 So, we have thus far seen the following result that if you have this operator from V to itself and suppose the minimal polynomial of this operator is split up into two co-prime factors such as then what did we see? We saw that if you have a very special representation in terms of what? Suppose you look at okay let us not make the claim right away. Let us just go through the steps. Suppose you have kernel of ta and kernel of qa here. So, if you obtain bases for these two then what do you have? What can you do with those bases? You can concatenate those two bases to get a basis for V itself, the entire vector space V right. That is only provided you have split it up into co-prime factors that is very important because we use that areyabhattabazoo identity to show that there is indeed a direct sum. So, that is we saw that V is then a direct sum of this and this. We have proved this result in the previous lecture right. What is even more interesting is that both of these happen to be a invariant subspaces right. So, if we say that consider the basis for kernel ta union the basis for kernel qa and define this to be a basis for V then corresponding to this representation what will we get? We will get the a represented under this basis that look like a1100a22 yeah. This is based on whatever work we have done earlier we have seen that if you have an a invariant subspace and if you represent an operator in terms of such a basis then of course the fellows that do not belong to that a invariant subspace because if you hit a fellow in the a invariant subspace with the operator a then the resultant will also live inside the a invariant subspace and after all what is this? If I were to use a different color I would just say this is nothing but the action of a on each individual element of this basis. So, suppose we define this basis to be V1 V2 till Vn then this is nothing but a V1 represented in terms of the basis BV then a V2 represented in terms of the basis BV so on till a Vn represented in terms of the same basis and because we know that each of these will only involve the representation of each of these the action of a on V1 is represented only in terms of the basis elements of the first one similarly for the last few. So, this has already split up right the first few of them form a basis for kernel PA the remaining vectors form a basis for kernel QA that is in this set right. So, it is an ordered basis. So, subject to that ordering these fellows are nothing but the representations of the first few fellows the action of a on the first few fellows which happen to be basis for kernel PA the last few are the representations of the action of a on kernel QA the basis elements of kernel QA and therefore, it will look like this. So, now there are two interesting things that you have to observe to keep this going first that now that this has become a matrix ok. So, here is the first observation now that this has become a matrix if I raise this matrix. So, I am dropping the subscript now if I raise this matrix to some arbitrary power in terms of the operator it is the action of the operator art times composition, but in terms of matrix it is just multiplication what do you think this is going to look like just the individual block diagonals going to be raised to the same power nothing else right. So, this is going to look like a 1 1 to the r 0 0 a 2 2 to the r. Therefore, if I want to cook up any polynomial such as summation alpha i i going from 0 to some k hat let us say a to the i of course, by a I mean the matrix what is that going to look like I will just write out the steps in two steps that is I could have written it in one step itself. So, what will this look like this is nothing but summation the same alphas I going from 0 to k hat alpha I now this is just a 1 1 the matrix a 1 1 to the I this is 0 this is 0 and this is also I going from 0 to k hat alpha I a 2 to to the I right any doubts about this so far we just raising a to different powers that entails raising the block diagonals to those same powers then adding them conformally. So, that is just addition of those block diagonals only and nothing else nothing else changes. Now, what is this form is not this the typical form of a matrix polynomial yeah. So, can I not say that if you take any polynomial and pass on the matrix as its argument then what you will essentially have is that same polynomial in terms of the small block diagonal matrices which is to say that if I now say that f of a is basically a matrix polynomial that is nothing but the same f on a 1 1 0 0 f on a 2 to because that is after on what this is my k hat is arbitrary. So, by choosing different arbitrary k hats and different choices of alpha I's I can get any polynomial. So, essentially for any arbitrary polynomial. So, for f belonging to the ring of polynomials right not for any arbitrary function of course, it is just polynomials because for polynomials we know this to be true right any questions so far this is clear ok. So, let us try and investigate what happens when we make that special choice of polynomial as one of these factors that is p ok. So, consider p of a remember what p is. So, I am omitting the box yes ok it is not visible ok perhaps this will be. So, it is visible ok. So, consider so again I am omitting the box it is just too cumbersome, but you understand what this means ok. It is just the matrix I have in mind where of course, this p is nothing but the minimal polynomials factorization into two co-prime factors. So, one of those co-prime factors I have picked out let us say p a what is this is going to look like on the one hand I know this p a is going to look like a 1 1 the action of p on a 1 1 0 0 the action of p on a 2 2 right straight forward from our build up. So, far further if I now want to look at this ok maybe yes let us retain this right maybe just it is a good idea. So, because it is a matrix and I am going to use another way of looking at the same object which is p a represented in terms of that basis let us say again even though I tried to cut down on these notations they somehow keep chasing me. So, this is so what is this again from very fundamental principles this is nothing but p a is action on the first fellow in the basis with that coordinate representation followed by p a is action on the second fellow until p a is action on the last fellow, but what do I know about the first few fellows this p a v 1 p a v 2 what can you say about them where does v 1 belong kernel of p a. So, this must identically be 0 right. So, at least these first ok wherever these first few blocks there must be 0 is it not please see if you understand this is very critical. So, a 1 1 is just the restricted operator acting on the kernel of p a yeah. So, a 1 1 is the action of a on the kernel of p a and its representation in terms of not the entire basis because that would entail considering the whole thing, but on the basis for the kernel of p a that is what a 1 is a 1 1 is. So, it is the action of the operator restricted to the kernel of p a that is what a 1 1 is. So, subject to this choice of basis we get this block diagonalization that is essentially the crux. We have defined this earlier in the previous lecture also if you recall that is how we called it a 1 1 this part is very important because otherwise if you did not do this if you had chosen b v instead then this entire thing would have had to be considered because that means you are using the representation in terms of the basis of v. So, each of them have to be n tuples, but a 1 1 is not n n n cross n right it is some r cross r. So, this is some r cross r this is some n minus r cross n minus r yeah a is an operator from v to v yeah that is what I am representing in terms of a smaller basis that is how I am cutting down on the size if I. So, if I had not done this instead then I would have gotten this entire padding with 0 as well. So, I would have gotten this a 1 1 padded with this 0 is a representation of a restricted to p kernel of p a represented in terms of b v, but I am not interested in this padding with 0 I am just interested in this r cross r. So, I am only going to represent it in terms of the basis elements of kernel p a not all the basis elements. If I did represent through all the basis elements I would have had to consider a rectangular matrix like so, because then I would be picking up objects in a smaller vector space smaller dimensional vector space which is r dimensional that is a kernel and they would have been mapped to a bigger vector space, but really that bigger vector space is after all of no consequence because this is an a invariant subspace. So, I can just focus my attention on the smaller dimensional subspace itself after hitting a vector in the kernel of p a I am not getting anywhere outside kernel p a because kernel p a is a invariant the same thing here you can also write yeah exactly otherwise I would know this might escape the kernel the subspace from which I am picking an object up it is not guaranteed that after the action of the operator it is still going to live inside it it is only going to live inside it if it is an a invariant subspace. So, by the same token in fact, this is also the action of a restricted to kernel q a and represented in terms of the basis of kernel q a and because of the direct sub I know that these two dimensions would add up to the dimension of v right ok. But now this is one representation of course is one representation we have gotten from this right please follow the line of reasoning very closely and do ask if you have any doubts at any point it is very important it is not very difficult to see, but if you lose the thread in one step it might be difficult to follow the subsequent steps right. So, what is happening here one representation tells me it must be like this p a 1 1 0 0 p a 2 2 from a second perspective I see that any operator now this p of a is also an operator. So, any operator represented in terms of a basis linear operator is completely characterized by what it does to every element in a basis. Now, the first r elements of this basis also form a basis for kernel p a it is not just the first r elements of the basis for v, but it is also all the basis elements of kernel p a and if any object in kernel p a is operated on by p a it must take it to 0. So, the first r fellows must vanish. So, in fact this must look like 0 0 0 p a 2 2 wouldn't you agree yeah which means comparing these two I might now conclude that p a 1 1 must be 0 yeah please take time to absorb this it is very important yeah because we are building up towards the Kelly Hamilton theorem shortly ok. What does this tell you from this step can I not conclude that p belongs to the annihilating ideal of a 1 1 yeah because obviously that is the definition any object any polynomial in which if you pass the argument of the matrix or the operator and it takes it to 0 then such a polynomial must be a member of the annihilating ideal of that particular operator therefore, the minimal polynomial of the annihilating ideal that is the generating polynomial of the annihilating ideal there is a minimal polynomial of a 1 1 which is this it must divide p right that is the definition the minimal polynomial is the generator of the annihilating ideal by definition. So, every other polynomial in the annihilating ideal must be a multiple yeah it has to be generated by the minimal polynomial that is 1 ok. Next consider f x is equal to mu a 1 1 x times q x ok just another polynomial just bear with me for a while what am I doing if I had kept p x here this would have just been the minimal polynomial for a, but now I am just changing this one bit here ok just consider this just indulge me for a while. Now f of a is what exactly this is going to be mu a 1 1 a and q a agreed right what is this going to be like mu a 1 1 acting on a 1 1 0 0 mu a 1 1 acting on a 2 2 times q acting on a 1 1 0 0 what is q acting on a 2 2 by very similar arguments as this if p acting on a 1 1 vanishes by exactly the same line of reasoning should not q a 2 2 also be 0. So, it is very important observation let me box it with the different color because I am not writing it out separately, but you should still understand that this is a very important observation it is exactly for the same reason as we argued that this fellow must be 0 that p a 1 1 must be 0 let me write it here similarly q a 2 2 is 0 and that is what I am using here. So, I have written that with pink any doubts clear ok what about mu a 1 1 a 1 1 has to be it is the minimal polynomial of a 1 1. So, this must also be 0. So, if you multiply these two what happens just carry out the usual multiplication like 2 by 2 matrices it is just that it is a block what happens can you check this acting on this is 0 this acting on this obviously is 0 this acting on this is 0 and this acting on this again 0. So, it is just 0 sorry so far yeah what does this mean that f x belongs to the annihilating ideal of a right. Therefore, mu divides f what is mu p q right. Therefore, p q divides what is f mu a 1 1 q right. So, mu a 1 1 q, but q is common. So, let us get rid of q what does it mean that p divides mu a 1 1 what did I have earlier combining these what can we say mu a 1 1 is obviously, monic p is also a factor of the original minimal polynomial for a. So, it is also monic right. So, therefore, based on these two mu a 1 1 is equal to p and similarly mu a 2 2 is equal to q isn't it, but the same line of reasoning is just that I am biased towards 1 1 then towards 2 2 if I wanted to be biased towards 2 2 I would have started with q and all those proofs, but it is fine. So, what is this telling us eventually you have a minimal polynomial you have split it up into two co-prime factors, but there is nothing special about 2. You can go ahead and split it up into as many co-prime factors as you like the largest possible number of co-prime factors is equal to the number of number of Eigen values are you sure or something missing there in that statement number of distinct Eigen values right. So, for each distinct Eigen value what do you have you have one co-prime factor. So, you stack them up together. So, I hope I can erase this part and here is the result that is suppose a is an operator from v to itself mu a 1 1 is equal to mu a 1 1 is equal to x is equal to ok let me just first write the characteristic polynomial ok. So, the characteristic polynomial of a x it is also a monic polynomial we know that. So, this is product of x minus lambda i i going from 1 through k and each x minus lambda i raise to the power d i where d i is the algebraic multiplicity of lambda i yeah. So, that summation d i is equal to n no doubts about this what can be a minimal polynomial for a we know that the roots of the minimal polynomial and the characteristic polynomial must be exactly the same except for the multiplicities. So, we definitely know i is equal to 1 to k x minus lambda i the only thing that can vary is this power 2 which it is raised nothing else is different between the minimal and a characteristic polynomial we prove this any root of minimal polynomial is an Eigen value any Eigen value is a root of a minimal polynomial root of the minimal polynomial right excuse me. So, what should I call this let us say this is F i right right clear so far now what can we say beyond this point based on what we have just understood can I not split up this a according to a choice of basis. So, that this looks like a 1 1 a 2 2 until a k k which choice precisely I consider p 1 x is equal to x minus lambda 1 to the power F 1 p 2 x is equal to x minus lambda 2 to the power F 2 so on till the p k x is equal to x minus lambda k to the power F k and then I look at the kernels right so I side by side look at the kernel of p 1 a the kernel of p 2 a and then the kernel of p k a and from this I get a basis for kernel p 1 a from this I get a basis for kernel p 2 a this until finally I get a basis for kernel p k a and if I stack these together if I take the union I am guaranteed to get a basis for v here. So, this b v is nothing but the union of the basis for the kernel of p i a i going from 1 through k right. So, a constructive method I have a characteristic polynomial a minimal polynomial of course so far we have not used the characteristic polynomial we will see shortly what its significance is. So, the moment I have an operator I get its minimal polynomial the moment I get its minimal polynomial I just split it up into factors like so at these lambdas are distinct right. So, corresponding to distinct I get values I get such factors such monic factors the moment I get such monic factors I can look at these vector spaces the moment I get these vector spaces since these are finite dimensional I can cook up a basis for them right. Once I have a basis for each of these I stack them up all to stack them up together all of them and I have a basis for entire v that is exactly what our work so far has guaranteed that we will get ok. Now, look at the individual blocks like a i i what do you think is the size of a i i very very important question what is the size of this fellow n cross n. So, the sizes of these fellows must sum to n this is where our minimal polynomial says nothing about. However, what are the Eigen values of a 1 1 we know that the Eigen values are exactly the roots of the minimal polynomial this has only repeated roots there is no other Eigen value possible apart from lambda 1 similarly this fellow can have no Eigen value other than lambda 2. Look at that beautiful result we had derived earlier remember neither the characteristic polynomial nor the minimal polynomial depend on a basis. So, it does not matter which basis you are looking at it from I am given any operator I have just conveniently casted in this form subject to this particular choice of basis, but the choice of basis is immaterial characteristic and minimal polynomials are both invariant to choices of basis. So, this is a basis under which it very transparently reveals everything, but in any other basis also after all the minimal polynomial and the characteristic polynomial would still be these two fellows which means exactly what. If you just straight away apply determinant to this what do you think is going to be like lambda i minus a the determinant thereof it must also be equal to this, but x minus lambda i factors can they come from anywhere other than the i ith block. So, how many lambda 1's must a 1 1 generate d 1 right because no other fellow if you take lambda minus a lambda i minus a then this is just lambda i of appropriate size minus a 1 1 lambda i of appropriate size minus a 2 2 so on and because this is a block diagonal structure the determinant is nothing, but the products of these block diagonals. So, no fellow other than a 1 1 can generate a factor in the characteristic polynomial such as x minus lambda 1. So, this generates all the lambda 1 that you need which is exactly d i copies d 1 copies this generates all the lambda 2's that you need which is d 2 copies this generates all the lambda k's that you need which is d k copies. So, what must be their sizes have to be d 1 d 2 d 3 and indeed they sum up to be this right. So, this then justifies that a i i belongs to d i cross d i right. Therefore, the characteristic polynomial of a i i is equal to x minus lambda i if I it is the other way round because I know that this must be true. Therefore, this must be of size d i cross d i and the minimal polynomial of a i i is what based on our work just recently derived yeah it is x minus lambda i to the power f i ok. So, with this observation we will bring this particular module to a close in the next module we will immediately see why indeed f i can be no bigger than d i which by my claim then would mean that the Cayley Hamilton's theorem is true ok that is what we shall see in this next module. But if you have any questions on this module so far please ask I hope this part is clear right take your time to absorb this and see if you understand this completely the arguments that we have presented as to why each of these must be of size d i cross d i d 1 cross d 1 d 2 cross d 2 d k cross d k so on and so forth that is clear yeah and also about these why their individual characteristic and monic polynomials should look like this ok ok. So, we will move over now to the next module.