 So, now it is time to try and prove this result. Yes, any questions? Yeah, yeah, but we do not know that. We do not know that about the degree of the minimal polynomial. As we saw with the factor with the vector e 2, you did not have to go up to the third stage, right. You just hit it with i, then you hit it with a and you lost linear independence. What if there is an operator that loses linear independence for every element of the basis at a degree less than the order, the degree of the size of the matrix, then it cannot be, right. So, that is the point, right. Yes? They have exactly the same roots. Yeah, the multiplicities will be different. We will talk about that later, like till what degree you might have to go. So, so far we have seen that the roots must be the same, right. But what if you need to repeat the root more number of times in one of them than in the other? We do not know whether a particular root appears more number of times in the carpoly or in the minpoly yet. We have not spoken about that. We have kept silent on that so far. We will later on see that, we will actually prove the Kelly Hamilton theorem without using the determinant based proof that you might find in many texts. We will see when we have done all this in a few lectures from now, that indeed the minimal polynomial is actually the, as we already proved the generator for the annihilating ideal of A and the characteristic polynomial belongs to the annihilating ideal of A. And therefore, the characteristic polynomial must vanish when you pass on the argument A in it, which is Kelly Hamilton's theorem, right. So, it actually matrix satisfies its characteristic polynomial. So, now, if it belongs to the annihilating ideal of A, then of course, so if it is divisible, then of course, got to be divisible by the minimal polynomial. So, therefore, the overall degree of the minimal polynomial is at most equal to the characteristic polynomial and no more than that. So, all of that we will prove. We will see when we get it to a very nice form like a Jordan canonical form and so on and so forth, then we will see it very transparently. The proof becomes very obvious. For that, we need a little bit more work to bring it out or to flesh it out a bit more, right, yeah, yeah. But so far we have not said that this characteristic polynomial, we have not given you any reason to believe that this characteristic polynomial belongs to the annihilating ideal. Only if you know that it belongs to the annihilating ideal, then it must be divisible by mu. Then you could have concluded. I have not told you that yet. Now of course, if independent of this, if you use the determinant based proof of Cayley Hamilton's theorem, which we are trying to avoid and then you will know that this belongs to the annihilating ideal of A and therefore, it must be divisible by mu. And then with this result, you can definitely say that the minimal polynomial cannot have a degree more than n. Then you can say, but somehow you have to invoke the Cayley Hamilton in between, but we are going to do it the opposite way. We are going to prove the Cayley Hamilton theorem later. So, we will not make any assumptions about this being the member in the annihilating ideal of A yet. They are sharing the same roots. No, but you do not know how many times they are sharing it, right. If this shares the root more number of times than it is here. So, the multiplicity of a root here is less than or equal to the multiplicity here. Then you can be sure that this fellow divides this and therefore, this belongs to the ideal. But so far we do not know if the multiplicity here is called actually the algebraic multiplicity. So, we do not know that the number of times a root appears in the minimal polynomial is strictly less than or equal to at most the algebraic multiplicity of that root in the characteristic polynomial. We have not proved that yet. So, that is why we have not given you any reason to believe that this is a member in the annihilating ideal as it turns out it will be, yeah, but that comes later. Of course, if you as I said, if you indirectly invoke Cayley Hamilton theorem from your past knowledge of determinants and the proof therein, then you can immediately conclude that. But we are going to try and avoid that determinant based argument here, right. So, what are we going to assume? Since these are two sets, equality of sets after all it is the same as like we use for subspaces because subspaces were also in the end sets themselves at the end of the day, right. So, we will show that this set is contained inside this set and also this set is contained inside this set and we shall be done. So, consider lambda belonging to, let us say, let us say we denote this by R chi A, the set. And let us say we denote this set as R mu A. So, let lambda belong to R chi A, which is to say that lambda is an eigenvalue, right. If lambda is an eigenvalue, then we must have the condition that A minus lambda I acting on V is equal to 0 for some V in V. Of course, this V is non-zero, agreed. That is the basic definition of an eigenvalue and eigenvector. So, it must satisfy this equation. If this is so, construct a basis for V given by B is equal to the first element you choose as this eigenvector, V. And then of course, you can extend it to a basis, you know that, right. So, then choose V1, V2, Vn minus 1. Now, if you want to get the minimal polynomial for A, through this basis, what should be our strategy? We have to find out mu V, mu A V, mu A V1, mu A V2, all the way till mu A Vn minus 1 and then take the LCM. What do you think is mu A V? What is mu A V? Yes, I am waiting for the answer. So, x minus lambda, right. But mu A V is equal to x minus lambda. We use a different color and use the dangerous term clearly. Why clearly? Of course, we have written that from the equation of this eigenvector. This is a polynomial, right. Matrix polynomial. If you look at the corresponding polynomial, it is x minus lambda. So, you need not go any further than I V, A V. And you have minus lambda times I V plus A times A V, 1 times A V and you have linear independence lost. Therefore, if you now look at this, what can you say that x minus lambda divides mu A or in other words, x minus lambda is a factor of mu A, yeah. Therefore, lambda belongs to R mu A. So, anything that belongs to R chi A has to also belong to R mu A. So, any root of the characteristic polynomial must also be a root of the minimal polynomial. In other words, every eigenvalue must be a root of the minimal polynomial. But in order to show this equality, we have to establish that the minimal polynomial cannot have any root other than the eigenvalues. In other words, if anything is a root of the minimal polynomial, it must necessarily also be a root of the characteristic polynomial or be an eigenvalue in other words, okay. So, that is one part of the proof, right. This essentially means that R chi A is contained in R mu A. That containment is shown. You started with something in R chi A, you saw that it belongs to R mu A. So, the set R chi A must be contained inside R mu A, okay. Now, conversely, suppose lambda, of course, lambda does not necessarily mean eigenvalue. So, we are just using the variable lambda. Lambda belongs to R mu A. This essentially means that mu, excuse me, mu A x, the polynomial is something x minus lambda times some q x, of course, where q x belongs to the ring of polynomials, all right. What do you know about the minimal polynomial? It is the smallest degree polynomial in the annihilating ideal of A. So, we can be sure that if this is our candidate for the minimal polynomial or the generating element for that annihilating ideal, can q x be a potential contender for a minimal polynomial? No, because q x has a degree that is strictly less than the degree of mu. So, if q x would identically reduce to 0 when you pass on A instead of x, then q would have been the minimal polynomial, but it is clearly not, right. You had to go up to that higher degree, right. So, clearly q of A is not identically equal to 0. Do you agree with this? If q of A had identically vanished, then that would have been the minimal polynomial because it has a degree less than or claimed minimal polynomial. And we know that minimal polynomial is the generating element and therefore, the lowest degree polynomial sitting inside the ideal, right. So, q clearly cannot be a member of the annihilating ideal. It means that, so if you have a non-zero matrix, let us think of matrices only. If you have a non-zero matrix, at least its rank is 1. Can you have a non-zero matrix whose rank is 0? Of course, not. Similarly, you cannot have a non-zero operator, which pulverizes every element, yeah. So, if it has a non-trivial rank, there exists v such that q A v is not equal to 0. Please focus on this. What I am saying is, if you fail to find any v for which q A v is not equal to 0, it means that q A pulverizes every v. If q A pulverizes every v, then q A must have been 0. But if q A is 0, then q A would have been the minimal polynomial. But we know that we had to go up to a higher degree to get our minimal polynomial. So, q A definitely, there is at least one vector. If not more, such that q A v is not equal to 0. Now, look at mu A A acting on this given vector. For this v, look at or I shouldn't say this, it gives the impression that there is only one. For such a v inside v, consider mu A acting on v. What is that equal to? That is equal to q A. Let me just write that one first. It does not matter really. The order can be flipped, right? When you are taking a univariate polynomial and you are passing on matrix argument, the order of the multiplication does not matter. So, what I am going to write is A minus lambda i times q A acting on v. But what is this? The left hand side is 0, right? Because it is a minimal polynomial. So, mu A of A is identically 0. So, therefore, 0 acting on v is 0, which means that I have A minus lambda i acting on q A v equal to 0. Let us call this fellow as sum w, which means that A minus lambda i w is equal to 0 for w not equal to 0 implying lambda is an eigenvalue. That is the eigenvalue eigenvector equation. Therefore, lambda belongs to r chi A, right? So, we started with something in mu A, r mu A, it ends up being in r chi A. Earlier, we started with something in r chi A, it ends up belonging to r mu A, thereby showing both sided containment. And from this box result here and this result I am about to box now, what we know is that r chi A is equal to r mu A. So, therefore, the roots of the minimal and the characteristic polynomial, if you discount the multiplicities are going to be 1 and the same. If there does not exist such a v, then it would mean that q A when it acts on any vector in the vector space v takes it to 0. But we know that only the 0 operator does that. But q A cannot be the 0 operator, because then it would be the, then q x would have been the minimal polynomial. But we know that the degree of the minimal polynomial is 1 greater than q by the nature of the construction. Therefore, there must be at least one vector. Yeah, yeah, that is one of the fundamental flawed proofs of the Cayley Hamilton theorem. When you take determinant, okay, yeah, that is what elucidating on a bit. Because unlike what we are doing here with matrices and the matrix polynomials, if you are talking about passing on the argument, you will see that it is like apples and oranges. Because the characteristic polynomial results from is equal to determinant x i minus a. Now, if you are passing on x as an argument, what is this x? Look at this identity structure. It is x x x x x like this. So, what you are saying in terms of this multiplication? It is something like a chronicle product, what we have in matrices. You are saying, oh, I am going to take this a a a like this. That is crazy, right? You cannot do such a thing. So, you cannot just plug in a i minus a i and then take its determinant like this. It does not even make sense, okay? So, it is one of the famous flawed proofs of Cayley Hamilton theorem that, well, you know, if you have a matrix, just replace it with a matrix a times i. But a times i is not the same as this operation we are doing here, yeah? It does not even make sense. Because you are passing on, you are operating using the determinant. It is not just a i minus a. You are using determinant. What is the determinant of something times a matrix times something? A scalar times some matrix and you take its determinant. You know that the scalar to the power n gets raised. But what do you mean? You mean by a matrix times something have then difference between two matrices. I mean, products of matrices you can still take determinants. But sums of matrices and their determinants, there is no such relation. It does not even make sense, as I said. It is a completely different operation. That is a very illegitimate proof of the Cayley Hamilton theorem. Do not even think about it. That is not how it works, okay? All right. No, no, no. I mean, you remove all the degrees by that. I mean, you are assuming that you are taking all of them as simple roots. No, that is not true. That is not true. The minimal polynomial does not unfortunately result from just taking one times every factor or first-degree factors like x minus lambda 1, x minus lambda 2. Sometimes within the minimal polynomial, you will have to take higher degrees. Yeah? Yes, yes, yes, exactly. So, we just ended up with x squared minus 2x plus 2. What is x squared minus 2x plus 1? But there is no guarantee that x plus 1 or x minus 1 would have been a minimal polynomial. In fact, x minus 1 might not have vanished, but the moment you add another, the scaled version results in a repetition. So, there is no reason to believe that just taking one copy of each will add up to my minimal polynomial. But that is a good question, because these are very deeply and intricately related to our fundamental question of when we are able to diagonalize and get the block diagonals to their smallest possible sizes. We will see the interconnections. It is quite interesting. So, next we are going to just see a very simple kind of a result, follows very straight forward. We have seen that for this chi A x is going to be equal to chi T inverse At x, right, which is to say that the characteristic polynomial is invariant under a change of basis. We use that property of determinants, the determinant of product is equal to product of determinants and we prove this. So, this was there. Now, we will see that even the minimal polynomial remains invariant under a change of basis. Why do you think that is going to be the case? Well, I mean how did we cook up the minimal polynomial? Just like we use the determinant for this. How did we cook this one up? Just think about the way it transpired. Rather, we are going to actually say a lot more than this. You see, what we are going to say is that the annihilating ideal of A or rather A is the same as the annihilating ideal of T inverse At. So, if we can manage to establish this, then the former follows from there, right. If it is the same annihilating ideal, then of course it has the same generating element, unique monic polynomial that generates the ideal. But why should this be true? So, these are sets of polynomials, remember. Everything that belongs here is a polynomial. So, anything that annihilates A, what can you say about it? So, suppose let f belong to the annihilating ideal of A, okay. It means that f of A vanishes. What about, so look at f on T inverse At. What is this action? After all, this is a polynomial. That is the one saving grace. It is not just some arbitrary function. And we have seen earlier that you take A and raise it to any power, okay. This is what? So, T inverse this is equal to T inverse At, T inverse At, like so, T inverse At. Because the fellows inside get pulverized. So, if I call this T inverse At at some A bar, this is what? A bar to the r is for A bar defined as T inverse At, right. So, what is this going to be equal to? Is this law going to be the same as T inverse pulled out from the left and f A from here and T pulled out from the right? We do not even need to go both ways. You start from f belonging to the annihilating ideal of T inverse At. It is the same, just the same. So, I am not even going to write that. So, of course, when this fellow is 0, this fellow is 0. Therefore, any member f that belongs to the annihilating ideal of this, yeah. Therefore, f belongs to the annihilating ideal of T inverse At. So, similarly, let G belong to and I leave it to you as an exercise to complete this T inverse At. If it has got to be 0, so if this fellow is 0, this is non-singular. If this is 0, you can hit the 0 with T on the left and T inverse on the right, because this is invertible. So, you will be led to concluding that G, I mean, I am leaving three dots for you to complete those steps and prove that G ends up belonging to the annihilating ideal of A. Therefore, the two annihilating ideals must be equal. So, this is a proof. I have left just three lines, three or four lines here for you to write, which means that the annihilating ideals are the same. If the annihilating ideals are the same, we are after all talking about the same object and the polynomial does not care about the change of basis then. So, whatever is the generating element of this is the same as the generating element of this. So, this then follows naturally. So, change of basis matters not, right? It is going to be very critical, because we are going to make choices of basis that endow the matrix or the operator with some beautiful structure, right? Which one? This part? This. So, if F A is 0, then this is 0. If this is 0, then T inverse 0 T is also 0. Therefore, F acting on T inverse A T is also 0. Therefore, F is an annihilating polynomial, right? If annihilates T inverse A T. Therefore, F belongs to the annihilating ideal of T inverse A T. But we started with an F that belongs to the annihilating ideal of A and it belongs to the annihilating ideal of T inverse A T. Therefore, we can say that the annihilating ideal of A is contained inside the annihilating ideal of T inverse A T. Then you start from the reverse direction and show that G that belongs to T inverse A T is annihilating ideal also answer belong to the annihilating ideal of A. So, therefore, anything that belongs here must belong here anything that belongs here must belong here therefore, those two sets must be one and the same yes sorry where we took this T out because after all F A is of what does it look like F A has a structure that looks something like this right F A because it is a polynomial it is a matrix polynomial. So, this has the form alpha i A to the i i going from 1 through some let us say some m. So, every individual term yeah every individual term is some A to the power r right. So, we have seen that this T inverse A T T inverse A T T inverse A T. So, this T inverse A T to the power r is nothing but T inverse A to the r T right that is why we could pull it out because when we are representing it in terms of the powers of T inverse A T which we call let us say A bar here as we have said here yeah. So, the powers of T inverse A T is nothing but T inverse times that same power of A times T. So, eventually from the whole polynomial term by term we are filtering out from alpha 0 i we are filtering out T inverse from the left T from the right from alpha 0 A i we are filtering out T inverse from the left T inverse from the right from alpha 0 A bar to the power 5 we can also filter out T inverse from the left and T from the right. So, the general polynomial has the same form as this right ok. So, this then shows that it is invariant to this change of basis it does not care about what basis we are choosing and this is good because now we can just deal with it like basis free give me any operator in terms of any basis I do not care whatever minimal polynomial I have once obtained in terms of some basis your friend is going to obtain the same minimal polynomial in terms of any other basis that he or she may have chosen ok just like the characteristic polynomial the minimal polynomial also remains invariant completely alright ok. So, what is the next thing we are going to do and what is the motivation for all this study again to remind you we are going to try and get this thing down to certain A invariant subspaces we want to get the basis for the whole vector space V in terms of the basis for certain A invariant subspaces and if you can write the whole vector space V as a direct sum of certain A invariant subspaces and not just that if those A invariant subspaces happen to be the smallest possible A invariant subspaces you can think of that is the best possible outcome we have when we had diagonalization we had precisely that each of those eigenvectors we had sufficient number of eigenvectors and each of those eigenvectors were the basis for the one-dimensional eigen spaces those were A invariant subspaces. So, we had this entire diagonal structure short of that when we cannot diagonalize we want to block diagonalize. So, that is the key now that we have this minimal polynomial we are going to try and harness this properties of this minimal polynomial to push towards that same goal can we somehow split up the minimal polynomial in some special ways so that the kernels of these factors of the minimal polynomials somehow allow us to cook up the best possible A invariant subspaces and then from thence we construct this diagonal structure for our operator or the block diagonal structure for our operator that is our goal. So, with that in mind we will next try and investigate how we can split up the minimal polynomial and towards that end we will have to encounter just like we encountered the LCM of polynomials we will now have to define the GCD the greatest common divisor or the HCF whichever you call it whatever you learnt in school like really middle school right highest common factor or the greatest common divisor that is what we call it and we will have to define that object which is what we are going to do in the next module.