 So we will continue where we left off the previous time, but just to recap, we were discussing about similarity and we say that a matrix A is similar to matrix B if there exists a non-singular matrix S such that B can be written as S inverse AS. This is an equivalence relation, so which in turn means that the relationship is reflexive, symmetric and transitive. Reflexive meaning every matrix is similar to itself and if A is similar and transitive means, symmetric means if A is similar to B then B is similar to A and transitive means that if there is a triangular, it holds that is if C is similar to B and B is similar to A then C is similar to A. We saw that similar matrices have the same characteristic polynomial and the corollary to this is that if two matrices are similar then they have the same eigenvalues counting multiplicities and we also defined the notion of diagonalizability and a matrix is diagonalizable if it is similar to a diagonal matrix and the similarity divides the space of all n cross n matrices into equivalence classes and if A is similar to a diagonal matrix then any other matrix belonging to the same equivalence class is diagonalizable into the same diagonal matrix and any matrix which is not in this same equivalence class will not be diagonalizable into this particular diagonal matrix and we saw that the condition for diagonalizability is that it should have n linearly independent eigenvectors and but we haven't discussed when a matrix will have n linearly independent eigenvectors. One such condition is that if A has n distinct eigenvalues then A is diagonalizable. When it has n distinct eigenvalues it will have n linearly independent eigenvectors and it will be diagonalizable. So one thing is that if you perturb a matrix by a small amount the eigenvalues also get perturbed by a small amount and we'll see that result a little bit later and so even if A is not diagonalizable you can always perturb it by a very small amount and obtain a matrix that is diagonalizable. We also defined the notion of simultaneous diagonalizability which is possible if there exists a single matrix so there's a common diagonalizing matrix S such that both S inverse A S and S inverse B S are diagonal matrices and one interesting property is that two matrices commute if and only if they are simultaneously diagonalizable. A and B commute meaning that A times B is the same as B times A for the last time and at the end of the previous class we stated the following theorem which says that if you have two rectangular matrices A which is of size m by n and B which is of size n by m with m less than or equal to n meaning that A is a fat matrix and B is a tall matrix but because we have defined them as m by n and n by m it is possible to consider what happens to BA and AB. Both multiplications are kosher they're both possible and these two matrices BA and AB have the same eigenvalues counting multiplicities but the matrix BA which is of size n by n and it's the bigger matrix in size than AB which is of size m by m the extra eigenvalues BA has n minus m extra eigenvalues compared to AB and those extra n minus m eigenvalues will be equal to zero. So in other words the difference between the characteristic polynomial of BA and the characteristic polynomial of AB is just this factor t to the n minus m you can see that this has n minus m zeros equal to zero t equal to zero and further if m equals n that is both these matrices are square and at least one of them is non-singular then the two matrices AB and BA are similar. So clearly if m equals n then they will have the same eigenvalues but remember that having the same eigenvalues or having the same characteristic polynomial is not sufficient for two matrices to be similar to each other but in this case if one of them is non-singular then the two matrices are indeed similar to each other. So let's see how to show this. So this is a slightly clever proof. There may be other ways to show it but this way is simple but nonetheless it is a clever proof. So the way it goes is you consider the following this times the matrix okay so consider this product of these two matrices now AB is of size m by m and B is of size n by m so basically there are this this zero here has m rows and n columns and this zero here has n rows and n columns so overall this matrix has a size m plus n by m plus n okay it's a square matrix and I'm multiplying it by this matrix i m a zero i n so basically since this is i m this has m columns and since this is i n this has n rows and so this matrix is also of size m plus n by m plus n so this matrix here is of size m by n so it has the same rows as this and the same number of columns as this identity matrix so if I multiply these matrices i m times AB which is AB and so this times this gives me A B A and this times this gives me B this times this gives me B A okay so once again this AB is of size m by m A B A is of size m by n this B is of size n by m and this B A is of size n by n and so overall this matrix is also of size m plus n by m plus n as it should be same size as this okay similarly if I consider i m a zero i n times the matrix zero zero B B A what do I get I get AB I get B I get BA so this is the same as this so these two matrices are the same and so and and further this block matrix here i m zero this is a block upper triangular matrix in fact these two blocks are both diagonal and so this is actually an upper triangular matrix with ones on the diagonal all its eigenvalues are equal to one and so this matrix is non-singular all its eigenvalues this is actually not required for the proof but this is just a side observation since this matrix is non-singular and these these two products are coming out to be the same matrix what we have is that if I consider if I pre-multiply this by the inverse of this matrix right then I will get this matrix here so that what I mean is i m a zero i n inverse times AB zero B zero times i m a zero i n will be equal to this matrix here zero zero B B A so basically what this means is that this matrix and this matrix are similar to each other and this matrix yes I say what did you mean by block when you say block upper triangular matrix so an upper triangular matrix is like this all the elements above the main diagonal are non-zero and everything below the diagonal are zero and a block upper triangular matrix consists of blocks which which go down like this and all these all the blocks have so let me let me raise here so a matrix which you can divide into blocks so there's a block here a block here a block here and a block here and if this block is always zero you call such a matrix a block upper triangular matrix it is arbitrary you can divide it in whichever way you like but if you can divide it into blocks where everything below those blocks is zero then you call it a block upper triangular matrix sir individual blocks they could be non upper triangular right correct correct but in this case you see that the individual blocks on the diagonal are identity matrix and therefore they are actually upper triangular in fact they have ones on the diagonal so this matrix contains only ones on the diagonal and so it is non singular okay so these two matrices are similar they're both m by n sorry m plus 1 n cross m plus n and they are similar so basically if I call this matrix say c1 and this matrix c2 they are similar to each other and the eigenvalues of c1 is now this is a block lower triangular matrix so and the size of this zero here was n cross n so the eigenvalues of c1 are the eigenvalues of ab plus n zeros similarly the eigenvalues of c2 is the eigenvalues of ba plus this matrix and this matrix is of size so this matrix b is of size n by m so this matrix is of size m by m okay so it's the eigenvalues of ba plus m zeros thank you so the eigenvalues of c1 are the eigenvalues of ab plus n zeros eigenvalues of c2 are the eigenvalues of ba plus m zeros and so this means that just comparing these two the eigenvalues of ba so basically this n is greater than or equal to m so the eigen the the the eigenvalues of ba are the same as the eigen they have same eigenvalues they are similar matrices so we have that this is the eigenvalues of ab plus n minus m zeros and as a consequence the characteristic polynomial pba of t is going to be t to the n minus m times pab of t now to close out the proof if m equals n and say a is non-singular then ab equals a times ba times a inverse which means that ab is similar to ba okay now so one one small remark before i start discussing the next thing is that how are the eigenvalues of a related to the eigenvalues of a transpose and the eigenvalues of a Hermitian so eigenvalues of a transpose are the same as eigenvalues of a counting multiplicities also eigenvalues of a Hermitian complex conjugate of the eigenvalues of a counting multiplicities okay so so basically a and a transpose have the same eigenvalues a and a Hermitian have the same eigenvalues except for the complex conjugate operation so we'll just see this very quickly so if i consider determinant of ti minus a transpose okay the solutions or the zeros of this equation are the eigenvalues of a transpose and this is equal to determinant of ti minus a transpose which is equal to the transpose operation doesn't change the determinant ti minus a and so they basically ti minus a and sorry a and a transpose have the same characteristic polynomials and so they have the same polynomial same eigenvalues similarly determinant of if i take t complex conjugate i minus a Hermitian that is equal to determinant of ti minus a whole Hermitian and when i take the Hermitian every element gets the conjugate operation so the determinant of this you can think of it as this transpose and then taking the complex conjugate and determinant of this matrix transpose is the same as the determinant of this matrix and the complex conjugate simply complex conjugates every element of the matrix and so this is equal to determinant of ti minus a whole complex conjugate so this means that p a Hermitian of t complex conjugate is equal to p a of t complex conjugate and so this means that if you find zeros of this and then you take its complex conjugate you will get the zeros of this matrix and so the eigenvalues of a and the eigenvalues of a Hermitian are related through the complex conjugate operation now we know that if two matrices are similar then they'll have the same characteristic polynomial and the same eigenvalues counting multiplicities but what we've shown here is that a transpose and a have the same eigenvalues and the same characteristic polynomial counting multiplicities obviously we cannot conclude from this that a is similar to a transpose okay this is not I mean so if two matrices are similar then they have this same eigenvalues but if they if two matrices have the same eigenvalues it does not mean that they are similar so do you think a and a transpose are similar to each other in other words if I give you any matrix a will you be able to find an invertible s such that a transpose equals s inverse a s what is your guess sir a is similar to itself so if we take a transpose of that the a transpose will be similar to a transpose okay from that you cannot conclude that a is similar to a transpose sir if we take transport in the definition in the definition of similarity so but the definition of similarity is that it says that they're similar if there exists an invertible s so it goes to I mean if you take the transpose in that it it simply gives you the reflexive relation that is if it is true that a is similar to a transpose it will just say that then it means that a transpose is similar to a but it won't really give you a proof that a is similar to a transpose right so this is one of those results which is you know in matrix theory which makes this this subject very intriguing it turns out that a and a transpose are actually similar to each other you will be given whatever a you would you can find an invertible matrix such that a transpose equals s inverse a s so they are actually similar to each other but to show that we have to use this something called the Jordan canonical form which we'll cover a little later in the course and but using that form we can show that a is actually similar to a transpose that will come later