All right. All right you guys. So last week we talked about diagonalization.  And what did we see?  We saw that, we saw the following, first of all, diagonalization for a linear operator acting on a vector space V over over field.  We say that several definitions of diagonalizable operator, One of them is that there exists a basis in such that the matrix of this operator is a diagonal matrix.  But another way to say it is that is diagonalizable if and only if the 22 equivalent formulations that V has a basis which consists of eigenvectors of T.  Here I want to remind you that e is called an eigenvector is called an eigenvector if two conditions are satisfied is non zero. The result of applying to V is that it gets multiplied by number for some lamb de nef.  That's one way to describe diagonalizable apparators, those for which there is a basis consisting of eigenvectors of this operator eigen vector of.  Okay, the second way to formulate it.  Equivalent formulation is that V can be written as a direct sum of subspaces denoted by lambda IT, which are called the eigenspaces.  By definition, the eigenspace consists of all vectors V which satisfy the equation T applied to V is lambda V. This is a lambda I here.  In other words, it consists of all vectors which are eigenvectors of, with eigenvalue lambda I plus the zero vector we are joining the zero vector, which technically is not an eigenvector according to this definition, to all eigen vectors with a given eigen value so as to obtain a subspace, it is an eigenspace. It's a subspace is diagnosable. A direct sum is equal to the direct sum of the subspaces which are called the eigenspaces.  Now we proved the theorem discussed at the end is that is diagonalizable if it's minimal polynomial, which we denote P of Z is equal to a product of linear factors and there are no multiplicities. Okay, this is a very essential point. No multiplicities means that each factor I appears exactly once. There are no terms, there are no factors, let's say of the form z minus lambda I squared, or cube or et cetera, only z minus lambda n squared n cube. And so that's a very strong condition. Armed with this condition, we can now look more deeply which operators are and are diagonalizable. The simplest example is two by two example, where t is given in a particular basis by a matrix like so. Here, to simplify shorthand, I'm going to use the same notation for an operator and its matrix, even though technically this is not the same thing. In this case, we ask what is the minimal polynomial? Okay? Here we can use another criterion that we've learned, which is that the zero another theorem. If has an upper triangular form, then the zero of P of Z are the diagonal entries. Right here is a matrix which is an upper triangular form. The only entry which appears on the diagonal is lambda because it appears twice, right? In our case this means that of z has to be z minus lambda to some power. But we also know that regardless of whether matrix diagonal triangular and so on, for any operator, the degree of the minimum polynomial is less than equal to the dimension of the vector space on which the operator acts. In our case, this means that M is less than equal to three. Right? That leaves the possibility 32 in my head. I'm already jumping to the next example. Two. That leaves two possibilities, z minus lambda or z minus lambda squared. We can see that this, this can happen on that, if you substitute t into this ponomal result is minus lambda, we should get zero, which is equivalent to saying that is given by the diagonal matrix because that's what this is. We look at the matrix and we see that it's not of this form. If we subtract this guy for us in our case minus lambda I is non zero. It's a matrix which has one in the upper right corner. So that means that this is not the mean polynomial, Right, so therefore we go to the next case. So there's only one possibility left, So we conclude that P of Z is z minus squared. This is a polynomial which has a multiplicity. It is not of the form that is dictated by this theorem. Polynomials appearing in this theorem are the ones in which each linear factor appears in exactly degree one could be z minus one minus m one times z m two or lamp one is not equal to lamb two, and so on. Z minus lamp is squared. It is not of this form. This implies that our operator is not diagnosizable. Okay, so one more comment on this to illustrate to, to u, address the issue how rare this situation is. We have unicorn, We have found something doesn't fit. In the case that we were the opposite of unicorn, we were looking for something and there was something else that doesn't fit the profile. How often? Rarely does this occur? For that, we can try to perturb this a little bit. Instead of considering this, we can shift one of them a little bit. Let's say lambda plus epsilon here. Lambda, where epsilentmly could be very small. Okay, so think about this way, plot the two diagonal entries on the real line. Here we have a situation like whereas previously in that example I have a situation where the two points coalesced, you have lambda, but twice, I'm trying to illustrate that it appears twice. Now, this is definitely a generic situation. Two parameters, you can move lambda or lambda plus epsilon, you can move them apart, and so on. Here, there's only one parameter, this is a special case. Only one parameter versus two parameters. This is generic, we call this generic two parameters. Right now, it turns out that by a very similar analysis, we will find out that if epsilon is non zero, this one is diagnoizable. Why? Because in this case, if absolute is non zero, it could be very small, look like 1 billion or 1 trillion or whatever. As long as it's non zero, these two numbers are different. Therefore, by the argument we used, we find that the zeros of the minimal pallynomial have to be both of them because they're distinct. Distinct entries appear on the diagonal. The minimal polynomial has no choice but to be the product of these two factors which are distinct. Therefore, the operator is diagnoizable by the first theory, right? Please ask me if it's not clear. That means that generic case is diagnosable, is a diagnalyzable case, or the case of diagonalizable operator. What we discussed on these two blackboards is really a special case, but it is a legitimate case. Nonetheless, a natural question arises how to deal with this case. This is just a two by two example, but what if we consider three by three, or four by four? And in this respect, if we start thinking about the general case, what can we say about the general case? One natural approach would be to try to generalize this example and see what happens when we do that. For example, for the three by three, a natural generalization of this would be a matrix like soul. It has lambda on the diagonal and it has ones just above the diagonal and then zeros everywhere else. What can we say about this case? In the same way as in the two by two case, we find that P of Z can be either z minus lambda or Z minus lamda squared or z minus lambda cubed, right? Because the degree of the minimal pomal now has to be at most three, It's only zero is lambda, therefore there are three choices. This is a case when it's diagonalizable, These two cases when it's not diagonalized. But again, this is not the case because this t minus lambda I is not equal to the zero matrix, right? There are entries above the diagonal. Next you ask what about this? This one also doesn't work because if you take minus lambda I, what you get is empty. The diagonal, you drop lambdusrom the diagonal, right? I want you all to very clearly understand this, because in our discussion today and the next couple of lectures, this expression will appear continuously multiple times. What is the meaning of t minus lambda for a matrix of this kind? It simply means replacing lambdus on the diagonal with zeros. It becomes cleaner a little bit. There was a parameter and now we get rid of it. You think also about this case as the case of the original matrix, but with lamb equal zero. In a sense, this is sufficiently representative case for all of all the possible matrix of this kind, but it's easier to work with this. Now, this minimal pallynomal is equal to Z minus lumber squared if and only if the square of this matrix is zero, right? That's the definition of minimal polynomial. But what is the square of this matrix? When we calculate, we see that we get this which is still non zero. Therefore it cannot be Zmn slander squared. That leaves only one possibility, and we find that P of Z indeed is equal to this, in particular has multiplicity. In fact, it's multiplicity. In this case, the operator would not be diagnoizable because there was already a second degree. But here it's in some sense even worse, it's like this phenomenon is even more accentiated. Multiplicity means it's non diagonalizable, right? Yet, it looks nice. And such matrices are easy to extrapolate to an arbitrary size. What would actually, let me keep this for now. This matrix clearly has an analog for an arbitrary, as an by a matrix. Namely, I'm now going to draw it schematically. So we're talking about by and matrices right by N, it's a square. Metrix on the diagonal will have lambdas just above the diagonal. We'll have ones, and then zero is everywhere else. That's what the shape of this matrix is. Clearly, it specialized to this one for two by two and to this one for three by three. This matrix I will denote J of lambda. What we need to know is the size of the matrix, that's n. And what is that entry's? The same entry, repeating as it did in the previous two cases. It's called the Jordan Block. Now what can we say about this case? By arguing exactly the same way as here, that we have to find the minus lambda n. This is T minus lambda I is going to be a square matrix where you will only zero entries will be just above the diagonal. We can call it like the first upper diagonal, one step removed from the main diagonal, so we have one here. And then you have zeros everywhere else, right? That's what subtracting lambda does. It empties the diagonal or sets all entries equal to zero. Then let me call this matrix A. If you take easy calculation, shows that if you raise it to the J power, you're going to get a matrix in which non zero entries are going to be J steps removed from the diagonal like this. Well, it doesn't look like a more like this. We have one and this is J, the distance is one. But here is j squared. It will start not here, but here. And it will go like this for cup will be here, here, and so on. That's easy to calculate. You see in particular that t minus lambda I to the power j is non zero for all j from one to n minus one. The last one, when you have the distance, is n minus one, is when it appears here. This is like this matrix that's a square for n equals three, that's two is n minus one. The resulting matrix still has one remaining non zero entry and it appears in the upper right corner. Okay? Is the same phenomenon will appear for general n, all the powers of this matrix up until n minus one non zero. Which means that the crystal inomial cannot be equal to, be equal to minus lambda to the j. But the rise panomenal has to be of degree or less. Which leaves out only one case, just like in the two previous examples. We find from this analysis that the minimal polynomial is z minus lambda to the N, which fits nicely into the previous two examples. For three by three, we found z minus lamb, the cubed, for equal two. I have raised it, We found minus lamb squared. It's a very interesting example because you can think of this as precisely transversal to the diagonal situation. Diagonal situation is when the minimal polynomial is a product of linear factors where each factor appears exactly once. What's the opposite situation? The opposite situation, minimal polynomial, is the power of a single linear factor. Now we see we have found a concrete realization of an operator with such minimal polynomial. It is realized by this Jordan Block. Okay, Now our natural question is, in general, you will have a minimal polynomial which is a product of linear factors, some of them degree one, some of them degree two. With multiplicities, could it be that the operators a basis in our Rector space so that the matrix of the separator with respect to the basis can be written in a block form where each block is a joint block. The answer is yes. This is actually the best we can do. I have to preface it in the minimal paloma can actually be split into linear factors. We know that for the F is real numbers, this may not be so. Therefore, in the next few lectures, we'll focus on the case one is a field of complex numbers, then any P of Z can be written as a product of linear factors. Um, but with some multiplicities in general, we call, maybe call it, This suggests that the generalization is a mix of the diagonal case and the Jordan case. In fact, we'll be able to prove this probably the first lecture next Tuesday, two weeks from today, first lecture after the break, we will. That V, where V is a finite dimensional vector space over the field of complex numbers. For every, there exists a basis of such that the matrix has a block form. Let me write it here such that the matrix of T with respect to beta looks like this. You will have J one of lambda one. Note by the way that in this notation, J one of lambda is actually diagonal matrix one by one matrix which has lambda. We are not losing the diagonal case because diagonal case can also be thought of as is the case of a matrix written in a block form where each block is strictly on the diagonal. There's nothing about the diagonal, but now we are allowing the possibility that block there exists blocks of larger size. It could be several blocks for a given lambda. So on the, after you list all the blocks responding to lambda one, could be, I don't know, yes, could be some other numbers. I don't want to overload it with indices, but maybe I should so that there's one here, then you have the same lambda two and so on. I hope it's clear there would be several blocks for each lambda, one. For each lambda, there could be different sizes. Of course, the sum, sum of the sizes has to be n. All are blocks of n by and matrix where n is the dimension and then zero is everywhere else. So every operator will have a basis in which the matrix looks like. So this will be lambda m some J with some index. Okay, Now this includes a diagonal case. It's not like it's a departure from diagonal keys. It's a generalization of the diagonal. The diagonal corresponds to a situation where all sides, all of these are actually one by one. It is diagomatris. Yes, that's next. Now you start thinking, what is key? K is the largest of the Jordan blocks corresponding to lumber. Size of Jordan of lambda present. I don't know, bliinkbet. Try to think of which operalyhichlynomialf insert into the polynomial. You'll get something which applied to every basis vector will give you zero. You will see that it follows from the previous consideration that you need the largest power Jordan block of size and you need to set en power. But if you have 22 jordan blocks of the same size, you don't need to take double this power because it will work equally well for the vectors servicing each of the two blocks. In particular, if all KI is are equal to one, that corresponds to the diagonal case, because that means that the largest size of a Jordan block is one, that means there are no jordan blocks. Okay, that's what we'd like to do. Note also that each of them, each of the blocks is upper triangular because it only has entries on the diagonal and above the diagonal. Therefore, this matrix is upper triangular in fact. This is in agreement with the second theory which I have raised about patrangulal matrices, because we see that all the entries on the diagonal are precisely lambda one, lambda m, that are the zero of the minimal ponomal. But a general patrangular matrix is a very complicated thing, right? Because it has a lot of entries. It has diagonal entries, right? And it also has times in Swanivid by two entries above the diagonals and a lot of parameters here by reducing the general case to upper triangular matrix of a very special kind, that these matrices can be written as block form where most entries are zero and the only non zero things appear just above the diagonal. Not everywhere, but in some places, that is our goal. Now, how do we achieve this goal? How do we prove this? For that, let's look more closely at the case of this matrix, how to prove this. This will be our goal this week and perhaps also probably one lecture after the break. The crucial point when we talk about diagonalization, a crucial notion is the notion of an eigenvector. An operator is diagonalizable if it has a basis consisting of eigenvectors, if at least one of these blocks has a size greater than one, which is equivalent to saying at least one of these powers is greater than one is certainly not diagonalizable as we have seen even in the simplest possible examples of two by 2.3, by three. What should replace the notion of an eigenvector? This is the first crucial idea that is introduced here. That's the notion of a generalized eigenvector. In fact, I spoke about it briefly a few lectures ago. But let me explain it more precisely. Let's look more closely what, how this aperator acts. Let me reproduce again this. And then zero is everywhere else. If we apply to the vector 100, we will get it multiplied by lambda one. Lambda, there's no lambda one, this is number one Lambda, This is an eigenvector. It is easy to show that every agen vector is a multiple of this one. That means that of lambda one. Lambda is the spin of this vector. Now let's call the elements of the standard canonical basis of N. In this case, let's denote them by V is the one which number one in the position and the zeros everywhere else this is V one, this is span of one. You can check that there are no other Eigen vectors. In other words, the space itself has dimension n. V has dimension n. But the dimension of this space, this subspace, is one. But what about other vectors? What can we say about other vectors V? One satisfies this property and we like to write it in this way because it's very intuitive, right? Tv is the V, but know that we can rewrite this formula in a way that will be more amiable to generalization. This is an equivalent way to write this formula because we simply take lambda V to the left hand side and express it as lambda times identity times V applied to V V one. In other words, satisfies minus lambda v one equals zero. Now we are presenting everything not in terms of but in terms of minus lambda I. That means removing this lambda. Things become a little bit cleaner. What can we say about V two? We find that minus lambda squared V two is equal to zero. V two is not annihilated by t minus lambda I, for otherwise it would be Ni in vector and I just told you that all invectors are multiples of V one. However, you can check easily that if you apply t minus lambda I squared to V two, you'll get zero. This pattern continues minus lambda to the power J applied to V j is equal to zero. Then finally, the last one is anhilighted by minus lambda to the n. But you see that means that every vector is annihilated by the n's power minus lumber to the applied to all V JS is equal to zero because, for example, one is already annihilated by T minus lumber, for sure. It's also annihilated by T minus lambda squared cube, and so on up to, and likewise for V23 and so on. Now we come across this equation, which is what we will replace this equation in the theory of eigenvectors. A vector which satisfies this equation will be called the Generalized Eigenvector. In other words, it is not annihilated by the first power of t minus lumber, but it is annihilated by some power of t minus lumber. We see that every vector in the vector space V here with this operator J N lambda is annihilated by the N's power of t minus slender. This leads us to the following definition. This is a definition of an Eigen space, but now we will have, we will call generalized Eigen space. It will consist of all vectors v such that minus m n for n. And so we will prove, we will prove that for any operator acting from V to V, we have this decomposition. Substitutes the decomposition of a diagonalizable operator for a diagnoizableperator decomposition of the vector space into a direct sum of eigenspaces. In general, we see that's not possible because not every operator is diagonalizable. But what will replace it in the case when V is defined over complex numbers is that we can write as a direct sum of subspaces in which this condition is relaxed. I have motivated this condition, this relaxed condition, by what happens for the Joan Blocks. Okay, I'm showing you the logic of it, of what is going to unfold because basically we'll get some interesting technical results which will enable us to prove this. In particular, in particular we will, this will be one of the theorems almost equivalent. A formulation statement is that given an operator from V to V, there is a basis of V consisting of generalized eigenvectors. You see how things generalize. T is diagnoizable if it has a basis consisting of eigenvectors. Eigen vectors are the ones which satisfy this equation, m V equals zero. We realize that not every operator is diagonalizable. What's the next best thing? What else can we do? This doesn't work in general, even though the cases when it. We saw kind of special cases not generic. Still they're interesting, it turns out that everything works out. The key to reformulating the statement so that it applies universally, and not just for dialyzable operators, is to replace the notion of an eigenvector, which is this equation, with the notion of a generalized eigenvector, which is this equation that the N's power of mind anilvector I have shown you what it means practically. Practically, this appears precisely for this matrix. This operator, the operator given by this matrix does have a basis of vectors which are all annihilated by some power of t minus lamb. It has a basis of generalized like in vectors. Any questions? Yes, because yeah, this is a slightly weaker statement. It could be. It doesn't rule out the possibility that you have a vector which is generalized in vector for two different lambs. Then even if you have such a basis, it's not clear that you can write, can separate different eigen values. Just that it's a small technical difference. We'll see it more clearly later today or Thursday. It's almost equivalent in this sense, but once we get, our strategy will be to prove this and then to derive from this, this statement. Okay? Now I'm going to prove some lemmas which we will use to establish the results. Okay? I hope you will see why I decided to start with this kind of a bird's eye view of an overview of the subject. I wanted to show you why it is interesting in this study to consider powers of apparator like Timanlmdy. Why it is interesting to consider all spaces of the apparat which are powers of a given apparator. See because now this is to appear, since you have seen now the kind operators we are talking about, you will not be surprised by what's going to come next. Here is what? Here is the first result that we're going to prove. We are going to take an operator, we are going to take an operator, some operator, from V to V. Again, V is a finite dimensional vector space over complex numbers. We're going to study so we know the notion of null space, right? Let me remind you the null space, in fact we will consider the operator minus lambda minus lambda squared, so on. That's what we want to do. We want to consider the null spaces of the Aperator because that's what generalized eigenspaces from this discussion. It is clear that it's interesting to consider the null space, which is actually eigenspace, right? Then to consider the null space of T minus lambda squared and so on. Different powers considered in all spaces of the separators and to try to understand how they relate to each other. This is space of, this will consist of all vectors V which are annihilated specifically by the second power of slam general. In this definition we don't say what is, we just say that exists for some, but it includes in particular the case when this number is two or three, or four and so on. That's why it's interesting to consider all of these powers to see how they relate to each other. This will be important for us in this analysis. However, without loss of generality, it will be convenient for our notation to remove lambda from this. Because in fact, without loss of generality, we can call this. Then what are we talking about? It will be this newly minted tea, the yellow A T squared, T so on, and we will be considering the null space of t. The space of t squared and so on, and the null space of ten. I hope you see now why we are considering just ten because we don't care what lambda is. We just, we previously denoted lambda. I will just denote it by, with a different color. Okay. So the first question is how do these two relate to each other? That's the subject of the first dilemma that we have. Natural inclusions of t is included inside null space of T squared, et cetera, space of t to the K, which in turn, is included in null space of t k p one. And thus an ad infinitum. For every K there is a notion of k to the K, it may well become zero. But in case it's a well defined peratoropace, well defined as a subspace of V, we have a nested family of subspaces you can think about like this is inside this nested right now, a comment on that nation. Some people use this notation, some people use this notation. This could be confusing. I'm using this as the greatest inclusion, which includes equality. For me, it means subspace, including the case when it's equal, which in the book this notation is used for me is primarily because it simplifies writing. If I want to say that it's a strict inclusion, so to say, a subspace which is not equal, I will comment and I will say included but not equal. But if I write like this, I allow the possibility that they are equal, I allow the possibility these are equal, and so on. I hope it's clear this nation does not confuse you. Okay, so let's prove this. This is obvious if you think about it. Let's start from this. Suppose V is in the null space of that means it's equivalent to V equals zero. But then t squared is equal to t of V. This is already zero. This is t of 0.0 right? If an element is already killed by t, it will definitely be killed by t squared. Which means that every element in the null space of t is automatically in the null space of t squared. But more generally, this means, okay, let me just write it, that the space of t is inside no space of t squared. But now what about t to the K? Suppose suppose that V is in a null space of t to the k. That means that's equivalent to saying that t to the V is equal to zero. But then if you take k, one of which the plus one, you can write as t times t to the right. This is already zero. You'll get t zero. For every linear operator, this is still zero. This means that the null space of t, t k is inside the null space of two k plus one. You get the nested property because you have something for, it has to be inside the null space of t is inside the no space which is squared inside the no space of cube inside the four. That proves this sequence of inclusions. Please ask me if it's not clear. Any operator and any K here. I prove that no space of k is inside, no space of k plus one. But it's about two specific powers. But it's nice to think about it as a collection of inclusions, not just one single in which I was trying to illustrate here. Because then it's also clear that next we prove that two and so on. It's actually a growing nested family of subspaces. Cannot grow too far. In some sense, the vector space is five dimensional, but there are the inclusions. Okay, that's the answer. Okay, great. So now the next question is, before I prove the next lemon, I want to make it a little bit nicer because I want to put call to one so that it's more uniform. You have to 123 and so on. Then I also want to recall that to zero is identity, That's our convention. What is an all space of identity? Just the zero element which is also subspace. Therefore, I could include include one more inclusion, namely one corresponding to zero. I can write all space of t zero, which is actually the smallest possible subspace simply by virtue of t to zero being in our convention, just the ident separator. This notation is not random. This notation conforms the general property that T t k plus one is t times t to the k. T is equal to t times t to zero, right? T times 20 is t times identity or composition with identity. Taking composition with identity doesn't change anything, therefore it is. This will be a slightly nicer formulation for us. That's Lema one. Okay? So so far it's almost trivial. Here is the first non trivial statement to motivate it. Let me create this diagram. I'm going to app the dependence of the dimensions of, of the no space to the K on K, right? If K is zero, the dimension is zero because just the zero vector. And then it could grow, something happens. But we know that it cannot be bigger than N, which is the dimensional vector space, because any subspace of end dimensional vector space has dimension less than equal to N, right? It's not like it can go to infinity. It has to, but the question is, how exactly does this process happen? Let's look at a couple of examples. Example one, suppose that is invertible, then the null space of t is zero. If I try to plot this, I will have the dimension of null space of t. For one, it will be at a point here. Then you'll also see that if t is invertible, then to the K is also invertible simply because K inverse is T inverse to the K. If T inverse exists to the inverse exists, just take K times composition of inverse, that would be the inverse of Tt. Therefore, the null space of Tk in this case is zero. The plot is very simple. Is just the horizontal line or the integer points, if you will, on the horizontal line. Okay, next. As our next example can see that this guy, that's now, what is the dimension of the, what is the null space of it is a span of one. Which we discussed. It's exactly the statement that there is only one dimensional space because the no space of the operator itself. Remember this is the operator in which lamb is equal to zero. We don't have to take t minus lambda I. We do what was used to be minus lamb. We do not buy the no space of t is the Eigen space for this operator and it's one dimensional. How interesting in this case, we also, we start here because if k is equal to zero, then we're talking about the null space of t to the zero, and that's the identity operator. For the identity operator, it's no space, is just a zero subspace has dimension zero. But now we are considering this is a case of the null space of it's like this. I want to connect them to, to make it look like a graph even though the values only makes sense for the integer case you calculate what t squared is. Remember I told you what t squared is. T squared will have on this, for, in the spots which once removed from the diagonal t squared on next diagonal. So on this one. Which vector space? The span of the first one and the second one. Because the first one will be here, here will be zero for the square. The null space of this is spin of V one and V two. Then you can easily check that to the J will have the J diagonal. And then the first J vectors will be in the null space or span the null space. The null space of the J will be the spin of V 1v2j, that's J dimensional. That means that the power n space of t to the K has dimension equal to t k until we get to N, in this case, anicle four. And then it stabilizes because the whole space is the entire space V. In this case, when you get to the n plus one, when it gets to n plus one dimension of the null space, to the entire space is the span of v one up to V. You see what happens. The graph is a linear function. Each time you raise the degree the power square, the dimension goes up by one. There are more and more vectors killed by a higher power. You see if you are killed by t, you're definitely killed by t squared. But the opposite is not true. We have seen that V two is killed by t squared. But it's not killed by t, right? Because if you apply to it, you get this vector, you get V one. That's why the space can grow. The null space of t to the K can only grow. The question is, how does it grow? In? These two examples show that the growth can be not growth at all, but we could just stay at zero. That's actually equivalent to the case when the null space of t. It is trivial. It just consists of zero vector. That's equivalent to being invertible. For invertible apperators, this graph is just the k line. But now, this is the opposite case where the dimension grows by one all the way up to the maximum. Because it cannot go higher, dimension is bounded by n, it goes diagonally up to n and then it stabilizes. The question is, what about the general case? Clearly, it can only the graph in principle, it could go like this, then it could be like this. We don't know apar, the only condition is that it grows and also at some point it has to it because that's the maximum. But for instance, it's not clear that we will always hit. Maybe it could remain at minus one. That's the next. Lemma shows that that's impossible, then it will stay. You see that's very interesting. So that's the first line of surprise that it stabilizes the first time. You don't go higher here, It did not go higher already in the first step and therefore you'll stay forever at this level. Here you go up, you have a possibility of either stabilizing at this level or go higher for that particular patter. It keeps going higher each time, but somewhere in the middle it would stay there. That is a statement of the next lemma that is surprising result in this section them to if the no space of t to the m is equal to the no space of t t m plus one for 01, then I referenced this P m, then it stays there. That is to say the space of t to the m. In particular, dimensions stay the same. If dimensions stay the same, the fact that they included it into one another shows that they're equal. You cannot have a subspaces of a given subspace which has the same dimensions but not equal to it. Stabilization of the dimensions is equivalent to the stabilization of the subspace is equal to the null space or to the plus I plus one for all I means stabilizes after M steps. Let me just draw, this is an example, but maybe it's not such a good example. First of all, this is not correct. Because here, if you go like this, you will stay there. Every time there is a horizontal segment, it will stay there. The only way is to go up or to stay there forever, okay? The only possibility between this and this scenario, a few steps then stabilize. These are the only possibility. Here's another go up, go up, stabilize. Or you could go up just once and then stabilize. Nothing else is possible. Maybe a little bit more nice, sir. Okay. So let's prove this. Please ask me if the statement is unclear, if you're lost what we're talking about, because then I will try to explain. We need to show that for any V in the null space, assuming this, assuming this, we need to show that if v is in the null space of t to the, that's, let's put it this way. For any V and V, V being here is equivalent to being in the null space of t to the m plus one plus I plus one plus I is the same as plus I plus one. It doesn't matter in which water. But maybe not to confuse you, let me just w one plus. Now we know that that's what it means to say that two subspaces are equal is to say that if you have a vector in the first, then it is in the second, and vice versa, so that it goes both ways. This statement therefore could be written as two. This implies, that is what it means for these two subspaces to be equal, but we already know. From lemon one. As you increase the power of the, any vector in the nose space of the smaller one will be in the nose space of a bigger one, right? This will already know, we need to show this. Let's suppose that is in an null space of t to the plus one plus I. That means that to the plus one plus I of V is equal to zero. Right? And that's what it means to be in the no space of the ride. But this is equivalent, let me actually equivalent to saying to the M plus one of T to the V is equal to zero right here. You have to be, I hope by virtue of practicing with various homework problems and so on, that you are proficient with powers of, for instance, for you, it comes as a second nature. This is the type of equation that to the a plus B is to a composed with t to the B. In other words, the usual arithmetic that we do with monomials applies to operators for the simple reason that this is nothing but T d d A plus B times. Right? Where if you wish we could put little circles to indicate what is going on. What do we mean you multiply with itself? It's just a composition which materializes as a matrix multiplication When we replace apparatus by the matrices, right, it's just TT as B times. But if so, you can break the sequence into two subsequences, one a of them and the other having of them, and that's a times composed with B. This is what I'm doing here. I have a property of that. It is annihilated by t to the plus one plus I, t to the power plus one plus I. But that's the same as t plus one applied to t to the V equals zero, right? Which is simply to say that t to the, belongs to the null space of t to the m plus one. However, and this is where we are going to use our assumption. The null space of t to m plus one is equal to the nu space of t to the m. This is our assumption I meant here. This guy not only belongs to the north space plus one, it also belongs to the north space of t to the M. What does that mean? It means that to the M of t to the IV is also equal to zero. But that's to say that to the M plus I is equal to zero. Which is to say that belongs to the nuspace of t to the m plus I. Which is which completes the proof. We started out with the vector in the no space of t to the m plus one plus I. We have shown that it is necessarily in the no space of t to the plus I. That proves the implication going up. The implication going down comes from lemma one. We're done. Okay? Any questions that graphically, you really can think about it in this way. In the case when n is equal to four, there are only 12345 choices for these graphs. Once there is a segment which goes horizontally, it goes horizontally all the way to infinity. Therefore, there are only these options. You either keep going up until you hit the wall, so to speak, of the erector space itself, which we have seen happens for the Jordan block. Or you make the torn, so you fold sooner. Although in a sense, the longer it goes on the weirder, the matrix, the weird matrix of all which is a Jordan block is the one for which it goes all the way up. The most vanilla matrices, so to speak, the invertible matrices are the ones which it never goes up. Okay? But now you see what happens. That what is the largest possible. What is the latest moment when it stabilizes? The latest moment is the dimension, right? Because it has to stabilize somewhere before. And that's elem of three, the latest point. Test me. No space, the largest. Okay. M, such that the no space of T to the M is equal to the no space of T to the M plus one where stabilization commences has to be less than or equal to the dimension of V. Because if it hasn't commenced yet, you have increased the dimension exactly n times, you can only increase up to the dimension of. This is the latest point at which it occurs, which means it happens at the power equal to the dimension. Technical statement. More precise technical statement. This is a colloquial expression of this. The more precise statement is that the null that stabilizes at the level of dimension for sure. Could be sooner, like here, here, here. But not later because there's no place to go. Right? There's a roof, there's a top. You cannot have a subspace or dimension greater than your space dimension of your space is N. That's the next Lemma three, which is that the null space of t to the dimension is equal to the null space of t to the dimension of V plus one. For sure the stabilization will already happen by then. Therefore, then it's also equal to the null space of t to the dimension of V plus I for all I equal to 12, et cetera. Okay, so far all of this looks little abstract. We're playing some game. What is this for? This is all for is the first theorem you see. I have named them lema one, lemur, lam three, because they are technical statements. But we have finally come to have accumulated enough material to prove something that is substantial. That is the first theorem for our set up. V is n dimensional vector space, is an operator from V to V. V is equal to the direct sum of the known space of t to the power dimension of an of department. That's mind boggling, if you think about it. There's absolutely no a prior, it looks very surprising. There's no reason why for any operator V can be written as a sum of its no space plus range. In generally, that's not the case. Obviously, we know that the dimensions add up. But this is not just for every apparator proof. For any operator from V to V, the dimension of V is equal to the dimension of the null space of plus the dimension of the range of. That is the fundamental theorem for linear maps, right? But if you have two subspaces whose dimensions add up to the dimension of the space, it doesn't mean that the sum is a direct sum. For instance, suppose that your space V is the classroom, One subspace is a stage there subspace is aligned in this stage. Okay, the dimension of the first is two, The dimension of the second is one. They add up to the dimension of the whole space. Three, but the sum of the plane and any line which goes, which is in this plane is going to be this plane. It's not equal to the whole space. It is equal to direct sum. In general, v is not equal to null space of S plus direct sum, the range of. But in fact, this if and only if these two intersect at zero. Yes. If you have a plane and you have a line which is not contained in it so that the intersection is zero, then yes, the fact that the dimension set up tells you that the sum is a direct sum and the covers the entire space. What we need to show therefore, is that the intersection of these two subspaces to show that the intersection of these two sub spaces is equal to zero, right? And I have just enough time to do it. So suppose that you have an element which belongs to both. Suppose let's sit belongs to the null space of. Now, let me simplify nation and write n for the dimension of v belongs to the null space and to the range. The first condition means that to the V is equal to zero. The second condition means that is equal to t to the n of W. Substitute this. So how do I have to show yeah, so to the N now apply, apply to the N, to the second formula. You get to the NV is to the two times, right? But to the V is equal to zero. To the two, n is equal to 02. To the n is equal to zero. It means that the n is equal to zero because of stabilization. Let me prove. This is where it becomes useful. Three, because we have shown that the dimensions of the No space is stabilized. After n, the double is equal to zero but is V. This equation is equivalent to the equation is zero. If v belongs to both of this, we have shown V is equal to zero. Hence, the intersection is equal to zero. Therefore, we do indeed have a direct. This completes the proof. Next time we will use this result to prove that the statement that I wrote on this board, that the has a basis consisting of generalized in vectors. And then we will get to the Jordan Canonical form. Okay, so we'll see on Thursday.