 So, we will define the GCD of polynomials. We will take 2, but you can easily generalize it to multiple polynomials in the same manner, all right. So, the GCD or you can write the H, C, F whatever you are comfortable with, okay. So, suppose F1 and F2 belong to the ring of polynomials, then D in Fx is the GCD of F1 and F2 if it satisfies the following properties. What is it? What should be it? Fi divides, sorry, not Fi, D divides Fi. Of course, i is equal to 1 to 2 in this case. Again, you can generalize it. We are just considering 2 polynomials here. So, D divides this. So, that is just a common divisor. But if it has to be a greatest common divisor, then it must additionally satisfy some other property. What is that other property? If H divides Fi for i is equal to 1 to 2, then H also divides D. So, that is the definition of the GCD of polynomials. Now, why are we interested in this? Because it turns out that the GCD of two polynomials much like numbers happens to be unity. Then what do we call such polynomials? Co-prime, fantastic. So, if GCD F1, F2 is equal to 1, then F1 and F2 are co-prime. Sorry, which one? So, I am not here talking about the uniqueness. If you want to make it unique, you can make it the monic. But yeah, because we are not interested in the scalings as such. As I said, when we talk about just scalers being the generators and all it is just the whole ring itself. So, if you want uniqueness, we will impose that it is monic. This is just the general definition. You can have multiple GCDs as your friend suggested by using different scaling factors of the same divisor. So, why is this interesting to us? Because we are going to take a close look at the minimal polynomial and eventually we are going to try and split it up into its co-prime factors. And that is where our interest in this object called GCDs lies. But before that, we have to use this Bezou's identity that we have proved for integers and show that it holds for polynomials as well. So, what is the Bezou's identity version for polynomials? In fact, sometimes also called the Aria Bhatta Bezou identity. So, this time we are restricting ourselves to just two polynomials with GCD of F1, F2 being given by D. Of course, then D also belongs to the ring of polynomials, goes without saying. This is not numbers that we are dealing with anymore. So, these are polynomials themselves too. Then, there exists AB also in the ring of polynomials such that what can you write? If you remember Bezou's identity, what was it? A F1 plus B F2 is equal to D. Can you suggest a way in which I will go lightly on this proof because we have already invested time behind this same proof for integers? It is only one catch or rather one difference, no catch really. So, this is easier because we do not have to deal with negative numbers. Unlike for integers where we dealt with both negative and positive. So, we had to refine this set. Here, we are just going to look at positive degree polynomials. So, the proof will rely on the following construction that is consider S, set S of polynomials given by A F1 plus B F2 where of course F1, F2 are given. So, if you like, you can just write F1, F2 here such that AB belong to. The claim is that this is an ideal. Is it an ideal? So, S F1, F2 is an ideal of univariate polynomials. Univariate means just a single variable X. Is it? You add any two fellows A1, F1 plus B1, F2 and A2, F1 plus B2, F2. It is just A1 plus A2 times F1 plus B1 plus B2 times F2. So, the sum definitely lives inside this set. What about the product of arbitrary fellows inside the ring with this? So, A F1 plus B F2 times G is just A G times F1 plus B G times F2. So, call A G as your A hat, B G as your B hat, then it is just A hat F1 plus B hat F2. So, this is an ideal. I am not even going ahead and proving this. If you want to fill in the blanks, you can just check. Is this clear that why this is an ideal? So, I am not writing this because we have already discussed ideals of polynomials in some length. So, if this is an ideal, what can we say? It is a polynomial ideal. What can you say about polynomial ideal? We have just beaten this point to death almost. So, it is a principle because the ring of polynomials is a principle ideal domain. This ideal must also be a principle ideal. So, it must be generated by a single element or a single element. If you want this single element, then you have to choose again the eumonic polynomial. So, consider D belonging to S F1 F2 such that S F1 F2 is equal to the ideal generated by D. The claim is D is the GCD of F1 F2. Why should this be true? Just cast your attention back if you are using the same exercise book. I doubt it. You might have run out of pages. If you are using the same exercise book where we proved the Bezu identity for integers, how do we proceed towards showing or what did we do for integers to show that this is indeed the GCD? So, this is because we are talking about ideals. Now, that time we had not defined ideals. So, now by our very definition, isn't it the same that we are doing here? This fellow generates every fellow inside this. Do you have any doubt that this is a non-empty ideal? If you have any doubts, just choose a is equal to 0, b is equal to 1 or a is equal to 1, b is equal to 0. So, F1 and F2 themselves definitely belong to this ideal. So, sub-proof may be of this claim that F1 and F2 definitely belong to S F1 F2. So, this ideal is definitely non-empty. So, what does this immediately tell you? That F1 and F2 belong to S F1 F2. What must be true? This is an ideal generated by D. This means, why is this true? Because choose a is equal to 0, b is equal to 1 and b is equal to 0, a is equal to 1. So, that is the hint, how you see that F1 and F2 definitely belong to this. If F1 and F2 definitely belong to this and this is an ideal that is generated by D, then F1 is equal to D times some Q1 and F2 is equal to D times some Q2. But Q1 and Q2 are just some polynomials in the polynomial ring. No, it is a ring of polynomials of course. That is the ring, the ring of polynomials over the field. Yeah, of course, of course, everything is dwelling inside. This is an ideal in the ring of polynomials. So, it is a polynomial ideal. That is why we are able to say that it is the principal ideal domain that we are living inside. And therefore, there is a unique monic polynomial or there is a polynomial of the lowest degree in that sitting inside that ideal which generates the entire ideal. So, therefore, this is definitely true. So, therefore, D, yeah, not necessarily in the ideal because the ideal is generated by the single element means you take that element from the ideal and multiply arbitrary fellows in the ring with that element that generating element and that also must live inside the ideal, right. D should be in the ideal. D should be in the ideal, but D is by my definition I am choosing. I have already claimed that this is an ideal. If it is an ideal, it is a principal ideal domain, the ring of polynomials. If it is a principal ideal domain, then every ideal has a unique monic polynomial that generates it or if you get rid of the uniqueness, you do not care about it then at least there is a polynomial of the lowest degree sitting inside it, lowest degree not of 0, ok, which generates it. So, if this generates it then it can be written in this form where q 1 and q 2, ok. Let me just write it for completeness q 1 and q 2 come from the ring of polynomials. So, therefore, this means that D is a common divisor at least, agreed? That D is a common divisor at least, ok, ok. What do I have to show next? The next part is that it is the greatest common divisor. So, suppose H belonging to f x is another common divisor f 1 and f 2 implies f 1 is equal to some, ok, let us say p times H or rather p 1 times H and f 2 is equal to p 2 times H with p 1 and p 2 sitting inside the ring of polynomials, right. Now, what do we know about D apart from the fact that it is a common divisor? D comes from this ideal. So, there must exist some A and B such that D is equal to A f 1 plus B f 2, ok. Now, since D by my choice because it is the generating element of the ideal. So, D belongs to S f 1 f 2, there exists A B in the ring of polynomials such that D is equal to A f 1 plus B f 2, no doubts about this. You see if you go back and reflect on what we have done for integers, you will see the uncanny resemblance, ok, but repetition does not hurt. It reinforces learning is what I am told. So, let us try that, we will see that in the exams if it has helped. So, A now what is f 1? I am going to replace this with p 1 H plus B p 2 H which is nothing but A p 1 plus B p 2 times H. What does that mean? H divides D. So, the common divisor you have found from this ideal as the generating element of this ideal is indeed going to be the greatest common divisor, all right. So, this is the proof of the Aray-Bhatta-Bazoo identity that D is not just the common divisor, let us just write another line. D is the GCD of f 1 and f 2, any doubts on this sofa? Is this clear? We have just replicated the steps for GCD of integers that we had done earlier, right. So, now since we know what the common, the greatest common divisor is or how it is representable and we know that when two polynomials are co-prime, then the greatest common divisor is unity, we will now consider the case. So, consider mu A x written as p x times q x such that p x and q x are co-prime, all right. Because we want to this, no we are not yet proving anything, this is just a premise I am just setting up something. So, this is a new construction, this we have proved, ok. So, we are going to just use the fact that the GCD is representable as A f 1 plus B f 2 and then if two polynomials are co-prime like we are positing here, then we will say that there exists A and B such that A p plus B q is equal to 1. Because if these are co-prime, then the greatest common divisor is 1, right. So, that is exactly what my next sentence is going to be. Then there exists A x comma B x in the ring of polynomials such that again I am dropping the x for brevity, but you understand what it means A p plus B q is equal to 1, right. So, let us say we start with the minimal polynomial, we split up into two factors such that there is nothing in common between these factors, ok. And then we will have the situation that there will we will always be able to find some A and B because of the Bezos identity we have just proved such that A p plus B q is equal to 1, ok. This also means does it not that A A times p A plus B A times q A is equal to identity because now it is an operator, ok. What do we know about kernels of polynomials A invariant great? Recall that any polynomial no special polynomial as such, right. Recall that kernel of p A or kernel of q A are A invariant subspaces, yeah. I mean p and q could be any arbitrary polynomial for that matter, we have proved this. You take any arbitrary polynomial constructed out of the operator or the matrix A, whatever lives inside that kernel is not just a subspace of course kernels are subspaces, but it is an A invariant subspace. That is if any vector V belongs to that subspace then A V must also belong to that subspace, ok, alright. So, this is all a setup for what we are now about to do. So, goes without saying that A is an operator from V to itself where V is finite dimensional, yeah. Maybe I should have put it right up top, but ok you understand what is going on, yeah. The claim is that V is equal to kernel p A direct sum with kernel q A. Imagine the consequences if this is true, just I will like you to without putting chalk to board, we will discuss this in further detail in the next module, but without putting chalk to board if this is true, what does it immediately suggest to you? You have this minimal polynomial, first you split it up into two co-prime factors, alright and then you cook up a basis for the first kernel and you cook up a basis for the second kernel, concatenate these two basis and you have the basis for V. In terms of this basis if you now represent A because these are both A invariant subspaces, what sort of a structure do you expect? Is it going to be a block diagonal 2 by 2 structure? Now if it is true for 2 look further into p, again split it up. So, you keep reducing it until you land up with a situation where you can no further split up a polynomial into co-prime factors, when will that be the case? So, what will be the co-prime factors of the minimal polynomial, the maximum possible partitions that you can split it up into? If you have x minus lambda 1 raise to the k 1 times x minus lambda 2 raise to the k 2, suppose there are let us say L distinct eigenvalues. So, it is x minus lambda 1 to the k 1 times x minus lambda 2 to the k 2 times so on till x minus lambda L to the k L that is your minimal polynomial. So, what are the co-prime factors? The maximum splitting that you can do. So, it is all the one co-prime factor is x minus lambda to the k 1, the other co-prime is x minus lambda 2 to the k 2, the other is. So, you have exactly L co-prime factorizations and each of them would then give you a basis of an A invariant subspace and you can write the entire vector space as a direct sum of these kernels of these polynomials from that come from a minimal polynomial right, these factors of the minimal polynomial and then you have that many block diagonals. So, you have exactly one block diagonal for every eigenvalue, every distinct eigenvalue right. So, of course, if you have just one eigenvalue then it is just the same thing and you might say that is where the Jordan form comes in more handy, but at least before we even dive into Jordan form if this is true in principle what we are trying to do is this. If we once we have proved this our next goal would be to show that look you will now be able to block diagonalize this A matrix into as many block diagonals as the number of distinct eigenvalues what will be the sizes well the sizes will depend on multiplicity algebraic multiplicity yeah, because it has to be equal and that is where you can also argue about we will see later why the power to which every factor is raised in the minimal polynomial cannot be more than the algebraic multiplicity ok, because we will split it up into a matrix and a nilpotent matrix and we will see that the nilpotence of that matrix cannot be more than the algebraic multiplicity all in good time, but I am just saying that this is what we are trying to get towards ok. So, eventually we will be able to show and that is why I said we will show you an indirect proof of the Cayley Hamilton theorem yeah closed yes yes yes. So, f is algebraically closed that is the first assumption. So, we are going to talk about Jordan forms and all there is a real Jordan form we will not deal with that we will talk about the complex Jordan form. So, because the field of complex numbers is algebraically closed. So, everywhere that is a standing assumption we had stated that earlier also the field is algebraically closed. So, that you can split this up into first degree polynomials because otherwise like s square x square plus 1 you cannot split it up into x plus i n x minus i without going into complex numbers. So, we always go for the complex domain which is algebraically closed not the real field ok. So, this is going to be what we shall prove in the next module ok.