 So now we still have to construct this polynomial. So if we can construct a polynomial p i of t with so with degree at most n minus 1 such that p i of j n i of okay so let me write this out then it will be clear. So if we can construct p i of t with degree at most n minus 1 such that p i of j n j j n j of lambda j. So bear with me I am just hypothesizing some things and then I will show that this is something desirable for us and then I will show how to construct these polynomials. So this is equal to 0 for every i not equal to j and when i equals j p i of j n i of lambda i equals d i then p of t which I will define as p 1 of t t plus etc plus p k of t k is the number of distinct eigenvalues of this matrix a. So if I consider p p 1 of t through p k of t will fulfill the requirements of the theorem is the polynomial just say the requirements the theorem what do I mean by that I mean that if I consider p of j this will turn out to be equal to b why because if I consider p of j this is equal to p 1 of j plus etc plus p k of j which is equal to p 1 of now j is a matrix which is of this block diagonal form j 1 through j k plus etc plus p k of j 1 through j k and a polynomial of a block diagonal form you can apply the polynomial inside this block for each of these block diagonals and that's exactly equal to the polynomial applied to the entire block diagonal form what I mean is that this is equal to p 1 of j 1 p k of j k plus etc plus p k of j 1 p k of j k okay this is in general not true you cannot apply you cannot push the polynomial inside each element of a matrix but for a block diagonal matrix you can push the polynomial into each of the blocks now we already said that p is a polynomial such that this is equal to b 1 and all these other terms in this block diagonal form are equal to 0 and here all these things will be equal to 0 p k of j k will be b k and so this is equal to p 1 of j 1 actually I should have said j n 1 but anyway j 1 up to all the other things are 0 plus etc plus here everything is 0 except p k of j k and this is equal to b 1 and so all these are non-overlapping blocks and so I'll get b 1 b k on the diagonal and 0 everywhere else so which is equal to my matrix b so basically if I can find these polynomials p i of t with degree at most n minus 1 such that these two properties hold one p i of j and j of lambda j equals 0 for all i not equal to j and p i of j n i of lambda i equals b i that is when i equals j then I'm all set so now we just need to figure out how to construct these polynomials so here it is q i of t is equal to the product from j equals 1 to k as a product except the ith term of t minus lambda j power n j okay then the degree of this polynomial q i of t equals what it's the sum of all these n j's except the ith term okay that would be the degree of this polynomial but the sum of all these n j equals n because that's the sum of the sizes of all the Jordan blocks and that should be of size n so this is equal to n minus n i okay so now one thing we can note immediately is that q i of j n j of lambda j if I compute this this is actually going to be equal to 0 for every i not equal to j why is that it's because this has this kind of form so if I substitute j n j of lambda j I'll get j n j of lambda j so the j term so this is a kind of bad notation because j is also the index of the summation here but whatever this j is for example for a moment think of it as l so I'll just write it as l so that it is not confusing j n l of lambda l equals 0 for all i not equal to l okay and now if I consider this thing one of the terms here will be the okay I'm delivering the point but there is a lambda l term j n l of lambda l minus lambda l times the identity matrix and then I'm raising that to the power n l but this difference is just going to be that nil potent matrix of size n i cross n i raise to the power n l sorry n l cross n l raise to the power n l and the nil potent matrix when you raise it to the power n l you will get the all zero matrix and so this is always equal to 0 j n l of lambda l minus lambda l times the identity matrix power n l is equal to 0 so one of the terms in this product will be equal to 0 which will make the whole product equal to 0 this is just the nil potent matrix of index n l okay so now now the so this satisfies one part of what I want q i of lambda j q i of j n l of lambda l equals 0 for i not equal to j but I also need that q i of j n i of lambda i should be equal to b i I need that property also but q n i of j q i of j n i of lambda i need not equal b i but one thing we can say about it is that because the i equal to j or j equal to i term is not included here um it is the all its eigenvalues will necessarily be non-zero because these lambda j's are all distinct so it is non-singular and furthermore um if you examine this matrix it will actually be so each of these matrix matrices when I take j n i of lambda i minus lambda j so when I substitute j n i of lambda i in here one of these terms that is the j term will be j n i small j term will be j n i of lambda i minus lambda j times the identity matrix and this matrix will have non-zero entries on the diagonal and non-zero entries on the first super diagonal and it has been raised to the power n j and when you raise it to the power n j those super diagonal terms may get fully occupied but it will remain upper triangular and it will also retain this topulate structure that it has that is the diagonal entries are the same the first super diagonal entries are the same the second super diagonal entries are the same and so on so it is of the form star that is the end of the form that is it is upper triangular and toplates so basically the point is that if I take an upper triangular toplates matrix its inverse is upper triangular toplates and the product of upper triangular toplates matrices is also upper triangular toplates so here is you are seeing a product of such upper triangular toplates matrices which will also remain upper triangular and toplates and in fact its inverse is also upper triangular and toplates so we will use that property next okay so this implies that if I consider the matrix j sorry q n 9 of j n i of lambda i inverse this is also this times if I multiply this by bi bi is also upper triangular and toplates so this is upper triangular and toplates so now I am pretty close so basically a matrix so so so basically so what I am trying to say is the following so we have bi I can write that as b 1 of i this is the first diagonal entries in this matrix bi times j n i of lambda i minus lambda i times the identity matrix power 0 plus d 2 of i this is the first super diagonal entry in the matrix bi times j n i of lambda i minus lambda i times the identity matrix power 1 plus and so on plus b n i of i times j n i of lambda i minus lambda i identity matrix power n i minus 1 so when I raise this to the power 1 this this will have once only on the first super diagonal entry and those ones are getting multiplied by b 2 of i and so they're placing b 2 of i in the first super diagonal entry of this bi and similarly when I raise it to the power n i minus 1 I'll get a matrix which has zeros everywhere else but one at the top left entry and that is getting multiplied by b n i of i and then that is getting placed at the top right entry of this bi so I can I can always write it like this so this implies that there exists now this itself is a polynomial of degree at most n i minus 1 so there exists a polynomial degree at most n i minus 1 such that q i of so j n i of lambda i inverse times bi which is an upper triangular topolitz matrix this can be written as this particular polynomial that I've written here a similar polynomial but using the entries of this matrix instead of b 1 i to b n i i some polynomial of degree at most n i minus 1 r i I'll call that of j n i of lambda i so I have to do all this circus because whatever I used earlier this q and my n i of this thing this need not equal bi and I have to do some circus to make this to find another polynomial such that this property continues to hold but q n i of j n i of lambda i will be equal to bi that's why I'm doing all this all these steps here so now I am almost done with the proof what I'll do is now set bi of t equal to q i of t which is what I defined earlier this q i of t which is of degree at most n minus n i okay so the degree is at most n i minus 1 plus n minus n i n minus 1 okay so then this the the claim is basically that this if I did pi of j this is that that's all I need so this matrix satisfies n j of lambda j is equal to q i of j n j of lambda j times r i of j n j of lambda j and this is equal to 0 by our construction so 0 times r i of j n j of lambda j which is equal to the all-zero matrix for all i not equal to j and pi of j n i of lambda i is equal to q i of j n i of lambda i times r i of j n i of lambda i and r i of j n i of lambda i is this matrix here and q i of j n i of so so I get q i of j n i of lambda i times q i of j n i of lambda i inverse times bi so these two just cancel with each other and this is equal to bi which is all that we were looking for so so what we just showed is that if the matrix a is non-derogatory then any other matrix b commutes with a if and only if there exists a polynomial p of degree at most n minus 1 such that b equals b of a of course if b equals p of a then a commutes with b that is trivial but showing the other way around was a little bit more involved we had to do quite a few steps to show that if a and b commute and a is non-derogatory then there must exist a polynomial of degree at most n minus 1 such that b can be written as a polynomial of a in fact the converse is also true namely that a in c to the n cross n is non-derogatory if and only if every matrix that commutes with a can be written as a polynomial in a okay so so that is all we have time for today I wanted to talk I wanted to also talk about convergent matrices and their properties um so we've already seen that a matrix a is convergent if all the elements of a power m go to 0 as m goes to infinity and if the matrix a is diagonal it means that it's convergent if and only if all the diagonal entries in magnitude are less than one which means that all the eigenvalues of this matrix a are less than one in magnitude and this directly extends to diagonalizable matrices because a power m can be written as v lambda power m times v Hermitian where v is the main or v Hermitian lambda power m times v where v is a matrix containing the eigenvalues of this matrix a and so basically we have also seen this before that diagonalizable matrices are convergent if the magnitude of all the eigenvalues of the matrix are less than one now what we'll discuss the next time is the extension of this idea to non diagonalizable matrices also which of course we will do through the Jordan canonical form so that's it for today and we'll continue next on continue in the next class