 So, the first part of this proof will entail us showing that for this fellow that I have here right it must be identically the 0 map which means what that if it acts on any vector in the vector space it must pulverize it yeah. So, for any V i we have what do we have mu a of a acting on V i a very careful attention to this is equal to mu a V i of a times some q i of a acting on V i check if you agree with this if you agree with this it is done almost this mu a is a common multiple of each of the mu a V i's therefore it must be representable as mu a V i times some quotient q i of course the remainder is 0 right. So, this object I have put an under brace on are identical you agree now that is the case this is equal to of course I can play around with the order as before polynomials of matrix a the order is a material a times V i is equal to q i a on 0 which is nothing but 0. So, what have I now seen ok hang on hang on hang on hang on it is a bit of nasty notation, but let me for a good enough reason let me just call this psi, psi a because we are we cannot up front assume that the LCM is equal to the minimal polynomial right. So, we should use a different notation for the LCM. So, let us define this psi as the LCM of mu a V 1 until mu a V n you realize. So, that is what I am doing here we realize why I made that correction I hope it is not confusing. Because here this is a symbol I have reserved for the minimal polynomial I cannot just say probably use that symbol for minimal polynomial for this right. So, I should I should use a different symbol for the psi which is the LCM. So, then what what can I say about this psi a acting on V i is equal to 0 for all i. So, if I know that the psi a hits every basis and sends it to 0 what does it do to any arbitrary vector this implies that psi a acting on i going from 1 through n alpha i or maybe I should use beta in this case just to distinguish from the alphas we used earlier. This is going to be also 0 because each term individually is 0 right which means that psi x belongs to the annihilating ideal of a of course, it must be identically the 0 operator otherwise it cannot send any arbitrary vector to 0 which means it is sitting inside the annihilating ideal wherein I know the minimal polynomial is also sitting in and it is the smallest degree harmonic polynomial sitting inside the annihilating ideal. Therefore, any object any other object sitting inside the annihilating ideal must be divisible by the minimal polynomial. So, I must have that mu a divides I should not use psi sorry sorry about that confusion yeah I should just write it as psi only when I am passing this because psi is after all a polynomial yeah sorry about that yeah I know a bit of confusion with the notation here, but I hope you get the message yeah because I want the subscript really for the operator and the vector the argument should be the x any questions I realize I messed up the notation a bit. So, if you have any questions please clarify it is ok right I can put a subscript if I want, but I am not putting it for the time being just call it psi it does not matter right I am just saying that you have these that are defined by a and v i's and you take the L same of these fellows just call it psi. So, this is psi of x it is a polynomial inside the polynomial ring we have not yet proven it is mu of a from this we have just shown that mu of a definitely divides psi you can give a subscript a or you may not it is a matter of taste, but you should not give it like the argument a that is only reserved when you are passing on the argument of the a as the argument instead of x. So, it is just a bit of that confusion, but I hope it is cleared up yeah. So, we have at least at least shown that mu of a divides psi now what should we do in the next part to prove that mu of a is equal to psi we should also show that psi divides mu of a ok. So, when this fail to be the case that psi fails to divide mu of a what is psi psi is by definition the least common multiple of these fellows. Now, if each of these fellows divide mu then can psi fail to divide mu it cannot right. So, the only way that psi can fail to divide the minimal polynomial mu is if at least one of these fellows mu a v 1 mu a v 2 or till mu a v n fails to divide mu a. So, let us investigate the possibility that one of these fellows mu a v 1 mu a v 2 till mu a v n fails to divide mu a and if you can rule the possibility out then we would have shown that psi also divides mu a right. So, suppose for some i we have mu a x is equal to mu a v i x times q x plus r x with mu a v r x not equal to 0. If this is the only way this is the only way in which because in each of these individual fellows divide this yeah if each of these individual fellows divides this then the least common multiple must also divide this because the least common multiple must contain at least those fellows in some factored form or the other just think of numbers whenever you run into such doubts and think of numbers and the LCMs of integers right. So, when can a number be not divisible by the LCM of a bunch of numbers? At least one of the fellows in that factorization in that LCM fails to be a factor of that number that is the only way. So, suppose we are positing that this is true then what will be the case? Let us see the action of mu a x sorry mu a a. So, look at mu a a acting on v i what is this? This is equal to mu a v i a times q a acting on v i plus r on a acting on v i right. What can you say of this? Of course, this fellow vanishes mu a is supposed to be the ideal is 0 operator identically 0 operator. So, this is 0 the 0 vector what about on the right hand side? What about this? You can flip the order and at least mu a v i definitely pulverizes v i that is the way we have cooked up mu a v i haven't we. So, this term is also going to be 0 plus r a v i which means that r of a acting on v i pulverizes it, but is that a contradiction or not? I am saying we already have a contradiction. What do you think I am saying it is a contradiction? Exactly you see the way I have written it out here this is the divisor this is the quotient this is the remainder right. So, this is sorry this is the dividend this is the divisor. So, a degree of this fellow must be less than this which means that this fellow then becomes yeah right. So, this fellow will then pulverize v i right. So, this is the contradiction right. So, the argument is this that if this is true then r a would divide every element inside right. So, the r a would be the generated for the annihilating ideal of v i or annihilating ideal of a with respect to v i, but that cannot be denied because we have already claimed that this fellow mu a v i is that unique monic element which sort of sits as the annihilating ideal and I mean the generator for the annihilating ideal of a with respect to v i. So, it is a contradiction. So, this cannot be true and because this cannot be true. So, we must have the fact that each of these mu a v i's must divide mu a yeah. If they do divide mu a then there is no way around the fact that what what did we prove earlier where yeah mu a divides psi. So, therefore, now psi should also divide mu a and therefore, indeed has claimed mu a can be obtained through this procedure, following procedure. Take a basis look at that polynomial which pulverizes or rather in whose kernel that particular member of the basis is sitting. How do you go about seeking the polynomial raise a to different powers higher powers until you come up with a set that becomes linearly independent linearly dependent. So, first you take a basis element v 1 look at v 1 a v 1 a squared v 1 a cubed v 1 at most you will have to go up to a to the n v 1, but you may not have to go that far alright. So, at some point you will find a linear dependence that point find the manner of linear dependence that is the linear combination of those fellows which leads to 0 translate that to a polynomial that becomes the minimal polynomial of the annihilating ideal of a with respect to v i. Once you have obtained these n polynomials corresponding to the n elements in the basis one for each of the elements in the basis take their least common multiple that gives you the minimal polynomial for the linear operator a right. Any questions about the steps I am willing to redo that again if you have any doubts? Yeah, because it is the I have already said the mu a by definition is the unique monic generator for the annihilating ideal of a and we know that the polynomials is a principal ideal domain the ring of polynomials the integral domain of polynomials is a principal ideal domain. In a principal ideal domain every ideal can be generated by a single element. Now, in case of polynomials if you do not restrict the search to just monic polynomials you can have multiple elements that are each serving as a generated for that ideal, but if you restrict it now as we have shown that if you restrict it now to a search over monic polynomials then that generating element becomes a unique element. So, this is all true of any polynomial ideal now we are casting our attention towards the special ideal of polynomials which is the annihilating ideal of a the way we have defined. So, because this is an ideal of the ring of or the ring of polynomials or the integral domain of polynomials it must have a unique monic generator and by definition itself that unique monic generator is nothing but the minimal polynomial yeah. So, it must be unique once you found it and now we have told you a constructive way of finding it. So, if you found it then it is you cannot help but be unique yeah. As for the fact which may not have received too much attention as to why each of these must be the unique generator because you cannot have. So, when you are see it is inherent in the construction maybe I have not highlighted that. If you are fixing up V1 and you are hitting it with a successively, where are we actually stopping? The first instance when we have a linearly dependent set. So, if you take any other polynomial of lower degree than that polynomial and if your polynomial indeed turns out to have Vi in its kernel then that would have been giving you the linearly dependent set right. So, you cannot do with a polynomial of degree smaller than this do you understand? I thought that might have been obvious but maybe now I feel like may not have been clarified the way we constructed it I acting on V, A acting on V, A squared acting on V and at every step when you are augmenting it by one additional action of A you go ahead and check is this linearly independent, is this linearly independent. The point at which it fails to be linearly independent is where you stop. Now, if you are saying no no I have stopped after raising A to fifth power but perhaps the minimal polynomial for the annihilating ideal of A with respect to V was of fourth or degree 4. That means, you should have had some I, A, A squared, A cubed and A to the power 4 acting on V giving you a linearly independent linearly dependent set but clearly that was not possible the way we went about with this construction. So, in the next lecture what I shall do is I shall take a small 3 by 3 example and you will begin to see how with numbers with actual numbers I will take a simple example and we will see how constructively you can actually obtain the minimal polynomial and then there will be absolutely no doubt that the way we have constructed this mu A V i that has to be the smallest degree polynomial sitting inside the annihilating ideal of A with respect to V i and if it is the smallest degree polynomial, smallest degree monic polynomial it is D generator. So, it is the minimal polynomial of A with respect to V i. Any questions? Otherwise we are done for today, thank you.