Understanding Matrix Diagonalization

Hi Engineering Math folks! I am Rasika and I'm back with 8.12 which is section 12 on chapter 8 diagonalization and this will be the last section on chapter 8. So there is a definition to give you an idea what is the word diagonalizable means. An n by n metric a that means a square metric a is diagonalizable if there is some invertible n by n metric that means another square metric that we're going to call as P and a diagonal metric D such that this relationship holds here so all these a P D are in by n square matrices same side and if this relation occurs so D is diagonalized the diagonal metric that means only the diagonal have elements other than zeros everything else is zeros When this occurs, we say that A and D are similar matrices. So the middle A here and D are going to be called similar matrices. So the first theorem, let A be an n by n metric. If A has n distinct eigenvalues, that means A is diagonalizable. That means for sure if you have n distinct eigenvalues, then A for sure is going to be diagonalizable. But... A is diagonalizable if and only if A has n linearly independent eigenvectors. So the first example saying we are both theorem can be visualized in the following way. So we have A is diagonalizable. and the other thing what we have is it's going to have n distinct eigenvalues and then the other condition we have is n linearly independent eigenvectors n linearly independent eigenvectors So the relation is n distinct eigenvalues going to imply that A is going to be diagonalizable. So this implies that but here this is like if and only if that goes ways. A is diagonalizable and N is independent eigenvectors so that means this having the double arrow and then what does this saying for us? If there's n distinct eigenvalues, that means you're going to have n linearly independent eigenvectors. So we can connect this one with one arrow. So this is what the relation between being a diagonalizable with the eigenvalues with the eigenvectors. Okay. So there are a couple of remarks. The process of diagonalizing a vector A involves two main steps. steps ok so the first step you have to find the eigenvalues including the multiple roots because sometimes if for example if it is like lambda minus 1 to the square equals 0 that means you are having the multiplicity of 1 to be twice that means lambda is 1 1 twice so that's what here means by including the multiple arrange them as the entries along the main diagonal of the metric called D so the diagonal metric once you know the eigenvalues you can create your own capital D metric all your eigenvalues going to be in the diagonal the rest is 0 that is your first step and then second step find the eigenvectors of A which means for every lambda you can find the eigenvectors make these the columns of P so that going to create your metric P all the icon vectors going to be on your columns and then that will satisfy this relationship the way that you creating your d and p you know how to find the p inverse multiply everything together on the left you will end up with the diagonal okay if the it is saying that here vi is an eigenvector corresponding to lambda i that means if you find the eigenvector corresponding to lambda 1 that should be in your first column if you have lambda n in like third then that Vn like corresponding eigenvector has to be on the third column of P if there aren't enough eigenvectors to fill up then of course A is not going to be diagonalizable okay so we have first example given the fact that the eigenvalues of A are 0 and 6 diagonalizable diagonalize A if possible okay so let's see so So it's giving that the eigenvalues already. So that's kind of little easy because one step have been done. Okay. So I will put a note here. This is a 2 by 2 metric, right? So the note I'm going to tell here, the theorem is going to guarantee that A is diagonalizable. because can someone think of why the reason it's a two by two and they have provided two distinct eigen values so the theorem says if you have n distinct eigen values that for sure is going to be diagonalizable because it is a two by two metric Okay, so once you know the eigenvalues, what you can do is you can create your d. So your d is going to be 0 in the 1 of the position and then the 6 is going to be the other diagonal element. The rest is going to be 0, so this is going to be your d. Now we need to create our p. For that, you need the eigenvectors corresponding to 0 and 6. So for lambda equal... 0 we need a minus 0 times I 2 which is going to be A that means negative 4 negative 5 8 10 that's the same thing as A and then I need to solve A minus 0 times I2 V equals to 0 for V equals V1 V2 I hope you guys know know how to get this in one step because we were doing over and over in the previous sections. So I need to create an augmented metric. So the augmented metric gonna be negative 4 negative 5 0 0 8 10 I need to use the row operations to make it to the row echelon form let's see so what I'm gonna do is I'm gonna add our 2 with plus r1 so negative 4 negative 5 0 0 this is gonna give me 0 it's gonna give me 0 so this is going to be the row echelon form that's gonna give me some values so v1 is going to give me negative 5 over 4 V2 and V2 is free that gives me the parametric form that v1 v2 is equal to v2 times negative 5 over 4 then 1. ok so you will see that v2 is a free variable so v2 can be anything and I am going to put this column vector when I'm creating my P. So the best choice to pick V2 in such a way that I can cancel out the denominator here. So I'm going to choose V2 to be 4. Why not? So that will give me to get V equals negative 5 and 4. because when i multiply by four this is gonna be four times one so it will give me so now we have to keep the order so on the first like the p metric your first column going to be negative 5 4 because I put 0 in the first position on D okay so let's go ahead and do the same thing for our lambda equals 6 so that's going to give me a minus 6 I 2 that's going to give me what a minus 6 let me go ahead and write it so I can see it which is going to be negative 10 negative 5 8, 4. This is what I'm going to get and then I have to solve for a minus 6i2, v equals 0 for v. See I write everything because when i'm doing an exam i want to get all the points that i can get and you can convince the instructor that you have you know you know everything you have put on everything on the paper okay so i have to solve this so negative 10 negative 584 I have 0 0 again row operations hmm now what I'm gonna do is I'm gonna replace my R2 first dividing by 5 and then multiplying by 4 so dividing by 5 will give you 2 so I have to multiply by 4 to get 8 so this is going to be negative 10 negative 5 if you don't feel comfortable like doing the fraction directly you can you know one time divide everything by two and then you know use the row operations and you know the row operations by now and feel free to do the algebra you can take as many as steps to like do it okay so this is going to give me v1 equals negative one half v2 and v2 is free that will give me the parametric Formed to be v1, v2 equals v2 out negative half and 1. Again, I'm going to choose my v2 in such a way that I can get rid of this fraction here, so 2. So I'm going to choose v2 to be 2 to get... V equals negative 1 and 2. Now since we have all the two independent eigen vectors so I'm going to do is we write d equals to 0 0 0 6 then we have p equals now belonging to 0 the eigenvector was negative 5 4 belonging to 6 it was negative 1 2 so that's going to be your p then it is true t equals to p inverse ap now even though i mentioned here it is true you better check at least once in your life okay so at least check and make sure that when you create P inverse that's a good practice for you guys to learn or refresh back again how to find the inverse of a metric you already know A you already know P multiply them together a good practice to see how to multiply three matrices and then see whether you're going to end up with D okay so Let's see the next example. We have a remark before of example. So the remark is saying there is a lot of freedom in order to, in the order of the eigenvalues as well as the choices for the free variables in the eigenvectors. That's what you saw in the previous example here. so I pick 0, 6 what's going to happen if you pick 6 to be here and 0 to be here then you have to switch these columns because 6 corresponding to 6 should be the first column if I move 6 to be the first position so you have more freedom you have freedom to do the row operations you have freedom to pick whatever your free variables depending on you know whatever the fractions you have and what's the best choice the reason for picking like something that you can cancel the fraction here that the denominator part is if you really have to multiply the things it's nice to have more numbers rather than you dealing with the fractions so that that's why it says you have a lot of choices ok so the next example diagonalize if possible ok now from the previous example to here the difference is we do not know what are the eigenvalues now so you have to do one more step than your previous problem so let's go ahead and find the eigenvalues so for eigenvalues you need the determinant of a minus lambda i2 so that's going to give you negative 2 minus lambda negative 1 1 negative 4 minus lambda you need to find the module the determinant value that's going to give you negative 2 minus lambda times negative 4 minus lambda minus 1 times negative 1 let's see what's going to happen so 8 2 lambda 4 lambda 6 lambda plus lambda squared plus 1 so that's going to give you lambda squared plus 6 lambda plus 9 you can factor this guy lambda plus 3 everything to the square okay so this is the characteristic polynomial so we can find so from here so the eigenvalues gonna be negative 3 with multiplicity 2 because I know that this is the 2 by 2 but I'm just only leaving one right so it has to be twice counted that the same thing okay now we have a 2 by 2 which have one eigenvalue what does that the previous theorem says the previous theorem guarantees only if n distinct eigenvalues then you will have a diagonalizable so we are not so sure when we use negative 3 we gonna get to diagonalizable a so you have to you know work and see what's gonna happen so there's a node Note that the theorem does not guarantee that A is diagonalizable. More info needed. ok let's see what are the moving force so at least for these negative three if I can get two independent vectors column vectors then you can form your P and then they are is sort of a guarantee that hey I will have a diagonalizable a so sometimes you will get like two independent vectors out of these eigenvectors sometimes no so you have to work on and see so for lambda equals negative 3 we need to find first a minus minus 3 times I2 first that means a plus 3 I2 that gives you 1 negative 1 and then 4 negative 4 did I do correctly plus three times one negative one but here i think i messed up right a so you have a plus so this is one because only the diagonal works and this is negative one okay negative 1 and this is the a minus minus 3 i 2 now i need to solve a plus 3 i 2 times v equals 0 for v which is v 1 v 2 that means i need the augmented metric so plus 1 negative 1 1 negative 1 0 0 i will do the row operations i have 1 negative 1 0 i need to get rid of this guy so r 2 minus r1 that gives you zero zero zero so this is going to give you v1 equals to v2 and v2 is free so the parametric I'm going to be V equals V1 V2 that means V211 okay so what happened we do is free and the only eigenvector I'm going to get is 1,1 but I have two eigenvalues it's going to be negative 3 comma negative 3 because the multiplicity is 2 but I'm ending up with 1 that means I cannot create my P by using only one vector that means a is not going to be diagonalizable so any eigenvector of a for lambda equals negative 3 must be a multiple of one more. As a result there is no way to fill second column of a potential P. ok so there is not enough linearly independent eigenvectors so A is not diagonalizable so in a three by three case on the same question sometimes you may have the multiplicity tool but you may have two free variable so in that way in that case for one eigen value that's going to create two independent eigenvectors and then for the other leftover one you will have one so in such a case you still may get a diagonalizable a so it depends on like the question but in this two by two case if you don't get two then you are done you have to stop from there okay so we have another example good to do more examples so in our next example diagonalize a if possible okay so let's see what's going to happen so just minute ago i said like you know sometimes in three by threes you will still get diagonalizable let's try out so we have a metric a you have to diagonalize there if possible okay so we don't know first we need the eigenvalues for that you need the determinant of a minus lambda i 3 now because it's a 3 by 3 metric that gives you what hmm i have to do more work now okay so one minus lambda five zero zero two minus lambda zero zero three one minus lambda I need to find the determinant now how are we going to do the determinant now it's a three by three so you have to go for the general formula remember that you have to the easiest path so I see two zeros in this column or this column so you can either pick the first column or the second the third column in order to do the determiner that will be the easiest path so I am going to choose the first column so I know that with the sign like the sign values that this is going to be plus alternate negative plus but really who cares because it's going to give you zero zero so this will give me one minus lambda positive sign and then you remove the first row remove the first column so you will have two minus lambda zero three one minus lambda so once you do this one you have to find the determinant of this two by two again you have one minus lambda it's going to give you 2 minus lambda 1 minus lambda minus 0 three times times 0. So you really have 1 minus lambda squared 2 minus lambda. So this will give you the characteristic polynomial. So the eigenvalues are going to be lambda equals 1 with multiplicity. 2 because I have the power 2 and then the other one lambda equals to 2 multiplicity 1 because it just have the power 1 there right okay so I have two eigenvalues for a 3 by 3 now I have to go ahead and find eigenvectors belonging corresponding to this values okay so we have a lot of work to do lambda equals 1 first so you need a minus i3 basically that's going to give you 050 010 if you don't see how that the answer is coming here you better write one more step okay and then I have to solve a minus i3 b equals 0 for b now we have v1 v2 v3 okay so i have to create the augmented metric 0 5 0 0 1 0 0 3 0 have 0 0 0 I need to create row reduced form row echelon form what gonna happen I don't have ones in my first column at all so I have no choice let's see what's gonna happen when I try to get rid of these three yeah let's try to get rid of these three and one so what i'm gonna do is i'm gonna do r2 minus r2 is going to get to dove let's see what's the best choice r2 i'm going to divide the first row by 5 and then going to subtract so negative 1 fifth of r1 and then r3 i will do three fifth r1 so dividing by five subtracting from here so that does not change my first row but these will be oh zeros so what's that gonna give us so it basically saying what I have 5 V 2 equals to 0 that means v2 equals 0 but the rest I just have three variables they all 0 so v1 and v3 are free so when I get into the parametric form I will have V to be V1, V2, V3. That's going to give me... v1, 0, v3. So you have v1 times 1, 0, 0 plus v3 times 0, 0, 1. Okay. So this is what's going to happen actually when you pull v1 and v3. So two free variables. v1 and v3 so we can choose whatever we want there's no luckily fraction so we don't have any problem to get rid of the fractions so we can pick like the simplest ones so let's see so I'm going to choose first v1 to be 1 and my v3 to be 0 so in that way I can just discard this guy and take out the first vector so this will will give me 1 0 0 and the next round I'm going to choose v1 to be 0 and v3 to be 1 so these are the simplest numbers now imagine if I pick v1 to be 1 and v3 to be 1 then I have to add them together in order to get this guy but the simplest you just multiplying by 0 everything disappeared multiply by 0 so you have to be little smart when you are picking values for free variables so this will give me 0 0 1 so they are like linearly independent so I had the multiplicity tool on lambda equals 1 and I have two linearly independent eigenvectors so it's like giving some kind of a promise that if I can find eigenvector For lambda equals 2, then I may have a diagonalizable A. But for that, you have to do a few more steps. So for lambda equals 2, I need to do A minus 2I3. that's going to give me negative 150 and then this line will be 0 0 0 this is going to give me 0 3 negative 1 okay and solve a minus 2 I 3 we equal 0 for B that means you need the augmented metric negative 150 000 0 3 negative 1 i need to do some row operations so what's the best i have a negative one on the first stop so i will get rid of what what's the best I will replace r1 first divide by r3 on 3 and then multiply it by 5. Let's see what's going to happen. So I'm going to do r1 minus 5 over 3 r3. so I have a zero here that will end up negative 1 this is divided by 3 so you're going to have 1 minus 5 gonna make it this guy 0 and this one negative 1 so that will end up with 5 over 3 0 this I didn't touch this I didn't do anything either I think that's enough because V2 is going to be free. Is it? No, I don't think so. V3 is going to be free. And then the rest you can find some values. Okay, let's see. Negative V1. v1 equals to 5 over 3 v3 actually and then you have from this guy v3 equals to 3v2 so v1 v3 or in other words i can write v2 as one third of v3 so v1 and v2 are both with respect to v3 therefore v3 is free so the parametric form gonna give you v equals what V3 out. So this is going to give you 5 over 3, 1 third and 1. And now you will see that here 3 is like annoying. It's in the denominator. So let's choose to get rid of that 3. So V3 to be 3. So that's going to give me 5, 1 and 3 to be an eigenvector. So this is going to give you, now you're happy. D I found lambda equals 1 first that has to be on the first two spots because that's the rule of the theorem and then I have lambda equals 2 on this part and then my P will be belonging to 1 I have two linearly independent vectors so either you can write this one first and this one second or vice versa it doesn't do anything so 1 0 0 0 0 0 1 belonging to 2 it became 5 1 3 so as a result we found our D and P you have to go ahead and check D equals to P inverse AP so you have to do the checking part there okay so if we go to our next page there are a couple of remarks so the first remark is saying that note that diagonalizing a metric with complex eigenvalues work the same way okay so we agreed that on the previous section that sometimes complex eigenvalues happen they come as a pair and then you will have a pair of eigenvectors as well still this theory am going to work and then the second remark what's the point in doing this someone can ask why the hell we do this much of algebra okay so for that we will actually give the answer on the next couple of remarks so we are doing this one to make our life little simpler okay so this is what gonna happen as a reminder we're gonna say that in a regular number if you need a square that means you are multiplying two things together right x squared means x times x so a squared is going to be a times a you're going to multiply a twice right they are matrices note that a squared is not just the square of each individual entry of a that's the important part because a if a is two by two and if you're multiplying that doesn't mean like you are squaring each of the component you have to do in a special way of multiplying the matrices so in more general quantity a to the k for any positive k is going to be defined similarly so this is what going to happen when you have the diagonal metric that means you're only the diagonal values are non-zeros that gives us the opportunity if i want to do the d to the power k any real number k that means i just have to blindly like take the power whatever only on the diagonal so that's make our life simpler so that's the whole point like doing all that process like checking whether a is diagonalized but if since a and D are similar according to our theorem because we promise that we will have such a P such a that a and D going to be similar so if I need a to the K that means the same thing you just take D to the k. Let's see. We will see some problems here. And then the last remark, it is possible to define concepts of matrix square roots as well as exponentials and logarithms. We will see them in future. An example, if a is an n by n diagonalizable metric, find a squared and use your answer to guess what a to the k. I just gave you the answer in my explanation but let's see what's going to happen. So, A is going to be A times A, right? But I know that... A out of this one. See P inverse AP equals to D. Right. This is coming from the theorem. But if I want to isolate my A what am I going to do is I have to multiply from the right side of both sides by P inverse. and then i have to multiply the left side by p to get cancel of this guy so i will like have pp inverse a and then pp inverse like this but that will give you pdp inverse this is the identity because you're multiplying with the inverse of that you have a this guy gonna give you identity what does that tell us so you can write a because multiplying with the identity gonna give you the same thing a is pdp inverse so we're going to replace this a that here a squared i forgot the square i'm going to replace this a by pdp inverse now so 1 a is p dp inverse i have again pdp inverse that's gonna give me matrix multiplication it's going to be commutative and the associative property holes so p d i'm going to multiply first p inverse p together and you have dp inverse this going to give you the identity so basically you have p d dp inverse what that gives you p d squared p inverse so when you have to find a squared you can actually find if you know the P and P inverse you just square the diagonal and the diagonal much easier you just square that the diagonal values that is going to be zeros so these can be generalized that a to the any k power is going to be just your p and d to the power k which is easy to handle the diagonal metric and p inverse. So this This is one main reason why we learned the diagonalizing business. Okay so let's see. Use the fact that these given matrices to come. compute 10th power of this given A. So let's see. Now if you consider this one to be A, this is the distance of P. This is my diagonal metric. This is my P inverse. Right? I think we just downed this problem here a little while ago. Okay. So we have this. We know what is. are PD and penis I need to find the 10th power of a right a to the 10th which is negative 4 negative 5 8 10 to the 10th power that's going to give you P t to the 10th and then p inverse this is possible because d is diagonal only the diagonal have non-zero values everything else zero so that gives you p is negative five negative one four two these zero to the tenth zero zero six to the tenth we're gonna see big numbers and in was luckily given if you haven't given the inverse you need to find so you should know how to find the inverse okay so this is negative 5 negative 1 4 2 this is going to give you 0 0 0 6 to the 10th what 6 to the 10th i'll leave it like that and then i have this guy negative 1 third negative 1 6 2 third 5 over 6 let's multiply so we know that we can't multiply three matrices together you have to do two at the time so i'm gonna do the first two negative five times zero is zero negative one times zero is zero so the first component zero negative five times zero and then negative six to the tenth four times zero zero four times zero then two times six to the tenth that's going to be the first two and then I still have negative 1 third negative 1 sixth 2 third 5 over 6 now I have to multiply these two again I'm lucky because two things are here 0 so 0 times this and then this guy so negative 6 to the 10 two-third and then zero negative six to the tenth five over six zero two to the sixth to the tenth times two-third and then zero two to the sixth to the tenth five over six you guys can use a calculator and you will see that big numbers coming up and I will put the answer here so negative 403 I guarantee I'm not going to give this much of a big numbers on you quizzes or exams so eight two six two one five six eight and then seven seven six nine six zero okay so that's the tenth power of your a given here okay guys so that's the end of the section last section on your chapter again there are practice problems you have to do odd problems and i will be posting on for these odd questions okay see you guys in my next chapter

So the middle A here and D are going to be called similar matrices. So the first theorem, let A be an n by n metric. If A has n distinct eigenvalues, that means A is diagonalizable.

That means for sure if you have n distinct eigenvalues, then A for sure is going to be diagonalizable. But... A is diagonalizable if and only if A has n linearly independent eigenvectors.

So the first example saying we are both theorem can be visualized in the following way. So we have A is diagonalizable. and the other thing what we have is it's going to have n distinct eigenvalues and then the other condition we have is n linearly independent eigenvectors n linearly independent eigenvectors So the relation is n distinct eigenvalues going to imply that A is going to be diagonalizable.

So this implies that but here this is like if and only if that goes ways. A is diagonalizable and N is independent eigenvectors so that means this having the double arrow and then what does this saying for us? If there's n distinct eigenvalues, that means you're going to have n linearly independent eigenvectors.

So we can connect this one with one arrow. So this is what the relation between being a diagonalizable with the eigenvalues with the eigenvectors. Okay. So there are a couple of remarks.

The process of diagonalizing a vector A involves two main steps. steps ok so the first step you have to find the eigenvalues including the multiple roots because sometimes if for example if it is like lambda minus 1 to the square equals 0 that means you are having the multiplicity of 1 to be twice that means lambda is 1 1 twice so that's what here means by including the multiple arrange them as the entries along the main diagonal of the metric called D so the diagonal metric once you know the eigenvalues you can create your own capital D metric all your eigenvalues going to be in the diagonal the rest is 0 that is your first step and then second step find the eigenvectors of A which means for every lambda you can find the eigenvectors make these the columns of P so that going to create your metric P all the icon vectors going to be on your columns and then that will satisfy this relationship the way that you creating your d and p you know how to find the p inverse multiply everything together on the left you will end up with the diagonal okay if the it is saying that here vi is an eigenvector corresponding to lambda i that means if you find the eigenvector corresponding to lambda 1 that should be in your first column if you have lambda n in like third then that Vn like corresponding eigenvector has to be on the third column of P if there aren't enough eigenvectors to fill up then of course A is not going to be diagonalizable okay so we have first example given the fact that the eigenvalues of A are 0 and 6 diagonalizable diagonalize A if possible okay so let's see so So it's giving that the eigenvalues already. So that's kind of little easy because one step have been done.

Okay. So I will put a note here. This is a 2 by 2 metric, right?

So the note I'm going to tell here, the theorem is going to guarantee that A is diagonalizable. because can someone think of why the reason it's a two by two and they have provided two distinct eigen values so the theorem says if you have n distinct eigen values that for sure is going to be diagonalizable because it is a two by two metric Okay, so once you know the eigenvalues, what you can do is you can create your d. So your d is going to be 0 in the 1 of the position and then the 6 is going to be the other diagonal element.

The rest is going to be 0, so this is going to be your d. Now we need to create our p. For that, you need the eigenvectors corresponding to 0 and 6. So for lambda equal... 0 we need a minus 0 times I 2 which is going to be A that means negative 4 negative 5 8 10 that's the same thing as A and then I need to solve A minus 0 times I2 V equals to 0 for V equals V1 V2 I hope you guys know know how to get this in one step because we were doing over and over in the previous sections.

So I need to create an augmented metric. So the augmented metric gonna be negative 4 negative 5 0 0 8 10 I need to use the row operations to make it to the row echelon form let's see so what I'm gonna do is I'm gonna add our 2 with plus r1 so negative 4 negative 5 0 0 this is gonna give me 0 it's gonna give me 0 so this is going to be the row echelon form that's gonna give me some values so v1 is going to give me negative 5 over 4 V2 and V2 is free that gives me the parametric form that v1 v2 is equal to v2 times negative 5 over 4 then 1. ok so you will see that v2 is a free variable so v2 can be anything and I am going to put this column vector when I'm creating my P. So the best choice to pick V2 in such a way that I can cancel out the denominator here.

So I'm going to choose V2 to be 4. Why not? So that will give me to get V equals negative 5 and 4. because when i multiply by four this is gonna be four times one so it will give me so now we have to keep the order so on the first like the p metric your first column going to be negative 5 4 because I put 0 in the first position on D okay so let's go ahead and do the same thing for our lambda equals 6 so that's going to give me a minus 6 I 2 that's going to give me what a minus 6 let me go ahead and write it so I can see it which is going to be negative 10 negative 5 8, 4. This is what I'm going to get and then I have to solve for a minus 6i2, v equals 0 for v. See I write everything because when i'm doing an exam i want to get all the points that i can get and you can convince the instructor that you have you know you know everything you have put on everything on the paper okay so i have to solve this so negative 10 negative 584 I have 0 0 again row operations hmm now what I'm gonna do is I'm gonna replace my R2 first dividing by 5 and then multiplying by 4 so dividing by 5 will give you 2 so I have to multiply by 4 to get 8 so this is going to be negative 10 negative 5 if you don't feel comfortable like doing the fraction directly you can you know one time divide everything by two and then you know use the row operations and you know the row operations by now and feel free to do the algebra you can take as many as steps to like do it okay so this is going to give me v1 equals negative one half v2 and v2 is free that will give me the parametric Formed to be v1, v2 equals v2 out negative half and 1. Again, I'm going to choose my v2 in such a way that I can get rid of this fraction here, so 2. So I'm going to choose v2 to be 2 to get... V equals negative 1 and 2. Now since we have all the two independent eigen vectors so I'm going to do is we write d equals to 0 0 0 6 then we have p equals now belonging to 0 the eigenvector was negative 5 4 belonging to 6 it was negative 1 2 so that's going to be your p then it is true t equals to p inverse ap now even though i mentioned here it is true you better check at least once in your life okay so at least check and make sure that when you create P inverse that's a good practice for you guys to learn or refresh back again how to find the inverse of a metric you already know A you already know P multiply them together a good practice to see how to multiply three matrices and then see whether you're going to end up with D okay so Let's see the next example.

We have a remark before of example. So the remark is saying there is a lot of freedom in order to, in the order of the eigenvalues as well as the choices for the free variables in the eigenvectors. That's what you saw in the previous example here.

so I pick 0, 6 what's going to happen if you pick 6 to be here and 0 to be here then you have to switch these columns because 6 corresponding to 6 should be the first column if I move 6 to be the first position so you have more freedom you have freedom to do the row operations you have freedom to pick whatever your free variables depending on you know whatever the fractions you have and what's the best choice the reason for picking like something that you can cancel the fraction here that the denominator part is if you really have to multiply the things it's nice to have more numbers rather than you dealing with the fractions so that that's why it says you have a lot of choices ok so the next example diagonalize if possible ok now from the previous example to here the difference is we do not know what are the eigenvalues now so you have to do one more step than your previous problem so let's go ahead and find the eigenvalues so for eigenvalues you need the determinant of a minus lambda i2 so that's going to give you negative 2 minus lambda negative 1 1 negative 4 minus lambda you need to find the module the determinant value that's going to give you negative 2 minus lambda times negative 4 minus lambda minus 1 times negative 1 let's see what's going to happen so 8 2 lambda 4 lambda 6 lambda plus lambda squared plus 1 so that's going to give you lambda squared plus 6 lambda plus 9 you can factor this guy lambda plus 3 everything to the square okay so this is the characteristic polynomial so we can find so from here so the eigenvalues gonna be negative 3 with multiplicity 2 because I know that this is the 2 by 2 but I'm just only leaving one right so it has to be twice counted that the same thing okay now we have a 2 by 2 which have one eigenvalue what does that the previous theorem says the previous theorem guarantees only if n distinct eigenvalues then you will have a diagonalizable so we are not so sure when we use negative 3 we gonna get to diagonalizable a so you have to you know work and see what's gonna happen so there's a node Note that the theorem does not guarantee that A is diagonalizable. More info needed. ok let's see what are the moving force so at least for these negative three if I can get two independent vectors column vectors then you can form your P and then they are is sort of a guarantee that hey I will have a diagonalizable a so sometimes you will get like two independent vectors out of these eigenvectors sometimes no so you have to work on and see so for lambda equals negative 3 we need to find first a minus minus 3 times I2 first that means a plus 3 I2 that gives you 1 negative 1 and then 4 negative 4 did I do correctly plus three times one negative one but here i think i messed up right a so you have a plus so this is one because only the diagonal works and this is negative one okay negative 1 and this is the a minus minus 3 i 2 now i need to solve a plus 3 i 2 times v equals 0 for v which is v 1 v 2 that means i need the augmented metric so plus 1 negative 1 1 negative 1 0 0 i will do the row operations i have 1 negative 1 0 i need to get rid of this guy so r 2 minus r1 that gives you zero zero zero so this is going to give you v1 equals to v2 and v2 is free so the parametric I'm going to be V equals V1 V2 that means V211 okay so what happened we do is free and the only eigenvector I'm going to get is 1,1 but I have two eigenvalues it's going to be negative 3 comma negative 3 because the multiplicity is 2 but I'm ending up with 1 that means I cannot create my P by using only one vector that means a is not going to be diagonalizable so any eigenvector of a for lambda equals negative 3 must be a multiple of one more. As a result there is no way to fill second column of a potential P. ok so there is not enough linearly independent eigenvectors so A is not diagonalizable so in a three by three case on the same question sometimes you may have the multiplicity tool but you may have two free variable so in that way in that case for one eigen value that's going to create two independent eigenvectors and then for the other leftover one you will have one so in such a case you still may get a diagonalizable a so it depends on like the question but in this two by two case if you don't get two then you are done you have to stop from there okay so we have another example good to do more examples so in our next example diagonalize a if possible okay so let's see what's going to happen so just minute ago i said like you know sometimes in three by threes you will still get diagonalizable let's try out so we have a metric a you have to diagonalize there if possible okay so we don't know first we need the eigenvalues for that you need the determinant of a minus lambda i 3 now because it's a 3 by 3 metric that gives you what hmm i have to do more work now okay so one minus lambda five zero zero two minus lambda zero zero three one minus lambda I need to find the determinant now how are we going to do the determinant now it's a three by three so you have to go for the general formula remember that you have to the easiest path so I see two zeros in this column or this column so you can either pick the first column or the second the third column in order to do the determiner that will be the easiest path so I am going to choose the first column so I know that with the sign like the sign values that this is going to be plus alternate negative plus but really who cares because it's going to give you zero zero so this will give me one minus lambda positive sign and then you remove the first row remove the first column so you will have two minus lambda zero three one minus lambda so once you do this one you have to find the determinant of this two by two again you have one minus lambda it's going to give you 2 minus lambda 1 minus lambda minus 0 three times times 0. So you really have 1 minus lambda squared 2 minus lambda.

So this will give you the characteristic polynomial. So the eigenvalues are going to be lambda equals 1 with multiplicity. 2 because I have the power 2 and then the other one lambda equals to 2 multiplicity 1 because it just have the power 1 there right okay so I have two eigenvalues for a 3 by 3 now I have to go ahead and find eigenvectors belonging corresponding to this values okay so we have a lot of work to do lambda equals 1 first so you need a minus i3 basically that's going to give you 050 010 if you don't see how that the answer is coming here you better write one more step okay and then I have to solve a minus i3 b equals 0 for b now we have v1 v2 v3 okay so i have to create the augmented metric 0 5 0 0 1 0 0 3 0 have 0 0 0 I need to create row reduced form row echelon form what gonna happen I don't have ones in my first column at all so I have no choice let's see what's gonna happen when I try to get rid of these three yeah let's try to get rid of these three and one so what i'm gonna do is i'm gonna do r2 minus r2 is going to get to dove let's see what's the best choice r2 i'm going to divide the first row by 5 and then going to subtract so negative 1 fifth of r1 and then r3 i will do three fifth r1 so dividing by five subtracting from here so that does not change my first row but these will be oh zeros so what's that gonna give us so it basically saying what I have 5 V 2 equals to 0 that means v2 equals 0 but the rest I just have three variables they all 0 so v1 and v3 are free so when I get into the parametric form I will have V to be V1, V2, V3.

That's going to give me... v1, 0, v3. So you have v1 times 1, 0, 0 plus v3 times 0, 0, 1. Okay.

So this is what's going to happen actually when you pull v1 and v3. So two free variables. v1 and v3 so we can choose whatever we want there's no luckily fraction so we don't have any problem to get rid of the fractions so we can pick like the simplest ones so let's see so I'm going to choose first v1 to be 1 and my v3 to be 0 so in that way I can just discard this guy and take out the first vector so this will will give me 1 0 0 and the next round I'm going to choose v1 to be 0 and v3 to be 1 so these are the simplest numbers now imagine if I pick v1 to be 1 and v3 to be 1 then I have to add them together in order to get this guy but the simplest you just multiplying by 0 everything disappeared multiply by 0 so you have to be little smart when you are picking values for free variables so this will give me 0 0 1 so they are like linearly independent so I had the multiplicity tool on lambda equals 1 and I have two linearly independent eigenvectors so it's like giving some kind of a promise that if I can find eigenvector For lambda equals 2, then I may have a diagonalizable A. But for that, you have to do a few more steps. So for lambda equals 2, I need to do A minus 2I3.

that's going to give me negative 150 and then this line will be 0 0 0 this is going to give me 0 3 negative 1 okay and solve a minus 2 I 3 we equal 0 for B that means you need the augmented metric negative 150 000 0 3 negative 1 i need to do some row operations so what's the best i have a negative one on the first stop so i will get rid of what what's the best I will replace r1 first divide by r3 on 3 and then multiply it by 5. Let's see what's going to happen. So I'm going to do r1 minus 5 over 3 r3. so I have a zero here that will end up negative 1 this is divided by 3 so you're going to have 1 minus 5 gonna make it this guy 0 and this one negative 1 so that will end up with 5 over 3 0 this I didn't touch this I didn't do anything either I think that's enough because V2 is going to be free.

Is it? No, I don't think so. V3 is going to be free. And then the rest you can find some values.

Okay, let's see. Negative V1. v1 equals to 5 over 3 v3 actually and then you have from this guy v3 equals to 3v2 so v1 v3 or in other words i can write v2 as one third of v3 so v1 and v2 are both with respect to v3 therefore v3 is free so the parametric form gonna give you v equals what V3 out.

So this is going to give you 5 over 3, 1 third and 1. And now you will see that here 3 is like annoying. It's in the denominator. So let's choose to get rid of that 3. So V3 to be 3. So that's going to give me 5, 1 and 3 to be an eigenvector. So this is going to give you, now you're happy.

D I found lambda equals 1 first that has to be on the first two spots because that's the rule of the theorem and then I have lambda equals 2 on this part and then my P will be belonging to 1 I have two linearly independent vectors so either you can write this one first and this one second or vice versa it doesn't do anything so 1 0 0 0 0 0 1 belonging to 2 it became 5 1 3 so as a result we found our D and P you have to go ahead and check D equals to P inverse AP so you have to do the checking part there okay so if we go to our next page there are a couple of remarks so the first remark is saying that note that diagonalizing a metric with complex eigenvalues work the same way okay so we agreed that on the previous section that sometimes complex eigenvalues happen they come as a pair and then you will have a pair of eigenvectors as well still this theory am going to work and then the second remark what's the point in doing this someone can ask why the hell we do this much of algebra okay so for that we will actually give the answer on the next couple of remarks so we are doing this one to make our life little simpler okay so this is what gonna happen as a reminder we're gonna say that in a regular number if you need a square that means you are multiplying two things together right x squared means x times x so a squared is going to be a times a you're going to multiply a twice right they are matrices note that a squared is not just the square of each individual entry of a that's the important part because a if a is two by two and if you're multiplying that doesn't mean like you are squaring each of the component you have to do in a special way of multiplying the matrices so in more general quantity a to the k for any positive k is going to be defined similarly so this is what going to happen when you have the diagonal metric that means you're only the diagonal values are non-zeros that gives us the opportunity if i want to do the d to the power k any real number k that means i just have to blindly like take the power whatever only on the diagonal so that's make our life simpler so that's the whole point like doing all that process like checking whether a is diagonalized but if since a and D are similar according to our theorem because we promise that we will have such a P such a that a and D going to be similar so if I need a to the K that means the same thing you just take D to the k. Let's see. We will see some problems here. And then the last remark, it is possible to define concepts of matrix square roots as well as exponentials and logarithms.

We will see them in future. An example, if a is an n by n diagonalizable metric, find a squared and use your answer to guess what a to the k. I just gave you the answer in my explanation but let's see what's going to happen.

So, A is going to be A times A, right? But I know that... A out of this one.

See P inverse AP equals to D. Right. This is coming from the theorem. But if I want to isolate my A what am I going to do is I have to multiply from the right side of both sides by P inverse.

and then i have to multiply the left side by p to get cancel of this guy so i will like have pp inverse a and then pp inverse like this but that will give you pdp inverse this is the identity because you're multiplying with the inverse of that you have a this guy gonna give you identity what does that tell us so you can write a because multiplying with the identity gonna give you the same thing a is pdp inverse so we're going to replace this a that here a squared i forgot the square i'm going to replace this a by pdp inverse now so 1 a is p dp inverse i have again pdp inverse that's gonna give me matrix multiplication it's going to be commutative and the associative property holes so p d i'm going to multiply first p inverse p together and you have dp inverse this going to give you the identity so basically you have p d dp inverse what that gives you p d squared p inverse so when you have to find a squared you can actually find if you know the P and P inverse you just square the diagonal and the diagonal much easier you just square that the diagonal values that is going to be zeros so these can be generalized that a to the any k power is going to be just your p and d to the power k which is easy to handle the diagonal metric and p inverse. So this This is one main reason why we learned the diagonalizing business. Okay so let's see.

Use the fact that these given matrices to come. compute 10th power of this given A. So let's see.

Now if you consider this one to be A, this is the distance of P. This is my diagonal metric. This is my P inverse.

Right? I think we just downed this problem here a little while ago. Okay.

So we have this. We know what is. are PD and penis I need to find the 10th power of a right a to the 10th which is negative 4 negative 5 8 10 to the 10th power that's going to give you P t to the 10th and then p inverse this is possible because d is diagonal only the diagonal have non-zero values everything else zero so that gives you p is negative five negative one four two these zero to the tenth zero zero six to the tenth we're gonna see big numbers and in was luckily given if you haven't given the inverse you need to find so you should know how to find the inverse okay so this is negative 5 negative 1 4 2 this is going to give you 0 0 0 6 to the 10th what 6 to the 10th i'll leave it like that and then i have this guy negative 1 third negative 1 6 2 third 5 over 6 let's multiply so we know that we can't multiply three matrices together you have to do two at the time so i'm gonna do the first two negative five times zero is zero negative one times zero is zero so the first component zero negative five times zero and then negative six to the tenth four times zero zero four times zero then two times six to the tenth that's going to be the first two and then I still have negative 1 third negative 1 sixth 2 third 5 over 6 now I have to multiply these two again I'm lucky because two things are here 0 so 0 times this and then this guy so negative 6 to the 10 two-third and then zero negative six to the tenth five over six zero two to the sixth to the tenth times two-third and then zero two to the sixth to the tenth five over six you guys can use a calculator and you will see that big numbers coming up and I will put the answer here so negative 403 I guarantee I'm not going to give this much of a big numbers on you quizzes or exams so eight two six two one five six eight and then seven seven six nine six zero okay so that's the tenth power of your a given here okay guys so that's the end of the section last section on your chapter again there are practice problems you have to do odd problems and i will be posting on for these odd questions okay see you guys in my next chapter

Transcript for:Understanding Matrix Diagonalization

Transcript for:
Understanding Matrix Diagonalization