All right. All right you guys. So today we hopefully we'll finish the discussion of the canonical representation of general operators on complex vector spaces.  Let me remind you where we left off before the break. The set up is that we have a finite dimensional vector space over the field of complex numbers.  The reason why we want to do it for the field of complex numbers is that in this case, every polynomial can be split into a product of linear factors.  Finite dimensional vector space over complex numbers.  And the reason is that in this case, we don't have to worry about the possibility that our polynomials, such as minimal polynomials or the Crits polynomial don't split into a product of degree one factors then the ideal situation and we have an operator on the ideal situation. Is the situation which we call in which we say that the Opiator T is diagonalizable.  To illustrate, it considers the basic example where acts on two in the two dimensional or space. It acts by multiplication by a two by two matrix.  Just to save time, in my notation I will not distinguish between an operator and a matrix which represents it. For now, I'll just write t is equal to, say, lamb one, lam two, lambda one and lamb two are two complex numbers. Let's suppose that they are distinct in this case.  This is a case of diagonalizable operator, obviously, because in this case there is a basis of eigen vectors, just the standard basis. Let's call this one and two we have applied to one is lambda one, one applied to two is lambda two, two. In fact, in this case we could easily consider the real counterpart of the situation. So we could consider if m one lamb two are two real numbers, we could consider the corresponding oppiator on two rather than two. The convenience of that is that we could then think about the opiator geometrically as an operator acting on the two dimensional plane in which we have chosen a coordinate system. This coordinate system is generated by the vector C one and two. Now let's suppose for simplicity that lambda one is one. Lambda two is two. What does this tell us? It tells us that for every vector along this line multiplies by two. All right, So it's very easy to know what the action of the operator is for every vector along this line. Whereas for every vector, let's see if I have a different color. For every vector along this line, the operator simply access the identity as identity. This information is a tremendous simplification of the, gives us a great sense of what operator does. Now say this vector twice this vector, this vector will go to itself. Which means that, for example, if you take a vector somewhere in the middle, so to speak, one plus two. Well, what does when it acts on one plus two? It takes one to two, two and takes to this vector, goes to this vector. You see on the generic vector, the operator looks complicated, but we have been able to find two pieces to break our vector space, as it were, the two dimensional space, into two, one dimensional pieces. Each of them is preserved by this operator or the term we use. Each of each of the subspaces is invariant, right? These are classic examples of invariant subspace. Both the horizontal and the vertical line. It is invariant because every vector it's in a horizontal line goes to a vector again, in a horizontal line under the actionless separator. And likewise for the vertical line. That is not the case, for example, for vectors along the bicector. Here is a vector along the bicector, one plus two. When we apply, we don't get a vector along this bicector line, we get a vector on a totally different line. We get two. One plus two. That means that this is not preserved. This is not invariant. It is not a invariant subspace, but this is invariant. And this is invariant as well. We have broken the two dimensional space into two pieces. Each of them is invariant. Now, when I say broken into two pieces, what do I mean? Technical term for this is that we have represented the two dimensional space as a direct sum of the two subspaces. Two one dimensional subspaces now is a linear operator. It takes to the sum. For instance, if you want to know how acts on one plus two, it is enough to know how it taxes on one, how it takes on two, since every vector can be written as a sum like so, a multiple of V one plus a multiple V two. This gives us complete information about how the operator acts. That's why diagnization is such a powerful tool. It allows us to compute the action of the operator in very simple terms. The key to it is breaking your space into pieces on which the operator acts in a very controlled and very simple way. In this case, just as a multiple of identity, lamb one times identity and lamb two times multipation by two along the horizontal line. Identity along the vertical line. In general, it's very similar, is diagonalizable, can be expressed as we saw before the break, actually even two weeks before, can be expressed in a variety of equivalent forms. But one of them is that there is a basis of beta such that the matrix of relative to this basis is diagonal. Okay? In this case, I already gave you the operator as operator of multiplication by a diagonal matrix. So the standard basis is such a basis, we often call this eigenbasis. That's one way to think about it. Another way is to say that V can be written as a direct sum of the eigenspaces corresponding to two distinct eigenvalues. This is called eigenspace corresponding to lambda I is equal to, it consists of all vectors in V, such as the t minus lambda V is equal to zero, or equivalently V is equal to lambda V. This is what I just illustrated in this example. These are the two eigenspaces. This is going to be of two and this is of one. The whole two dimensional space is direct sum of these two lines of two dimensional subspaces. Finally, the third one was that the minimal polynomial p of z is equal to a product of linear factors, each with multiplicit one, all distinct, no multiplicities. So, this is an ideal situation, and it would have been nice if every appeator could be represented. This form? Yes. Oh, yes. Yes, you're right, I'm sorry. So you're paying attention. Thank you. Yes, that's right. The first one is response to the one, so therefore is equal to two. Unfortunately, it's not always the case. The first counter example is not always the case. The first counter example is when the operator has the form like or is represented by a matrix like at first glance is not very different because it's almost diagonal. There is just this little thing above the diagonal. Yet this little thing changes everything dramatically because in this case, the criteristic polynomial is minus lamb squared. Which we have discussed a number of times. I will just leave it at that. Which shows you that we are in the case when T is diagnosisable. Because if we were in the case that is diagonalizable, the critics polynom minimal polynomial would have to be z minus lambda and Z minus lambda squared. Compare to when t is actually lambda zero, zero lambda. We get rid of this one above the diagonal. In this case, indeed P of Z, Z minus lambda, and clearly the operator is diagnosizablejt. What happened is that the two eigenvalues, lambda one and lambda two have collided. That's a counter example that gives us an example of an operator which is not diagonalizable. Then the question is, can we still do something in this case? Is there a similar decomposition of our vector space into a direct sum of subspaces labeled by the eigen values? There's still a notion of eigen value for such an operator. Lambda is a unique eigen value. Eigen values, I remind you, are precisely the roots of the minimal pallynomial. There's only one root of this pallynomial, and that's lambda itself. It's just that it appears with multiple Tom which creates all these complications. But one could still hope that in general one could generalize subspaces and still have a possibility to break our space V into smaller subspaces labeled by the eigen values, each of them invariant on each of them. The action of perhaps not being a multiple of the identity by something similar, something close enough. That's what we're going to establish today. This is called Jordan Canonical Form or Jordan form. Before I get to this, I want to give a bunch of definitions. The idea is that, in fact, everything can be an operator, will always be represented by a matrix which has a block form, which is a union of blocks. Just like that. I think it's touching my shirt. That's what happens to study. It's an engineering problem. So anyway, are you hearing it? Yeah. I don't know what's I think the microphone doesn't want to go back to work, you know, so after the spring break. Okay. It's not my third. Okay, Let's don't move, okay? Okay. So let's generalize this from two by two to n by n. We have this matrix which I have introduced before, which is called the Jordan Block, which is a natural generalization of this. It has lambdas on the diagonal, the same lambda everywhere, and it has just above the diagonal everywhere, there is a main diagonal and there is like the next shifted diagonal just above. Okay. The idea is that we put on everywhere, skipping any place that's the Jordan block. In this case, the minimal polynomial is, as we already discussed, the s power of Z minus lumber. In fact, what does it mean? It means that every vector is annihilated by the operator minus lambda I to the n. In other words, we're substituting for z in the minimal polynomial. This is equal to zero for any V in the n dimensional space. That gives us an idea how to generalize the condition that we have imposed that corresponds to Eigen vectors. Eigen vector is one, is a vector which is annihilated by t minus lambda times identity, or Tmalmda times identity. Here we see that every vector in the sentimental space is annihilated by the Ens power. This leads us to the notion of generalized eigenvector that we discussed just the week before the break. But in fact there is more. I think that it's very useful here to keep have in mind the following. Diagram, the following geometric description of this matrix. You see, if I take J, this is okay. If I take minus lambda I, then I will get a matrix in which we'll have zero everywhere actually, except just above the diagonal where it have ones. Okay? So it's convenient to do so so we don't have to carry this lambda with us, but we still capture the essential property of the operator. This equation will be this operator to the N's power is equal to zero. But let's look how this operator actually acts on the standard basis. You have the standard basis on which is one, et cetera, et, and then two N which is zero, et cetera, one. Now we know that for every matrix, the meaning of the ice colon is how this appiator acts on the ice basis vector. Right from this we can read off how this appert on the standard basis. We see that I want to put them like this. N N minus one, minus two and so on. Two e one is a vector which has one in the first position. If you apply the operator minus lumber to it, you will get the first colon of the matrix. And guess what? The first colon of the Smtrix consists of zeros on it hasn't yet gotten to the one one appears as the second colon. In the first colon, there is nothing that means that if I apply minus lambda to one, I go to zero. Okay. What about two? Two has one. If I apply minus lambda to it, I will get the second colon, and the second colon has one on the top. It's one. Therefore, if I have this, do you see this? Do you see that? Two goes to one. Likewise, three goes to two and so on. And minus one goes to minus two. You have the standard basis of CN forms this a chain from the perspective of the operator minus lambda I. It forms the chain in the sense that each vector goes to the preceding one. N goes to the minus first in first, to minus second, and so on. And then we get to one, and then we fall off the cliff, so to speak. Now, what does it mean? It means that one is annihilated by the first power of T minus lambda. What about two? Two is not annihilated by T minus lambda. We get to one, but if you apply it one more time, you get zero. Two is annihilated by T minus lambda squared. Likewise, EK is annihilated by T minus lambda to the K. Okay, That gives us a good understanding of what the meaning of the Jordan Block is. First of all, Jordan Block is a matrix. It corresponds to a particular opperatorf, it's a size n matrix. It's an operator acting on dimensional vector space. If we take t minus, if t is represented by a Jordan block, like minus lambda I is represented by a matrix like. So that the action of that matrix on our vector space is captured very nicely by this diagram. Okay, we can think of a Jordan Block as, as an operator acting on anti meenal vector space such that its action is given by a very simple procedure relative to the standard basis is not just multiplying something by lambda, but if you take t minus lambda I, it's a very controlled procedure which is just basically sends each basis vector to the preceding one until you reach one and then you go to zero. Let's keep this picture in mind, it's still relatively simple. The statement that we're going to prove today is that The theory that we're going to prove today is that a linear every operator on a finite dimensional vector space over the field of complex numbers can be represented in terms of hesomticx representation, which is built from the Jordan blocks. Let's give a couple of definitions. I will say that at a matrix has Jordan form. A matrix, I don't know, matrix A form. If a can be represented in a block form, in other words you can, you can break the matrix which is after all and by table can break it into pieces. There will be a few square pieces on the diagonal, okay, Which have non zero values, potentially have non zero values. And then there are zeros everywhere else. And each I I is is a Jordan Block, like so. Okay, to illustrate it, let me give you a couple of examples. Well, first of all, if it's just one Jordan block, then surely it's an operator. It corresponds to an operator which has a Jordan form, which is presented by a matrix having Jordan form. But more generally, let's say a block two by two. Three by three, let's say with different eigen values. Okay, so that's an example. This number, this was one, this lamb two. Now this numbers in principle could also be the same. You could have two Joan blocks of different sizes or equal sizes that have the same numbers on the diagonal. So in other words you could have lambda one, lamb one, here also lambda one. Then you could have a block, another block of size two with another Eigen value. Then you could also have a diagonal part. Now a diagonal part would say would be lambda three and then lambda four. Notice that one by one Jordan block is just a number. A diagonal matrix isn't a Jordan form. It's just that every block in it has size one by one. As soon as you get a size two by two or more, you start getting things which are not on the diagonal, which are off diagonal. Because for two by two Jordan block, there is already one above the diagonal, three by three, four by four and so on. You have things above the diagonal, but Jordan form does not preclude a diagonal part. It's a special case when the blocks are actually of the size one by one. How do you tell what are the sizes? For example, you see, I want you to recognize the difference between this example, this example. Let's suppose I put here one. By the way, in this picture, if I don't put, if I don't put numbers is zero. But what if I put one here? You see that changes the situation. Before I put one, I have two blocks, lambda one of size two and another lamb, one block of size three. Three by three. But if I put one here, this is no longer in block form. Because block form means that outside of those squares on the diagonal, it's on the zeros. If I put just one non zero element, is already not a block decomposition. The smallest block here is five by five. That's how you distinguish between them. Another way, combinatorially to think about this, is that a matrix has Jordan form if it has diagonal entries, and then some of the entries above the diagonal are equal to one. Some of them are zero, some of them are one. Not all of them are on. The way the ones appear gives you a unique way to break it into block diagonal form. For instance, if you don't have one here, then you have two separate blocks. But if you have one here, you don't have two separate blocks. You have a five, five combined block. But suddenly here you get zero. But you see between different, if you have lambda one and lambda two, it will always be zero. The only way that you have one is in a chain which combines the same lambdas. Yes, you could have distinct gas in different blocks like this. Yes, you could have that too. You cannot put one if there are different Lambdas, that's the point. You can only have one linking the same lambdas. In other words, this is not allowed. This is not a German block, it's not in German form because if it were in German form, it would have to either not to have one here or to have lamda one equal lambda two. This is not good, but this is. This is okay. This is okay. A few weeks ago, I did a calculation showing you that if you have a matrix like this, it is actually diagonalizable. We now actually know enough to understand why this is so, because we know that the minimal polynomial is a product of linear factors which correspond to diagonal entries. Therefore, the minimal pent for this matrix is Z minus lambda one times Z minus two. If lambda one is not equal to lambda two, we are back to the situation and therefore there is a basis of eigenvectors. This basis is not the standard basis one zero and zero one. In fact, one zero is an eigenvector with eigenvalue lamda one. But I did a calculation showing you that a linear combination of one and two is an eigenvector with eigenvalue lamda two. However, the formula for that eigenvector involves dividing by lambda one minus lamb two. So this matrix is diagonalizable if and only if lamb one is not equal to Lamberto. Yes, sure this is like this. There's no one here like this. This is zero. I emphasized that this is not allowed. You cannot have two distinct diagonal entries and one above them. If you look at this picture there, none of that is happening here. Do you see that? It's very important. I suggest you go back to your notes or video of my lecture or to do it calculation yourself to see, to convince yourself that this mats has two linear independent vector lambda one and lambda two. The first one is obvious, one, zero. The second one is not obvious, is not zero one, but it's a linear combination of one, zero and zero one. However, that combination only makes sense when lambda one is not equal to lambda two. Because in the process of calculation, you have to divide by lamb one minus lambda two. Therefore, if lambda one is equal to lambda two, that's not diagnoizable anymore, which is no on general grounds but because in this case the mineral polynomal is not Z minus lambda one is Z lamda one squared. Therefore, we're out of the diagalizatione. I suggest you carefully look at all of these examples to familiarize yourself with the combinatorics of all of this. This is not random, it's a very, very structured thing. You see this once they appear, but they don't appear randomly. They only appear when you have multiplicities so that they create blocks. Okay, So that's the definition of a matrix that has Jordan form or is presented in Jordan form. Now what yeah, we could have here lambda one also, is that what you mean here? Zero and here you have an option to put another one, but then the block will be like this here, yes, you can do both. If things are equal, you can have either one or zero. If things are distinct, you have to have zero. That's basically the rule. Yes. Let if lambda one, it's a union of two Jordan blocks, one by one block and one by one block. So it's kind of a degenerate Jordant block where there are no ones, there's no rule for one, because it's one by one, it starts with something like poetry. Anyway. Okay, So I hope it's clear now. Now, what does it mean from the perspective of what I described here? Diagrammatically, it can be expressed by saying that there is a basis of V, which has a very particular form. Okay, so. The corresponding basis. So what does it mean for the basis for the standard basis? The action of on elements of the standard basis has Jordan form. You see here the basis is like 123 and so on. The basis can be broken into union of chains like so. In some textbooks they call cycles. But I think it looks more like a chain than a cycle because you're not cycling around, you're just moving in one direction. In this particular case, you're going to have two goes to one minus lumber one and then falls off the cliff by minus lumber one, right? The 345, they have to remember that you go from larger one to smaller one. Not 1-2 but 2-1 as we saw when we zoomed in on one specific Jordan block. Here you have 543 and then we go down minus lambda two. For simplicity, I will just put it over the first arrow. Next we have six and 776, I put lambda one here. Also, let's return to lambda two so that we're writing them in sequence minus lambda here it was lambda one, lambda two. Then finally we have a chain. Could be the general chain consisting of a single, that's the case when there's an eigenvector and nothing comes to it. In this case, it will be eight and nine. This is minus lambda three. Minus lambda four. Okay? That's what the basis looks like, or rather not what the basis looks like, that's what the action of the operator looks like. That the basis is such that the action can be broken into these pieces and each piece has its regular structure of a chain corresponding to specific lambda. Note that we are not writing the action of, we're writing action of minus lambda one or lambda two and so on. Whatever is appropriate for the particular block. Now I decided to change this, Put one here instead of zero. What would happen? These two chains will have to combine, because now we unfortunately will combine, but in a weird way. It will be two will appear here, then one not combining like so. Because you see it should go 5-4 to three, then to two to one. This is the red. This one is responsible for the jump 3-2 Don't have, if you have zero in that position, three goes to zero, right? Three is going to minus lamb one times three. Without one would be zero colon. Because you remove the only non zero entry in the third colon in this way. But if I put one here, it means that minus lambda one, three is two. Right? That's what I mean here by this diagram. That you have two chains of size three and size two get combined together on. We can play with this, but I just wanted to give you the different angles to see what this Jordan form means. This kind a basis such that the action of the operator can be described on the basis of vectors. In this way, by a union of chains is called a Jordan basis. In other words, a Jordan basis is a basis that If we represent our operator with respect to this basis, we will get a matrix which has Jordan form. This is called Jordan Basis. Now, the main theorem that you can always do that, that's what replaces diagonalization for a general operator. That should okay, let me put it here again. You have T from V to V, where V is a finite dimensional vector space over the field of complex numbers. For any such, there is a Jordan basis. That is to say, let's say beta. That is to say the matrix of relative to the basis has Jordan form. Okay, Now let me connect this to something we discussed just before the break. Just before the break, we have seen that in general this equation can be replaced by a more sophisticated equation. Where we replace the eigenspace given here by what's called the generalized Eigen space. In general, this direct sum of decomposition becomes the decomposition into the direct sum of Generalized eigenspaces. Okay, where the generalized eigen space contains the eigenspace, it is defined by an equation where you put some I'm like I'm being a little bit combining different equations together, color coding them. The yellow corresponds to the general case when it's not necessarily diagonalizable. Diagonalizable means that generalized eigenspace is equal to the honest to goodness eigenspace, right? But in general, this is bigger and it's defined by this equation for some k. Now observe that all elements in a Jordan basis satisfy these equations. Right here, you have minus lambda one, I two squared is equal to zero, for example, whereas T minus lambda one I one is equal to zero. Likewise here, T minus lambda one of five cubed is zero because if you have a chain of length n, you have to apply minus lambda, cresponing lambda times identity and times. And you will get to zero because you will travel through the entire chain and then go to zero. In other words, all of these basis vectors are actually generalized like in vectors from this perspective. Not surprisingly, this theorem dovetails very nicely with theorem one of existence of, of the Jordan basis. Namely, moreover, lamb I of t has a basis consisting or appearing in the Jordan blocks. The basis appearing in the blocks blocks with lambda eye on the diagonal. You see, in other words, this theorem is a more refined statement then the theorem we proved last time about decomposing a vector space and to direct some generalized against spaces. Here, there was no basis given here. You see what we only knew that the space decomposes into those pieces. But what happens in each piece, we didn't know. The best we could say that operator acts as an upper triangular matrix. That every vector satisfies this equation for some k. But it's still too amorphous, not very rigid. Now rigidifying it for instance. What happens here in this case, Let's go back to the case when this is zero, so I have fewer colors, then remove this. What will be in this case in values are lambda one, lambda two, lambda four equal to four, there are four pieces. What is lambda one? Lambda one, It is the span of all the vectors appear which are relevant to the jordan blocks corresponding to lambda one. And those are 12345, Even though there are two different blocks for lambda one, all the vectors which correspond to them, three, four, and five, how do I know? I know because whichever vector from this five I take, I know that T minus lambda one I applied to it is zero. For example, the first one is killed by T minus lambda one. The second one is killed by Timanus lambda one squared. The third one is killed by T minus lambda one. The first one by Tim one squared the fifth one by time. For example, the last one I know from this diagram, if I apply it once, twice, three times, I go to zero for five, the number k which appears in the definition of generalized. Again, space is equal to. That's right. Thank you. Likewise for each of them, this number is precise response to its position in its Jordan block. The first one in the Jordan block is an honest wooden invectexte. The number is two for the next one, number three and so on. That's how they are distributed. But all of them are generalized invectors because for some K they will be annihilated. Is that clear? Please ask me if it's not clear of lambda one is the five dimensional subspace which on which our operator at just by the five matrix cris point to this two blocks. It's nothing to do with those blocks because we are talking about the action on the span of the first five vectors. Then lambda two is a span of six and seven. Each of them is a generalized like invect but for lumber two, six is annihilated by T minus lambda two, seven is annihilated by T minus two squared. When I say Ti minus Lm two, I mean minus lumber two times identity. But I'm just trying to yes, they all spend because this is the basis of the whole space long. I don't know. Obviously, for the guys I cannot get this equation with lambda one. If I subtract lambda one, I will get some matrix which is not. I need to get the matrix which has once above the diagonal so that I get this structure there. Things which correspond to blocks with other lambdas have nothing to do with generalized vectors for lambda one. On the blocks with lambda one will contribute, the basis vectors appearing in the blocks for lambda one will contribute to lambda one, T, likewise for lambda two, and so on. You see that indeed in this case 5789, it's a nine dimensional space and it breaks into 1234 generalized in spaces in general. For example, lumber two there could be extra blocks, lumber three could be extra blocks, and so on. But all of them will come the basis elements which correspond to those blocks, they will all combine to give a basis of that subspace. But that's roughly how it looks. I just wanted to give one representative example so that you see what's going on. This is therefore this theorem. This theorem one. But the existence of Jordan Jordan basis is how to say gives additional information. It's clarification of this, it gives more structure to this. Here. I'm just saying that there is a five dimensional subspace, There is a two dimensional subspace, the one dimensional subs is one dimension subspace. But how the operator acts on them is not clear from this. The new result gives us a much more precise information about how the operator acts. It acts by a block form where the blocks are the Jordan canonical are the jordan blocks. Which means that There is a basis such that the basis for each Lambda can be broken into pieces. Here there is a subdivision. The fact that the lambda one parts correspond not a single block of size five, but to two blocks of size two and three shows that this subspace for lambda one actually has a subdivision as a direct sum of two smaller subspaces on one of them, the operator xs, a joint block of size two by two. On the other one of size three by three. Yes. Well, the span is just the five dimensional part. In general, if you have a matrix like this, let's think about how this matrix acts. What does it mean that a matrix has this form? It means that if you look at the vectors which are labeled by the rows of the first block, the subspace, which is invariant. The fact that there, the fact that all of this is zero means that if I take the first colon and apply my operator, I will get a linear combination of only the first few here. Nothing from here contributes. That means that the first, if I apply to it at my operator, I will get an element in the spin of the guys. Likewise for the second. Likewise for the third, it means that the spin of the first, let's say the first five, the spin of the first five is invariant. That's exactly what it means. By the way, if you're wondering about the notion of T invariant, it is the case when a matrix, an operator has a matrix form like so with blocks means that your vector space breaks into direct sum of pieces, each of them being T invariant, then the matrix acting on that piece is the corresponding block. That's a general fact even before Jordan block decomposition in general, if a matrix has this form, this is equivalent to the statement that the vector space is a direct sum of subspaces and each of them is t invariant. Then if it's invariant, it means that we get a well defined apparator just on VI. Therefore, it has its own matrix. Once I choose a basis in VI, that matrix is exactly the Crisponi block in this block decomposition. But it turns out that we have a lot of control also that we can adjust our basis in such a way that these blocks are not just matrices and by matrices, but they always have this form. So it's really powerful result, you see. All right, so now I have explained how it fits into the previous result. It gives us a more precise picture of what those generalized agen spaces are. Not only they're just amorphous generalized agen space with some abstract vector skill annihilated by some powers of Tim can actually be represented as a union of pieces of subsets. Or they can be spanned by vectors such that in each of those subsets, the action of the operator is given by a specific Jordan block. You see? So that's what the statement is. Okay, So now let's prove it unless there are more questions. Okay, So the idea is to prove it when all lambdas are equal to zero. To prove it, we will consider the special case is what's called Neil Potent. This was actually introduced already in section eight point B, but actually I didn't have time to talk about it, but it was in your assignment anyway. Is Neil potent, What does it mean? That is nilpotent. It means the to some power is equal to zero for some K. Now in retrospect, we're trying to prove that can always be represented by a matrix like this. If you start raising this matrix to power power. If you have a block matrix and you want to raise it to power, it corresponds to raising each block separately. Well, unfortunately it's called K, but let's call it S. Okay, now under what conditions can a block like so become zero when you raise it to a particular power lambda? Let's look at this three by three block. If lambda one is non zero, you can see that in the square there will be lambda one squared, in the cube there will be lambda one cubed, and so on. So if lambda one is not zero, no power is equal to zero. Every power of this matrix is non zero. The only way how this matrix raised to some power can become identically zero is that lambdas are equal to zero. In fact, those matrices are going to be the ones where all diagonentris are equal to zero. That's a special case. It turns out that this case is representative enough that if we can prove it, then everything will follow. This corresponds to the case when all lambdas are zero. Okay, So let's state the theorem in this case as theorem two. Suppose T is nilpotent under the conditions of V is finite dimensional over complex numbers. And suppose you have an operator which is nilpotent, then there is a Jordan basis. Okay? So we're considering, we're tackling the simplest case when there is only one lambda and that lambda is equal to zero when I wrote here, corresponds the case when all lambda zero. Better to say there is only one, It is equal to zero. Because as we said, usually when we write lambda one, lamb two, and so on, we mean distinct numbers. If they are equal to zero, there is only one distinct number, that number is zero. That's the case that we're considering right now. Vivi, the general case. In the general case, there will be many blocks with different lambda. Now we'll have only blocks with lambda equal zero. We want to prove that indeed we can always represent any impotent operator, the one for which certain power is zero as an operator. Indeed, for such an operator, there exists a joint basis in which then the matrix will appear like so, but with all diagonal entries identically zero, only once, some once above the diagonal, right? Maybe none. If it's diagonalizable, it could be that there are no ones. The case when there are no ones above the diagonal in the join caanical form is exactly the case when the diparators diagonalizable. That's a generic case. Actually what we are doing now, we are trying to, it's like we know 90% of the zoo, 95% of the zoo, the animals in the zoo. And we try to knock cages of the exotic animals. These are exotic animals which have Joan Blocks of size greater than one. But we love them anyway, right? So that we want to include them. The proof is going to be by induction one dimension of the vector space if you dimension is equal to one. There's nothing to prove because if you have a one dimensional vector space and an important operator, first of an operator is just a number scalar multiplication by scalar. And it's impotent. If only the scalar is zero, your operator is zero. Zero operator has joganeical form. In higher dimensions, certain dimension one. Nothing to prove. Suppose we know Now let's in the case when the dimension is greater than one, assuming we have already proved for smaller dimensions, all smaller dimensions, that's what the induction means. The thing is the operator is potent. There exists some K such that to the K is zero as an operator on, which means that the is to zero is zero. Vector for all. That's what it means. There is some to the K. Apply to any vector zero. Okay. Pick vector. Actually, let me say m, there is this key. Oh yeah, sorry. I actually, let me leave it like this, but I want to explain. Then there is the smallest let M be the smallest number number such that to the m is zero, but t to the minus one is non zero. If some power is zero, there will be a smallest power. But then there exists a vector u such that t to the minus one is actually non zero, right? We're going to use this to reduce this case to a smaller dimension. The first step is to observe. Let us define the following subspace U t squared u t to the minus one, okay? No need to take because u is zero, t minus one. U is non zero is zero. Because M is zero for everybody, I claim that the vectors are linearly independent. In other words, the vectors form a basis of you. This is actually an exercise I think from last homework, but let me just give you a quick explanation. These vectors form a basis. The statement is equivalent to these vectors being linear independent because the space, I mean spin spin. If they were linearly independent, linearly dependent, then we would have the equation zero plus one plus, et cetera, plus minus one, minus one equals zero, right? But then let's suppose, could be that zero is non zero. Then we'll just look at this formula. If zero is zero, we'll raise it. It will start with some r to the R I before are equal to zero. R is the smallest non zero number in this equation, right? We will, we will write the equation in such a way that there are no I which are equal to zero. We will start with the first non zero element. This is non zero, but all the other ones are equal to zero. Okay? Suppose there is such an equation where this guy is non zero. Then how to get contradiction? You have to get rid of all of them. And you do that by applying t to the power minus r minus one. You see it's a very typical argument in this type of situation, apply. I want to be sure that this guy will not disappear. If I apply minus R, I will get to the two. M will kill it. But I will do just. I will take one smaller, minus r, minus one to this, you see, All of them will disappear to M is zero if I apply it to each of the terms. Let me write it more carefully. This expression becomes to the r t to the minus one u. We have assumed that this is non zero, and this is a contradiction you get by applying t to the minus R minus one to this linear combination, yet just a non zero multiple of this vector. We have agreed, we have picked it with this condition and we assume that CR is non zero. Therefore, this is non zero economical to zero as a contradiction. Therefore, these vectors are linearly independent and therefore they form a basis in the span. Okay? We have the subspace of dimension m is m. We have a specific basis in there, consisting of U and powers of applied to it. Now we need to find, the idea is to find and observe that is environed to see that U is invariant, is equivalent, that if I apply to any of those vectors, I will get the linear combination of those vectors because those vectors form a basis. A subspace is invariant if it has a basis such that you apply to every L basis element, and you get a linear combination of that basis. That is to say a vector of that subspace to is to the R plus one, which is still in our subspace. It is invariant. Now the idea, and that's going to be step two, is find another subspace W, V, which is actually, sorry, strike that. The first possibility is that you could actually be equal to step two, two possibilities. The first one is V is equal to, that is to say U already is the whole space. The second possibility is that V is not equal to you. It could happen that this is a whole space because exactly in the case when the appiator is given by Jordan, a block with lamb equal zero. That's exactly the case. Now in this case we are done if I write the basis in the opposite order, because if I write the basis in the opposite order, M plus one, minus one u t to the minus two, et cetera. And this will be my M minus 121. Right? I'm confusing you. Let me just write it simply. I have maps to squared, maps to t to the minus one U, and the maps to zero. This is a chain, this is a Jordan basis, Just the way it's constructed, it is already Jordan basis of if U is equal to V, we're done. Now, suppose that, therefore, suppose step three, suppose V is not equal to you, then the idea is to find another subspace, which you will call W, which is invariant. Then by induction it has a smaller dimension. Because this one has positive dimension, it contains at least one vector U, non zero vector U, the dimension subspace which is the invariant. And such that V is a direct sum of U and W, then the dimension of W is smaller than the dimension of V. By inductive assumption or by induction, W has the Jordan basis, which we'll call beta. You have a Jordan basis here, right? By induction, but we also have a Jordan basis here, That's this guy. This ones, we have constructed the subspace U as a span of UTU and so on. But just by virtue of this construction, it's clear that those vectors actually form a Jordan basis because the action of on them is given precisely by the chain that gives us a Jordan block, right? So we have a basis in U, which is a joint basis with one block. Actually we have a basis in W which could have many blocks. We don't know. Here we exploit the inductive assumption because it has a smaller dimension. We combine them together, we got a joint basis for the whole space, and we're done. The question is to find this W here. There is a very nice trick where we use dual spaces. I have to say, I previously taught the scores from another book by Friedberg, Insul, and Spence, and their proof of Jordan Anicalfourmu. More difficult, much more convoluted. So I'm actually impressed. It's a very elegant and concise. You're lucky. If you think it's complicated, you should see the proof. The other book. Okay, here's how we look constructed. We know in general, if you have a vector space, then we have a dual space, which is the space of all linear maps. I write C because we are really over complex numbers here, right? We know that for any vector in there exists a functional in V prime such that the value of on U is non zero. This is one of the basic facts about dual spaces. The way you prove it is, for example, by extending to non zero vector. If you have a vector in V, you can extend it to a basis, then take the dual basis, then the dual element to U will have the properties will have its value on the vector. U will be non zero. We don't care at this point what will be the value on other vectors. All we need is that there exists a linear functional whose value on this vector U is non zero. That applies in any situation, No regard to Jordan canonical forms or whatever finite dimension a space and any non zero vector in it, there exists linear function which takes non zero value on it. Okay? In this key, in our case, let's catch you call it V. We will take V s m minus one, where this is our U. Remember we have this U with the property that minus one U is non zero, but MU is zero. And we'll find a functional Fi which takes non zero value on this guy. Now define W as follows. It consists of all vectors in V such that f of v is equal to zero. Five of v is equal to zero, and so on. Five of t to the minus one, V is equal to zero, okay? So I use this Fi as a pivot to define a subspace by the property by considering all vectors on which is on the contrary, fi is non zero on the property, is that fi on t to the minus one is non zero. But now I define W. Okay? Now I've run out of space, but I guess I don't need this anymore. I claim that it satisfies the necessary condition. Namely that V is equal to U direct. What does it mean? How to prove this? We have done this already a bunch of times. We need to prove that U intersect is zero. And the dimension of U plus dimension of W is dimension of V. That's all it takes. Let us show that intersect consists of just a zero element. Maybe first I should say is indeed invariant because if you have a vector which satisfies this equation, if you apply to it, it will also satisfy this equation, because this is going to be the same equations, essentially it is invariant. And now we need to show that it's a direct sum. Suppose that intersect. We'll argue by contradiction that it is non zero. Then v can be written as zero. It's an element of U. It can be written zero plus one. Et plus M minus one, minus one. Again, in the previous discussion, we will, if the zeros, like zero is equal to zero, one equal to zero, we'll remove them and start with the first non zero element, j to j plus, et cetera, M minus one to the M minus one, where j will assume to be non zero. Then just as before, we will apply to the power minus j minus one to it. It's actually argument very similar to what we did showing that they are linear independent. When we apply this, only this guy will survive. This will be to the minus G minus one u. Now, this vector, it belongs to W on the one hand, because it is equal to t, to the minus j minus one applied to V, V is assumed to be in and T invariant. If we apply to this power, we'll still get a vector in. But on the other hand, this vector is non zero multiple of t minus one. By our construction of this C J t to the minus one, u is non zero. And that's a contradiction, right? Because for this vector to be in W, it means that i of this vector is zero. But five of this vector is non zero because it's equal to j times five of t minus one. We have constructed i with this property. Okay, that shows that the intersection is equal to zero. But also next, we use dimension theorem to show that the dimension by a fundamental theorem of linear maps, we can see that the dimension of W is greater than equal to n minus m and the dimension of v dimension.