Transcript for:
Vector Spaces and Subspaces

good evening everyone good evening ma'am good evening ma'am okay so I have add the uh live streaming uh Link in the chat box so let us start uh the revision session for week three and week four topics today okay so in week three we start with what are vector spaces okay so for Vector spaces first of all you should have a set and also you should consider uh set V along with two functions which is a vector addition and scalar multiplication okay so first of all this means that your this operation means that this set is closed under vector addition okay what does this mean this means that this set is closed under vector addition with the with respect to this operation and same here uh with respect to scalar multiplication this set is closed okay so closed with respect to vector addition Clos with respect to scalar addition scalar multiplication that is the first condition what does that mean so it means if you take two vectors V1 V2 from V then V1 V2 should be in V so this is what is called closed under vector addition and what is mean by closed under scalar multiplication if you take any scalar and you multiply with any vector v then the scalar multiplication Alpha V should be in V so this if this happens then such a set is called closed under vector addition and scalar multiplication okay so that is the first condition you check you should check for Vector space okay after that you have the following aums so in this axum the first four axum corresponds to vector addition okay so the first one checks the commutativity next associativity here you have the identity vector and here the inverse vector okay and then this two is for scalar multiplication so 1 * V = to V and then ABV equal to a * of BV the associative property and these two are distributive properties ma'am can you repeat please from starting see for a vector space suppose you are given a set along with two functions uh first one is the addition function and second one is scalar multiplication first thing you should check is that you have to check whether V1 + V2 is in v and also for random vectors V1 V2 in v and also for any Alpha in r and v in V Alpha V should be in V if these two conditions are satisfying you call it closed under vector addition and scalar multiplication once this condition is verified then you go for all the eight aums so the first four axum deals with the vector addition this two is for scalar multiplication and this two is for distributive property so if all the aums are satisfied all the 10 properties are satisfied then only you say that this set ve uh V is a vector space over R along with the given operations of addition and scalar multiplication even if one of this fails out of this 10 even if one of them fail then it cannot be a vector space okay so remember for verification of a vector space if it is really a vector space then all the 10 aium will satisfy if it is not a vector space so you just have to show that one of the property fails that is sufficient okay fine so let us see some examples for this so the very basic examples you know that R and the ukan space the set of all M by and matrices so the ukian space is this okay and set of all M byn matrices Solutions of the homogeneous system ax equal to Z okay so this is in fact a subset of RN and hence is a Subspace of RN what is a Subspace you already know anyway we'll revise it in the next slides coming slides okay fine so and what is cancellation law for vector addition if you have some addition of vectors like this where you have a common Vector on both sides then you can cancel them out okay so from this what you can get so the identity ve the zero Vector is unique and also the inverse Vector is unique okay so these are some properties of vector spaces okay now here is an exercise where you have given where you are given different sets along with different addition and scalar multiplication we are going to see if it satisfies the vector space axioms or not okay fine so first observe the set V1 what is the set V1 can you say V1 is R3 R3 right and addition is the usual addition given and scalar multiplication it is given in a little different way okay so uh since the addition is usual so all the properties of addition you can verify it is easily satisfied it is closed under vector addition and all the first four XMS will be verified for the vector addition because it is the usual addition this is what is called the usual addition coordinate wise addition so there is nothing special about this so let us look at the scalar multiplication part is it closed under scalar multiplication no ma'am vors in R3 yes or no yes or no are these Vector in R3 yes ma'am they are in they are in r3ed so they are in R3 means so you know that it is closed under scalar multiplication so after SC uh the closed property then you have to check what is the next property one times of any Vector B is equal to V for any V in this set V so if one means it is a nonzero scalar so you know that it is going to be again X Y and Z only so this property will be satisfied and what is the next property after this a b * of V = to a * of B right so this is the next property right where is that yeah AB * of V is equal to a into BB so this also will not have any issues so if you check uh since this numbers are multiplied here so let us check whether this property is satisfi so a b times of XY Z is equal to ABX AB Y and Z by AB okay assuming both A and B are non zero and uh yeah so a times of BV if you take so a * of BV again it is going to be the same yes or no yes yes ma'am you see B is so this satis this also is satisfied this also is satisfied you can easily verify that okay what is the next property then uh a plus b into V isal A plus b right so let us check whether this property is satisfied or not so maybe I will do it in the next page so let us say V is X comma y comma Z okay so what is a + BV addition is the usual addition right so you have a x a + b * of X and A + B * of Y and a + b * of Z since it is a usual addition that is what you this is what you will get and what about a + b * of V so it is A + B * of X A + B * of Y and Z by A+ B are they equal no ma'am so you see that they are not equal then for the first one A + B into Z how we will get that we should get something like Z by a + b which one uh the AV plus BV that you did this is usual addition right okay okay here I made a mistake instead of making the um scalar multiplication I did the usual scalar multiplication yeah so you have to do ax b y and z by uh C sorry Z by a all of this are a and then here you will have BX b y and z by B so you will have a + b * of X and A + B * of Y and Z by a + z by B so the first two coordinates are matching but what about the third coordinate is z by a plus b same as no Z by a plus Z by B it's not same right they are not equal so you see that this particular axum distributive property for this uh set V so V1 is not a vector space okay V1 with respect to this addition and scalar multiplication is not a vector space okay let us look at V2 what about B2 so here the scalar multiplication is the usual one but the addition is different first of all is it closed under addition yes it is okay okay so what about the properties of the first four properties ma'am commutativity associativity yeah ma'am uh if it's Z1 Z2 then why should it not be closed under addition it is not the usual addition right usual addition you have Z1 plus Z2 right yes ma'am yeah so this is not the usual addition so this is something different what's your question here so it does so V2 is not closed under addition is it m no I did not say it is not closed under addition so I asked you is it closed in addition because you see that X1 X1 + X2 y1 + Y2 Z1 Z2 all these are going to be real numbers only so this is impact closed under addition only ah okay what about the second uh first axium then is it commutative with respect to addition when you change so let us say this is U and this is V so U plus v is this what about V plus u it will be set set one it is going to be the same the first property is satisfied so U + V equal to v+ U so that is satisfied and what is the second property then U + V + W so that will also be satisfied yes or no yes ma'am yeah ma'am so this will also be satisfied what about the next one what is the identity element here zero you add which Vector you will get the same element here it fails here yeah ma'am it will fail because the last element we will get zero instead of Z1 okay let us see whether there is an identity element or not so you have X1 y1 Z1 okay the first two coordinates the uh addition is the usual addition so let us put z0 okay because the first two part the addition property is the usual one so let us see what about the third part what happens to the third part we will find whether such a vector exist or not so here it is X1 y1 and here you need Z1 so here according to this addition what is this X1 y1 and Z1 Alpha right so what should be this Alpha then so that this are equal one yeah so it means given any Vector if you add it with the vector 0 01 then you get the same Vector again so this is the identity here zero Vector here this is the zero Vector not identity let me see Zero Vector this is the zero vector okay for this Vector space with respect to this addition and scalar multiplication you have a you have a zero Vector so if you have a zero Vector can you also find find an inverse yes yes you can find an inverse right so what will be the inverse of an element X1 y1 Z1 with respect to this addition minus X1 - y1 0 - X1 - y1 and zero yes ma'am okay so let's see X1 y1 Z1 plus - X1 - y1 here what should be the element you have to find is equal to 0 Alpha so 0 0 1 actually so which element if you take you will get one here 1 by Z 1 by Z so so only when Z1 is non zero you will get uh one here what about if Z1 is zero what happens so you will get 1 by Z1 if Z1 is non zero if Z1 is zero then then also we will get then if Z one is zero this part is actually okay let me show you what happens let me put some um value here maybe something like a c and see what is happening so this is actually X1 first term is zero second is zero and the third is Z1 C okay so Z1 C isal to 1 implies C is = 1 by Z1 when Z1 is non zero and when Z1 equal to 0 you are getting 0 is equal to 1 so it means are there inverse entries inverse elements for vectors of the form X Y comma 0 with respect to this addition yes ma'am yes what is the inverse entry uh for X we can put minus X for y we can put minus y and for Z we can put any element but then this entry will become zero right yes but here you are having one how will they be equal me it is not necessary that we have to get 0 00 0 no identity as the identity and see this is this is the definition that X so this is the vector X and this is the inverse of the vector that should be equal to the zero Vector right that is the definition ma' zero Vector is is it that zero Vector is not necessarily always zero yes that I have told you multiple times zero Vector is here itself I have written that this is the zero Vector so if it is 0 0 1 How can there be an inverse for this element because when Z one equal to Z this part is zero and this cannot be equal to this right yes or no so you see that there is no inverse for elements of this form you agree yeah okay no inverse for elements of this form so this is also not a vector space okay what about V3 so again you are in R3 the addition is the usual addition scalar multiplication again for the first two variables it is the same and only for the third one it is changing so I uh I directly uh jump the scalar multiplication part because all the addition properties you can verify them to to satisfy and scalar multiplication 1 V equal to V yes or no yes ma' okay and what is the second property a b * V equal to a * BB is that satisfying here yes ma'am it is yes what about the next property what about a + b * of B taking this to be the vector B will not be satisfi so now you will get a + b * of X A + B * of Y and this is z so again what is a plus BV is it going to be the same no ma'am it will be two said two is that yeah so you are getting a plus BV means you are getting a + b x a + b y and 2 Z so you see that these two are not equal so this is also not a vector space okay you understand how we are verifying the properties here so if one property is like usual addition or something then you skip that and move to the one where it is different so that way you can quickly identify what is the problem okay fine so what about V4 V4 set of all X comma y comma Z such that X comma y comma Z and R and X+ y + zal to 1 so before even uh so this is the usual multi addition and scalar multiplication so before even verifying or doing any calculation can anyone justify it without making any calculation first of all what does this represent V4 earlier here what we have seen all the set were representing R3 now what does R4 represent sorry V4 represent R2 no R3 ma'am it is a subset of R3 but what does it represent X+ y + z plus uh Z equal to 1 yesterday also I wrote The General equation of a plane what is this plane plane right equation of a plane so now this is a plane and is this passing through origin yes is it passing through origin not passing through origin this plane is not passing through origin so here we are considering useful addition in scalar multiplication and this does not pass through origin which is the uh zero Vector so if it is not passing through the zero Vector it means what the zero Vector is not there in this set not a vector so you can say that it is not a vector space I don't understand M how is it possible so if 0 comma 0 comma 0 belongs to V4 then 0 + 0 + 0 should be equal to one is that possible no ma' so that is why we are saying this zero Vector is not in V4 if the zero Vector is not there in a set then it cannot be a vector space right with respect to usual addition and scalar multiplication okay so this is one way of saying if you want to look at look at the uh General ways like first of all the vector addition itself or scalar multiplication if you take let us take a vector X comma y comma Z in V4 then what do you know x + y + z is equal to 1 now let us do scalar multiplication Alpha times of X comma y comma Z does this also belong to V4 what happens to Alpha X Plus alpha y + Alpha Zed Alpha it is Alpha and it need not be one right yes ma'am so this does not belong to V4 so you can see that it is not closed under scalar multiplication you can also show that it is not closed under vector addition so there are several ways to justify that this is not a vector space okay in the previous cases the sets were R three but the uh addition or scalar multiplication was defined differently here you see the addition and scalar multiplication are the usual definitions but the set is different okay fine what about the set V5 but ma'am I have a doubt that zero Vector which you have I can see that4 is this the condition for Subspace I think so but how you can just finalize true that zero Vector is not belongs to the vector space of whether it is a vector space or Subspace zero Vector is necessary without zero Vector how will you how will the third axm get will be satisfied okay so zero Vector will be performed both for Vector space as well as sub yes yes exactly okay okay fine okay and yeah okay and uh this set any doubts ma'am for V4 I didn't got this totally the things you explained see for V4 I have taken let us say x comma y comma Z is any Vector in V4 so now Alpha times of this Vector should be in v4s or no yes ma'am but what happens to Alpha times of alpha X Plus alpha y + Alpha Z it is equal to Alpha and it need not be one always yeah so it is not in V4 okay means for since the condition is given X Plus y + z equal to 1 that means uh the resultant is should be always be equal to one yes okay any element in this set should always satisfy this condition okay understood thank you ma' okay so now what is the set V5 can you say what is the set V5 passing through the origin line is a line given by y = - x x right and is this a line passing through origin yes ma'am yes is a line passing through origin in R2 or R3 will that be a vector space every Subspace itself is a vector space so you know that it is a sub space so in fact it is a vector space with respect to usual addition and scalar multiplication yes or no yes ma'am okay so you know that so every every Subspace is also a vector space okay so we know that any line passing through origin in R2 is a Subspace that is why therefore V5 is a vector space is the justification clear for everyone ma'am yes m' could you just please explain also with the help of identity expain which part this one only V5 if we need to prove that whether it is a vector space or not can be used with the of addtive identity also your doubt is see first of all every Subspace is a vector space is that fine for you yes ma'am okay so every Vector space is a sub every Subspace is a vector space so up to this it is fine now the next part is that if we what are the subspaces of R2 we have already seen so for R2 the subspaces are for R2 R2 itself and then you consider the origin and any line passing through origin so these are the subspaces of R2 okay and any line passing through origin is of the form Y is equal to MX yes ma'am okay ma'am actually I'm asking that whether we can prove it also with the help of additive identity I'm asking that question you're asking whether it has an additive identity yes you tell me what is the additive identity here tell me what happens to the usual 0 0 yeah you are having us addition and scalar multiplication so what is the issue here so if you take 0 0 it is there in VY and it also satisfies the vector addition property so what is the issue here okay all right Madam I have a difficulty 3.1 3.1 four question second part uh please uh ask your doubts at the end of the session otherwise it will interrupt the flow okay excuse me ma'am yeah M can you just explain V4 I I understood but I want to like see if you want to consider vector addition also you can do or scalar multiplication also you can do if you know what is scalar multiplication by now you should have understood this okay anyway I will explain again see X comma y comma Z you are taking any random element V in V4 now what you should prove for was closed under scalar multiplication Alpha times of V should also be in V4 but what happens to the vector Alpha V for Alpha V Alpha X Plus alpha y + Alpha Z is equal to Alpha and if you take any Alpha which is not one then this is not going to be in the set V4 right so it is not closed under scalar multiplication is it fine yes ma okay so whatever I am teaching at that moment if you have doubts in that definitely you can ask but if it is something different so please wait till the end of the session okay fine okay so now what are subspaces so for subspaces so again you have to verify closeness of vector addition and scalar multiplication so actual definition a nonempty subset W of a vector space V is called a Subspace if W is a vector space under the operations of addition and scalar multiplication defined in V by definition itself every Subspace is a vector space right so now you see that to show that a non-empty subset W is the vector space you don't have to check all the vector space aums it is much simpler you simply have to check only this two condition what is the first condition closed under vector addition and the second condition is closed under scalar multiplication and also the zero Vector also should be there something it is a you can say that it is a result of either you can say it as a result of one or result of two because you see that it is a shortcut to sometimes prove that something is not a Subspace you cannot use that as a uh property to prove something is a Subspace because what happens here Alpha times of W1 should be in W so if you take Alpha to be zero then 0 * of w equal to zero right and you know that this should also be in W so you know that always the zero Vector is there in the W so this is a shortcut to sometimes say that it is not a Subspace but you cannot prove something is Subspace using this property okay fine every Vector space v/ r has two trivial subspaces one is the set itself and the other one is the zero Vector so by zero Vector it is not always the 0 00 0 as I gave you an example it can be something different depending on the vector addition and scalar multiplication given to you okay fine yeah the subset consisting of yeah so these are the trivial subspaces so already I have mentioned uh the subspaces of R2 can you say what are the subspaces of R3 plane passing through origin through origin yes and all lines passing through origin line passing through origin and and plane passing through orig plane plane passing through origin so these are the one subspaces of R3 okay subspaces of R3 okay now let us look at the next example here I have considered the various Matrix uh examples okay V1 V2 V3 V4 V5 so all are different subsets of uh 2 by2 matrices we are considering only 2 by two matrices okay so uh first thing here you know that uh if we can verify that it is a Subspace then if you have to verify it is a Subspace you just have to check closeness under addition and scalar multiplication so is the sum of two symmetric matrices also a symmetric Matrix and not necessarily why not so let us take a a comma B in V1 so you know a transpose is equal to a and b transpose is equal to B now what about a plus b whole transpose B transpose plus a transpose a transpose so you see that a transpose plus b transpose means so you know that again this entries are nothing but a transpose is a and b transpose is B so a plus b whole transpose is a plus b so it means your a plus b is also in V V1 is that correct yes so it is closed under vector addition what about closed under scalar multiplication Alpha take any Alpha in R and A in V1 so Alpha times of a whole transposes when you're taking transpose you can simply take out the scalar right so it is simply Alpha * of a so this implies Alpha a belongs to V1 so it is both closed under addition and scalar multiplication so is V1 a Subspace yeah okay so V1 is a Subspace what about V2 scalar Matrix so scalar matrixes so let us take a comma B in V2 how how do you represent scalar matrix by the way when the diagonal entri arex entri are you can actually represent it as Alpha I in general okay so let us take B is equal to let us say beta times of 0 0 beta now A + B you know that it is simply Alpha plus beta times of I so again it is a scalar Matrix only obviously same way Al a is going to be maybe I'll use some c c * of a is C Alpha * of I so this is also in V2 so V2 is also a Subspace so V2 is also a Subspace what about the set of all diagonal matrices so you have a comma B in B3 so a is equal to A1 A2 B is equal to B1 B2 so A + B is equal to A1 + A2 B1 + B2 so this is also same way you can yeah what's the difference scal and this is set of all diagonal matrices the previous one was scalar matrices and this is diagonal matrices this is scal me the values are equal diagonal yes for scalar matrices the diagonal entries are all equal so you see V3 is also a Subspace now what about V4 and V5 triangular matrices it is easy to verify actually so I will do for upper triangle you can verify it for lower triangle so you say A1 A2 a m it should be it should be A1 Plus B1 no in a plus b uh A2 plus B2 oh sorry sorry yeah I WR WR yeah correct thank you so A1 + B1 A2 + B2 yeah okay and here yeah so now you see a + b is again A1 + B1 A2 + B2 0 A3 + B3 so obviously this is a upper triangular Matrix same way you can show that Alpha a is also an upper triangular Matrix similarly for B5 so all of them are actually subspaces so only V1 is a Subspace incorrect only V4 is a Subspace incorrect both V2 and V3 are subspaces yes all subspaces all are subspaces of M2 uh 2 by2 of R yes okay fine so we will see more examples using matrices when we look at Dimension okay so we will also see it at the later part of this session okay so now consider this set consider the following set of R2 with usual addition and scalar multiplication you have V1 and V2 so what is V1 representing x-axis the x-axis is it a line passing through origin yes ma'am and what about V2 y AIS and that is also line passing through origin so are V1 and V2 subspaces yes ma'am yes ma'am okay so V1 and V2 are subspaces what about V1 intersection V2 what is V1 intersection V2 where does xaxis and y y axis intersect origin origin origin right so you see that it is actually just 0 comma 0 and is this a Subspace of R2 yes yes ma yes ma'am okay and what about the union V1 Union V2 is it a Subspace of R2 yeah yes okay so what is V1 Union V2 the whole XY plane how do you say it is whole X comma plane it is just x- axis and y axis right just no just the x-axis and y axis so all the in between these entries are missing so does this form a Subspace for example I'm taking 1 comma 0 it is there in V1 Union V2 so 0 comma 1 this is also there in V1 Union V2 so what about the sum of this two vectors is it so one comma one is it there in the union of these two sets no no ma'am okay so you'll see that the sum is not here okay so V1 Union V2 is not is not a Subspace okay fine so V1 is a Subspace of V R2 but so definitely these two are incorrect In general so leave about this problem in general what can you say about intersection of two subspaces z z subspaces of a vector space v u intersection W is always aace always a Subspace and U Union W is not be a Subspace okay please remember this okay fine okay so here in general it is given so if you are taking two lines passing through origin they are subspaces but they are union it is not a Subspace because if you take the sum of these two vectors it is a vector in between this two line so it is not there in U Union U Dash so it is saying that the union of two subspaces is not NE necessarily A Subspace okay let U1 uh U2 uh subset of V be subspaces of we what is U1 plus U2 what is U1 + U2 it is set of all U1 plus U2 U1 and U2 are varying in U1 and U2 right now always U1 plus U2 is a Subspace of V so what are the different combinations we have seen the intersection is always a Subspace Union need not be a Subspace and the sum is always a Subspace so now for these two examples given here V1 and V2 can you say how the sum is Subspace always okay you can prove that so the proof I'm not doing so it is very simple if you want to prove it so you have to do the usual addition and scalar multiplication so you take two vectors from here so let us say U1 + U2 so another Vector V1 1 + V2 the sum you see that it is U1 + V1 plus U2 + V2 so this is in U1 this is in U2 so it is closed under vector addition same way you can show that it is closed under scalar multiplication okay so it is always a Subspace of we okay so why not a Subspace not be a sub someone talking with was like discussing among yourself for asking a question here I'm asking a question that the earlier one U Union W need not be a Subspace it's a little confusing for me there is nothing confusing about it so you see that you see that this you look at this figure okay so now this is a Subspace you U is a Subspace U Dash is a Subspace by parallelogram law you know that if you take this Vector U U and this vector v the sum U + V is this Vector okay and this is only this what is your U Union U Dash U Union U Dash is only this line and this line apart from this nothing is included in U Union U Dash but this Vector U plus v is outside this two line so it means it is not closed under addition right okay okay got it got it okay fine so now coming back to this what is V1 + V2 for this example here what is V1 + V2 here X Y X Y okay set of all X comma y such that X comma Y in R suppose I give another set V3 which is X comma X now can you tell what is V1 + V3 uh 2x comma X okay fine okay so here U1 U2 are these two sets so you are in R3 so what is you1 U do x x y z x comma Y in R okay suppose you have u3 is this what is U1 + u3 x + y y+ y 2 y comma 0a y comma 0 right it should be 2 y right why it is 2 a because U1 is X+ y X+ y comma comma Z right okay fine now I'm writing another set let us say U which is set of all X comma y comma 0 such that X Y belongs to R what is a relation between this set and this set what is the relation between this set and this set here Z AIS is zero both they lie in same plane they are the same plane right yes ma'am yeah they are actually the same XY XY plane both are representing the same XY plane actually both are the same set only both are equal so this x + y you can replace with with some other dummy variable maybe something like X Das comma y comma 0 where x d comma Y is in R so these are just change of variables so it does not matter so these two sets are same right okay now in this case what is U1 intersection U2 what is U1 intersection U2 in this case origin yeah it is the origin right what about U1 intersection u3 origin again origin y AIS no again origin right what are the intersecting element yes yes origin it's just origin right yes okay so this is the x-axis and this is a line passing through 1 comma 1 comma 0 and origin so these two will intersect only at the origin right okay yes ma'am okay fine okay so the next part the linear combination of vectors so let V be a vector space and V1 V2 VN are vectors in V so what is mean by linear combination linear combination means Sigma I = to 1 to n a i da so this kind of scalar multiplication and vector addition combined together is called linear combination okay so for example if V1 is 1 comma 0 and V2 is 0 comma 1 can you write any Vector X comma Y in R2 as a linear combination of these two vectors yes yes so you what is the coefficient here what are the coefficients Z zero X and Y X and Y okay so now for this example if you want to write 5 comma 6 as a linear combination of 1 comma 2 and 1 comma 1 what are the coefficients how do you find let us put Alpha and beta so how do you find Alpha and beta Alpha plus beta betaal 5 and 2 Al plus beta so we have to solve the system to find this answer so in this case it will Al 1 Al = 1 and beta right so you see 1A 2 + 4 * of 1 comma 1 is equal to 5A 6 yes right the same way you can do this also so I will not do this you complete this part find Alpha and beta for this example okay okay so this I leave it as a home so note that if V1 V V1 V2 are Vector such that V is equal to A1 V1 + A2 V2 then each of the vector is a linear combination of the other two vectors here this expression is actually saying that V is a linear combination of V1 and vs2 but in general if this happens then each Vector is a linear combination of the other vectors how here one condition is missing if A1 A2 non zero this condition is needed here because see if you want to write V1 as a linear combination of v and V2 how do you do that you write A1 V1 is equal to first of all V minus A2 V2 then you can write B1 as V minus A2 by A1 V2 okay so you call this as some scalar and call this as another scalar okay so let us say this is your Alpha 1 and and this is your Alpha 2 okay same way you can write V2 as a linear combination of the other two vectors given this A2 is non zero okay fine further zero can be written as a linear combination of V V1 V2 can you say uh tell how what are the coefficients Z Zer only zero only yes why maybe negative also - one if I take A1 V1 A2 V2 then yeah ma'am then also it will be zero yes so you see that from here only the definition of linear dependence we saw if you forgot so you see that in this case Zero coefficients also is a possibility you can also have zero coefficients here but nonzero coefficients is also a possibility that is where linear dependence comes right so what is the definition of linear dependence if you have such a linear combination equal to zero and if you have some of them to be non zero if not all of them are zero then you know that these vectors are linearly dependent and you know that if one vector can be written as a linear combination of the other vectors then the vectors are linearly dependent yes or no yes ma'am right so that is why you see that this expression tells you that this vectors V V1 V2 are linearly dependent okay so you know that what is this what is special about this set of vectors V1 V2 V3 what is larpent standard yeah it is linearly independent and in fact it is the standard basis for R3 yeah right okay what about this V1 V2 V3 sorry V1 V2 V so can we write 3 comma 7 comma 2 as the linear combination of 0 comma 2 comma 1 yes ma'am what are the coefficients Al Bet 2 Beta = 3 beta is equal 3 by okay let us take 3 by2 then this part will actually become 3 comma 3 comma Z and what about Alpha Al equal to two alpha is equal to two Okay so 2 * of 2 plus four * of 3 7 and here two yeah fine so Alpha is two fine okay Alpha is equal to 2 so let B Be any linear combination of the vectors V1 and V2 does v lie in the plane containing V1 and V2 assuming V1 and V2 are linearly independent does v lie in the plane containing V1 and V2 no' no no you are given that V is a linear combination of Av V1 plus a V2 so does v lie in the plane containing V1 and V2 so V1 and V2 are lying in some plane you know that yes what about the vector v so if it is a plane passing through origin if this condition is added okay plane passing through origin then you can definitely say the linear comination will also lie in that otherwise you know that it need not be a Subspace right any plane need not be a Subspace but a plane passing through origin if you are considering you know that every linear combination will also be a Subspace right see sorry in that Subspace so here uh in fact um I wanted to say that I have told it in our regular session so these two conditions can be combined and written as Alpha V plus beta W belongs to W for Alpha comma beta in W and V comma W in W so this one condition is satisfied then you can say that W is closed under vector addition and scalar multiplication so here you are seeing that this is a scal this is actually uh the linear combination means it is exactly satisfying that condition right so uh will that line a plane passing through origin yes or no okay so that if V1 and V2 are lying in a plane passing through origin Z in this yeah then we can say V is in that plane yeah then V is also in W okay now let W equal to 1 comma 2 comma 0 does this Vector lie on the plane 2x - 2 y + 4 is z what is your answer no no ma'am no so let us see 2 * of 1 minus 2 * of 2 plus four * of 0 what is the answer 2 - 4 + 0 is-2 so it is not there in the plane so it means can we write w as a linear combination of V1 and V2 no no if not V1 and V2 are linearly independent vectors in that plane okay so it means what is happening if you are having a plane passing to origin and if you are having two linearly independent vectors what does this form for this space w it forms a this bis this is so it means you know that span of V1 V2 is w what do mean by span all the vectors combinations right all possible linear combinations is your span this is W so any linear combination of we 1 and B2 is inside the set W any element outside W will not be there in this set so the answer is no you don't have to calculate in this way and check using this argument you can say the answer is no okay fine so this is the definition of linearly dependent vectors the set of vectors V1 V2 VN from a vector space V is said to be linearly dependent if you you have such a linear combination where at least one of the coefficient is non zero then such a set of vector is called linearly dependent vectors okay and linear independence mean if all the coefficients are zero then you call that set to be linearly independent okay fine now for how many values so here is an exercise so for how many values of a is this set linearly independent so you are having four vectors from R3 so for which values of a is the set linearly independent how will you find first of all how many vectors are there four Vector four four if four vectors are there in R3 what is the condition for linear Independence it cannot be linearly independent independent right so there is no for for any value of a this set is linearly dependent right so how many values empty set only no values for no values this set is linearly independent what is a condition actually I have written it somewhere um yeah so a set with K vectors in RN where K is greater or equal to n + one is always linearly dependent in our case you have four vectors in R3 so it means four is greater than or equal to this four so it is a linearly dependent set So based on this property you can say that this set is always linearly dependent okay fine choose the correct set of options Union of two distinct linearly independent sets may be linearly dependent yes ma'am example quickly one Z is this a linearly independent set yes sir yes so maybe I'll for counter example maybe I'll Cho other example so let us take one comma one here so now what about the union of this two said will it be linearly independent or dependent dependent larly dependent okay Union of two distinct linearly independent set must be linearly dependent true true why no false considering not necessarily this Union is linearly independent only right yes ma'am so both options are incorrect it may or may not be linearly independent so so in this case if it is May then it is the correct option but here it first one is correct then then First Union of two distinct L yeah in that case maybe linearly dependent yeah it is correct sorry may be linearly dependent so it is correct we have given an example it can be linearly dependent and here it is saying must be linearly dependent so it is incorrect nonempty intersection of two linear independent sets must be linearly independent what is the answer true true right because any sub intersection means it is going to be a subset and you know that any subset of a linearly independent set is also a linearly independent set and hence this is true and so this will be always be true yeah this will always be true option three yes because you know that any just like like for linearly dependent set every Super set is linearly dependent same you can write for this any subet of a linearly independent set linearly independent any subset in the sense non empty subset okay it's uh they are opposite in dependency I mean independent sets and dependent yeah so here you see that intersection of two set is going to be a subset of each of the set so you see that they are going to be linearly independent okay fine yes okay so now what is the relation between linear Independence dependence and solutions of a system of linear equation yes before answering this question what what do you know about it so let me ask the question in this way suppose you are having three vectors V1 V2 V3 okay in R3 all these are from R3 okay and you have a matrix with The Columns of V1 V2 V3 columns as V1 V2 V3 now if the set is linearly independent let us say this set is some s independent what do you know about the determinant of a not equal to zero and if linearly dependent equal equal to Z this is the first criteria you should know so in general it is not necessary about three vectors in R3 if you choose any n vectors from RN forming a matrix like this and you know that the determinant is non zero if they are linearly independent and lineally dependent means the determinant is zero now look at this question assume that the set A1 D1 1 C1 A2 B2 C2 A3 B3 C3 so these are the columns so each of this is a column okay they are given to be linearly dependent so what do you know about the coefficient Matrix a here they are given to be dependent so what you yeah so if determinant of a is equal to zero for a system a is equal to B what do you know about the solution it can be no solution no solution or exactly either no solution or infinitely many solution it can never have a unique solution right when determinant is zero it can never have a unique solution and so here also unique solution option is there so this is also incorrect this is incorrect so determinant is zero means the Matrix is non invertible right not invertible yeah okay so based on the linear dependence and Independence you can tell about the determinant and also the solution of a system of linear equations okay fine so sorry what are the ways to check a given set is a linearly independent set or not ma'am can you explain uh what is invertible Matrix how invertible yeah see if determinant is zero then you call that matrices to be not invertible right yes yeah that is what is given here okay okay okay so what are the different ways of checking linear Independence so in R2 you know that what what do you know about the zero Vector linear dependent linearly dep take any nonzero Vector it is linearly independent right you forgot if there are two vectors and one is a multiple of the other what do you say dependent linearly dependent and the determinant if you take the two vectors V1 and V2 just now I told you the determinant is non zero you know that linearly independent dependent and zero means lar right and if there are three or more vectors from R2 then yeah same things for R3 for any zero vector or a set containing zero Vector is linearly dependent and any nonzero Vector is always linearly independent so if there are two vectors you check whether one is a multiple of the other if one is a multiple of the other then it is linearly dependent dependent and if there are three vectors you know the determinant method determinant non zero means linearly independent and zero means linearly dependent four or more vectors nothing to verify straight away you write linearly depend okay set containing a zero Vector is always [Music] dependent a super set of a linear dependent set is dep dep a subset of a linearly independent set is independent okay if s is a linearly independent set in Vector space V equal to set all X comma y such that y = 2x with usual addition and scalar multiplication find the maximum possible cardinality of s first of all uh we have not yet started the dimension part but since you already know what is dimension can you say what is the dimension of the space we one so the dimension is one how many linearly independent vectors can be there from B one only one the maximum possible cardinality of s is one one dimension of V is one how is the dimension one so first of all it is a straight line passing through origin so any straight line passing through origin will have Dimension One or precisely if you want to say you can say that X comma 2X x in r so basis is given by 1 comma 2 so only one element in bases so Dimension is one so this way also you can say that Dimension is one so since the dimension is one any linearly independent set will have maximum only one okay yes ma'am so ways of checking linear Independence using homogeneous equations so if if you are having a set of vectors V1 V2 VN and if you want to identify suppose up to R3 you know if it is R4 or something more than that how do you check in that case you can use a system of equations to check whether it is linearly independent or not so you write the given vectors okay so as Columns of this V this Matrix V okay and if VX is equal to Z reduces to a system which has only the trivial syst solution then the set is linearly independent okay and if we get a nonzero solution then the set is linearly dependent here sometimes what happens suppose you have like two vectors and you are or maybe three vectors from R4 then this method is also useful okay three vectors from R4 means you cannot go to the determinant method so that in that case you can use this kind of a technique to identify whether the given set is linearly independent or okay fine so already we know that for square Matrix you know the determinant technique for non-square matrices only we will go for this technique right okay span of a set the span of a set s denoted by span of s is a set of all finite linear combinations of the elements of the set s so that is all possible linear combination and always this set will form a Subs this now you are given with two vectors can you say what is the span of s what is the span of s it's a plane right which plane XY XY XY plane so here you see that what is this uh span first first of all span if you write it precisely then it is Alpha times of 1 comma 1 comma 0 plus beta * of 1 comma 0 comma 0 so now what is Alpha comma beta here now it is set of all Alpha plus beta comma Alpha comma Zer where Alpha comma beta are in R so which plane does this represent next okay this represents the XY plane okay now let V be a vector space and S be a subset of v s is said to be a spanning set if s is the SP if span of s is equal to B right for example if you see that any basis is a spanning set so any basis is a spanning set so for R2 we can say 1 comma 0 0a one yeah you can take one zero the standard basis you can definitely take or you can take any other basis okay similar way for R3 also you can take any basis okay is the spanning set of a space unit no ma'am no ma'am ma'am no okay why we are having different spanning sets yes if S1 and S2 are two spanning set of a space does S1 and S2 contain the same number of vectors yes yes ma no yes ma'am so for example what is the span of this X1 R3 it is R2 R of this set that is also R2 so are these two sets having the same number of ENT no no no right okay so find the value of C so next question find the value of C for which the vector will be in the spanning set of the vectors with usual addition so please uh try this problem uh we'll have just one minute break just try this problem excuse me Mom c equal to Z for so did you try what is the answer here Cal to Z Cal to z c is z is zero okay so how does the typical element of uh span of 1 comma 0 comma 1 and 0 comma 1 comma minus one how does it look like Alpha comma Alpha comma betaa alha minus beta so this is how a typical element will look like right now you are having Alpha is equal to p and beta is equal to Q so what should be cus q q pus Q okay fine now let us look at this example yeah I need to ask a question whether we will consider a general General case uh for the spanning set if it contains RN elements sorry if it is a vector space of RN then it will contain only n elements yes okay so the cardinality of that RN will be n so you are asking about you linearly independent set or what is that n elements mean what do you mean by n elements here ma'am you have just shown me about that spaning set okay spanning can have n element or more than n element see suppose you are having a vector space where is the spanning and definition here right suppose dimension of V is equal to n so any spanning set of V will contain minimum n elements minimum n so it can be more than n also here in this case you see there are three Els and here one of it is a linear combination of the other so this is actually not contributing much so any spanning any spanning set of V will contain minimum n elements it can contain more than n also okay okay ma'am yeah ma'am yeah tell me this spanning set is super set of this bases yes spanning sets always contain a base okay fine okay now coming back to this problem here let V and W be Vector spaces which are defined as follows V is set of all X comma y such that yal to MX where m is non zero and X comma y comma m is in r with usual addition and scalar multiplication and W equal to set of all X comma y x is non zero with usual addition and scalar multiplication so the set 1 comma M here it is 1 by m comma 1 is linearly independent set in B do you agree given m is non zero is this set linearly dependent or independent independent how is it linearly independent 1 m 1 by M1 what is the determinant 1 - one then how can it be linearly independent it will be dependent so this way you can argue or very easily you can argue that M * of 1 by m comma 1 what is this one comma one comma so you see that one is a multiple of the other so they are linearly dependent so the next so this is incorrect so next option is the set one comma M comma 1 by m comma 1 is a spanning set for B is this correct yes ma'am yeah we can say that okay how is this a spanning set s Y8 satisfy y = to MX yeah you see that so V is actually the set of all X comma MX where X M comma X is in r and M is non zero so it means so 1 comma m in fact forms a spanning set or in fact a basis or spanning set you can say so any set containing the spanning set will also be a spanning set any set any subset of V which is also containing this will be a spanning set for V so you see that this this is a spanning the set 1 comma m is linearly independent set in B yes or no yeah yes we can say that what about this yeah for this one too we can see both of them are non zero vectors in V if you if there are nonzero vectors in V you know that they are always linearly okay the set0 comma 1 comma 0 comma 2 is linearly independent set in w no not not independent definitely this is a multiple of this so it is linearly dependent right so 0a 2 is two times of 0 comma 1 and the deterant is also zero the set 0 comma 1 is the linearly independent set in W yeah yeah the set 0 comma 5 is a spanning set for w yes yes so any element of the form 0 comma Alpha will be a spanning it right yes okay up to this is everything fine for everyone yes ma'am and can we say that for RN if a set containing less than n element then it would be larly independent any any set with less than n elements yeah for RN any set less than n element would be linearly independent no no for example you take r itself in R3 if I'm taking one 0 2 0 0 are they linearly independent dependent this is a linearly dependent set right the number of elements does not guarantee Independence here okay fine okay so what is the basis of a vector space a basis of a Vector space is a linearly independent set that also spans B right so your B should be linearly independent and span of B should be equal to V if these two conditions are satisfied then you call such a set B to be a basis for p okay so for R2 and R3 you know that you know that standard bases are there standard bases for both of this are there and any linearly independent set of size two and here size three will be a basis so in general for RN for RN any linearly independent set with n vectors forms a basis okay so in RN if you take any n vectors which are linearly independent then that will form a basis okay okay so for MN of R for example M2 of R what is the basis standard basis one 0 right so this is one of the bases same way for M3 M4 you can write right okay is every spanning set of v a basis for V what is the answer no no so for example we consider this set for R2 so this is a spanning set for R2 but this is not linearly independent hence not a basis right so is every linearly independent set a basis for V yes yes so you see 1 comma 0 is linearly independent in R2 does this form a basis no' no' no it's not spanning R2 so not a basis so it should be both linearly independent and also it should be a spanning set then only it will be a basis so what are the equivalent conditions for basis this is the definition the first one so if you know that the given set is a maximum linearly independent set it means if you take any other linearly independent set it cannot be bigger than this set then such a set is called the maximum linearly independent set so if it is a maximum linearly independent set then that will form a basis same way minimal spanning set means suppose you have let us say B1 is a subset of B such that span of B1 is also equal to V okay then this should imply that B1 is equal to B so if this happens then such a set B is called a minimal spanning set it means there is no smaller set which is the span of the entire set okay fine is the basis of a uh space unique already we have given some examples here itself you see that uh this is a basis for R2 this is also a basis for R2 there can be infinite number of bases for a given Vector space so bases is definitely not unique okay answer is no suppose B1 and B2 are two bases for V then the number of vectors in B1 and and the number of vectors in B2 what is the relation should be they should be equal should be equal is there any doubt no ma'am no ma'am this is not true for spanning set sorry what is the question this uh Forest spaning set this is not true spaning set this is not true if s is a subset of R7 such that span of s is equal to R7 what is the minimum number of elements possible in s seven seven seven right so a spanning sh set for RN should contain at least seven entries minimum there should be n entries okay so the dimension of a vector space is nothing but the number of in the bases of yeah in the previous end maximum number of Entry maximum it can be anything okay the dimension of a vector space b is the size or cordal that is the number of elements in the bases okay and it is denoted by DM now for any bases for RN has how many elements excuse me ma' element ma'am Dimension can be less than the number of I mean cardinality of bases isn't it no it is exactly the cardinality of the bases always always that is a definition okay okay any basis for RN has exactly how many elements n elements and and so if the basis contains n elements the dimension of RN is always n n yeah yeah what is the doubt okay okay let W be a Subspace of R3 spanned by these vectors what is the dimension of w two so of all how many vectors are here three vectors so you can say dimension of w is less than or equal to three this much at least you can say okay after that can we say a precise answer yeah we can say it is two two Dimension is two yeah Dimension is two how why because the third is always zero so this is a linear combination of one and two R2 so let us say this is V1 V2 so V1 and V2 so that is why this is negligible so the spanning set of let us say this is V3 V1 V2 V3 is span of V1 and V2 okay and these two are linearly independent so you see that this is the dimension of the space then okay ma'am can we say that the Z is zero in all these three vectors then it is dimension is two that does not tell you anything if that is zero does that does not tell you anything okay yeah ma' I have doubt in one question which is from Dimension can you take it okay uh is it from activity question or something or it is from previous year yeah previous year uh then please ask at the end of the session only half an hour left you can ask after half an hour okay okay okay let me cover all the topics and then we can discuss the Dos okay steps to obtain bases from a spanning set so if you are having a spanning set for example like this if you're having a spanning set what is the way to obtain a basis from this okay so for simple examples like this you can simply eliminate the linearly dependent Vector here but it is not always this much simple so in that case what you do you write down the vectors as rows of the Matrix apply row reduction on a so that you get the row eilon form the nonzero rows of the reduced Matrix forms the basis for w okay so for example the same Matrix if you're considering 1 0 0 0 1 0 350 so if you are performing this row operation R3 going to R3 minus 3 R1 and R3 going to R3 - 5 R2 then the Matrix you're are going to get is 1 0 0 0 1 0 0 0 0 so the non Z Z rows what are the nonzero rows of this Matrix first and second so that that will form a basis for w okay fine so not column what is the question again nonzero columns we'll look for or so that is the next case that is Method two if you are writing the vectors as the vectors are arranged as Columns of a matx and you perform row reduction then The Columns of a corresponding to The Columns of R containing pivers forms the uh basis for the given space this is another way where you write the vectors as columns okay fine ma'am Rank and diamension always will be same yes Rank and dimension in the sense that suppose you are having V1 V2 VN for an in an N byn Matrix okay so what is the dimension of span of V1 V2 VN or you can even consider so this is same as the rank of this Matrix suppose this is a so this is same as rank of a okay so we will come to this definition I have given this definition in the later part we will see it there okay fine okay ma'am If U we U and W are subspaces of a finite dimensional Vector space then dimension of U plus v is dimension of U plus dimension of w minus dimension of U intersection W so sometimes you don't even have to calculate what is U plus W straight away by applying this formula you can see what is the dimension of u+ v u plus W okay so for example let us consider U to be this set and W to be this it now what is U inter what is the dimension of U first of all it is the XY plane what is the dimension two two two yeah two right and what is the dimension of w one so dimension of U is two and dimension of w is one now what is U intersection W one one so this Dimension is again one right okay so now what is without even calculating what is u+ w can you say what is dimension of u+ w two 2 + 1 - 1 so 2 so now if you calculate U + W what is U + w x such that X and Y are in so you see that the dimension of this SP is to okay so using this formula without calculating the sum you can easily identify the dimension sometimes when you have complicated subspaces U and W this technique is useful okay because in some subspaces uh finding this U plus W may not be very straightforward like this it may be something complicated in that case you can use this formula okay fine okay find the D dimension of the vector space V is equal to set of all X comma y comma Z such that x + y + z equal to Z and Z equal to Z under usual addition and usual scalar multiplication how do you find the dimension of this set di coordinate what do you know about the Z coordinate Z coordinate is always zero it is always zero so now this equation then reduces to xal what do you know about y then- so x - x zero where X is in R so your set is actually only this much so you can write any Vector X comma - x comma 0 as x * of 1 comma -1 comma 0 so it means what the vector 1 comma - one comma 0 this forms a basis for V so what is the dimension of V then one one one so what about this element maybe I will say 1 by minus 1 by 5 1X 5 0er does this form a basis for V yes yes ma'am yes yes ma'am SC M yeah so you see that any scalar multiple of a vector will also be a basis for that particular space okay fine now find the basis for this set V is equal to set of all x y z 0 z x y0 0 such that x + 2 y = to Z and x y z r and r so first of all X and Y we cannot reduce but Z we can replace with X and Y so let us rewrite the set V is equal to X Y instead of Z I can write x + 2 y again here 0 x + 2 y x y 0 0 x y in r R now any Matrix in this set how can you write if you split the uh if you have to remove the unknown variables like x times of what is this Matrix can you tell x * of one one y z uh one 1 0 0 0 y * of z0 one two two two 1 okay so now can you tell me what is the basis for V yes basis I'm asking not dimension 1 0 1 0 1 1 0 so these two right 1 0 1 0 1 1 0 0 0 and 0 1 2 02 0 1 0 0 this will form a basis for we is that correct yes yes ma'am okay now V is a diagonal matrix whose sum of the diagonal entries are zero so it is a 3X3 diagonal matrix dimension of this so if the basis contains two elements tell me the dimension two two two right okay okay so now for the next problem you can write any Matrix a in this set as first of all it is a 3X3 diagonal matrix what is the other condition given sum of diagonal matrix is zero diagonal entri is z so D1 D2 d3al Al to0 so if it is given as trace of a equal to Z it is saying the same thing don't forget what do mean by Trace okay so now D1 + D2 + D3 equal to Z so from this can you say what is your [Music] D3 D2 minus D1 minus D2 right so you can write this Matrix as D1 0 0 0 D2 0 0 0 - D1 minus D2 now what is the dimension of this Matrix how many unknown variables are there two unnown variables so D1 * 1 0 0 0 0 0 0 0- 1 plus D2 times 0 1 1 one so you see that this two will form a basis for what v v so what is the dimension of V then two two okay so given a matrix space or given a set like this now you know how to find the dimension and bases right yes ma'am okay yes ma'am so one or two more problems from this kind of problems so yeah last question could please repeat one second last one the previous question yeah yes any doubt in this what we are finding what is I'm finding a find a basis this is a this is B we are finding bases okay this is also finding bases we are finding bases for V excuse me fine let B be a VOR space yeah I'm for calculating the dimension uh in the notes that you have given you have written that for nonzero rows is equals to the dimension but you are explaining that um unknown variables is equal to the dimension of V do the difference you have to understand if you are talking about dimension of one single Matrix then what your saying is correct but if you are considering a matrix space this is Matrix space collection of matrices forming a vector space then this is what is a method to find the dimension what you are talking is the dimension of a matrix like this single Matrix right so this is what is the rank of a matrix but if you are having a space of matrices then this is how you find the dimension for the space of matrices it is not one single Matrix it is a space of matrices you understand the difference yes ma' okay let B be a vector space which is defined as follows B is equal to set of all X comma y comma Z comma W such that x + z is equal to y + W so now find the basis for we how do you find so we have to again try to reduce the number of variables how many variables can we reduce two dimens how many how many can we r two I guess two okay let's say I'm keeping X and Y how do you replace that no ma'am we we can only reduce one we can reduce only one we can reduce only one right because only one equation is here so we only one so here from this what you can say w is equal to x + z y so x y z in R now any element x y z comma x + z - y can be written as x * of 1 0 0 1 plus y * of 0 1 0 minus one yes and Z times of 0 0 1 1 okay so now these three vectors and the last element will have two one oh here that will be one right that times of so this three Vector will form a basis for V okay so what is the dimension of V then three is this clear okay yes ma'am okay now instead of in all the scenarios we were having only one equation suppose you are having two equations then how do you solve so here you are having 2x + 3 y = 0 and y + y z is = 0 5 Z we can substitute that in equation one so you take y isal to Minus 5 Z you substitute here so 2x minus 3 * of 5 Z equal to Z so either you do like this or you have to look at these options and see what is the answer okay now which of the foll following set forms a basis for we first of all uh in this uh set how many variables can we reduce two two we can reduce two so how many unknowns are there three if you are reducing two out of three one unknown only one unknown right so now yeah so uh you say that suppose you know what is z suppose let us that is equal to one because we already know that only one is the possible answer let us start with the one Element Case suppose Z is equal to 1 What is the value of y- 5 minus 5 and what is the value of x of 15 by right so you see that this element is in fact in V but uh just like the previous problem let us try to write this uh set in terms of one single variable so the first is X or I can write it as Min - 3x2 y y and minus uh 1 by 5 y right so either you can write it in this way where Y is in R you can write it in this way or here if you are writing in terms of three is right yeah or if you want to write the same set in terms of Z then how do you write so here 2x uh - 15 Z is equal to Z is that correct Y is equal to - 5 Z so - 15 Z = 0 so X is = to 15 by 2 Z comma and Y is equal to minus Pi Z comma z z r so in either way you can write okay so now if you are considering this format then you will get this as a basis element so this is correct but can these form a basis for the set without even verifying can you say the answer yes ma'am they are linearly depend they will not form a basis because the dimension is one the Bas bases should have only one element you know that the dimension is one dimension of V is one so the bases should have only one element so these two set canot from a basis so this is fine what about this yeah ma'am it is also yes because you see it is just a multiple of two when you take that is equal to two then you will get this element so any non-zero element here will form a basis because the dimension of V is one so is this clear how we did this problem yes okay yeah excuse me ma'am yeah ma'am here we took the Jal to one yeah for option three you have taken Z equal to one we can take any how you choose you can take any any number to form a basis if the dimension is just one here you can just form a take any that and write this answer so usually what to option yeah we take according to option yeah we can take according to option if not no option is given then we do it in the previous method to take out the variables and try to find the basis okay thank you why option first and second incorrect sorry option first and second are incorrect what is the reason yeah as I told you Dimension is equal to one that is obvious because you are having only one unknown variable so if the dimension is one then the number of elements in the bases is equal to one one one and how many elements are there in this set two there are two elements so it cannot form a basis okay okay is it fine yes ma'am okay okay so we are there almost at the for the at the last topic rank of a matrix okay let a be and M byn Matrix the rank of the Matrix a is the number of linearly independent Columns of a okay this is one way of saying that it is the number of linearly independent Columns of a the Subspace spanned by The Columns of a is the column space of a and the dimension of the column space is the column rank of a similarly the Subspace Spann by the row of a is the row space of a and the dimension of row space is the row rank so it is always true that row rank is equal to column rank is equal to the rank of The Matrix so the number of linearly independent columns or the number of linearly independent rows of a matrix will always be the same okay irrespective of the size of the Matrix that will always be the same okay and now you are having a matrix here so you are having three columns here what is the column space of a according to the definition it is a Subspace panned by The Columns of it so can you say what is this it is the pan of which vectors 1 -1 4 2 0 - 2 3 -1 2 okay so this is the column space now what is the basis for the column space how do you identify a basis how do you identify your bases we'll take variables a b and c for example uh then we will uh do the addition is there a simpler way in this case for example 3us what you say about the relation between these two vectors and this in this Cas yes or no yes now you know that this is a linear combination of this two what about these two vectors are they linearly independent yes yes ma'am so what can you say about the basis 1us 4 2 - 2 so this will form a basis because this is just a linear comination so this set is actually linearly dependent so this is a linearly dependent set so you have to rce use this set to a linearly independent set to get a basis okay fine so now what is the column rank of a column rank is the dimension of the column space so what is the dimension of the column space how many elements are there in the bases of the column space two it is two okay now what is the row space of a 1 - 1 4 tan of 1 1 2 3 - 1 - 1 2 3-- 0 - - now what is the basis for the row space what is the basis for the row Space 4 - 2A 2 is equal to minus of 1A 2A 3 - 5 of -1a 0-1 so you can verify that this is actually a linear combination of these two vectors so what is Alpha and beta here Alpha should be equal to if you calculate you can verify that Alpha is equal to Minus 5 and betaal to Minus 5 sorry one and betaal to Minus 5 is that correct does it satisfy if you substitute min-1 comma -2a -3 + 5 comma 0a 5 4 Min -2 and two F so you see that this is again a linear combination of these two vectors so what is the basis here 1 2 3 - 1 0 - one 1 2 3 - one 0 minus one okay if for yeah if for some Matrix we don't get the linear combination I mean the three vectors which we get are not linear combination of each other then what should we do see usually this in simple cases you can find like this if it is not simple what you have to do is that write this vectors first consider this matx find the determinant find the determinant if the determinant is zero then only you have to guess the dimension is lesser than this if the determinant is non zero then all the three will be linearly independent right okay so based on the determinant you can guess whether you have to check this or just straight away write this as a basis okay do you understand the technique I'm telling yes ma'am means if the determinant is zero then we have to check or else just declare that they are all linear yeah if the determinant is zero definitely rank of a will be strictly less than three okay it cannot be equal to three okay in that case definitely one of this is going to be a linear combination of the other two okay by checking the determinant of minor of that Matrix you will get able to get the checking determinant of minor why why do you want to do that means if determinant of this Matrix is zero then we will find the determinant of minus like 2x2 matx and if you get the that you don't have to do for a 3X3 Matrix you don't have to do so if the determinant of the Matrix is zero it is just a 3X3 Matrix so you only are going to be left with two vectors so two vectors dependence or Independence you can easily check by looking at whether it is a multiple of the other or not so in this case looking at a minor all the principal minors you have to look so that Technique we are not seeing here but there is one technique where you can find the rank by looking at all the determinants of all the miners of the Matrix so that part we are not covering here so forget about it ma'am yeah okay uh writing these vectors in column reducing to row row form that the next technique so the some people find it convenient just by looking at this Matrix itself they tell or if you want you can go to the row eulon form so by the way here the row rank is two so sometimes you can uh find the row Kon form find the number of nonzero rows in the row Kon form that will give you the rank of it so this is another way to calculate the rank so either you can use the colum space or you can use the row space or you can use the row eulon form there are three ways to find the rank of a given Matrix is that fine all of this is going to give you the same answer ma'am in reduced toon form it will also give us the basis right for the yes it will also give a basis the nonzero rows will form a basis okay by the way whatever Elementary row operations you are performing it is not going to the number of linearly independent rowes yes ma' okay fine now find the value of a if the rank of the Matrix a is two so first of all this is a 3X3 Matrix and if the rank is less than three what can you say about the determinant determin is zero is not equal to zero oh yeah zero determinant is equal to zero right so now what is the determinant of this Matrix can you compute a + 6 2 * of a - 6 + 3 * of 2 + 4 * of -1 is that correct yes 2 a - 12 + 6 - 4 so 2 a - 6 - 4 - 10 so a is = to five is that correct yes yes okay so this just guarantees that determinant rank of a is less than three but uh is the rank uh equal to two when a is equal to five is the rank equal to two so if you have to check the rank is equal to two you have to just see whether you can uh see the two uh either two rows or columns are not multiples of each other definitely this column is not a multiple of this column yes or no yes ma'am so in that way you can see that obviously the rank is two okay determinant is zero and you have two linearly independent column so rank is two or linearly independent rows either way you can check so rank is two so determinant zero only guarantees the rank is less than three less than three means it can be one two or three for a non-zero Matrix so in this case it is definitely two okay fine is this clear for everyone yes ma'am okay choose the set of correct options from the following row Rank and column rank of a matrix is always the same yes or no yes ma'am yes ma'am yes ma'am same the rank of a zero Matrix is always zero yes okay let me write it in a different way the only Matrix for which rank is zero is the zero Matrix is that correct yes ma'am yes ma'am okay so you will have rank to be zero only when the Matrix is zero otherwise you will never have a zero Matrix even if there is one nonzero entry then the rank will be one okay fine the rank of a matrix all of whose entries are the same nonzero real number Must Be One do you agree zero so it is it says a is equal to Alpha Al alpha alpha all of them are alpha how do you how do you justify we'll reduce to only one one the form it will be actually one one one one all the other things are z0 right yes ma'am yes so you see that rank is one rank of the N byn identity Matrix is one for any N in N or false what is the rank of an identity Matrix of size 10 right so this is incorrect so rank cannot exceed the number of columns or the number of rows of a matrix so if you are having a m byn Matrix then rank of a is less than or equal to minimum of M comma n it means what suppose you are having a 2x4 Matrix what can you say about the rank of a can rank of a be three yeah yes yeah how can it be three it is less than the minimum of this two what is the minimum of this two two two two so minimum of 2 comma 4 is two so rank of a is less than or equal to two it cannot be three okay okay so rank of a is less than or equal to minimum of M comma n okay same way rank of a into B product of two matrices is less than the minimum of rank of a into rank of B okay consider the following upper triangular I guess this is the last problem here yeah this second one sir uh second one second if two Matrix are multiplied uh then uh the rank will be minimum of uh the rows and column of those two A and B of that rank will be what suppose two Matrix are multiplied then what the resultant Matrix we are getting the rank will minimum of that row column not row or column rank of a and rank of B okay so suppose you are having a matrix a of size 3x two and you're having a matrix of size uh 3 by uh 2 by one so what can be the rank of B one one one okay you forget about the rank of now what is the uh let us say this is a non-zero matrix by the way what is the minimum of rank of a comma rank of B rank rank B it is less than or equal it is equal to one okay so what do you know about the Matrix a Ab is of size 3x 1 so the rank of ab is less than or equal to one yes yes okay you understand yes yes okay so let us look at the last problem here consider the following upper triangular Matrix to choose the correct options where a b c d e f r and r so if f is equal to zero the rank of the Matrix must be less than or equal to two cor yes ma'am correct because DET will be Zero have only two nonzero rowes so it will be either two or less than two right yes ma'am if uh f is equal to zero then the rank of the Matrix must be exactly two false is that correct no no ma'am so for example d and e both are zero then it is possible that can rank is one so it is incorrect if a b c d e f all are non zero then the rank of the Matrix must be three yes yes why because when you reduce it to the row Kon form you will actually get the matx of this form here some entries are here so you will get three nonzero entries if since a d f are non zero this is the rilon form you will get okay fine option four if a DF are non zero then the rank of the Matrix must be three true true true correct so if a DF are non zero then the rank of the Matrix must be three so actually the rank of a triangular Matrix depends on the number of nonzero entries in the diagonal diagonal so the nonzero diagonal uh entri of a triangular Matrix tells you the rank of the Matrix okay so this option is in correct only right yeah and what about the dimension equal to rank that thing Dimension ma'am uh that thing that one point someone said earlier that Dimension is always equal to rank is it true oh here if you're considering this as a one single Matrix then the dimension whatever we are talking is same but if you are now considering all upper triangular Matrix a b B C 0 d e 0 0 f where a b c d e f are in R so if you are considering one single Matrix and finding what is this rank of this Matrix then whatever we discussed is fine but this is a different problem this is a different question if you are considering this space now you are considering a collection of matrices now what is the dimension of this it is a different question what is the dimension of this what is the answer how many un here six dimension of V is six this is a completely different question and this is a different question okay okay ma'am here one single Matrix here collection of matrices okay okay so now you can ask whatever doubts you have so far I have cover I try to cover all the topics from week one to we four was it fine uh yes ma'am okay then so now you can ask your doubts yes ma'am ma'am I have doubt uh from Dimension I have one problem uh sum of the diagonal uh sum of the diagonal inas of a is zero and the sum of each rows is zero okay then what will be the dimension of the Matrix okay so what is the size of cross okay yes ma'am so you are having a matrix a b c d e f g h i where a + e + i = z a + b + C = to 0 D + e + f is equal to z g + H + I equal to Z this is your question right yes ma'am sum of diagonal is zero sum of each row is zero what is the dimension of such matrices correct yes ma'am yes or no yeah so now uh again here we are trying to reduce the number of variables how many variables are originally here variables are here so we are trying to reduce the number of variables so first in this case we can reduce one variable I you can write as minus a minus B okay so this we have reduced okay now C can be reduce minus a minus B and F can be reduce yesus tus now coming to this already I is reduced right already I is reduced so you can reduce one more variable here so let us take galus I which is plus a okay so you have reduced four variables here so this Matrix you can rewrite as a B minus a minus B and D you can write as d e we have not reduced F you can write as minus D minus E G you have reduced as minus H + a + e h and minus a minus C so now from the this how many unknown variables are there five so now you know the answer right five five right a b d e and H five variables are there so Dimension is and you know how to B right yes okay yes ma'am yes ma'am yeah uh one general questions uh regarding the paper how many questions will be there and what will the weightage for that no fixed rule like that at least I don't know such rule exist or not I think yeah any other questions ma'am what is proper subset proper sub space can you yeah uh tell me Amit Kumar yes yeah ma'am proper sub space or proper subset what the means proper proper Subspace or proper subset yeah yeah what is the difference you are asking uh I'm asking ma'am proper the word of proper okay what is proper okay okay okay fine so usually usually suppose you are having uh some set let us say x a which is having a b c so what are all the subsets of X all subsets of x if you are writing then you will write start with empty set then you will write all the singl T sets and then you will write all this pad sets and then you will write the original set okay so now except mty set and except X whatever is in between these are called the proper subsets the same goes for so the same goes for Subspace also if you are having a vector space V so this zero and V they are called the except this except these two subspaces other subspaces got it m got it thank you are proper yeah' okay yeah okay one question is there in 3.1 I couldn't get if uh 3.1 uh six question if a comma B comma C is inverse element of one sorry can you say from where is the question 3.16 question activity question 3 3.16 question can you please hold on I will some I didn't get it right that's what I was wondering uh 3.16 question it is something like inverse element of 1 comma 2 comma 5 yes exactly exactly yes a comma B is inverse element of 1 2 5 yeah so it is a very simple question right so you are having uh 1 comma 2 comma 5 in R3 what is the inverse element they are asking so 1 comma 2 comma 5 plus which element will give you 0 comma 0 comma 0 what is a bcus that's all right from this you find a b c and then write a minus B plus C sorry what is your doubt no no answer there it was evenus so wondering how okay so here what is the answer so it is -1 -2 - 5 and you are asked to find a - b + C a - b + C is -1 + 2 - 5 so - 6 + 2 - 4 what's wrong here okay I don't know what looks fine right yeah yeah this looks fine okay ma'am yeah ma'am disjoin Subspace means what disjoin subspaces yes ma'am there are no subspaces which are disjointed always there is a uh zero Vector in the intersection this joint means intersection empty right so intersection of any two subspaces will always contain the zero Vector so there are no disjoint subspaces okay ma'am hello ma'am yes yes ma'am is BAS is your Subspace or proper Subspace or it is the entire space what is Basis a Subspace no no basis is not a Subspace it is just a set if you are adding two elements of a basis it will not be in the bases right then it is not basis is just a set it cannot be a space first of all remember a vector space will never have finite number of elements unless it is a zero space okay' okay thank you okay yes any other questions ma' uh yeah spanning span uh whenever uh ma'am can you uh take a example and can explain the span spanning Set uh okay I have little spanning set and basis how do I teach okay see spanning set means here you see all the possible linear combinations you are considering so you are considering V1 V2 V3 here so all possible linear combination means Alpha V1 plus Alpha V2 plus Alpha V3 so now your V3 is dummy what do you mean by dummy here it means what if Alpha is equal to Z sorry alpha 1 Alpha 2 Alpha 3 if Alpha 3 is equal to zero still you will have and 1 comma 0 + 0 comma 1 one times of this right is equal to 1 comma 1 still this this is going to be there in the span of V1 V2 V3 so this V3 is not really contributing anything because you are able to get the V3 from V1 and V2 okay so this V3 is kind of a dummy Vector in the spanning set only the linearly independent vectors are contributing to the span of that set okay this Vector is not creating any difference so you see that even if you have this it is not making any uh any big difference but if you are having this set then you see that this is a linearly dependent set so your spanning set is not a basis you know that if the spanning set is linearly independent then only it is a basis here it is not linearly independent so it is not a basis so a spanning set which is linearly independent will form a basis is that fine okay yeah fine and between the spanning set and span if any any set of s that is have the vector V1 V2 uh V3 V5 and uh how to define uh span of s and spanning Set uh we have conf I have confusion both of them so you are asking suppose you are having span of no no ma I have this s that is the set of V1 V2 V3 V then how to write the span of s and uh spanning set yeah I understand so first of all you should understand so span of let us say this is uh V1 V2 V3 V4 V5 and this set is your s this is what you are saying right this is what you're calling as the set s right yes ma yes okay now if V is equal to span of s then you know that s is the spanning set of V now what is V you want to find so first of all you see you try to reduce reduce this to a linearly independent set reduce to this to a linearly independent set if you are reducing this to a linearly independent set suppose you are for example you are in r four then how many maximum how many elements can be here four yes in s how many elements can be there if you reduce to a linearly independent set in R4 how many elements can be there in the four only four four four maximum four elements can be there right so you take that four elements and write Alpha V1 alpha 1 V1 Alpha 2 V2 Alpha 3 V3 and Alpha 4 V4 the reduced linearly independent set from that you write the linear possible linear combinations where each Alpha is in r that is the set B okay so if you directly use all the vectors it is going to give you too much of calculation So to avoid too much calculation first of all reduce the spanning set to a linearly independent set that is we are reducing the spanning set to a basis and from the basis you can tell what is the set B okay is it fine okay ma'am fine fine okay thank you okay any other doubts okay then so thank you Ma yeah thank you everyone wish you all the best for the exam thank you ma'am ma'am what will be the difficulty level for this it will be mixed you will have some easy questions some difficult questions everything will be there Madam please share the PDF for this sorry please share the PDF for today's of today yeah yeah definitely I will share definitely so right after this session I will put it in the master folder please check it thank you so much thank you ma'am okay