 So today we will discuss some more properties of the matrix of a linear transformation okay let me recall the important result that we proved last time towards the end, V and W are finite dimensional vector spaces over the same field let us say real field T from V to W is linear I have to order basis I have to order basis of V and W respectively I am using this notation for that B V B W then we had seen last time that for every X and V if you look at the matrix of T X, T X is in W the matrix of T X relative to V W this is the matrix of T relative to the basis B V B W into the matrix of X relative to B V. This is what we said is the converse of the statement that if A is a matrix then T X equal to A X is a linear transformation okay. Let us look at a few more properties of the matrix of a linear transformation especially we will look at first how it behaves what is the matrix of the composition of linear transformations what is the matrix of the composition of linear transformations what we will see is that this is the multiplication of the corresponding matrices and this is really the defining place for matrix multiplication something that we should always remember matrix multiplication the unique peculiar way it is defined comes really by looking at matrices as linear transformations okay. So the next result really defines matrix multiplication then we will also look at the question if T is invertible what how can you compute the matrix of the inverse transformation T inverse the natural answer is it will be the inverse the matrix of T inverse relative to the same basis will be the inverse of the matrix of the transformation T okay and also finally establish a relationship between matrices corresponding to different basis okay these three results we will discuss today so the first is composition so this is the framework for me the framework is I will use a slightly different notation for this theorem you will see it is because of its simplicity I have u v w finite dimensional vector spaces real vector spaces say with I will write down the basis for u so these are ordered basis I will write down the basis b u explicitly I need to talk about composition of maps I have two maps T is from u to v T is from u to v and S is from v to w suppose these are linear suppose T and S are linear then I am looking at the composition first I must define the composition but before that I want to write down the formula for the matrix of the composition operator remember S circle T S circle T is a map from see T is from u to v so T takes a vector x from u to v S takes that vector T x to w so this is a matrix from u to w so I must write b u b w okay this is a linear transformation this is a function in the first place from u into w what we would like to demonstrate is that this is S now S is from v to w so b v b w into T T is from u v b u b v where I have not yet defined the composition where S circle T of x the composition S circle T of x this is equal to S of T of x where x is in u this is the formula for the composition for x in u S circle T of x is S of T of x composition okay so let us prove this result and you will see that this really defines see if S and T are linear transformations we will now show that S it is easily seen that S circle T is a linear transformation so we know what this matrix is what this formula says is the matrix multiplication of the matrix of this transformation and the matrix of this transformation the product is defined by the left hand side matrix okay so this defines matrix multiplication really given two matrices I can always find linear transformations S and T such that the first matrix A is this the second matrix B is this then I would like to know what is A B that is given by the composition S circle T relative to these two bases okay so this is really the definition of matrix multiplication okay so let us prove this before proving this formula I must show that S circle T is linear but I am going to leave that as an exercise. To prove clearly composition of linear transformations is again a linear transformation we need to observe that S circle T is a linear transformation now from u into W okay remember this is a formula connecting to showing that two matrices are equal let us say I want to show matrix P is equal to matrix Q then I will have to show that the corresponding entries are the same a little more general I will show that the jth column of the matrix P is equal to the jth column of the matrix Q then it follows that P is equal to Q that is what I will do okay. So let me start with I have written down the bases I have written down the bases B u explicitly I am going to exploit that consider S circle T of u j S circle T of u j is a vector in W I want to look at the matrix of this relative to B W I want to look at this as a matrix relative to B W this is an element in W this is a vector in W I will appeal to this I will appeal to this result and also use the definition of composition so this is equal to let me write this as it is S of T of u j that is composition and then keep this B W I have simply expanded what is inside this bracket now I have S of some vector what is the matrix of S of some vector what is the matrix of T of some vector it is a matrix of T into that vector just write the appropriate bases this is a matrix of S or the matrix of S S is a function from B to W so the matrix of S will be B V B W into the matrix of T of u j okay now matrix of T of u j T of u j is from u into W so I must remember to write this as B W it is correct B V B V T is a function from u into V so this is B V T u j B V apply this formula once again so this is equal to matrix of T T is from u to V B u B V and u j is a vector in u B u is that okay this is u this is okay so what I have done is on the left I have a circle T of okay now I will expand the left hand side this will be this is another linear transformation I can call that R if you want so this is a circle T is now a matrix is now a linear transformation from u to W u j is in u B u this is my left hand side and the right hand side I will write it as it is to agree with this a circle T is a linear transformation so I can call that as R if you want then it is left hand side is R of u j R of u j I will appeal to the same formula R of u j is matrix of R into matrix of u j what is R a circle T this is because R is a circle T that is linear okay the only thing that you need to observe is what is a matrix of u j relative to the basis B u make that observation we are through ith component 1 all other entries 0 that is the column vector u j matrix of u j relative to B u how do you write that you must write u j as a linear combination of u 1 u 2 etc u n the only unique linear combination of u j is 0 times u 1 plus 0 times u 2 etc 1 into u j plus 0 u j plus in etc 0 u n so the matrix of u j relative to B u is the column vector e j this happens in the j th coordinate this is what we call as e j the standard the j th standard basis vector of R n the j th standard basis vector of R n this is a matrix of u j relative to B u relative to B u with respect to some other basis you would not get this are we through with the proof you need to make one more observation this observation was made much earlier this is a matrix on the left this is a vector it is really e j I have written down so this left hand side is a circle T B u B w e j is equal to s e j for all j this is to for all j as j varies from 1 to n what you observe is that if a is a matrix and e j is the j th standard basis vector a is of order m cross n then a e j is the j th column e j is the j th standard basis vector a is an m cross n matrix then a e j is the j th column of a j th column of a on the left is the j th column of this product if a e j equals B e j for all j then a is equal to B that is what we have so if you want you can call this m you can call this as n and I have m e j equals n e j for all j j th column of m is equal to j th column of n j arbitrary so m is equal to n so these two matrices must be the same and so I have this formula a circle T B u B w on the right B v B w s B u B v T okay. I told you this defines matrix multiplication and we know that matrix multiplication is associative that can be shown by using this result and one of the previous theorems so let me just give this as a corollary to fill up the details is an exercise for you matrix multiplication is associative is a corollary of this result okay. One of the consequences matrix multiplication is associative just a few lines of this proof given two three matrices matrix multiplication A B C given three matrices A B C such that multiplication is possible the product A B C is possible then product A B product B C will be possible so given three matrices A B C such that the product A B C is possible A into B C is A B into C that is what matrix as a multiplication associativity means this I want to show A into B C equals A B into C you are given A B C construct three transformations T A T B T C such that T A X equals A X T B X equals B X T C X equals C X what is the matrix of the transformation T A corresponding to the standard basis that will be A matrix of T B corresponding to the standard basis that will be B matrix of T C corresponding to standard basis that will be C standard basis in the appropriate spaces see product A B C must be defined so the number of columns of A must be the same as the number of rows of B the number of columns of B must be the same as the number of rows of C the order of A B C will be the number of rows of A times the number of columns of C so you need to choose appropriate basis in appropriate spaces K L M whatever then use the fact that this formula holds and show that you have A B into C is A into B C okay so just take matrices write down the obvious linear natural linear transformations define through these matrices and look at the matrices of these linear transformations in turn with relative to the standard basis and apply this theorem okay you can show that matrix multiplication has a set of okay. One of the other consequences is what is the inverse matrix of a linear transformation which is known to be invertible okay the answer has been given let us prove this result but before that let us look at the specific case see we are talking about inverse transformations in particular we will look at a linear transformation over a vector space that is from the vector space to itself consider T from V to V linear such a transformation will be called an operator if the domain and the core domain are the same then T will be called an operator if the spaces V and W are different then you have to obviously choose different basis but if the spaces are the same then it is comfortable to deal with only one basis okay so let us say script B I will not use V there is only one vector space let script B be a basis of V I want to use BV there is only one space here then I will use this notation TB there is only one basis I will use this notation to denote TBB see I know how to write down the matrix of a linear transformation when two bases are given in particular if the bases coincide then I know what this right hand side is instead of writing BB I will simplify the notation by writing TB now this I can do when I know that T is an operator the space from the space V to itself so I will use this notation this is just a notation okay this is just a terminology TB will be this matrix which we know how to compute let us look at the particular case what is the matrix of the identity transformation on V what is the matrix of the identity transformation on V can you see that it is identity matrix but if it is between two different bases then it is not the identity matrix why is this identity matrix that is because you must look at the first basis vector write it as a linear combination of the same basis the only choices first coordinate is 1 all other coefficients are 0 so the first column is 1 0 0 second column is 0 1 0 etc so the identity matrix sorry the for the linear transformation I on V with respect to a particular fixed basis B treating that as two different bases is the identity matrix the identity matrix of order the same as a dimension of V so for the identity transformation if the two bases are the same then it is identity matrix if the bases are different then it is not identity you can verify easily by simple examples also what is the matrix of the 0 transformation relative to a single basis that is a 0 matrix okay left hand side 0 transformation right hand side is a 0 matrix this is an equation involving matrices okay in this case let us go back to that formula that we derived just now I have T, S from V to itself the linear operators this I have proved earlier I have proved this earlier what happens with this in this example in this particular situation a circle T there is only one basis okay this is simplified formula when you are dealing with linear operators now what it actually means is that little more abstraction can be brought here I defined the function Phi from T L recall I defined a function Phi from on a linear transformation T so this is in L V W to R m cross n R m n or R m cross n by Phi of T equals the transformation T relative to two bases this time I will choose in the case when W is V I will have only one basis so this will be with respect to only one basis one basis that I started with L V V can be shortened to L V but I will leave it as it is we had observed that this is an isomorphism this is linear one to one and on to and so it is an isomorphism and we use this formula to compute the dimension of L V W if V is m dimensional W is n dimensional then we computed the dimension of L V W by using this isomorphism now if in the light of this formula what it also follows is that this Phi preserves products Phi preserves products what is the meaning of this Phi of S circle T equals Phi of S into Phi of T it is like f x y equals f x into f y that is called a multiplicative function Phi preserves products what is the consequence of this consequence of this formula really the formula that I told you for the inverse transformation when you know that the inverse exists so let us derive that next. So I want to show this result that let T from V to V be an invertible transformation be an invertible linear operator and a script B be a basis of V T inverse is also a linear transformation from this T inverse is also a linear operator we had seen this before what is the matrix of T inverse relative to B what we want to show is that this is equal to the matrix of T relative to B take the inverse of that. So I will introduce a bracket and write this minus 1 outside remember this is an equation again involving matrices equation involving matrices what is inside the bracket are linear transformations okay proof I will make use of what we had seen just now this formula and the fact that the identity transformation with respect to the matrix of the identity transformation relative to a fixed basis is the identity matrix. Since T is invertible there exists S from V to V such that T is invertible there exists S such that T S by which I mean T circle S so T circle S is a circle T equals the identity transformation this is a formula for transformations there are no matrices here S and T are linear transformations the right hand side I is the identity transformation so I am using the same notation the context must make it clear whether it is matrix or a linear transformation. So I will apply this formula to this equation so if you look at T circle S relative to the fixed basis B that I started with that will be equal to identity relative to B I am using the first equation T circle S equals I am using just that identity relative to B is the identity matrix this time it is an equation involving matrices the right hand side I is the identity matrix T circle S B I will invoke this formula that is a matrix of T relative to B into the matrix of S relative to B this is equal to the identity matrix I can do a similar thing for the second equation and just write down on the other side this is equal to S into T matrix of S into matrix of T this is matrix of S into matrix of T coming from this equation so I have two matrices let us say A and B A into B equals identity that is equal to B into A in fact one of them is enough because it is a square matrix rank nullity dimension theorem could be used okay in any case I have an equation like A into B equals identity A and B are matrices so since A is square and B is square A and B both must be invertible what it also means is that this S B is the inverse of this matrix this matrix must be the inverse of this matrix let me write this what in particular this means is that S this is a matrix remember this must be the inverse of this matrix by definition A B equals identity A and B are square so B equals A inverse that is what I have written down but S is T inverse and so I have this so you do not have to compute the matrix of the inverse transformation if you know the matrix of the original transformation you take the inverse of the matrix of the transformation that you started with that will be the matrix of the inverse transformation relative to the same basis that you started with okay if you change the basis then this changes okay that brings us really to the next probably the most crucial question how do matrices corresponding to different basis behave how do matrices corresponding to different basis behave okay so let us answer that question how do masses my matrices corresponding to different basis for the same transformation behave. The answer is given in the next result let me state this theorem so I have a single linear transformation a linear operator really I have two bases let T be linear let T be linear let B1 and B2 let B1 and B2 be bases of V look at the identity transformation and then look at the matrix of the identity transformation relative to these two bases let me call that as the matrix M I am looking at the identity linear transformation relative to I am computing the matrix of this identity transformation relative to these two bases we know that this is not I this is not the identity matrix I am calling that as a matrix M then we have the following for every x and v the matrix of x relative to B2 the second basis is M times the matrix of x relative to B1 this really gives us the other formula how are matrices of a particular linear transformation corresponding to different bases related let me write that formula here the matrix of T relative to B2 by which I mean B2 B2 is M into the matrix of T relative to B1 B1 that is matrix of T relative to B1 times M inverse okay this is the important probably the most important relationship remember that this involves M inverse so we must show that M is invertible okay but I am going to leave that little part as an exercise this is similar this is similar to what we did earlier this is similar to what we did earlier use the previous result composition rather okay use the earlier result to show that M is invertible so I am going to leave this part as an exercise. Exercise show that M is invertible remember M is a matrix to show that M is invertible one could for instance show that the system Mx equal to 0 for x in Rn if n is a dimension of V has x equal to 0 as the only solution Mx equal to 0 for x in Rn has x equal to 0 as the only solution now that you can use this idea to prove that M is invertible. So I will assume M is invertible and derive this formula but this formula is really a consequence of this formula so we need to prove this first okay so let us prove this let us start with the matrix of x relative to B2 I can write this as the matrix of identity x identity transformation x relative to B2 and then use this formula that we proved earlier let me recall that here matrix of Tx relative to B2 Bw actually that is a matrix of T relative to B1 B2 into matrix of x relative to B1 this is what we proved earlier we used Bv Bw this is Bw Bv Bw Bv I am using B1 B2 here vector space V is the same two different bases now so let me use this result here this is the matrix of I relative to B1 B2 into the matrix of x relative to B1 I have the first formula immediately this is what we are denoting by M so x B2 is M x B1 that has proved the first formula okay x B2 is identity transformation I am applying and then I am appealing to this formula matrix of I relative to B1 B2 and then matrix of x relative to B1 this is what we are calling as M so I have the first formula x B2 is M x B1 that is the first formula I need to prove this so let me now start with consider Tx B2 Tx B2 I am going to appeal to the previous formula let me call Tx as y then I am looking at y relative to B2 y relative to B2 is M times y relative to B1 y relative to B2 is M times y relative to B1 instead of Tx I have y y relative to B2 is M times y relative to B1 Tx B1 I will apply this formula I will apply this formula with B2 replacing B1 with B1 equals B2 really I will apply this formula with B1 equals B2 this formula holds for any two bases in particular B1 equals B2 so I am going to look at this formula tell me if this statement is clear I will write this as M do you agree with this that is I am looking at a single basis now I am looking at a single basis and then I am looking at this formula for a single basis M into T B1 into x B1 that is it is actually matrix of T B1 B1 into matrix of x B1 the formula for a single basis I am using that but matrix of T B1 B1 is what we are denoting as T B1 matrix of T relative to B1 on the left hand side I have Tx B2 I will use that formula for B2 so I have really got what I wanted on the right hand side so I want this I have sort of got of what I want really here I will now expand this Tx B2 for a single basis B2 will be T B2 B2 that is just T B2 is it okay same thing Tx B2 what I did here I have what I have done here I have used for B2 then invoke the first formula T B2 into the first formula is x B2 is M x B1 I hope it is clear I start with Tx B2 apply apply this formula for as though there is a single basis so as a single basis I get this formula B2 and then formula 1 x B2 is M times x Bn I proved it so I invoke that here so finally what do I have that is this is on the left T B2 M x B1 this is on the left that is the expanded form of this on the right I have this let me write this M T B1 x B1 I want to show two matrices equal I will show that their jth columns are equal this is true for all x in V that I started with this is true for all x in V in particular say I am looking at x B1 in particular take B1 write down explicitly let us say u1 u2 etc un and then look at x being replaced by u1 u2 u3 etc let us say x being replaced by uj if I replace x by uj and write the matrix relative to that same basis then this is a jth column this is something that we did just now instead of x replace instead of x replace them by the basis elements that are present in B1 then I will get T B2 M is M T B1 this is true for all x apply to the basis elements I get this again something like P E j equals Q E j so jth column of P is equal to jth column of Q so P is equal to Q this whole thing is P for me this is a matrix so remember M is a matrix this is a matrix this is a product I am calling P this whole thing is again a matrix I am calling that Q I will apply the basis elements in B1 to this equation then this will be the first column of the identity matrix second column of the identity matrix etc so these two matrices are the same invoke the fact the M is invertible post multiply by M inverse I get the required formula since M is invertible I can post multiply this equation by M inverse this is now an equation involving matrices so I have T B 2 equals M T B1 M inverse okay that is if P is okay let us say A is a matrix of a linear transformation corresponding to one basis and B is the matrix of a linear transformation corresponding to another basis then A and B are related by the formula A equals M times B times M inverse for some invertible matrix M if A is a matrix of a linear transformation corresponding to one basis B is a matrix of the same linear transformation corresponding to another basis then A and B are related by the formula A equals M B M inverse for some invertible matrix M how one could determine M is a different matter but there is an invertible matrix M that satisfies this equation such okay can you see that if A equal to M B M inverse then B is equal to M inverse A M that is M inverse A M inverse inverse instead of M inverse let us use N then if A is equal to M B M inverse then B is equal to N A N inverse okay so we can if A is related to B by means of this formula we say that B is similar to A if A is related to B by means of this formula we say B is similar to A by the observation that we have seen just now it follows that if B is similar to A then A is similar to B okay A is similar to itself A is equal to identity A identity inverse if A is similar to B B is similar to C product of inverses you can use to show that A is similar to C so this is an equivalence relation similarity of matrices is an equivalence relation what does it preserve that is something we cannot discuss right now it preserves what are called Eigen values it preserves Eigen values okay that is something we will discuss later let me stop here.