\nand = 0 1 60 2 12 0 4 24 \n\nand = 1 2 90 1 6 0 0 1.\n\nNow the change-of-basis formula 9.7 asserts that = t , which you can verify with matrix multiplication using the matrices above. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9A Bilinear Forms and Quadratic Forms 337 \n\n# Symmetric Bilinear Forms \n\n> 9.9 definition:\n\nsymmetric bilinear form, (2) \n\n> sym\n\nA bilinear form (2) is called symmetric if \n\n( , ) = ( , ) \n\nfor all , . The set of symmetric bilinear forms on is denoted by (2) \n\n> sym\n\n. \n\n> 9.10 example:\n\nsymmetric bilinear forms \n\n If is a real inner product space and (2) is defined by \n\n( , ) = , ,\n\nthen is a symmetric bilinear form on .\n\n Suppose is a real inner product space and () . Define (2) by \n\n( , ) = , . \n\nThen is a symmetric bilinear form on if and only if is a self-adjoint operator (the previous bullet point is the special case = ). \n\n Suppose () () is defined by \n\n( , ) = tr (). \n\nThen is a symmetric bilinear form on () because trace is a linear functional on () and tr () = tr () for all , () ; see 8.56. \n\n> 9.11 definition:\n\nsymmetric matrix \n\nA square matrix is called symmetric if it equals its transpose. An operator on may have a symmetric matrix with respect to some but not all bases of . In contrast, the next result shows that a bilinear form on has a sym-metric matrix with respect to either all bases of or with respect to no bases of . \n\n> 9.12\n\nsymmetric bilinear forms are diagonalizable \n\nSuppose (2) . Then the following are equivalent. (a) is a symmetric bilinear form on .(b) ( , ( 1, , )) is a symmetric matrix for every basis 1, , of .(c) ( , ( 1, , )) is a symmetric matrix for some basis 1, , of .(d) ( , ( 1, , )) is a diagonal matrix for some basis 1, , of .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n338 Chapter 9 Multilinear Algebra and Determinants Proof First suppose (a) holds, so is a symmetric bilinear form. Suppose \n\n 1, , is a basis of and , {1 , , } . Then ( , ) = ( , ) because \n\nis symmetric. Thus ( , ( 1, , )) is a symmetric matrix, showing that (a) implies (b). Clearly (b) implies (c). Now suppose (c) holds and 1, , is a basis of such that ( , ( 1, , )) \n\nis a symmetric matrix. Suppose , . There exist 1, , , 1, , \n\nsuch that = 1 1 + + and = 1 1 + + . Now \n\n( , ) = ( \n\n> \n> \n> = 1\n\n , \n\n> \n> = 1\n\n )=\n\n> \n> \n> = 1\n> \n> \n> = 1\n\n ( , )=\n\n> \n> \n> = 1\n> \n> \n> = 1\n\n ( , )= ( \n\n> \n> \n> = 1\n\n , \n\n> \n> = 1\n\n )= ( , ) ,\n\nwhere the third line holds because () is a symmetric matrix. The equation above shows that is a symmetric bilinear form, proving that (c) implies (a). At this point, we have proved that (a), (b), (c) are equivalent. Because every diagonal matrix is symmetric, (d) implies (c). To complete the proof, we will show that (a) implies (d) by induction on = dim .If = 1 , then (a) implies (d) because every 1-by-1 matrix is diagonal. Now suppose > 1 and the implication (a) (d) holds for one less dimension. Suppose (a) holds, so is a symmetric bilinear form. If = 0 , then the matrix of \n\n with respect to every basis of is the zero matrix, which is a diagonal matrix. Hence we can assume that 0 , which means there exist , such that \n\n( , ) 0 . Now \n\n2( , ) = ( + , + ) ( , ) ( , ). \n\nBecause the left side of the equation above is nonzero, the three terms on the right cannot all equal 0. Hence there exists such that ( , ) 0 .Let = { ( , ) = 0} . Thus is the null space of the linear functional ( , ) on . This linear functional is not the zero linear functional because . Thus dim = 1 . By our induction hypothesis, there is a basis 1, , 1 of such that the symmetric bilinear form | has a diagonal matrix with respect to this basis. Because , the list 1, , 1 , is a basis of . Suppose {1 , , 1} .Then ( , ) = 0 by the construction of . Because is symmetric, we also have \n\n( , ) = 0 . Thus the matrix of with respect to 1, , 1 , is a diagonal matrix, completing the proof that (a) implies (d). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9A Bilinear Forms and Quadratic Forms 339 \n\nThe previous result states that every symmetric bilinear form has a diagonal matrix with respect to some basis. If our vector space happens to be a real inner product space, then the next result shows that every symmetric bilinear form has a diagonal matrix with respect to some orthonormal basis. Note that the inner product here is unrelated to the bilinear form. \n\n> 9.13\n\ndiagonalization of a symmetric bilinear form by an orthonormal basis \n\nSuppose is a real inner product space and is a symmetric bilinear form on \n\n . Then has a diagonal matrix with respect to some orthonormal basis of .\n\nProof Let 1, , be an orthonormal basis of . Let = ( , ( 1, , )) .Then is a symmetric matrix (by 9.12). Let () be the operator such that \n\n( , ( 1, , )) = . Thus is self-adjoint. The real spectral theorem (7.29) states that has a diagonal matrix with respect to some orthonormal basis 1, , of . Let = ( , ( 1, , ), ( 1, , )) .Thus 1 is the matrix of with respect to the basis 1, , (by 3.84). Hence \n\n 1 is a diagonal matrix. Now \n\n( , ( 1, , )) = t = 1 ,\n\nwhere the first equality holds by 9.7 and the second equality holds because is a unitary matrix with real entries (which implies that 1 = t; see 7.57 ).Now we turn our attention to alternating bilinear forms. Alternating multilinear forms will play a major role in our approach to determinants later in this chapter. \n\n> 9.14 definition:\n\nalternating bilinear form, (2) \n\n> alt\n\nA bilinear form (2) is called alternating if \n\n( , ) = 0 \n\nfor all . The set of alternating bilinear forms on is denoted by (2) \n\n> alt\n\n. \n\n> 9.15 example:\n\nalternating bilinear forms \n\n Suppose 3 and is defined by \n\n(( 1, , ), ( 1, , )) = 1 2 2 1 + 1 3 3 1.\n\nThen is an alternating bilinear form on .\n\n Suppose , . Then the bilinear form on defined by \n\n( , ) = ()() ()() \n\nis alternating. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n340 Chapter 9 Multilinear Algebra and Determinants \n\nThe next result shows that a bilinear form is alternating if and only if switching the order of the two inputs multiplies the output by 1 . \n\n> 9.16\n\ncharacterization of alternating bilinear forms \n\nA bilinear form on is alternating if and only if \n\n( , ) = ( , ) \n\nfor all , .\n\nProof First suppose that is alternating. If , , then \n\n0 = ( + , + ) = ( , ) + ( , ) + ( , ) + ( , ) = ( , ) + ( , ). \n\nThus ( , ) = ( , ) , as desired. To prove the implication in the other direction, suppose ( , ) = ( , ) \n\nfor all , . Then ( , ) = ( , ) for all , which implies that \n\n( , ) = 0 for all . Thus is alternating. Now we show that the vector space of bilinear forms on is the direct sum of the symmetric bilinear forms on and the alternating bilinear forms on . \n\n> 9.17\n\n (2) = (2) \n\n> sym\n\n (2) \n\n> alt\n\nThe sets (2) \n\n> sym\n\nand (2) \n\n> alt\n\nare subspaces of (2) . Furthermore, \n\n (2) = (2) \n\n> sym\n\n (2) \n\n> alt\n\n.\n\nProof The definition of symmetric bilinear form implies that the sum of any two symmetric bilinear forms on is a symmetric bilinear form on , and every scalar multiple of any symmetric bilinear form on is a symmetric bilinear form on . Also, the zero bilinear form is in (2) \n\n> sym\n\n. Thus (2) \n\n> sym\n\nis a subspace of (2) .Similarly, the verification that (2) \n\n> alt\n\nis a subspace of (2) is straightforward. Next, we want to show that (2) = (2) \n\n> sym\n\n+ (2) \n\n> alt\n\n. To do this, suppose (2) .Define , (2) by \n\n( , ) = ( , ) + ( , ) 2 and ( , ) = ( , ) ( , ) 2 .\n\nThen (2) \n\n> sym\n\nand (2) \n\n> alt\n\n, and = + . Thus (2) = (2) \n\n> sym\n\n+ (2) \n\n> alt\n\n.Finally, to show that the intersection of the two subspaces under consideration equals {0} , suppose (2) \n\n> sym\n\n (2) \n\n> alt\n\n. If , , then 9.16 implies that \n\n( , ) = ( , ) = ( , ) \n\nand hence ( , ) = 0 . Thus = 0 . Hence (2) = (2) \n\n> sym\n\n (2) \n\n> alt\n\n(by 1.46). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9A Bilinear Forms and Quadratic Forms 341 \n\n# Quadratic Forms \n\n9.18 definition: quadratic form associated with a bilinear form, \n\nFor a bilinear form on , define a function by () = ( , ) .A function is called a quadratic form on if there exists a bilinear form on such that = .Note that if is a bilinear form, then = 0 if and only if is alternating. \n\n9.19 example: quadratic form \n\nSuppose is the bilinear form on 3 defined by \n\n(( 1, 2, 3), ( 1, 2, 3)) = 1 1 4 1 2 + 8 1 3 3 3 3.\n\nThen is the quadratic form on 3 given by the formula \n\n ( 1, 2, 3) = 12 4 1 2 + 8 1 3 3 32.\n\nThe quadratic form in the example above is typical of quadratic forms on ,as shown in the next result. \n\n9.20 quadratic forms on \n\nSuppose is a positive integer and is a function from to . Then \n\nis a quadratic form on if and only if there exist numbers , for \n\n , {1 , , } such that \n\n( 1, , ) = \n\n> \n\n\n\n> = 1 \n\n\n\n> = 1\n\n , \n\nfor all ( 1, , ) .\n\nProof First suppose is a quadratic form on . Thus there exists a bilinear form \n\n on such that = . Let be the matrix of with respect to the standard basis of . Then for all ( 1, , ) , we have the desired equation \n\n( 1, , ) = (( 1, , ), ( 1, , )) = \n\n> \n\n\n\n> = 1 \n\n\n\n> = 1\n\n , .\n\nConversely, suppose there exist numbers , for , {1 , , } such that \n\n( 1, , ) = \n\n> \n\n\n\n> = 1 \n\n\n\n> = 1\n\n , \n\nfor all ( 1, , ) . Define a bilinear form on by \n\n(( 1, , ), ( 1, , )) = \n\n> \n\n\n\n> = 1 \n\n\n\n> = 1\n\n , .\n\nThen = , as desired. \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler 342 Chapter 9 Multilinear Algebra and Determinants \n\nAlthough quadratic forms are defined in terms of an arbitrary bilinear form, the equivalence of (a) and (b) in the result below shows that a symmetric bilinear form can always be used. \n\n> 9.21\n\ncharacterizations of quadratic forms \n\nSuppose is a function. The following are equivalent. (a) is a quadratic form. (b) There exists a unique symmetric bilinear form on such that = .(c) () = 2() for all and all , and the function \n\n( , ) ( + ) () () \n\nis a symmetric bilinear form on .(d) (2) = 4() for all , and the function \n\n( , ) ( + ) () () \n\nis a symmetric bilinear form on .\n\nProof First suppose (a) holds, so is a quadratic form. Hence there exists a bilinear form such that = . By 9.17, there exist a symmetric bilinear form \n\non and an alternating bilinear form on such that = + . Now \n\n = = + = .\n\nIf (2) \n\n> sym\n\nalso satisfies = , then = 0 ; thus (2) \n\n> sym\n\n (2) \n\n> alt\n\n,which implies that = (by 9.17). This completes the proof that (a) implies (b). Now suppose (b) holds, so there exists a symmetric bilinear form on such that = . If and then \n\n() = ( , ) = ( , ) = 2( , ) = 2() ,\n\nshowing that the first part of (c) holds. If , , then \n\n( + ) () () = ( + , + ) ( , ) ( , ) = 2( , ). \n\nThus the function ( , ) ( +)()() equals 2 , which is a symmetric bilinear form on , completing the proof that (b) implies (c). Clearly (c) implies (d). Now suppose (d) holds. Let be the symmetric bilinear form on defined by \n\n( , ) = ( + ) () () 2 .\n\nIf , then \n\n( , ) = (2) () () 2 = 4() 2() 2 = (). \n\nThus = , completing the proof that (d) implies (a). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9A Bilinear Forms and Quadratic Forms 343 \n\n> 9.22 example:\n\nsymmetric bilinear form associated with a quadratic form \n\nSuppose is the quadratic form on 3 given by the formula \n\n( 1, 2, 3) = 12 4 1 2 + 8 1 3 3 32.\n\nA bilinear form on 3 such that = is given by Example 9.19, but this bilinear form is not symmetric, as promised by 9.21(b). However, the bilinear form on 3 defined by \n\n(( 1, 2, 3), ( 1, 2, 3)) = 1 1 2 1 2 2 2 1 + 4 1 3 + 4 3 1 3 3 3\n\nis symmetric and satisfies = .The next result states that for each quadratic form we can choose a basis such that the quadratic form looks like a weighted sum of squares of the coordinates, meaning that there are no cross terms of the form with . \n\n> 9.23\n\ndiagonalization of quadratic form \n\nSuppose is a quadratic form on .(a) There exist a basis 1, , of and 1, , such that \n\n( 1 1 + + ) = 1 12 + + 2\n\nfor all 1, , .(b) If = and is an inner product space, then the basis in (a) can be chosen to be an orthonormal basis of .\n\nProof \n\n(a) There exists a symmetric bilinear form on such that = (by 9.21). Now there exists a basis 1, , of such that ( , ( 1, , )) is a diagonal matrix (by 9.12). Let 1, , denote the entries on the diagonal of this matrix. Thus \n\n( , ) = {{ if = ,\n\n0 if \n\nfor all , {1 , , } . If 1, , , then \n\n( 1 1 + + ) = ( 1 1 + + , 1 1 + + )=\n\n> \n> \n> = 1 \n> \n> = 1\n\n ( , )= 1 12 + + 2,\n\nas desired. (b) Suppose = and is an inner product space. Then 9.13 tells us that the basis in (a) can be chosen to be an orthonormal basis of .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n344 Chapter 9 Multilinear Algebra and Determinants \n\n# Exercises 9A \n\n1 Prove that if is a bilinear form on , then there exists such that \n\n( , ) = \n\nfor all , .\n\n2 Let = dim . Suppose is a bilinear form on . Prove that there exist \n\n 1, , , 1, , such that \n\n( , ) = 1() 1() + + () () \n\nfor all , .\n\nThis exercise shows that if = dim , then every bilinear form on is of the form given by the last bullet point of Example 9.2. \n\n3 Suppose is a bilinear form on and also is a linear functional on . Prove that = 0 .\n\n4 Suppose is a real inner product space and is a bilinear form on . Show that there exists a unique operator () such that \n\n( , ) = , \n\nfor all , .\n\nThis exercise states that if is a real inner product space, then every bilinear form on is of the form given by the third bullet point in 9.2. \n\n5 Suppose is a bilinear form on a real inner product space and is the unique operator on such that ( , ) = , for all , (see Exercise 4). Show that is an inner product on if and only if is an invertible positive operator on .\n\n6 Prove or give a counterexample: If is a symmetric bilinear form on , then \n\n{ ( , ) = 0} \n\nis a subspace of .\n\n7 Explain why the proof of 9.13 (diagonalization of a symmetric bilinear form by an orthonormal basis on a real inner product space) fails if the hypothesis that = is dropped. \n\n8 Find formulas for dim (2) \n\n> sym\n\nand dim (2) \n\n> alt\n\nin terms of dim .\n\n9 Suppose that is a positive integer and = { () (0) = (1)} .Define by \n\n( , ) = 10\n\n .\n\nShow that is an alternating bilinear form on .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9A Bilinear Forms and Quadratic Forms 345 \n\n10 Suppose that is a positive integer and \n\n = { () (0) = (1) and (0) = (1)}. \n\nDefine by \n\n( , ) = 10\n\n .\n\nShow that is a symmetric bilinear form on .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n346 Chapter 9 Multilinear Algebra and Determinants \n\n# 9B Alternating Multilinear Forms \n\n# Multilinear Forms \n\n> 9.24 definition:\n\n \n\nFor a positive integer, define by \n\n = \n\n> times\n\n.\n\nNow we can define -linear forms as a generalization of the bilinear forms that we discussed in the previous section. \n\n> 9.25 definition:\n\n -linear form, () , multilinear form \n\n For a positive integer, an -linear form on is a function \n\nthat is linear in each slot when the other slots are held fixed. This means that for each {1 , , } and all 1, , , the function \n\n ( 1, , 1 , , + 1, , )\n\nis a linear map from to .\n\n The set of -linear forms on is denoted by () .\n\n A function is called a multilinear form on if it is an -linear form on \n\nfor some positive integer .In the definition above, the expression ( 1, , 1 , , + 1, , ) means \n\n( , 2, , ) if = 1 and means ( 1, , 1 , ) if = .A 1-linear form on is a linear functional on . A 2-linear form on is a bilinear form on . You can verify that with the usual addition and scalar multiplication of functions, () is a vector space. \n\n> 9.26 example:\n\n -linear forms \n\n Suppose , (2) . Define a function 4 by \n\n( 1, 2, 3, 4) = ( 1, 2) ( 3, 4). \n\nThen (4) .\n\n Define (()) by \n\n( 1, , ) = tr ( 1 ). \n\nThen is an -linear form on () .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9B Alternating Multilinear Forms 347 \n\nAlternating multilinear forms, which we now define, play an important role as we head toward defining determinants. \n\n> 9.27 definition:\n\nalternating forms, () \n\n> alt\n\nSuppose is a positive integer. \n\n An -linear form on is called alternating if ( 1, , ) = 0 whenever \n\n 1, , is a list of vectors in with = for some two distinct values of and in {1 , , } .\n\n () \n\n> alt\n\n= { () is an alternating -linear form on } .You should verify that () \n\n> alt\n\nis a subspace of () . See Example 9.15 for examples of alternating 2-linear forms. See Exercise 2 for an example of an alternating 3-linear form. The next result tells us that if a linearly dependent list is input to an alternating multilinear form, then the output equals 0. \n\n> 9.28\n\nalternating multilinear forms and linear dependence \n\nSuppose is a positive integer and is an alternating -linear form on . If \n\n 1, , is a linearly dependent list in , then \n\n( 1, , ) = 0. \n\nProof Suppose 1, , is a linearly dependent list in . By the linear depen-dence lemma (2.19), some is a linear combination of 1, , 1 . Thus there exist 1, , 1 such that = 1 1 + + 1 1 . Now \n\n( 1, , ) = ( 1, , 1 , 1 \n\n> \n> = 1\n\n , + 1, , )=\n\n> 1\n> \n> = 1\n\n ( 1, , 1 , , + 1, , )= 0. \n\nThe next result states that if > dim , then there are no alternating -linear forms on other than the function on that is identically 0. \n\n> 9.29\n\nno nonzero alternating -linear forms for > dim \n\nSuppose > dim . Then 0 is the only alternating -linear form on .\n\nProof Suppose that is an alternating -linear form on and 1, , .Because > dim , this list is not linearly independent (by 2.22). Thus 9.28 implies that ( 1, , ) = 0 . Hence is the zero function from to .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n348 Chapter 9 Multilinear Algebra and Determinants \n\n# Alternating Multilinear Forms and Permutations \n\n> 9.30\n\nswapping input vectors in an alternating multilinear form \n\nSuppose is a positive integer, is an alternating -linear form on , and \n\n 1, , is a list of vectors in . Then swapping the vectors in any two slots of ( 1, , ) changes the value of by a factor of 1 .\n\nProof Put 1 + 2 in both the first two slots, getting \n\n0 = ( 1 + 2, 1 + 2, 3, , ). \n\nUse the multilinear properties of to expand the right side of the equation above (as in the proof of 9.16) to get \n\n( 2, 1, 3, , ) = ( 1, 2, 3, , ). \n\nSimilarly, swapping the vectors in any two slots of ( 1, , ) changes the value of by a factor of 1 .To see what can happen with multiple swaps, suppose is an alternating \n\n3-linear form on and 1, 2, 3 . To evaluate ( 3, 1, 2) in terms of \n\n( 1, 2, 3), start with ( 3, 1, 2) and swap the entries in the first and third slots, getting ( 3, 1, 2) = ( 2, 1, 3). Now in the last expression, swap the entries in the first and second slots, getting \n\n( 3, 1, 2) = ( 2, 1, 3) = ( 1, 2, 3). \n\nMore generally, we see that if we do an odd number of swaps, then the value of \n\nchanges by a factor of 1 , and if we do an even number of swaps, then the value of does not change. To deal with arbitrary multiple swaps, we need a bit of information about permutations. \n\n> 9.31 definition:\n\npermutation, perm \n\nSuppose is a positive integer. \n\n A permutation of (1 , , ) is a list ( 1, , ) that contains each of the numbers 1, , exactly once. \n\n The set of all permutations of (1 , , ) is denoted by perm .For example, (2 , 3, 4, 5, 1) perm 5. You should think of an element of \n\nperm as a rearrangement of the first positive integers. The number of swaps used to change a permutation ( 1, , ) to the stan-dard order (1 , , ) can depend on the specific swaps selected. The following definition has the advantage of assigning a well-defined sign to every permutation. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 9B Alternating Multilinear Forms 349\n> 9.32 definition:\n\nsign of a permutation \n\nThe sign of a permutation ( 1, , ) is defined by \n\nsign ( 1, , ) = (1) ,\n\nwhere is the number of pairs of integers ( , ) with 1 < such that appears after in the list ( 1, , ).Hence the sign of a permutation equals 1 if the natural order has been changed an even number of times and equals 1 if the natural order has been changed an odd number of times. \n\n> 9.33 example:\n\nsigns \n\n The permutation (1 , , ) [no changes in the natural order] has sign 1.\n\n The only pair of integers ( , ) with < such that appears after in the list \n\n(2 , 1, 3, 4) is (1 , 2) . Thus the permutation (2 , 1, 3, 4) has sign 1 .\n\n In the permutation (2 , 3, , , 1) , the only pairs ( , ) with < that appear with changed order are (1 , 2) , (1 , 3) , , (1 , ) . Because we have 1 such pairs, the sign of this permutation equals (1) 1 . \n\n> 9.34\n\nswapping two entries in a permutation \n\nSwapping two entries in a permutation multiplies the sign of the permutation by 1 . \n\n> Proof\n\nSuppose we have two permutations, where the second permutation is obtained from the first by swapping two entries. The two swapped entries were in their natural order in the first permutation if and only if they are not in their natural order in the second permutation. Thus we have a net change (so far) of 1\n\nor 1 (both odd numbers) in the number of pairs not in their natural order. Consider each entry between the two swapped entries. If an intermediate entry was originally in the natural order with respect to both swapped entries, then it is now in the natural order with respect to neither swapped entry. Similarly, if an intermediate entry was originally in the natural order with respect to neither of the swapped entries, then it is now in the natural order with respect to both swapped entries. If an intermediate entry was originally in the natural order with respect to exactly one of the swapped entries, then that is still true. Thus the net change (for each pair containing an entry between the two swapped entries) in the number of pairs not in their natural order is 2, 2 , or 0 (all even numbers). For all other pairs of entries, there is no change in whether or not they are in their natural order. Thus the total net change in the number of pairs not in their natural order is an odd number. Hence the sign of the second permutation equals \n\n1 times the sign of the first permutation. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n350 Chapter 9 Multilinear Algebra and Determinants \n\n9.35 permutations and alternating multilinear forms \n\nSuppose is a positive integer and () \n\n> alt\n\n. Then \n\n( 1 , , ) = ( sign ( 1, , ))( 1, , )\n\nfor every list 1, , of vectors in and all ( 1, , ) perm .\n\nProof Suppose 1, , and ( 1, , ) perm . We can get from \n\n( 1, , ) to (1 , , ) by a series of swaps of entries in different slots. Each such swap changes the value of by a factor of 1 (by 9.30) and also changes the sign of the remaining permutation by a factor of 1 (by 9.34). After an appropriate number of swaps, we reach the permutation 1, , , which has sign 1. Thus the value of changed signs an even number of times if sign ( 1, , ) = 1 and an odd number of times if sign ( 1, , ) = 1 , which gives the desired result. Our use of permutations now leads in a natural way to the following beautiful formula for alternating -linear forms on an -dimensional vector space. \n\n9.36 formula for (dim ) -linear alternating forms on \n\nLet = dim . Suppose 1, , is a basis of and 1, , . For each \n\n {1 , , } , let 1, , , , be such that \n\n =\n\n> \n\n\n\n> = 1\n\n , .\n\nThen \n\n( 1, , ) = ( 1, , ) \n\n> ( 1,, ) perm \n\n(sign ( 1, , )) 1,1 , \n\nfor every alternating -linear form on .\n\nProof Suppose is an alternating -linear form on . Then \n\n( 1, , ) = ( \n\n> \n\n \n\n> 1= 1\n\n 1,1 1 , , \n\n \n\n> = 1\n\n , )=\n\n> \n\n \n\n> 1= 1\n\n\n\n> \n\n \n\n> = 1\n\n 1,1 , ( 1 , , )= \n\n> ( 1,, ) perm \n\n 1,1 , ( 1 , , )= ( 1, , ) \n\n> ( 1,, ) perm \n\n(sign ( 1, , )) 1,1 , ,\n\nwhere the third line holds because ( 1 , , ) = 0 if 1, , are not distinct integers, and the last line holds by 9.35. \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler Section 9B Alternating Multilinear Forms 351 \n\nThe following result will be the key to our definition of the determinant in the next section. \n\n> 9.37\n\ndim (dim ) \n\n> alt\n\n= 1 \n\nThe vector space (dim ) \n\n> alt\n\nhas dimension one. \n\nProof Let = dim . Suppose and are alternating -linear forms on with \n\n 0 . Let 1, , be such that ( 1, , ) 0 . There exists such that \n\n ( 1, , ) = ( 1, , ). \n\nFurthermore, 9.28 implies that 1, , is linearly independent and thus is a basis of .Suppose 1, , . Let , be as in 9.36 for , = 1 , , . Then \n\n ( 1, , ) = ( 1, , ) \n\n> ( 1,, ) perm \n\n(sign ( 1, , )) 1,1 , \n\n= ( 1, , ) \n\n> ( 1,, ) perm \n\n(sign ( 1, , )) 1,1 , \n\n= ( 1, , ),\n\nwhere the first and last lines above come from 9.36. The equation above implies that = . Thus , is not a linearly independent list, which implies that \n\ndim () \n\n> alt\n\n 1 .To complete the proof, we only need to show that there exists a nonzero alternating -linear form on ( thus eliminating the possibility that dim () \n\n> alt\n\nequals 0) . To do this, let 1, , be any basis of , and let 1, , be the linear functionals on that allow us to express each element of as a linear combination of 1, , . In other words, \n\n = \n\n> \n\n\n\n> = 1\n\n () \n\nfor every (see 3.114). Now for 1, , , define \n\n> 9.38\n\n( 1, , ) = \n\n> ( 1,, ) perm \n\n(sign ( 1, , )) 1 ( 1) ( ). \n\nThe verification that is an -linear form on is straightforward. To see that is alternating, suppose 1, , with 1 = 2. For each \n\n( 1, , ) perm , the permutation ( 2, 1, 3, , ) has the opposite sign. Be-cause 1 = 2, the contributions from these two permutations to the sum in 9.38 cancel either other. Hence ( 1, 1, 3, , ) = 0 . Similarly, ( 1, , ) = 0 \n\nif any two vectors in the list 1, , are equal. Thus is alternating. Finally, consider 9.38 with each = . Because ( ) equals 0 if and equals 1 if = , only the permutation (1 , , ) makes a nonzero contribution to the right side of 9.38 in this case, giving the equation ( 1, , ) = 1 . Thus we have produced a nonzero alternating -linear form on , as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n352 Chapter 9 Multilinear Algebra and Determinants \n\nThe formula 9.38 used in the last proof to construct a nonzero alternating -linear form came from the formula in 9.36, and that formula arose naturally from the properties of an alternating multilinear form. \n\nEarlier we showed that the value of an alternating multilinear form applied to a linearly dependent list is 0; see 9.28. The next result provides a converse of 9.28 for -linear multilinear forms when \n\n = dim . In the following result, the statement that is nonzero means (as usual for a function) that is not the function on that is identically 0. \n\n> 9.39\n\nalternating (dim ) -linear forms and linear independence \n\nLet = dim . Suppose is a nonzero alternating -linear form on and \n\n 1, , is a list of vectors in . Then \n\n( 1, , ) 0 \n\nif and only if 1, , is linearly independent. \n\nProof First suppose ( 1, , ) 0 . Then 9.28 implies that 1, , is linearly independent. To prove the implication in the other direction, now suppose 1, , is linearly independent. Because = dim , this implies that 1, , is a basis of (see 2.38). Because is not the zero -linear form, there exist 1, , such that \n\n( 1, , ) 0 . Now 9.36 implies that ( 1, , ) 0 .\n\n# Exercises 9B \n\n1 Suppose is a positive integer. Show that dim () = ( dim ) .\n\n2 Suppose 3 and is defined by \n\n(( 1, , ), ( 1, , ), ( 1, , )) = 1 2 3 2 1 3 3 2 1 1 3 2 + 3 1 2 + 2 3 1.\n\nShow that is an alternating 3-linear form on .\n\n3 Suppose is a positive integer and is an -linear form on such that \n\n( 1, , ) = 0 whenever 1, , is a list of vectors in with = + 1\n\nfor some {1 , , 1} . Prove that is an alternating -linear form on .\n\n4 Prove or give a counterexample: If (4) \n\n> alt\n\n, then \n\n{( 1, 2, 3, 4) 4 ( 1, 2, 3, 4) = 0} \n\nis a subspace of 4.\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9B Alternating Multilinear Forms 353 \n\n5 Suppose is a positive integer and is an -linear form on . Define an \n\n -linear form on by \n\n( 1, , ) = \n\n> ( 1,, ) perm \n\n(sign ( 1, , ))( 1 , , )\n\nfor 1, , . Explain why () \n\n> alt\n\n.\n\n6 Suppose is a positive integer and is an -linear form on . Define an \n\n -linear form on by \n\n( 1, , ) = \n\n> ( 1,, ) perm \n\n( 1 , , )\n\nfor 1, , . Explain why \n\n( 1 , , ) = ( 1, , )\n\nfor all 1, , and all ( 1, , ) perm .\n\n7 Give an example of a nonzero alternating 2-linear form on 3 and a linearly independent list 1, 2 in 3 such that ( 1, 2) = 0 .\n\nThis exercise shows that 9.39 can fail if the hypothesis that = dim is deleted. \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler 354 Chapter 9 Multilinear Algebra and Determinants \n\n# 9C Determinants \n\n# Defining the Determinant \n\nThe next definition will lead us to a clean, beautiful, basis-free definition of the determinant of an operator. \n\n> 9.40 definition:\n\n \n\nSuppose that is a positive integer and () . For () \n\n> alt\n\n, define \n\n () \n\n> alt\n\nby \n\n ( 1, , ) = ( 1, , )\n\nfor each list 1, , of vectors in .Suppose () . If () \n\n> alt\n\nand 1, , is a list of vectors in with \n\n = for some , then = , which implies that ( 1, , ) = ( 1, , ) = 0 . Thus the function is a linear map of () \n\n> alt\n\nto itself. We know that dim (dim ) \n\n> alt\n\n= 1 (see 9.37). Every linear map from a one-dimensional vector space to itself is multiplication by some unique scalar. For the linear map , we now define det to be that scalar. \n\n> 9.41 definition:\n\ndeterminant of an operator, det \n\nSuppose () . The determinant of , denoted by det , is defined to be the unique number in such that \n\n = ( det ) \n\nfor all (dim ) \n\n> alt\n\n. \n\n> 9.42 example:\n\ndeterminants of operators \n\nLet = dim .\n\n If is the identity operator on , then = for all () \n\n> alt\n\n. Thus det = 1 .\n\n More generally, if , then = for all () \n\n> alt\n\n. Thus det () = .\n\n Still more generally, if () and , then = = (det ) \n\nfor all () \n\n> alt\n\n. Thus det () = det .\n\n Suppose () and there is a basis 1, , of consisting of eigenvectors of , with corresponding eigenvalues 1, , . If () \n\n> alt\n\n, then \n\n ( 1, , ) = ( 1 1, , ) = ( 1 )( 1, , ). \n\nIf 0 , then 9.39 implies ( 1, , ) 0 . Thus the equation above implies \n\ndet = 1 .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9C Determinants 355 \n\nOur next task is to define and give a formula for the determinant of a square matrix. To do this, we associate with each square matrix an operator and then define the determinant of the matrix to be the determinant of the associated operator. \n\n> 9.43 definition:\n\ndeterminant of a matrix, det \n\nSuppose that is a positive integer and is an -by- square matrix with entries in . Let ( ) be the operator whose matrix with respect to the standard basis of equals . The determinant of , denoted by det , is defined by det = det . \n\n> 9.44 example:\n\ndeterminants of matrices \n\n If is the -by- identity matrix, then the corresponding operator on is the identity operator on . Thus the first bullet point of 9.42 implies that the determinant of the identity matrix is 1.\n\n Suppose is a diagonal matrix with 1, , on the diagonal. Then the corresponding operator on has the standard basis of as eigenvectors, with eigenvalues 1, , . Thus the last bullet point of 9.42 implies that \n\ndet = 1 .For the next result, think of each list 1, , of vectors in as a list of \n\n -by-1 column vectors. The notation ( 1 ) then denotes the -by- \n\nsquare matrix whose th column is for each = 1 , , . \n\n> 9.45\n\ndeterminant is an alternating multilinear form \n\nSuppose that is a positive integer. The map that takes a list 1, , of vectors in to det ( 1 ) is an alternating -linear form on .\n\nProof Let 1, , be the standard basis of and suppose 1, , is a list of vectors in . Let ( ) be the operator such that = for = 1 , , .Thus is the operator whose matrix with respect to 1, , is ( 1 ).Hence det ( 1 ) = det , by definition of the determinant of a matrix. Let be an alternating -linear form on such that ( 1, , ) = 1 . Then \n\ndet ( 1 ) = det = ( det ) ( 1, , )= ( 1, , )= ( 1, , ),\n\nwhere the third line follows from the definition of the determinant of an operator. The equation above shows that the map that takes a list of vectors 1, , in \n\nto det ( 1 ) is the alternating -linear form on .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n356 Chapter 9 Multilinear Algebra and Determinants \n\nThe previous result has several important consequences. For example, it immediately implies that a matrix with two identical columns has determinant 0.We will come back to other consequences later, but for now we want to give a formula for the determinant of a square matrix. Recall that if is a matrix, then \n\n , denotes the entry in row , column of . \n\n> 9.46\n\nformula for determinant of a matrix \n\nSuppose that is a positive integer and is an -by- square matrix. Then \n\ndet = \n\n> ( 1,, ) perm \n\n(sign ( 1, , )) 1,1 , .\n\nProof Apply 9.36 with = and 1, , the standard basis of and the alternating -linear form on that takes 1, , to det ( 1 ) [see 9.45]. If each is the th column of , then each , in 9.36 equals , . Finally, \n\n( 1, , ) = det ( 1 ) = det = 1. \n\nThus the formula in 9.36 becomes the formula stated in this result. \n\n> 9.47 example:\n\nexplicit formula for determinant \n\n If is a 2-by-2 matrix, then the formula in 9.46 becomes \n\ndet = 1,1 2,2 2,1 1,2. If is a 3-by-3 matrix, then the formula in 9.46 becomes \n\ndet = 1,1 2,2 3,3 2,1 1,2 3,3 3,1 2,2 1,3\n\n 1,1 3,2 2,3 + 3,1 1,2 2,3 + 2,1 3,2 1,3.\n\nThe sum in the formula in 9.46 contains ! terms. Because ! grows rapidly as \n\n increases, the formula in 9.46 is not a viable method to evaluate determinants even for moderately sized . For example, 10! is over three million, and 100! is approximately 10 158 , leading to a sum that the fastest computer cannot evaluate. We will soon see some results that lead to faster evaluations of determinants than direct use of the sum in 9.46. \n\n> 9.48\n\ndeterminant of upper-triangular matrix \n\nSuppose that is an upper-triangular matrix with 1, , on the diagonal. Then det = 1 .\n\nProof If ( 1, , ) perm with ( 1, , ) (1 , , ) , then > for some \n\n {1 , , } , which implies that , = 0 . Thus the only permutation that can make a nonzero contribution to the sum in 9.46 is the permutation (1 , , ) .Because , = for each = 1 , , , this implies that det = 1 .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9C Determinants 357 \n\n# Properties of Determinants \n\nOur definition of the determinant leads to the following magical proof that the determinant is multiplicative. \n\n> 9.49\n\ndeterminant is multiplicative \n\n(a) Suppose , () . Then det () = ( det )( det ) .(b) Suppose and are square matrices of the same size. Then \n\ndet () = ( det )( det ) \n\nProof \n\n(a) Let = dim . Suppose () \n\n> alt\n\nand 1, , . Then \n\n ( 1, , ) = ( 1, , )= ( det )( 1, , )= ( det )( det )( 1, , ),\n\nwhere the first equation follows from the definition of , the second equation follows from the definition of det , and the third equation follows from the definition of det . The equation above implies that det () = ( det )( det ) .(b) Let , ( ) be such that () = and () = , where all matrices of operators in this proof are with respect to the standard basis of . Then \n\n() = () () = (see 3.43). Thus \n\ndet () = det () = ( det )( det ) = ( det )( det ) ,\n\nwhere the second equality comes from the result in (a). The determinant of an operator determines whether the operator is invertible. \n\n> 9.50\n\ninvertible nonzero determinant \n\nAn operator () is invertible if and only if det 0 . Furthermore, if \n\n is invertible, then det ( 1 ) = 1 \n\n> det \n\n.\n\nProof First suppose is invertible. Thus 1 = . Now 9.49 implies that \n\n1 = det = det ( 1 ) = ( det )( det ( 1 )). \n\nHence det 0 and det ( 1 ) is the multiplicative inverse of det .To prove the other direction, now suppose det 0 . Suppose and \n\n 0 . Let , 2, , be a basis of and let () \n\n> alt\n\nbe such that 0 . Then \n\n( , 2, , ) 0 (by 9.39). Now \n\n( , 2, , ) = ( det )( , 2, , ) 0. \n\nThus 0 . Hence is invertible. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n358 Chapter 9 Multilinear Algebra and Determinants \n\nAn -by- matrix is invertible (see 3.80 for the definition of an invertible matrix) if and only if the operator on associated with ( via the standard basis of ) is invertible. Thus the previous result shows that a square matrix is invertible if and only if det 0 . \n\n> 9.51\n\neigenvalues and determinants \n\nSuppose () and . Then is an eigenvalue of if and only if \n\ndet ( ) = 0 .\n\nProof The number is an eigenvalue of if and only if is not invertible (see 5.7), which happens if and only if is not invertible, which happens if and only if det ( ) = 0 (by 9.50). Suppose () and is an invertible linear map. To prove that \n\ndet ( 1 ) = det , we could try to use 9.49 and 9.50, writing \n\ndet ( 1 ) = ( det 1 )( det )( det ) = det . \n\nThat proof works if = , but if then it makes no sense because the determinant is defined only for linear maps from a vector space to itself, and \n\nmaps to , making det undefined. The proof given below works around this issue and is valid when . \n\n> 9.52\n\ndeterminant is a similarity invariant \n\nSuppose () and is an invertible linear map. Then \n\ndet ( 1 ) = det . \n\nProof Let = dim = dim . Suppose () \n\n> alt\n\n. Define () \n\n> alt\n\nby \n\n( 1, , ) = ( 1 1, , 1 )\n\nfor 1, , . Suppose 1, , . Then \n\n 1 ( 1, , ) = ( 1 1, , 1 )= ( 1, , )= ( 1, , )= ( det )( 1, , )= ( det )( 1, , ). \n\nThe equation above and the definition of the determinant of the operator 1 \n\nimply that det ( 1 ) = det . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 9C Determinants 359\n\nFor the special case in which = and 1, , is the standard basis of ,the next result is true by the definition of the determinant of a matrix. The left side of the equation in the next result does not depend on a choice of basis, which means that the right side is independent of the choice of basis. \n\n> 9.53\n\ndeterminant of operator equals determinant of its matrix \n\nSuppose () and 1, , is a basis of . Then \n\ndet = det ( , ( 1, , )). \n\n> Proof\n\nLet 1, , be the standard basis of . Let be the linear map such that = for each = 1 , , . Thus ( , ( 1, , ), ( 1, , )) and \n\n( 1 , ( 1, , ), ( 1, , )) both equal the -by- identity matrix. Hence \n\n> 9.54\n\n( 1 , ( 1, , )) = ( , ( 1, , )) ,\n\nas follows from two applications of 3.43. Thus \n\ndet = det ( 1 ) = det ( 1 , ( 1, , )) = det ( , ( 1, , )) ,\n\nwhere the first line comes from 9.52, the second line comes from the definition of the determinant of a matrix, and the third line follows from 9.54. The next result gives a more intuitive way to think about determinants than the definition or the formula in 9.46. We could make the characterization in the result below the definition of the determinant of an operator on a finite-dimensional complex vector space, with the current definition then becoming a consequence of that definition. \n\n> 9.55\n\nif = , then determinant equals product of eigenvalues \n\nSuppose = and () . Then det equals the product of the eigen-values of , with each eigenvalue included as many times as its multiplicity. \n\n> Proof\n\nThere is a basis of with respect to which has an upper-triangular matrix with the diagonal entries of the matrix consisting of the eigenvalues of ,with each eigenvalue included as many times as its multiplicitysee 8.37. Thus 9.53 and 9.48 imply that det equals the product of the eigenvalues of , with each eigenvalue included as many times as its multiplicity. As the next result shows, the determinant interacts nicely with the transpose of a square matrix, with the dual of an operator, and with the adjoint of an operator on an inner product space. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 360 Chapter 9 Multilinear Algebra and Determinants\n> 9.56\n\ndeterminant of transpose, dual, or adjoint \n\n(a) Suppose is a square matrix. Then det t = det .(b) Suppose () . Then det = det .(c) Suppose is an inner product space and () . Then \n\ndet ( ) = det . \n\n> Proof\n\n(a) Let be a positive integer. Define ( ) by \n\n( 1, , ) = det (( 1 )t)\n\nfor all 1, , . The formula in 9.46 for the determinant of a matrix shows that is an -linear form on .Suppose 1, , and = for some . If is an -by- \n\nmatrix, then ( 1 )t cannot equal the identity matrix because row \n\n and row of ( 1 )t are equal. Thus ( 1 )t is not invertible, which implies that ( 1, , ) = 0 . Hence is an alternating \n\n -linear form on .Note that applied to the standard basis of equals 1. Because the vector space of alternating -linear forms on has dimension one (by 9.37), this implies that is the determinant function. Thus (a) holds. (b) The equation det = det follows from (a) and 9.53 and 3.132. (c) Pick an orthonormal basis of . The matrix of with respect to that basis is the conjugate transpose of the matrix of with respect to that basis (by 7.9). Thus 9.53, 9.46, and (a) imply that det ( ) = det . \n\n> 9.57\n\nhelpful results in evaluating determinants \n\n(a) If either two columns or two rows of a square matrix are equal, then the determinant of the matrix equals 0.(b) Suppose is a square matrix and is the matrix obtained from by swapping either two columns or two rows. Then det = det .(c) If one column or one row of a square matrix is multiplied by a scalar, then the value of the determinant is multiplied by the same scalar. (d) If a scalar multiple of one column of a square matrix is added to another column, then the value of the determinant is unchanged. (e) If a scalar multiple of one row of a square matrix is added to another row, then the value of the determinant is unchanged. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 9C Determinants 361\n> Proof\n\nAll the assertions in this result follow from the result that the maps \n\n 1, , det ( 1 ) and 1, , det ( 1 )t are both alternating -linear forms on [see 9.45 and 9.56(a) ].For example, to prove (d) suppose 1, , and . Then \n\ndet ( 1 + 2 2 )= det ( 1 2 ) + det ( 2 2 3 )= det ( 1 2 ),\n\nwhere the first equation follows from the multilinearity property and the second equation follows from the alternating property. The equation above shows that adding a multiple of the second column to the first column does not change the value of the determinant. The same conclusion holds for any two columns. Thus (d) holds. The proof of (e) follows from (d) and from 9.56(a). The proofs of (a), (b), and (c) use similar tools and are left to the reader. For matrices whose entries are concrete numbers, the result above leads to a much faster way to evaluate the determinant than direct application of the formula in 9.46. Specifically, apply the Gaussian elimination procedure of swapping rows [by 9.57(b), this changes the determinant by a factor of 1] , multiplying a row by a nonzero constant [by 9.57(c), this changes the determinant by the same constant ], and adding a multiple of one row to another row [by 9.57(e), this does not change the determinant ] to produce an upper-triangular matrix, whose determinant is the product of the diagonal entries (by 9.48). If your software keeps track of the number of row swaps and of the constants used when multiplying a row by a constant, then the determinant of the original matrix can be computed. Because a number is an eigenvalue of an operator () if and only if det ( ) = 0 (by 9.51), you may be tempted to think that one way to find eigenvalues quickly is to choose a basis of , let = () , evaluate \n\ndet ( ) , and then solve the equation det ( ) = 0 for . However, that procedure is rarely efficient, except when dim = 2 (or when dim equals 3 or \n\n4 if you are willing to use the cubic or quartic formulas). One problem is that the procedure described in the paragraph above for evaluating a determinant does not work when the matrix includes a symbol (such as the in ). This problem arises because decisions need to be made in the Gaussian elimination procedure about whether certain quantities equal 0, and those decisions become complicated in expressions involving a symbol .Recall that an operator on a finite-dimensional inner product space is unitary if it preserves norms (see 7.51 and the paragraph following it). Every eigenvalue of a unitary operator has absolute value 1 (by 7.54). Thus the product of the eigenvalues of a unitary operator has absolute value 1. Hence (at least in the case \n\n = ) the determinant of a unitary operator has absolute value 1 (by 9.55). The next result gives a proof that works without the assumption that = . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 362 Chapter 9 Multilinear Algebra and Determinants\n> 9.58\n\nevery unitary operator has determinant with absolute value 1\n\nSuppose is an inner product space and () is a unitary operator. Then |det | = 1 . \n\n> Proof\n\nBecause is unitary, = (see 7.53). Thus \n\n1 = det ( ) = ( det )( det ) = ( det )( det ) = | det | 2,\n\nwhere the second equality comes from 9.49(a) and the third equality comes from 9.56(c). The equation above implies that |det | = 1 .The determinant of a positive operator on an inner product space meshes well with the analogy that such operators correspond to the nonnegative real numbers. \n\n> 9.59\n\nevery positive operator has nonnegative determinant \n\nSuppose is an inner product space and () is a positive operator. Then det 0 . \n\n> Proof\n\nBy the spectral theorem (7.29 or 7.31), has an orthonormal basis con-sisting of eigenvectors of . Thus by the last bullet point of 9.42, det equals a product of the eigenvalues of , possibly with repetitions. Each eigenvalue of is a nonnegative number (by 7.38). Thus we conclude that det 0 .Suppose is an inner product space and () . Recall that the list of nonnegative square roots of the eigenvalues of (each included as many times as its multiplicity) is called the list of singular values of (see Section 7E). \n\n> 9.60\n\n|det | = product of singular values of \n\nSuppose is an inner product space and () . Then \n\n|det | = det ( ) = product of singular values of . \n\n> Proof\n\nWe have \n\n|det | 2 = ( det )( det ) = ( det ( ))( det ) = det ( ) ,\n\nwhere the middle equality comes from 9.56(c) and the last equality comes from 9.49(a). Taking square roots of both sides of the equation above shows that \n\n|det | = det ( ) .Let 1, , denote the list of singular values of . Thus 12, , 2 is the list of eigenvalues of (with appropriate repetitions), corresponding to an orthonormal basis of consisting of eigenvectors of . Hence the last bullet point of 9.42 implies that \n\ndet ( ) = 12 2.\n\nThus |det | = 1 , as desired. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Section 9C Determinants 363\n\nAn operator on a real inner product space changes volume by a factor of the product of the singular values (by 7.111). Thus the next result follows immediately from 7.111 and 9.60. This result explains why the absolute value of a determinant appears in the change of variables formula in multivariable calculus. \n\n> 9.61\n\n changes volume by factor of |det | \n\nSuppose ( ) and . Then \n\nvolume () = | det |( volume ). \n\nFor operators on finite-dimensional complex vector spaces, we now connect the determinant to a polynomial that we have previously seen. \n\n> 9.62\n\nif = , then characteristic polynomial of equals det ( ) \n\nSuppose = and () . Let 1, , denote the distinct eigenvalues of , and let 1, , denote their multiplicities. Then \n\ndet ( ) = ( 1) 1 ( ) . \n\n> Proof\n\nThere exists a basis of with respect to which has an upper-triangular matrix with each appearing on the diagonal exactly times (by 8.37). With respect to this basis, has an upper-triangular matrix with appearing on the diagonal exactly times for each . Thus 9.48 gives the desired equation. Suppose = and () . The characteristic polynomial of was defined in 8.26 as the polynomial on the right side of the equation in 9.62. We did not previously define the characteristic polynomial of an operator on a finite-dimensional real vector space because such operators may have no eigenvalues, making a definition using the right side of the equation in 9.62 inappropriate. We now present a new definition of the characteristic polynomial, motivated by 9.62. This new definition is valid for both real and complex vector spaces. The equation in 9.62 shows that this new definition is equivalent to our previous definition when = (8.26). \n\n> 9.63 definition:\n\ncharacteristic polynomial \n\nSuppose () . The polynomial defined by \n\n det ( ) \n\nis called the characteristic polynomial of .The formula in 9.46 shows that the characteristic polynomial of an opera-tor () is a monic polynomial of degree dim . The zeros in of the characteristic polynomial of are exactly the eigenvalues of (by 9.51). \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 364 Chapter 9 Multilinear Algebra and Determinants\n\nPreviously we proved the CayleyHamilton theorem (8.29) in the complex case. Now we can extend that result to operators on real vector spaces. \n\n> 9.64\n\nCayleyHamilton theorem \n\nSuppose () and is the characteristic polynomial of . Then () = 0 . \n\n> Proof\n\nIf = , then the equation () = 0 follows from 9.62 and 8.29. Now suppose = . Fix a basis of , and let be the matrix of with respect to this basis. Let be the operator on dim such that the matrix of (with respect to the standard basis of dim ) is . For all we have \n\n() = det ( ) = det ( ) = det ( ). \n\nThus is the characteristic polynomial of . The case = (first sentence of this proof) now implies that 0 = () = () = () .The CayleyHamilton theorem (9.64) implies that the characteristic polyno-mial of an operator () is a polynomial multiple of the minimal polynomial of (by 5.29). Thus if the degree of the minimal polynomial of equals dim ,then the characteristic polynomial of equals the minimal polynomial of . This happens for a very large percentage of operators, including over 99.999% of \n\n4-by-4 matrices with integer entries in [100 , 100] (see the paragraph following 5.25). The last sentence in our next result was previously proved in the complex case (see 8.54). Now we can give a proof that works on both real and complex vector spaces. \n\n> 9.65\n\ncharacteristic polynomial, trace, and determinant \n\nSuppose () . Let = dim . Then the characteristic polynomial of \n\ncan be written as \n\n ( tr ) 1 + + (1) (det ). \n\n> Proof\n\nThe constant term of a polynomial function of is the value of the poly-nomial when = 0 . Thus the constant term of the characteristic polynomial of \n\nequals det () , which equals (1) det (by the third bullet point of 9.42). Fix a basis of , and let be the matrix of with respect to this basis. The matrix of with respect to this basis is . The term coming from the identity permutation {1 , , } in the formula 9.46 for det ( ) is \n\n( 1,1)( , ). \n\nThe coefficient of 1 in the expression above is ( 1,1 ++ , ), which equals \n\n tr . The terms in the formula for det ( ) coming from other elements of \n\nperm contain at most 2 factors of the form , and thus do not contribute to the coefficient of 1 in the characteristic polynomial of .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9C Determinants 365 \n\nThe next result was proved by Jacques Hadamard (18651963 ) in 1893. \n\nIn the result below, think of the columns of the -by- matrix as ele-ments of . The norms appearing below then arise from the standard inner product on . Recall that the notation , in the proof below means the th column of the matrix (as was defined in 3.44). \n\n> 9.66\n\nHadamards inequality \n\nSuppose is an -by- matrix. Let 1, , denote the columns of . Then \n\n|det | \n\n> \n\n\n\n> = 1\n\n . \n\nProof If is not invertible, then det = 0 and hence the desired inequality holds in this case. Thus assume that is invertible. The QR factorization (7.58) tells us that there exist a unitary matrix and an upper-triangular matrix whose diagonal contains only positive numbers such that = . We have \n\n|det | = | det | | det | = | det | =\n\n> \n\n\n\n> = 1\n\n , \n\n\n\n> \n\n\n\n> = 1\n\n , =\n\n> \n\n\n\n> = 1\n\n , =\n\n> \n\n\n\n> = 1\n\n ,\n\nwhere the first line comes from 9.49(b), the second line comes from 9.58, the third line comes from 9.48, and the fifth line holds because is an isometry. To give a geometric interpretation to Hadamards inequality, suppose = .Let ( ) be the operator such that = for each = 1 , , , where \n\n 1, , is the standard basis of . Then maps the box ( 1, , ) onto the parallelepiped ( 1, , ) [see 7.102 and 7.105 for a review of this notation and terminology]. Because the box ( 1, , ) has volume 1, this implies (by 9.61) that the parallelepiped ( 1, , ) has volume |det | , which equals |det | .Thus Hadamards inequality above can be interpreted to say that among all paral-lelepipeds whose edges have lengths 1, , , the ones with largest volume have orthogonal edges (and thus have volume = 1 ) .For a necessary and sufficient condition for Hadamards inequality to be an equality, see Exercise 18. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n366 Chapter 9 Multilinear Algebra and Determinants \n\nThe matrix in the next result is called the Vandermonde matrix . Vandermonde matrices have important applications in polynomial interpolation, the discrete Fourier transform, and other areas of mathematics. The proof of the next result is a nice illustration of the power of switching between matrices and linear maps. \n\n9.67 determinant of Vandermonde matrix \n\nSuppose > 1 and 1, , . Then \n\ndet \n\n1 1 12 1 1 \n\n1 2 22 2 1 \n\n1 2 1 \n\n= \n\n> 1 < \n\n( ). \n\nProof Let 1, , , 1 be the standard basis of 1 () and let 1, , denote the standard basis of . Define a linear map 1 () by \n\n = (( 1), , ( )). \n\nLet denote the Vandermonde matrix shown in the statement of this result. Note that \n\n = ( , (1 , , , 1 ), ( 1, , )). \n\nLet 1 () 1 () be the operator on 1 () such that 1 = 1 \n\nand \n\n = ( 1)( 2)( )\n\nfor = 1 , , 1 . Let = ( , (1 , , , 1 ), (1 , , , 1 )) . Then is an upper-triangular matrix all of whose diagonal entries equal 1. Thus det = 1 \n\n(by 9.48). Let = ( , (1 , , , 1 ), ( 1, , )) . Thus = (by 3.81), which implies that \n\ndet = ( det )( det ) = det . \n\nThe definitions of , , and show that equals \n\n1 0 0 01 2 1 0 01 3 1 ( 3 1)( 3 2) 01 1 ( 1)( 2) ( 1)( 2) ( 1 ).\n\nNow det = det = \n\n> 1 < \n\n( ), where we have used 9.56(a) and 9.48. \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler Section 9C Determinants 367 \n\n# Exercises 9C \n\n1 Prove or give a counterexample: , () det ( +) = det +det .\n\n2 Suppose the first column of a square matrix consists of all zeros except possibly the first entry 1,1. Let be the matrix obtained from by deleting the first row and the first column of . Show that det = 1,1 det .\n\n3 Suppose () is nilpotent. Prove that det ( + ) = 1 .\n\n4 Suppose () . Prove that is unitary if and only if |det | = = 1 .\n\n5 Suppose is a block upper-triangular matrix \n\n = 1 \n\n0 \n\n\n\n,\n\nwhere each along the diagonal is a square matrix. Prove that \n\ndet = ( det 1)( det ). \n\n6 Suppose = ( 1 ) is an -by- matrix, with denoting the th \n\ncolumn of . Show that if ( 1, , ) perm , then \n\ndet ( 1 ) = ( sign ( 1, , )) det . \n\n7 Suppose () is invertible. Let denote the characteristic polynomial of and let denote the characteristic polynomial of 1 . Prove that \n\n() = 1(0) dim ( 1 )\n\nfor all nonzero .\n\n8 Suppose () is an operator with no eigenvalues (which implies that \n\n = ). Prove that det > 0 .\n\n9 Suppose that is a real vector space of even dimension, () , and \n\ndet < 0 . Prove that has at least two distinct eigenvalues. \n\n10 Suppose is a real vector space of odd dimension and () . Without using the minimal polynomial, prove that has an eigenvalue. \n\nThis result was previously proved without using determinants or the charac-teristic polynomialsee 5.34. \n\n11 Prove or give a counterexample: If = , () , and det > 0 , then \n\n has a square root. \n\nIf = , () , and det 0 , then has a square root (see 8.41 ).\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n368 Chapter 9 Multilinear Algebra and Determinants \n\n12 Suppose , () and is invertible. Define by \n\n() = det ( ). \n\nProve that is a polynomial of degree dim and that the coefficient of dim \n\nin this polynomial is det .\n\n13 Suppose = , () , and = dim > 2 . Let 1, , denote the eigenvalues of , with each eigenvalue included as many times as its multiplicity. (a) Find a formula for the coefficient of 2 in the characteristic polynomial of in terms of 1, , .(b) Find a formula for the coefficient of in the characteristic polynomial of in terms of 1, , .\n\n14 Suppose is an inner product space and is a positive operator on . Prove that \n\ndet = det . \n\n15 Suppose is an inner product space and () . Use the polar decom-position to give a proof that \n\n|det | = det ( ) \n\nthat is different from the proof given earlier (see 9.60). \n\n16 Suppose () . Define by () = det ( + ) . Show that \n\n(0) = tr .\n\nLook for a clean solution to this exercise, without using the explicit but complicated formula for the determinant of a matrix. \n\n17 Suppose , , are positive numbers. Find the volume of the ellipsoid \n\n{( , , ) 3 2\n\n 2 + 2\n\n 2 + 2\n\n 2 < 1} \n\nby finding a set 3 whose volume you know and an operator on 3\n\nsuch that () equals the ellipsoid above. \n\n18 Suppose that is an invertible square matrix. Prove that Hadamards inequality (9.66) is an equality if and only if each column of is orthogonal to the other columns. \n\n19 Suppose is an inner product space, 1, , is an orthonormal basis of ,and () is a positive operator. (a) Prove that det = 1 , .(b) Prove that if is invertible, then the inequality in (a) is an equality if and only if is an eigenvector of for each = 1 , , .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9C Determinants 369 \n\n20 Suppose is an -by- matrix, and suppose is such that | , | for all \n\n , {1 , , } . Prove that \n\n|det | /2 .\n\nThe formula for the determinant of a matrix (9.46 ) shows that |det | ! .However, the estimate given by this exercise is much better. For example, if \n\n> = 1\n\nand = 100 , then ! 10 158 , but the estimate given by this exercise is the much smaller number 10 100 . If is an integer power of 2, then the inequality above is sharp and cannot be improved. \n\n21 Suppose is a positive integer and , is a function such that \n\n() = () () \n\nfor all , , and () equals the product of the diagonal entries of \n\nfor each diagonal matrix , . Prove that \n\n() = det \n\nfor all , .\n\nRecall that , denotes the set of -by- matrices with entries in . This exercise shows that the determinant is the unique function defined on square matrices that is multiplicative and has the desired behavior on diagonal matrices. This result is analogous to Exercise 10 in Section 8D, which shows that the trace is uniquely determined by its algebraic properties. \n\nI find that in my own elementary lectures, I have, for pedagogical reasons, pushed determinants more and more into the background. Too often I have had the expe-rience that, while the students acquired facility with the formulas, which are so useful in abbreviating long expressions, they often failed to gain familiarity with their meaning , and skill in manipulation prevented the student from going into all the details of the subject and so gaining a mastery. \n\nElementary Mathematics from an Advanced Standpoint : Geometry , Felix Klein \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 370 Chapter 9 Multilinear Algebra and Determinants\n\n# 9D Tensor Products \n\n# Tensor Product of Two Vector Spaces \n\nThe motivation for our next topic comes from wanting to form the product of a vector and a vector . This product will be denoted by ,pronounced tensor , and will be an element of some new vector space called \n\n (also pronounced tensor ). We already have a vector space (see Section 3E), called the product of \n\n and . However, will not serve our purposes here because it does not provide a natural way to multiply an element of by an element of . We would like our tensor product to satisfy some of the usual properties of multiplication. For example, we would like the distributive property to be satisfied, meaning that if 1, 2, and 1, 2, , then \n\n( 1 + 2) = 1 + 2 and ( 1 + 2) = 1 + 2. \n\n> To produce in TEX, type \\otimes .\n\nWe would also like scalar multiplica-tion to interact well with this new multi-plication, meaning that \n\n( ) = () = () \n\nfor all , , and .Furthermore, it would be nice if each basis of when combined with each basis of produced a basis of . Specifically, if 1, , is a basis of \n\nand 1, , is a basis of , then we would like a list (in any order) consisting of , as ranges from 1 to and ranges from 1 to , to be a basis of \n\n . This implies that dim ( ) should equal (dim )( dim ) . Recall that \n\ndim ( ) = dim + dim (see 3.92), which shows that the product \n\nwill not serve our purposes here. To produce a vector space whose dimension is (dim )( dim ) in a natural fashion from and , we look at the vector space of bilinear functionals, as defined below. \n\n> 9.68 definition:\n\nbilinear functional on , the vector space ( , ) A bilinear functional on is a function such that \n\n ( , ) is a linear functional on for each and ( , ) \n\nis a linear functional on for each .\n\n The vector space of bilinear functionals on is denoted by ( , ) .If = , then a bilinear functional on is a bilinear form; see 9.1. The operations of addition and scalar multiplication on ( , ) are defined to be the usual operations of addition and scalar multiplication of functions. As you can verify, these operations make ( , ) into a vector space whose additive identity is the zero function from to .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9D Tensor Products 371 \n\n> 9.69 example:\n\nbilinear functionals \n\n Suppose and . Define by ( , ) = ()() .Then is a bilinear functional on .\n\n Suppose and . Define by ( , ) = ()() .Then is a bilinear functional on .\n\n Define by ( , ) = () . Then is a bilinear functional on \n\n .\n\n Suppose . Define () by ( , ) = () . Then is a bilinear functional on () .\n\n Suppose and are positive integers. Define , , by ( , ) = \n\ntr () . Then is a bilinear functional on , , . \n\n> 9.70\n\ndimension of the vector space of bilinear functionals \n\ndim ( , ) = ( dim )( dim ) .\n\nProof Let 1, , be a basis of and 1, , be a basis of . For a bilinear functional ( , ) , let () be the -by- matrix whose entry in row ,column is ( , ). The map () is a linear map of ( , ) into , .For a matrix , , define a bilinear functional on by \n\n ( 1 1 + + , 1 1 + + ) = \n\n> \n> \n> = 1 \n> \n> = 1\n\n , \n\nfor 1, , , 1, , .The linear map () from ( , ) to , and the linear map \n\nfrom , to ( , ) are inverses of each other because () = for all \n\n ( , ) and ( ) = for all , , as you should verify. Thus both maps are isomorphisms and the two spaces that they connect have the same dimension. Hence dim ( , ) = dim , = = ( dim )( dim ) .Several different definitions of appear in the mathematical literature. These definitions are equivalent to each other, at least in the finite-dimensional context, because any two vector spaces of the same dimension are isomorphic. The result above states that ( , ) has the dimension that we seek, as do \n\n( , ) and dim ,dim . Thus it may be tempting to define to be ( , ) \n\nor ( , ) or dim ,dim . However, none of those definitions would lead to a basis-free definition of for and .The following definition, while it may seem a bit strange and abstract at first, has the huge advantage that it defines in a basis-free fashion. We define \n\n to be the vector space of bilinear functionals on instead of the more tempting choice of the vector space of bilinear functionals on .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n372 Chapter 9 Multilinear Algebra and Determinants \n\n> 9.71 definition:\n\ntensor product, , The tensor product is defined to be ( , ).\n\n For and , the tensor product is the element of \n\ndefined by \n\n( )( , ) = ()() \n\nfor all ( , ) .We can quickly prove that the definition of gives it the desired dimension. \n\n> 9.72\n\ndimension of the tensor product of two vector spaces \n\ndim ( ) = ( dim )( dim ) .\n\nProof Because a vector space and its dual have the same dimension (by 3.111), we have dim = dim and dim = dim . Thus 9.70 tells us that the dimension of ( , ) equals (dim )( dim ) .To understand the definition of the tensor product of two vectors \n\nand , focus on the kind of object it is. An element of is a bilinear functional on , and in particular it is a function from to . Thus for each element of , it should produce an element of . The definition above has this behavior, because applied to a typical element ( , ) of \n\nproduces the number ()() .The somewhat abstract nature of should not matter. The important point is the behavior of these objects. The next result shows that tensor products of vectors have the desired bilinearity properties. \n\n> 9.73\n\nbilinearity of tensor product \n\nSuppose , 1, 2 and , 1, 2 and . Then \n\n( 1 + 2) = 1 + 2 and ( 1 + 2) = 1 + 2\n\nand \n\n( ) = () = (). \n\nProof Suppose ( , ) . Then \n\n(( 1 + 2) )( , ) = ( 1 + 2)() = ( 1)() + ( 2)() = ( 1 )( , ) + ( 2 )( , ) = ( 1 + 2 )( , ). \n\nThus ( 1 + 2) = 1 + 2 .The other two equalities are proved similarly. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9D Tensor Products 373 \n\nLists are, by definition, ordered. The order matters when, for example, we form the matrix of an operator with respect to a basis. For lists in this section with two indices, such as { } = 1 , , ; = 1 , , in the next result, the ordering does not matter and we do not specify itjust choose any convenient ordering. The linear independence of elements of in (a) of the result below captures the idea that there are no relationships among vectors in other than the relationships that come from bilinearity of the tensor product (see 9.73) and the relationships that may be present due to linear dependence of a list of vectors in or a list of vectors in . \n\n> 9.74\n\nbasis of \n\nSuppose 1, , is a list of vectors in and 1, , is a list of vectors in .(a) If 1, , and 1, , are both linearly independent lists, then \n\n{ } = 1 , , ; = 1 , , \n\nis a linearly independent list in .(b) If 1, , is a basis of and 1, , is a basis of , then the list \n\n{ } = 1 , , ; = 1 , , is a basis of .\n\nProof To prove (a), suppose 1, , and 1, , are both linearly independent lists. This linear independence and the linear map lemma (3.4) imply that there exist 1, , and 1, , such that \n\n ( ) = {{1 if = ,\n\n0 if and ( ) = {{1 if = ,\n\n0 if ,\n\nwhere , {1 , , } in the first equation and , {1 , , } in the second equation. Suppose { , } = 1 , , ; = 1 , , is a list of scalars such that \n\n> 9.75\n> \n> \n> = 1 \n> \n> = 1\n\n , ( ) = 0. \n\nNote that ( )( , ) equals 1 if = and = , and equals 0 otherwise. Thus applying both sides of 9.75 to ( , ) shows that , = 0 , proving that \n\n{ } = 1 , , ; = 1 , , is linearly independent. Now (b) follows from (a), the equation dim = ( dim )( dim ) [see 9.72], and the result that a linearly independent list of the right length is a basis (see 2.38). Every element of is a finite sum of elements of the form , where \n\n and , as implied by (b) in the result above. However, if dim > 1 \n\nand dim > 1 , then Exercise 4 shows that \n\n{ ( , ) } . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n374 Chapter 9 Multilinear Algebra and Determinants \n\n> 9.76 example:\n\ntensor product of element of with element of \n\nSuppose and are positive integers. Let 1, , denote the standard basis of and let 1, , denote the standard basis of . Suppose \n\n = ( 1, , ) and = ( 1, , ) .\n\nThen \n\n = ( \n\n> \n\n\n\n> = 1\n\n ) ( \n\n> \n\n\n\n> = 1\n\n )=\n\n> \n\n\n\n> = 1 \n\n\n\n> = 1\n\n( )( ). \n\nThus with respect to the basis { } = 1 , , ; = 1 , , of provided by 9.74(b), the coefficients of are the numbers { } = 1 , , ; = 1 , , . If instead of writing these numbers in a list, we write them in an -by- matrix with \n\n in row , column , then we can identify with the -by- matrix \n\n 1 1 1 \n\n 1 \n\n.\n\nSee Exercises 5 and 6 for practice in using the identification from the example above. We now define bilinear maps, which differ from bilinear functionals in that the target space can be an arbitrary vector space rather than just the scalar field. \n\n> 9.77 definition:\n\nbilinear map \n\nA bilinear map from to a vector space is a function \n\nsuch that ( , ) is a linear map from to for each and \n\n ( , ) is a linear map from to for each . \n\n> 9.78 example:\n\nbilinear maps \n\n Every bilinear functional on is a bilinear map from to .\n\n The function defined by ( , ) = is a bilinear map from to (by 9.73). \n\n The function () () () defined by ( , ) = is a bilinear map from () () to () .\n\n The function ( , ) defined by ( , ) = is a bilinear map from ( , ) to .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9D Tensor Products 375 \n\nTensor products allow us to convert bilinear maps on into linear maps on \n\n (and vice versa), as shown by the next result. In the mathematical literature, (a) of the result below is called the universal property of tensor products. \n\n> 9.79\n\nconverting bilinear maps to linear maps \n\nSuppose is a vector space. (a) Suppose is a bilinear map. Then there exists a unique linear map such that ( ) = ( , ) \n\nfor all ( , ) .(b) Conversely, suppose is a linear map. There there exists a unique bilinear map # such that \n\n #( , ) = ( ) \n\nfor all ( , ) .\n\nProof Let 1, , be a basis of and let 1, , be a basis of . By the linear map lemma (3.4) and 9.74(b), there exists a unique linear map \n\nsuch that ( ) = ( , )\n\nfor all {1 , , } and {1 , , } .Now suppose ( , ) . There exist 1, , , 1, , such that \n\n = 1 1 + + and = 1 1 + + . Thus ( ) = ( \n\n> \n> \n> = 1 \n> \n> = 1\n\n( )( )) =\n\n> \n> \n> = 1 \n> \n> = 1\n\n ( )=\n\n> \n> \n> = 1 \n> \n> = 1\n\n ( , )= ( , ) ,\n\nas desired, where the second line holds because is linear, the third line holds by the definition of , and the fourth line holds because is bilinear. The uniqueness of the linear map satisfying ( ) = ( , ) follows from 9.74(b), completing the proof of (a). To prove (b), define a function # by #( , ) = ( ) for all \n\n( , ) . The bilinearity of the tensor product (see 9.73) and the linearity of imply that # is bilinear. Clearly the choice of # that satisfies the conditions is unique. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n376 Chapter 9 Multilinear Algebra and Determinants \n\nTo prove 9.79(a), we could not just define ( ) = ( , ) for all \n\nand ( and then extend linearly to all of ) because elements of \n\n do not have unique representations as finite sums of elements of the form \n\n . Our proof used a basis of and a basis of to get around this problem. Although our construction of in the proof of 9.79(a) depended on a basis of \n\n and a basis of , the equation ( ) = ( , ) that holds for all and \n\n shows that does not depend on the choice of bases for and .\n\n# Tensor Product of Inner Product Spaces \n\nThe result below features three inner productsone on , one on , and one on , although we use the same symbol , for all three inner products. \n\n9.80 inner product on tensor product of two inner product spaces \n\nSuppose and are inner product spaces. Then there is a unique inner product on such that \n\n , = , , \n\nfor all , and , .\n\nProof Suppose 1, , is an orthonormal basis of and 1, , is an ortho-normal basis of . Define an inner product on by \n\n9.81 \n\n> \n\n\n\n> = 1 \n\n\n\n> = 1\n\n , , \n\n\n\n> = 1 \n\n\n\n> = 1\n\n , = \n\n> \n\n\n\n> = 1 \n\n\n\n> = 1\n\n , , .\n\nThe straightforward verification that 9.81 defines an inner product on \n\nis left to the reader [use 9.74(b) ].Suppose that , and , . Let 1, , be such that \n\n = 1 1 + + , with similar expressions for , , and . Then \n\n , = \n\n> \n\n\n\n> = 1\n\n \n\n> \n\n\n\n> = 1\n\n , \n\n\n\n> = 1\n\n \n\n> \n\n\n\n> = 1\n\n = \n\n> \n\n\n\n> = 1 \n\n\n\n> = 1\n\n , \n\n\n\n> = 1 \n\n\n\n> = 1\n\n =\n\n> \n\n\n\n> = 1 \n\n\n\n> = 1\n\n \n\n= \n\n> \n\n\n\n> = 1\n\n \n\n> \n\n\n\n> = 1\n\n = , , . \n\nThere is only one inner product on such that , = , , \n\nfor all , and , because every element of can be written as a linear combination of elements of the form [ by 9.74(b) ].\n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler Section 9D Tensor Products 377 \n\nThe definition below of a natural inner product on is now justified by 9.80. We could not have simply defined , to be , , (and then used additivity in each slot separately to extend the definition to ) without some proof because elements of do not have unique representations as finite sums of elements of the form . \n\n> 9.82 definition:\n\ninner product on tensor product of two inner product spaces \n\nSuppose and are inner product spaces. The inner product on is the unique function , from ( ) ( ) to such that \n\n , = , , \n\nfor all , and , .Take = and = in the equation above and then take square roots to show that \n\n = \n\nfor all and all .The construction of the inner product in the proof of 9.80 depended on an or-thonormal basis 1, , of and an orthonormal basis 1, , of . Formula 9.81 implies that { } = 1 , , ; = 1 , , is a doubly indexed orthonormal list in \n\n and hence is an orthonormal basis of [ by 9.74(b) ]. The importance of the next result arises because the orthonormal bases used there can be different from the orthonormal bases used to define the inner product in 9.80. Although the notation for the bases is the same in the proof of 9.80 and in the result below, think of them as two different sets of orthonormal bases. \n\n> 9.83\n\northonormal basis of \n\nSuppose and are inner product spaces, and 1, , is an orthonormal basis of and 1, , is an orthonormal basis of . Then \n\n{ } = 1 , , ; = 1 , , \n\nis an orthonormal basis of .\n\nProof We know that { } = 1 , , ; = 1 , , is a basis of [ by 9.74(b) ].Thus we only need to verify orthonormality. To do this, suppose , {1 , , } \n\nand , {1 , , } . Then \n\n , = , , = {{1 if = and = ,\n\n0 otherwise .\n\nHence the doubly indexed list { } = 1 , , ; = 1 , , is indeed an orthonormal basis of .See Exercise 11 for an example of how the inner product structure on \n\ninteracts with operators on and .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n378 Chapter 9 Multilinear Algebra and Determinants \n\n# Tensor Product of Multiple Vector Spaces \n\nWe have been discussing properties of the tensor product of two finite-dimensional vector spaces. Now we turn our attention to the tensor product of multiple finite-dimensional vector spaces. This generalization requires no new ideas, only some slightly more complicated notation. Readers with a good understanding of the tensor product of two vector spaces should be able to make the extension to the tensor product of more than two vector spaces. Thus in this subsection, no proofs will be provided. The definitions and the statements of results that will be provided should be enough information to enable readers to fill in the details, using what has already been learned about the tensor product of two vector spaces. We begin with the following notational assumption. \n\n> 9.84 notation:\n\n 1, , \n\nFor the rest of this subsection, denotes an integer greater than 1 and \n\n 1, , denote finite-dimensional vector spaces. The notion of an -linear functional, which we are about to define, generalizes the notion of a bilinear functional (see 9.68). Recall that the use of the word functional indicates that we are mapping into the scalar field . Recall also that the terminology -linear form is used in the special case 1 = = (see 9.25). The notation ( 1, , ) generalizes our previous notation ( , ) . \n\n> 9.85 definition:\n\n -linear functional, the vector space ( 1, , ) An -linear functional on 1 is a function 1 \n\nthat is a linear functional in each slot when the other slots are held fixed. \n\n The vector space of -linear functionals on 1 is denoted by \n\n( 1, , ). \n\n> 9.86 example:\n\n -linear functional \n\nSuppose for each {1 , , } . Define 1 by \n\n( 1, , ) = 1( 1) ( ). \n\nThen is an -linear functional on 1 .The next result can be proved by imitating the proof of 9.70. \n\n> 9.87\n\ndimension of the vector space of -linear functionals \n\ndim ( 1, , ) = ( dim 1) ( dim ).\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9D Tensor Products 379 \n\nNow we can define the tensor product of multiple vector spaces and the tensor product of elements of those vector spaces. The following definition is completely analogous to our previous definition (9.71) in the case = 2 . \n\n> 9.88 definition:\n\ntensor product, 1 , 1 \n\n The tensor product 1 is defined to be ( 1, , ).\n\n For 1 1, , , the tensor product 1 is the element of 1 defined by \n\n( 1 )( 1, , ) = 1( 1) ( )\n\nfor all ( 1, , ) 1 .The next result can be proved by following the pattern of the proof of the analogous result when = 2 (see 9.72). \n\n> 9.89\n\ndimension of the tensor product \n\ndim ( 1 ) = ( dim 1) ( dim ).Our next result generalizes 9.74. \n\n> 9.90\n\nbasis of 1 \n\nSuppose dim = and 1 , , is a basis of for = 1 , , . Then \n\n{ 1 1 } 1 = 1 , , 1; ; = 1 , , \n\nis a basis of 1 .Suppose = 2 and 11, , 1 1 is a basis of 1 and 21, , 2 2 is a basis of 2.Then with respect to the basis { 1 1 2 2 } 1 = 1 , , 1; 2 = 1 , , 2 in the result above, the coefficients of an element of 1 2 can be represented by an 1-by- 2 matrix that contains the coefficient of 1 1 2 2 in row 1, column 2. Thus we need a matrix, which is an array specified by two indices, to represent an element of 1 2.If > 2 , then the result above shows that we need an array specified by \n\nindices to represent an arbitrary element of 1 . Thus tensor products may appear when we deal with objects specified by arrays with multiple indices. The next definition generalizes the notion of a bilinear map (see 9.77). As with bilinear maps, the target space can be an arbitrary vector space. \n\n> 9.91 definition:\n\n -linear map \n\nAn -linear map from 1 to a vector space is a function \n\n 1 that is a linear map in each slot when the other slots are held fixed. \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n380 Chapter 9 Multilinear Algebra and Determinants \n\nThe next result can be proved by following the pattern of the proof of 9.79. \n\n> 9.92\n\nconverting -linear maps to linear maps \n\nSuppose is a vector space. (a) Suppose that 1 is an -linear map. Then there exists a unique linear map 1 such that ( 1 ) = ( 1, , )\n\nfor all ( 1, , ) 1 .(b) Conversely, suppose 1 is a linear map. There there exists a unique -linear map # 1 such that \n\n #( 1, , ) = ( 1 )\n\nfor all ( 1, , ) 1 .See Exercises 12 and 13 for tensor products of multiple inner product spaces. \n\n# Exercises 9D \n\n1 Suppose and . Prove that = 0 if and only if = 0 or \n\n = 0 .\n\n2 Give an example of six distinct vectors 1, 2, 3, 1, 2, 3 in 3 such that \n\n 1 1 + 2 2 + 3 3 = 0 \n\nbut none of 1 1, 2 2, 3 3 is a scalar multiple of another element of this list. \n\n3 Suppose that 1, , is a linearly independent list in . Suppose also that \n\n 1, , is a list in such that \n\n 1 1 + + = 0. \n\nProve that 1 = = = 0 .\n\n4 Suppose dim > 1 and dim > 1 . Prove that \n\n{ ( , ) } \n\nis not a subspace of .\n\nThis exercise implies that if dim > 1 and dim > 1 , then \n\n> { \n\n ( , ) } . \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\nSection 9D Tensor Products 381 \n\n5 Suppose and are positive integers. For and , identify \n\n with an -by- matrix as in Example 9.76. With that identification, show that the set \n\n{ and }\n\nis the set of -by- matrices (with entries in ) that have rank at most one. \n\n6 Suppose and are positive integers. Give a description, analogous to Exercise 5, of the set of -by- matrices (with entries in ) that have rank at most two. \n\n7 Suppose dim > 2 and dim > 2 . Prove that \n\n{ 1 1 + 2 2 1, 2 and 1, 2 } . \n\n8 Suppose 1, , and 1, , are such that \n\n 1 1 + + = 0. \n\nSuppose that is a vector space and is a bilinear map. Show that \n\n( 1, 1) + + ( , ) = 0. \n\n9 Suppose () and () . Prove that there exists a unique operator on that takes to for all and .\n\nIn an abuse of notation, the operator on given by this exercise is often called .\n\n10 Suppose () and () . Prove that is an invertible operator on if and only if both and are invertible operators. Also, prove that if both and are invertible operators, then ( ) 1 = 1 1 ,where we are using the notation from the comment after Exercise 9. \n\n11 Suppose and are inner product spaces. Prove that if () and \n\n () , then ( ) = , where we are using the notation from the comment after Exercise 9. \n\n12 Suppose that 1, , are finite-dimensional inner product spaces. Prove that there is a unique inner product on 1 such that \n\n 1 , 1 = 1, 1 , \n\nfor all ( 1, , ) and ( 1, , ) in 1 .\n\nNote that the equation above implies that \n\n> 1 = 1 \n\nfor all ( 1, , ) 1 .\n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n382 Chapter 9 Multilinear Algebra and Determinants \n\n13 Suppose that 1, , are finite-dimensional inner product spaces and \n\n 1 is made into an inner product space using the inner product from Exercise 12. Suppose 1 , , is an orthonormal basis of for each \n\n = 1 , , . Show that the list \n\n{ 1 1 } 1 = 1 , , 1; ; = 1 , , \n\nis an orthonormal basis of 1 .\n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler Photo Credits \n\n page v: Photos by Carrie Heeter and Bishnu Sarangi. Public domain image. \n\n page 1: Original painting by Pierre Louis Dumesnil; 1884 copy by Nils Forsberg. Public domain image downloaded on 29 March 2022 from \n\n> https://commons.wikimedia.org/wiki/File:Ren_Descartes_i_samtal_med_Sveriges_drottning,_Kristina.jpg\n\n.\n\n page 27: Public domain image downloaded on 4 February 2022 from https://commons.wikimedia.org/wiki/File:IAS_Princeton.jpg. \n\n page 51: Photo by Stefan Schfer; Creative Commons Attribution Share Alike license. Downloaded on 28 March 2022 from https://commons.wikimedia.org/wiki/File:BurgDankwarderode2016.jpg. \n\n page 119: Photo by Alireza Javaheri. Creative Commons Attribution license. Downloaded on 12 March 2023 from \n\nhttps://commons.wikimedia.org/wiki/File:Hakim_Omar_Khayam_-_panoramio.jpg .\n\n page 132: Statue completed by Giovanni Paganucci in 1863. Photo by Hans-Peter Postel; Creative Commons Attribution license. Downloaded on 14 March 2022 from https://commons.wikimedia.org/wiki/File:Leonardo_da_Pisa.jpg. \n\n page 181: Photo by Matthew Petroff; Creative Commons Attribution Share Alike license. 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Downloaded on 9 July 2019 from https://commons.wikimedia.org/wiki/File:Mathematik_Gttingen.jpg. \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler \n\n383 Symbol Index \n\n 1 , 91 \n\n , , 74 \n\n , , 69 \n\n , , 74 \n\n , 354 \n\n , 231 \n\n t, 77 , 375, 380 \n\n , 287 \n\n( 1, , ), 378 \n\n( , ) , 370 \n\n , 2 \n\n, 55 deg, 31 \n\n, 196 \n\ndet , 355 \n\ndet , 354 dim, 44 \n\n, 21 \n\n( 1 1, , ), 287 \n\n( , ) , 164 \n\n , 4 \n\n , 13 \n\n , , 72 \n\n , 6 \n\n , 13 \n\n( , ) , 308 \n\n , 52, 90 \n\n , 23 \n\nIm , 120 \n\n , 31 \n\n() , 52 \n\n( , ) , 52 \n\n() , 334 \n\n() , 69, 154 \n\n() , 88 perm, 348 \n\n () , 30 \n\n , 101 \n\n () , 31 \n\n() , 137 \n\n , 214 \n\n , 341 , 7 \n\n , 2 \n\nRe , 120 \n\n , 381 \n\n, 299 \n\n , 253 , 102 \n\n , 107 \n\n , 228 \n\n 1 , 82 \n\n() , 288 \n\n , 68 \n\n , 221 \n\n , 137 \n\n , 280 \n\n #, 375, 380 \n\ntr , 326 \n\ntr , 327 \n\n| , 133 \n\n/ , 142 \n\n , 211 \n\n 0, 109 \n\n , , 184 \n\n , 15 \n\n , 105, 204 \n\n/ , 99 \n\n , 15 \n\n 1 , 379 \n\n 1 , 379 \n\n (2) , 334 \n\n (2) \n\n> alt\n\n, 339 \n\n (2) \n\n> sym\n\n, 337 \n\n , 17 \n\n , 103, 346 \n\n () , 346 \n\n () \n\n> alt\n\n, 347 \n\n , 372 \n\n , 372 \n\n + , 98 \n\n|||| , 186 \n\n , 120 \n\n|| , 120 \n\nLinear Algebra Done Right , fourth edition, by Sheldon Axler \n\n384 Index \n\nAbbott, Edwin A., 6 absolute value, 120 addition in quotient space, 100 of complex numbers, 2 of functions, 13 of linear maps, 55 of matrices, 71 of subspaces, 19 of vectors, 12 of vectors in , 6 additive inverse in , 3, 4 in , 9 in vector space, 12, 15 additivity, 52 adjoint of a linear map, 228 algebraic multiplicity, 311 alternating bilinear form, 339 alternating -linear form, 347 annihilator, 109 Apolloniuss identity, 195 Artin, Emil, 80 associativity, 3, 12, 56 backward shift, 53, 59, 84, 140 ball, 287 Banach, Stefan, 227 basis, 39 of eigenvectors, 165, 245, 246, 250 of generalized eigenvectors, 301 Bernstein polynomials, 49 Bessels inequality, 198 bilinear form, 333 bilinear functional, 370 bilinear map, 374 block diagonal matrix, 314 Bunyakovsky, Viktor, 189 \n\n -algebras, 295 Carroll, Lewis, 11 Cauchy, Augustin-Louis, 189 CauchySchwarz inequality, 189 Cayley, Arthur, 312 CayleyHamilton theorem, 364 on complex vector space, 312 change-of-basis formula for bilinear forms, 336 for operators, 93 characteristic polynomial, 363 on complex vector space, 311 ChatGPT, 196, 279 Cholesky factorization, 267 Cholesky, Andr-Louis, 267 Christina, Queen of Sweden, 1 closed under addition, 18 closed under scalar multiplication, 18 column rank of a matrix, 77, 114, 239 columnrow factorization, 78 commutativity, 3, 7, 12, 25, 56, 73, 80 commuting operators, 138, 175180, 209, 235, 248249, 256 companion matrix, 152 complex conjugate, 120 complex number, 2 complex spectral theorem, 246 complex vector space, 13 \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n> 385 386 Index\n\ncomplexification eigenvalues of, 140 generalized eigenvectors of, 318 minimal polynomial of, 153 multiplicity of eigenvalues, 318 of a linear map, 68 of a vector space, 17, 43 of an inner product space, 194 conjugate symmetry, 183 conjugate transpose of a matrix, 231 coordinate, 6 cube root of an operator, 248 De Moivres theorem, 125 degree of a polynomial, 31 Descartes, Ren, 1 determinant of matrix, 355 of operator, 354 of positive operator, 362 of unitary operator, 362 diagonal matrix, 163, 274 diagonal of a square matrix, 155 diagonalizable, 163, 172, 176, 245, 246, 294, 307, 316 differentiation linear map, 53, 56, 59, 61, 62, 67, 70, 79, 138, 208, 304 dimension, 44 of a sum of subspaces, 47 direct sum, 21, 42, 98 of a subspace and its orthogonal complement, 212 of null dim and range dim ,299 discrete Fourier transform, 269 distributive property, 3, 12, 15, 56, 80 division algorithm for polynomials, 124 division of complex numbers, 4 dot product, 182 double dual space, 118 dual of a basis, 106 of a linear map, 107, 153, 162, 174 of a vector space, 105, 204 of an operator, 140 eigenspace, 164 eigenvalue of adjoint, 239 of dual of an operator, 140 of operator, 134 of positive operator, 252 of self-adjoint operator, 233 of unitary operator, 262 on odd-dimensional space, 150, 318, 367 eigenvector, 135 ellipsoid, 287 Euclidean inner product, 184 Fibonacci, 132 Fibonacci sequence, 174 field, 10 finite-dimensional vector space, 30 \n\nFlatland , 6 forward shift, 140 \n\nFrankenstein , 50 Frobenius norm, 331 Fugledes theorem, 248 fundamental theorem of algebra, 125 fundamental theorem of linear maps, 62 Gauss, Carl Friedrich, 51 Gaussian elimination, 51, 65, 361 generalized eigenspace, 308 generalized eigenvector, 300 geometric multiplicity, 311 Gershgorin disk, 170 \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Index 387\n\nGershgorin disk theorem, 171 Gershgorin, Semyon Aronovich, 171 Gram, Jrgen, 200 GramSchmidt procedure, 200 graph of a linear map, 103 Hadamards inequality, 365 Halmos, Paul, 27 Hamilton, William, 297 harmonic function, 196 Hilbert matrix, 256 HilbertSchmidt norm, 331 homogeneity, 52 homogeneous system of linear equations, 65, 95 hyponormal operator, 241 identity matrix, 90 identity operator, 52, 56 imaginary part, 120 infinite-dimensional vector space, 31 injective, 60 inner product, 183 inner product space, 184 Institute for Advanced Study, 27 invariant subspace, 133 inverse of a linear map, 82 of a matrix, 91 invertible linear map, 82 invertible matrix, 91 isometry, 258 isomorphic vector spaces, 86 isomorphism, 86 Jordan basis, 322 Jordan form, 324 Jordan, Camille, 324 kernel, 59 Khayyam, Omar, 119 Laplacian, 196 length of list, 5 Leonardo of Pisa, 132 linear combination, 28 linear dependence lemma, 33 linear equations, 6465, 95 linear functional, 105, 204 linear map, 52 linear map lemma, 54 linear span, 29 linear subspace, 18 linear transformation, 52 linearly dependent, 33 linearly independent, 32 list, 5 of vectors, 28 lower-triangular matrix, 162, 267 Lviv, 227 Lww, 227 matrix, 69 multiplication, 73 of bilinear form, 334 of linear map, 69 of nilpotent operator, 305 of operator, 154 of product of linear maps, 74, 91 of , 113 of , 232 of vector, 88 minimal polynomial and basis of generalized eigenvectors, 306 and characteristic polynomial, 312 and diagonalizability, 169 and generalized eigenspace decomposition, 316 and generalized eigenspaces, 317 and invertibility, 149 and upper-triangular matrices, 159, 203 \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler 388 Index\n\ncomputing, 145 definition of, 145 gcd with its derivative, 173 no direct sum decomposition, 325 of adjoint, 241 of companion matrix, 152 of complexification, 153 of dual map, 153 of nilpotent operator, 305, 324 of normal operator, 241 of quotient operator, 153 of restriction operator, 148 of self-adjoint operator, 244 polynomial multiple of, 148 zeros of, 146 minimizing distance, 217 \n\n> \n\n-linear form, 346 \n\n> \n\n-linear functional, 378 \n\n> \n\n-linear map, 379 monic polynomial, 144 Moon, v, xvii MoorePenrose inverse, 221 multilinear form, 346 multiplication, see product multiplicity of an eigenvalue, 310 nilpotent operator, 303, 322 Noether, Emmy, 332 nonsingular matrix, 91 norm of a linear map, 280 norm of a vector, 182, 186 normal operator, 235 null space, 59 of powers of an operator, 298 of , 111 of , 231 one-to-one, 60 onto, 62 operator, 133 orthogonal complement, 211 projection, 214 vectors, 187 orthonormal basis, 199 list, 197 parallelogram equality, 191 Parsevals identity, 200 partial differentiation operator, 175 Peabody Library, 181 permutation, 348 photo credits, 383 point, 12 polar decomposition, 286 polynomial, 30 positive definite, 266 positive operator, 251 positive semidefinite operator, 251 principal axes, 287 product of complex numbers, 2 of linear maps, 55 of matrices, 73 of polynomials, 138 of scalar and linear map, 55 of scalar and vector, 12 of scalar and vector in , 9 of vector spaces, 96 pseudoinverse, 221, 250, 255, 275, 279 Pythagorean theorem, 187 QR factorization, 264, 365 quadratic form, 341 quotient map, 101 operator, 142, 153, 162, 173 space, 99 range, 61 of powers of an operator, 306 \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler Index 389\n\nof , 112 of , 231 rank of a matrix, 79, 114, 239 real part, 120 real spectral theorem, 245 real vector space, 13 reverse triangle inequality, 129, 193, 294 Riesz representation theorem, 205, 210, 216, 224, 225 Riesz, Frigyes, 205 row rank of a matrix, 77, 114, 239 scalar, 4 scalar multiplication, 9, 12 in quotient space, 100 of linear maps, 55 of matrices, 71 Schmidt pair, 278 Schmidt, Erhard, 200, 278 Schurs theorem, 204 Schur, Issai, 204 Schwarz, Hermann, 189 self-adjoint operator, 233 Shelley, Mary Wollstonecraft, 50 sign of a permutation, 349 simultaneous diagonalization, 176 simultaneously upper triangularizable, 178 singular matrix, 91 singular value decomposition of adjoint, 275 of linear map, 273 of pseudoinverse, 275 singular values, 271, 362 skew operator, 240, 247, 269 span, 29 spans, 29 spectral theorem, 245, 246 square matrix, 91 square root of an operator, 248, 251, 253, 320 standard basis of , 39 of () , 39 subspace, 18 subtraction of complex numbers, 4 sum, see addition sum of subspaces, 19 Supreme Court, 210 surjective, 62 SVD, see singular value decomposition Sylvester, James, 181 symmetric bilinear form, 337 symmetric matrix, 269, 337 tensor product, 372, 379 \n\nThrough the Looking Glass , 11 trace of a matrix, 326 of an operator, 327 translate, 99 transpose of a matrix, 77, 231 triangle inequality, 121, 190, 281 tuple, 5 two-sided ideal, 58 unit circle in , 262, 269 unitary matrix, 263 unitary operator, 260 University of Dublin, 297 University of Gttingen, 332 upper-triangular matrix, 155160, 264, 267, 314 Vandermonde matrix, 366 vector, 8, 12 vector space, 12 volume, 292, 363 of a box, 291 zero of a polynomial, 122 \n\n> Linear Algebra Done Right , fourth edition, by Sheldon Axler\n\n# Colophon: Notes on Typesetting \n\n This book was typeset in LuaL ATEX by the author, who wrote the L ATEX code to implement the books design. \n\n The L ATEX software used for this book was written by Leslie Lamport. The T EXsoftware, which forms the base for L ATEX, was written by Donald Knuth. \n\n The main text font in this book is the Open Type Format version of T EX Gyre Termes, a font based on Times, which was designed by Stanley Morison and Victor Lardent for the British newspaper The Times in 1931. \n\n The main math font in this book is the Open Type Format version of T EX Gyre Pagella Math, a font based on Palatino, which was designed by Hermann Zapf. \n\n The sans serif font used for page headings and some other design elements is the Open Type Format version of T EX Gyre Heros, a font based on Helvetica, which was designed by Max Miedinger and Eduard Hoffmann. \n\n The LuaL ATEX packages fontspec and unicode-math , both written by Will Robertson, were used to manage fonts. \n\n The L ATEX package fontsize , written by Ivan Valbusa, was used to gracefully change the main fonts to 10.5 point size. \n\n The figures in the book were produced by Mathematica , using Mathematica \n\ncode written by the author. Mathematica was created by Stephen Wolfram. The Mathematica package MaTeX , written by Szabolcs Horvt, was used to place L ATEX-generated labels in the Mathematica figures. \n\n The L ATEX package graphicx , written by David Carlisle and Sebastian Rahtz, was used to include photos and figures. \n\n The L ATEX package multicol , written by Frank Mittelbach, was used to get around L ATEXs limitation that two-column format must start on a new page (needed for the Symbol Index and the Index). \n\n The L ATEX packages Ti kZ, written by Till Tantau, and tcolorbox , written by Thomas Sturm, were used to produce the definition boxes and result boxes. \n\n The L ATEX package color , written by David Carlisle, was used to add appropriate color to various design elements. \n\n The L ATEX package wrapfig , written by Donald Arseneau, was used to wrap text around the comment boxes. \n\n The L ATEX package microtype , written by Robert Schlicht, was used to reduce hyphenation and produce more pleasing right justification.\n","transcriptionSegments",[],"notesText","title","flashcards",{"_69":71},[72,84,90,96,102,108,114,120,126,132,138,144,150,156,162,168,174,180,186,192,198,204,210,216,222,228,234,240,246,252],{"_73":74,"_75":76,"_77":78,"_79":80,"_74":81,"_82":83},"answer","choice_d","choice_a","Emphasis on solving systems of equations","choice_b","Emphasis on computational techniques","choice_c","Emphasis on determinants and their properties","Emphasis on abstract vector spaces and linear maps","question","What is the main focus of 'Linear Algebra Done Right' compared to most introductory linear algebra courses?",{"_73":79,"_75":85,"_77":86,"_79":87,"_74":88,"_82":89},"Read quickly to get an overview","Read as you would a novel","Read slowly, verifying steps and working out details","Skip exercises and focus on theorems","According to the preface, what is the recommended approach to reading and learning from this book?",{"_73":75,"_75":91,"_77":92,"_79":93,"_74":94,"_82":95},"Jordan canonical form","Singular value decomposition","Gram-Schmidt procedure","Cauchy-Schwarz inequality","Which of the following is NOT listed as one of the author's top ten favorite results in the book?",{"_73":77,"_75":97,"_77":98,"_79":99,"_74":100,"_82":101},"A set with only addition defined","A set with addition and scalar multiplication satisfying certain properties","A set of numbers with multiplication","A set of points in space","What is the definition of a vector space as given in the book?",{"_73":77,"_75":103,"_77":104,"_79":105,"_74":106,"_82":107},"Commutativity of addition","Existence of a multiplicative inverse for every element","Associativity of addition","Distributive property of scalar multiplication","Which property is NOT required in the definition of a vector space?",{"_73":74,"_75":109,"_77":110,"_79":111,"_74":112,"_82":113},"A subset with no additional structure","A subset containing only the zero vector","A subset closed under multiplication only","A subset that is also a vector space with the same operations","What is a subspace of a vector space?",{"_73":74,"_75":115,"_77":116,"_79":117,"_74":118,"_82":119},"It is always equal to V","It is always a direct sum","It is always the intersection of U and W","It is the smallest subspace containing both U and W","Which of the following is true about the sum of two subspaces U and W of a vector space V?",{"_73":77,"_75":121,"_77":122,"_79":123,"_74":124,"_82":125},"The number of elements in the space","The length of any basis of the space","The number of subspaces","The number of vectors in the standard basis only","What is the dimension of a finite-dimensional vector space defined as?",{"_73":79,"_75":127,"_77":128,"_79":129,"_74":130,"_82":131},"A list is linearly independent if all vectors are nonzero","A list is linearly independent if it contains the zero vector","A list is linearly independent if no vector is a linear combination of the others","A list is linearly independent if all vectors are equal","Which statement about linear independence is correct?",{"_73":79,"_75":133,"_77":134,"_79":135,"_74":136,"_82":137},"dim V = dim null T × dim range T","dim V = dim null T / dim range T","dim V = dim null T + dim range T","dim V = dim null T - dim range T","What does the fundamental theorem of linear maps state for a linear map T on a finite-dimensional vector space V?",{"_73":77,"_75":139,"_77":140,"_79":141,"_74":142,"_82":143},"U ∪ W = V","U ∩ W = {0}","U + W = V","U = W","Which of the following is a necessary and sufficient condition for a sum of two subspaces U and W to be a direct sum?",{"_73":79,"_75":145,"_77":146,"_79":147,"_74":148,"_82":149},"Any polynomial that annihilates T","A polynomial with all roots equal to zero","The unique monic polynomial of smallest degree such that p(T) = 0","The characteristic polynomial of T","What is the minimal polynomial of an operator T?",{"_73":77,"_75":151,"_77":152,"_79":153,"_74":154,"_82":155},"It is computationally faster","It provides more insight and simpler proofs","It is only valid for real vector spaces","It avoids the need for linear maps","According to the book, what is the main advantage of the determinant-free approach to eigenvalues on complex vector spaces?",{"_73":79,"_75":157,"_77":158,"_79":159,"_74":160,"_82":161},"It is proved using determinants","It is only about invertible operators","It characterizes self-adjoint and normal operators via the existence of an orthonormal basis of eigenvectors","It applies only to real vector spaces","Which of the following is true about the spectral theorem as presented in the book?",{"_73":79,"_75":163,"_77":164,"_79":165,"_74":166,"_82":167},"To solve systems of equations","To find eigenvalues","To construct an orthonormal basis from a linearly independent list","To compute determinants","What is the main purpose of the Gram-Schmidt procedure?",{"_73":77,"_75":169,"_77":170,"_79":171,"_74":172,"_82":173},"Existence of additive identity only","Every nonzero element has a multiplicative inverse","Associativity of addition only","Commutativity of addition only","Which property distinguishes a field from a general set with addition and multiplication?",{"_73":75,"_75":175,"_77":176,"_79":177,"_74":178,"_82":179},"It is the vector from the origin to the sum of their coordinates","It is the intersection point of the two vectors","It is the product of their magnitudes","The sum is always the zero vector","In the context of vector spaces, what is the geometric interpretation of the sum of two vectors in R^2?",{"_73":79,"_75":181,"_77":182,"_79":183,"_74":184,"_82":185},"A group of vectors with no order","A collection of elements with infinite length","A collection of elements where order and repetition matter","A set of unique elements","What is the definition of a list in the context of linear algebra as described in the book?",{"_73":77,"_75":187,"_77":188,"_79":189,"_74":190,"_82":191},"There can be multiple additive identities","It is unique for each vector space","It must be a nonzero vector","It is not required to exist","Which of the following is true about the additive identity in a vector space?",{"_73":77,"_75":193,"_77":194,"_79":195,"_74":196,"_82":197},"Lists are always unordered","Order matters in lists but not in sets","Lists cannot have repeated elements","Sets are always infinite","What is the main difference between a set and a list in linear algebra?",{"_73":79,"_75":199,"_77":200,"_79":201,"_74":202,"_82":203},"Multiplying two vectors together","Adding a scalar to each coordinate","Multiplying each coordinate of a vector by a scalar","Multiplying a vector by another vector","What is the definition of scalar multiplication in the vector space F^n?",{"_73":75,"_75":205,"_77":206,"_79":207,"_74":208,"_82":209},"Distributivity of multiplication over addition only for real numbers","Commutativity of addition and multiplication","Existence of additive and multiplicative inverses","Associativity of addition and multiplication","Which of the following is NOT a property of complex arithmetic as listed in the book?",{"_73":75,"_75":211,"_77":212,"_79":213,"_74":214,"_82":215},"It is at most the dimension of the space","It is always equal to the dimension of the space","It is unrelated to the dimension","It is always greater than the dimension","What is the significance of the minimal polynomial's degree for an operator on a finite-dimensional vector space?",{"_73":75,"_75":217,"_77":218,"_79":219,"_74":220,"_82":221},"A quotient space consists of all translates of a subspace","A quotient space is always one-dimensional","A quotient space is always the same as the original space","A quotient space is the intersection of two subspaces","Which of the following statements about quotient spaces is correct?",{"_73":79,"_75":223,"_77":224,"_79":225,"_74":226,"_82":227},"To provide a set of vectors orthogonal to V","To define a new vector space unrelated to V","To represent all linear functionals from V to the field","To find the dimension of V","What is the main purpose of the dual space V* in linear algebra?",{"_73":74,"_75":229,"_77":230,"_79":231,"_74":232,"_82":233},"The intersection is never a subspace","The intersection is always the whole space","The intersection is always the zero subspace","The intersection of any collection of subspaces is a subspace","Which of the following is true about the intersection of subspaces?",{"_73":77,"_75":235,"_77":236,"_79":237,"_74":238,"_82":239},"A set of vectors that do not span V","A list of vectors that spans V and is linearly independent","A list of vectors that are all zero","A set of vectors that are all equal","What is the definition of a basis for a vector space V?",{"_73":77,"_75":241,"_77":242,"_79":243,"_74":244,"_82":245},"Every linearly independent list contains the zero vector","Every linearly dependent list contains a vector that is a linear combination of the previous ones","Every vector space is infinite-dimensional","Every list of vectors is linearly independent","Which of the following is a consequence of the linear dependence lemma?",{"_73":74,"_75":247,"_77":248,"_79":249,"_74":250,"_82":251},"dim U is unrelated to dim V","dim U = dim V for all subspaces","dim U > dim V always","dim U ≤ dim V","What is the relationship between the dimension of a subspace U and the dimension of the whole space V?",{"_73":77,"_75":253,"_77":254,"_79":255,"_74":256,"_82":257},"dim(U + W) = dim U + dim W + dim(U ∩ W)","dim(U + W) = dim U + dim W - dim(U ∩ W)","dim(U + W) = dim U - dim W","dim(U + W) = dim U × dim W","Which of the following is true about the sum of the dimensions of two subspaces U and W?","definitions","youtubeUrl","https://linear.axler.net/LADR4e.pdf","emoji","noteLanguage","errorMessage","Error creating notes from the 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