All right, let's talk about, let's finish our discussion of the operators which admit representation as upper triangular matrices. This is what we talked about last time.  Let me remind you the definition that if you have a linear operator on a vector space over some field F, it will say that. We will say, we say that has an upper triangular matrix representation representation if there exists a basis of v called beta such that the matrix of t with respect to this basis is upper. Triangular. Okay? So that's the definition.  Then we have the theorem which I stated at the end of last lecture and will now prove that has a per triangular matrix representation if and only if the minimal polynomial. Of z of t has the form the product of some z minus lamb I from one to m for some lambda one, lambda m in our field of definition. In other words, in other words I, Z splits into linear factors. We know that in general, if our field is a field of real numbers or another field which is not algebraically closed, then not every polynomial can be written in this form as a product of linear factors. It turns out that the ability to represent our linear operator as an upper triangular matrix is actually equivalent to the minimal polynomial decomposing into a product of linear factors. This gives a very nice criterion to know when it can have a matrix representation of this form just from looking at the minimal polynomial. If it splits, then if it doesn't split, then you cannot. Okay, let's prove it. It is if and only if statement we have implications going in both directions. Now suppose that has upper triangular representation. We discussed this last time in great detail. The matrix of relative to this basis beta as in this definition then is going to have this form where you have some entries on the diagonal which will lambda one, lambda. Then we have zeros below the diagonal, right? And then we can have anything above. That's what it means that the matrix is upper triangular as we did last time. We will denote the diagonal entries lambda one to lambda n, assuming that n is the size of the matrix, that is to say n is the dimension of our vector space time. Given an upper triangular matrix, we defined another polynomial, not the minimal polynomial, but the polynomial in which the zeros are precisely the diagonal entries. Last time we defined, define this polynomial specifically for triangular matrices, which we call of Tilda of Z. That's just a product of Z minus lambda I I from one to n. Note that here we are not claiming that this is a product of n factors. It could be which is different from N. Last time we proved that if we substitute our operator into the tilda. Tilda, we'll get zero. In other words, if you put minus lamb product, you get zero. This I explained last time. But we know that minimum polynomial will always divide. If you have any polynomial pitilda of z such that pitild of is zero, then pitilda must be divisible by p of z. P of Z has also this property of the smallest degree which has this property. Therefore, every other polynomial with this property is divisible by p of z. Pid is divisible by what does it mean? Is it has this form, Minus lambda one minus lambda, two minus lambda n. What are the penormous viivisibleyZ, what does it mean? It means that Z is a product of some of the factors, put this way for I in some set minus lambda. Here is a subset in a set from one to end. The only polynomials which are divisible by this product are the ones first of all itself but also you can remove one of the factors or you can remove two of the factors and so on. There are no other polynomials that bid, could be divisible. We conclude that the minimal pinomial, Minimal pinomial He is a product of linear factors. Now there's a slightly unfortunate thing, which is that here I wrote same lambda I from one to M. Maybe it's better to write like in some set S I, where I is is a subset, one to n. It's just a matter of labeling here. I label from one to m. But if we call lambda one, lambda the diagonal entries, then we better adjust the nation to say that the labels are not from one to M necessarily, but it could be like 1357 or some subset is the cardinality of S. Okay, so this way, this direction is obvious by comparing Tilda, coming from diagonal entries and the critisic polynomial and concluding that P of Z must be a product of some of the factors. A product of some of the factors. There is no other choice of PTO. Okay, so that's the proof in this direction. That's easy. This is a little bit harder. Now we say suppose of z is a product, like, since I want to go make the induction by m, let me put it like this, I from one to M, and here I will write z, but I will use a different notation. In other words, I is what in the statement of the theorem would be called the lambda, Lambda j. Okay, There is a slight discrepancy in my notation, because lambda one to lambda d in the diagonal entries. I also want notation for the zeros. In general, the zeros are going to form a subset of those lambdas. It's probably better to introduce a different notation for them then there is no ambiguity. Suppose the principle panoma is a product of linear factors where the zero is denoted by mu I. Let's that has a upper triangular Matx representation. Upper triangular for shorthand, I will just call it upper triangular representation. Okay? We'll do it by induction on M. The first case is equal one you have is minus one. Okay? So clearly that means that if P of t is minus one, I is equal to zero. That's the property of minialolynomal. If you stick into it, you'll get zero. But that means that t is equal to one. In fact, it's one by one matrixsilly to consider one by one matrix, all of them are upper triangular because any diagonal matrix is upper triangular. By definition, if you have one by one matrix, it only has one entry. It is on its diagonal, there is no other entries. In this case, it works indeed has upper triangular representation. I put in quotation marks because in this case, there is not much room to have taper triangular matrices. Nonetheless, it fits in our general definition. That's fine. We'll take it now. Suppose we prove that plus one of degrees up to minus one, this is a number of linear terms. The degree of p is a number of linear factors. Right? Let's prove it now for M. This is a very nice proof. Let's. 44p equal to this product. There are factors. Now let me write it like this. Minus lambda one minus lamda, minus one minus lambda M. How can I reduce it from factors to minus one factor? Okay, here's the trick, and we will use this trick many times, in what will follow the range of minus lambda m. This is a subspace of, in other words, we take this last factor. Of course, we're not claiming that this factor will annihilate if you substitute into it, but we say, okay, fine, let's see what we can get out of it. The range of T minus LumbderMI, the first property that we immediately see. Let's call this U. It's a subspace of a range of any linear operator is a subspace. I claim that it is invariant. What does it mean? It means that for any U, U, U is also in. How can we prove this? By observing that being in this U must be equal to minus lambda M I of V for some V V. All right, But then if you apply to it, which is what we want to do of U is minus lambda m I V here, we can switch the factors because they commute with each other. This is something we discussed before. The operator commutes with any polynomial in t. If you have any polynomial in like this one, then you can switch the order. That's what we're using here. We get minus lambda M T of V O. That means that it is in the range, it belongs to the range which is you. By switching t and t minus lambda I, we are able to show that if we apply operator to any element of U, we again get an element of U, which is the range of this appiator. That's good. This is invariant. Therefore, we can restrict, we can restrict to U in general, if you have a subspace, you talked about this. Imagine V is like this classroom and U is the stage. This is a zero point. You have rotation about the vertical axis going through this point. An example of viubspace stage, because every vector in the stage will just rotate to another vector on the stage. Therefore, this rotation, which is defined initially on the entire space, restricts to a well defined operator on the stage. If the stage were not invariant, for instance, if you take some line which goes through zero, it will rotate to another line. It's not going to be an operator from this line to itself, it will be an operator from this line to another line. But precisely when the subspace is invariant under your operator. If you restrict your operator to the subspace, that is to say you only take as argument elements of the subspace, the result will be in the same subspace. Therefore, it gives rise to a well defined linear operator, which traditionally we do not like this, where this vertical line and subscript U indicates restricting to is a linear operator from U to you. Okay, so now I claim that here, and this is where this is a crucial insight. The crucial insight that annihilates is not the original polynomial, but the originally, but only the first minus one factors. That's okay. Let me explain. Take P of t by definition will be the product of the first minus one factors. We are not including this guy. I claim that this one, if you substitute, let me simplify it, just otherwise, it too much notation. I claim that this guy, if I substitute into it, is equal to zero on. If I substitute sub, this is equivalent to saying that if I take of U of T, so what this means, minus lambda one, I'm now writing, but this is, I substitute for z minus lambda two and so on, M minus one, right? I claim that this is O for all, because is, it must be of this form. It must come by acting of twist minus lambda M I on for V and V. Therefore, this equation is equivalent to this equation. In other words, the fact that U lies in the range means that we can insert an extra factor which will complete this product to the minimal polynomial. But we know that this result is a minimal polynomial. Since we know that P of T is zero, we now get the entire minimal polynomial by restoring this last factor. Now we know that on U, this PU has the property that it has a product of the near factors, but the degree of U is minus one by inductive assumption. We assume that we have already proved in the case when minimal polynomial has degree minus one or less. Therefore, by inductive assumption assumption. What did we prove? Remember, we are now proving this way, that U has an Ppromtx representation restricted to U has upper triangular representation. That's great. Let's choose a basis in which it is a patrangular. Therefore, there is a basis u1um of such that the corresponding matrix is a triangular. But we discussed at the beginning of last lecture, what does it mean to be upper triangular. It means that if you apply it to the element of the basis, it's going to be in the span of the first k elements. For a k, for one, for one to capital. Now, extended, extend it. Let's call it beta prime to beta. Beta is going to be the union of this and some additional set of vectors which we'll call V1v. This is a basis of, there are additional elements we have to throw in to extend it to the basis of the entire space. Like you have a basis of the stage, you have two elements. You can extend it by taking a vector which is transversal to the stage. That's one in this case, but in general could be several. I claim that this basis, the matrix of the entire, will be upper. Triangular. Okay? Because I label them from one to capital. This will be, if you will, plus one. This is M plus n. I already checked the upper triangular property for the first M because from my inductive assumption, it remains to show that the same property remains for those vectors that we have adjoined. Here's another trick. If you apply to one of those vectors V, okay, what am I going to get? I'm, I can write it as t minus lambda m, lambda M. You can always subtract and add the same thing. Then it will not change the result. Right now apply to V. This is going to be minus lambda V plus lambda I. But this is in, because it's now in the range of this operator. Because this vector is obtained by applying this apart. This vector, it is in the range of this guy, remember is the range of t minus lamb MI. Therefore, it's in the span of the guys. Plus there is a component, there is an extra element which is a multiple of V. This is for sure in the span of u1um capital, right? But therefore, if it's in the spin of this, it can also be like this. I can say always I can put extra vectors. Now you see that not only UK satisfies this property triangular property, but this additional vector that we have adjoined satisfies the triangular property. With respect to the ordering on this entire set. Where the ordering goes like this. First you go through the basis of U and then you go through additional basis elements. In fact, only one of them will appear when we apply the operator to V I. But it means that you can think of it as a linear combination of all the U plus all the vs, up to, up to it is triangular in this basis. That's what this means. Okay. Any questions? This completes the result. This is really great because we now have a super nice criterion. When an operator has an upper triangular form or could have an upper triangular form with respect to some basis, all, all we need is the polynomial is a product of linear factors. Now if our field of definition is the field of complex numbers or other algebraticlose field, this will be always. We have a corollary to this. If F is equal to C, every operator has an upper triangular form representation, upper triangular matrix representation. Now this is good. Every app has a triangle representation, but triangular is not the optimal. We discussed that the optimal is diagonal for diagonal matris, for instance, easily raise them to powers and stuff like that. If you have two diagonal metris, they commute with each other and so on. Let's find out, is there an analog of this condition? If we actually want to claim that it has a diagonal representation. That's our next step. Under what conditions can an operator have a diagonal representation? We just found out that plgebrically closed field, every operator has a triangular form, which is great. The only abstraction to an operator not having triangular form for a general field is the inability to factor the minimal polynomial into a product of linear factors. Now we will ask a similar question in a case when T has a representation by a diagonal matrix. What we will see by the end of this lecture is that there is a similar theorem. And the only difference is that if and only if condition is not just that the minimum polynomial can be written as a product of linear factors, but that all of these linear factors are distinct, that there are no multiplicities here. That is a necessary and sufficient condition for a linear appert to be representable in diagonal form. Let's Let's expand this definition. We will say that has a diagonal matrix representation, but in fact we will call it diagonal a nicer term. It's not nice to say upper triangle, we say upper triangular representation. But for diagonal, we can say diagonalizable terminology. We will say no, I will put this in brackets. Is diagonalizable if there exists a basis such that this is a diagonal matrix. Before I start, I get to the theorem that I promised just now. I want to give you an example of diagonalizable matrix and application of this dit, how to diagonalize it and how to use this fact to compute something interesting. Okay, That has to do, actually, I think I should keep this in case you haven't copied yet. And I'll start here. This example has to do with something which is very famous. A lot of people have heard about. This is called Fibonacci numbers. I'm sure you have heard about this too. This is a very interesting example. This is actually is exercise number 21. It's not a good, okay? Exercise 21 in five D Fibonacci numbers. Leonardo Fibonacci di Pizza was Italian mathematician who wrote a book in 1202, more than 101,000 years ago, which was very influential, showed Varus tricks in various methods for calculations of Arius quantities. It was extremely influential for several centuries and some people have argued, there's a beautiful book by Keith Devlin that it was a precursor of the Renaissance. That it was so influential that in many ways, it was one of those things that pushed Western civilization to the new frontiers. The problem itself was motivated by mating rabbits. Okay, I'll spare you the details. Our time is limited. It's not a course about rabbits, it's a course about matrices and inter, operators. But if you can Google it, you'll find it. So the mathematical interpretation is the following. That we have certain numbers zero, f121n, there are some positive integers labeled by non negative integers, 01, et cetera. They are defined recursively as follows. That zero is equal to zero is equal to n minus one plus n minus one. If you follow this, then let's say n. Oh, yeah. Yeah, Yeah, that's right, You're right. I'm sorry, I skipped. I skipped Yes, absolutely. Yes. Because it's a kind of a second order equation. So we the initial condition has to have to correct. Yes. One. Sorry, I have too many numbers in my head right now and I'm confusing them, but please please correct me if you see this again. Okay. So, N and N012, 3456, and so on. So, here is 112 358-132-1304 And so on. Okay? I'm sure you've seen it. So the question is, can you write a close formula for it? Just a formula in terms of not recurrence without using this recurrence relation n times, but just like in one shot. Okay? And it turns out that yes, you can. And for that you need to diagonalize the two by two matrix. Here's how it works. You convert this equation into a matrix equation. How to do it? You introduce a vector VN, which is going to have two components. One of them is V n minus one and the minus on the other one is N. Then I claim that actually this equation, it is equivalent to the matrix equation, that V is equal to 01 11vn minus one. In other words, you can say that this is a second order equation because you find N not just from minus one, but from minus two. It's actually similar in a world of differential equations to the class of equations which involve second derivative. These are called difference equations, where you have a step, you are calculating something that corresponds to n. But it's an expression which depends not only on n minus one, that would, that would be a first order equation, but also n minus two. The way to deal with it, both for differential and for difference equations, is to convert it into a matrix equation. If you have second order equation, you got two by two matrix equation if it's a third order three by three and so on. Now let's check. You have by this definition this is And minus two. Minus one. It is minus one, right? Then and minus two plus and minus one. But this by our definition is F N. We find that this is exactly N. Indeed this is correct. Let's call this matrix matrix A. I can write N as it's a V minus one, Therefore it's A squared and minus two. Because V and minus one in turn is A times V and minus two, I can replace it, and the result will be A squared times V minus two. I continue this procedure and I get a n a n minus one times v one. Now one is well defined. One is 11 because it's 01n is 11. But actually I can also introduce zero, which will be 01. Then one actually is equal to eight times zero. The form will be slightly, it will be eight to the n, and then you put 01. You see, what does it mean? Let me rewrite this VN, which is and minus one N is equal to a to the n times 01. What it shows that a fan is just the lower. You have a matrix and you multiply first row and the second row, it's going to be the lower right corner of A to the N. That's what it is. But how to calculate a to the N, it's easy. Now, we have a nice formula, but the formula is matrix form. Is a matrix formula. And we want a scalar formula, we want a number, we don't want a matrix. But we have reduced the difficulty to the calculation of the N's power. If we could find power of a in some way, then all you need to do is take it up lower right corner because that's what will give you a fen when you multiply by 01, right? So this is where diagonalization helps tremendously, as I already explained, as you have seen before obviously in math 54 in the first course of linear algebra. I'm not saying anything particularly new. One element which is new is the usage of minimal polynomial because previously in the study of diagonalization, in the first course of linear algebra such as 54, you use characteristic polynomial. Which we are for now reluctant to use before we properly introduce the determinants, which we will soon. But for now our tool is not the polynomial, it is the minimal polynomial. Because minimal paloma gives us information about agon values as we know. Right? What is the minimal polynomial of this matrix of A? Which is again, well, this is actually a special case of the class of apparators that we have already discussed, right? It's a special case of this matrix which has one just below the diagonal, right? Then minus a zero, minus one, minus a minus one. In this case, we saw that the principle, the mini polynomial is z to the n. Well, a zero plus a one, Z plus etc. Plus a n minus one. Well, let's just try a n minus one to one plus z to the n. That's a special case. When n is equal to two. This is minus a zero, And this is minus a one. What is the critics pallynomial? It is z squared, I write it now. Minus z, minus one. In this way, right? The cofficient front of z is minus one in front of a constant term, minus one. And then the leading term is z squared. That's one way to find it. Indeed, you can calculate that T squared a squared minus a minus identity is zero. Let's calculate, if you multiply, if you take the square of this matrix, you're going to get 112, minus 0111, minus 1001. You see that all terms cancel out. But now you say, okay, well what a, a coincidence. This matrix just turned out to be of this form. What if it was some generic matrix? How would we find its minimal polynomial, even though it is not relevant to the question, to the problem at hand. I would like to comment on it because in fact one can easily find the minimal polynomial for any two by two matrix. Here is, is a remark. Suppose you're given a generic two by two matrix which is not of this form. That's your matrix A. There are two choices to two possibilities for the minimal polynomial. We know that the minimal polynomial has degree less than equal to the size of the matrix. In this case, when the size is two by two, it can either be of degree one or of degree two. There are no other choices. Well, I suppose if it's a zero matrix, you could say that it has to be positive degree. Yes, because it's monic polynomial has to start with positive degree. Strike that, two possibilities. The degree of, of the minimal polynomial is one. That means that the principal polynomial has to be of the form z minus lambda. Or lambda some number. But then if you subs into it, we should get zero minus lm is zero, which means that is a multiple of the identity matrix. This is actually true for any size minimal polynomial of a matrix, or an operator, if you will. In this case, I'm doing at the level of matrices, let me write a instead of t. But the same argument would work for a general linear operator. Minimal polynomial has degree one. And if your operator is a multiple of the identity operator, if that's not the case, which for example, for us, is not the case, obviously it has to be a degree two. It's a very special case, you can immediately spot that the operator is a multiple of the identity then, okay, you know what the mino panoma is, is just z minus lambda. Where lambda is that multiple for all other apparatus has to have degree greater than one. But for two by two, it has to have degree less than equal to two. Therefore, the only possibility is degree two. How to find it? You know that it has to be Munich P of Z has to be equal to Z squared plus a one. Z plus a zero. Here's what you do. You simply calculate a squared, You have a and you have identity. And you find the unique, there are unique a 0.1 such that this is satisfied. Well, by this I mean 0000. This is how you can do it. We could do it even if we did not. Remember that this is of the special form, we could easily calculate this and this. And we found a one is negative 1.0 is equal to negative one. For two by two matrices, you actually need to use the determinants, never need to use critisic polynomials. It's actually much easier to do it in terms of minimal polynomials. Okay, let's go back to our calculation. We know that the agen values of the matrix, or agen values of the cris point operator, are precisely the zeros of the minimal polynomial. So what are the zeros of Z squared minus minus one? You use the formula for the solution of a quadratic equation and you get one plus minus squared root 5/2 These are the 212 plus and minus. Okay, once you have found it, then a first step. First step is to find the polynomial itself. The second, strike that minimal polynomial. Second step, find zero. Third step, find the corresponding eigen vectors. Now you know which equation to solve. You solve the equation. A times x1x2 is equal to one plus square root of 5/2 times x1x2. Then this is plus minus. Then you have, I have already done this. It's easy. A system of two linear equations. You find the, so it's a homogeneous system of equations. Obviously, there is not a unique solution. There's a solution up to a scalar, obviously. Because if you have an eigenvector, you can always multiply by non zero number, you will still get an eigenvector. The answer is plus minus x. Two plus minus is one minus plus square root of 5.2 negative one. These are the eigenvectors confine them by solving this equation, the system of equations. Then you can build the matrix from these guys. You get one minus square root of 5/2 minus 11 plus square root of 5/21 Then matrix A is going to be Q inverse times lambda one, Lambda two. Did I call them lamb? 12? Let me actually call them lambda plus minus, maybe it's easier. Lamb plus lambda minus Q inverse, right? Then A to the N. As I explained already, multiple times is obtained because if you multiply, it is met by itself the intermediate Q inverses and Q's will cancel out. You will only have one Q on the left, one Q on the right. Here you have the power, but the power of diagonal matrix is obtained by simply raising to the power each of the entries on the diagonal. Lambda to the n, lambda minus to the n. Okay, now we have to apply it, we have to find its right corner. You could do it in general, but there is the simplest way to actually calculate it is to say that it's going to be a combination. It follows that N, which is the right lower corner entry of this is a combination alpha times one plus square root of 5/2 to the n. That's lambda plus to the n beta because does not depend on n. Q is just a matrix with numbers, it's written over there. Q inverse is also made numbers. When you calculate this product, you see that every entry of this two by two matrix is going to be a combination of the Ns power of lambda plus, which is one of the diagonal entries, n s power of lambda minus. Okay? And so instead of calculating precisely what this product is, you can just find alpha and beta from the initial conditions because the initial condition is at zero, is equal to zero, right? It follows that alpha plus beta is equal to O alpha, beta is equal to zero, right? Can I actually check? I want to make sure that I'm not making a mistake. I might be confusing and plus one. Mm, there it is. Yeah, I think I am confusing something. Mm hmm. That's one. So let's see. We have no, I think it's correct actually. So alpha is equal to 01 is equal to one, right? I know zero is equal to one. That's where I made a mistake. Zero. Okay. I'm I want to be consistent with the what's written in them. Zero is equal to zero. I want to be consistent with what is written in the book. And it may be slightly different from the convention that I've been using. So I just want to make sure that it is correct. We have this equation, okay, I think it works. One is equal to one, which means alpha times one plus square root of 5/2 plus beta times one minus square root of 5/2 is equal to one, right? These are two equations, two equations on beta, from which we find that. First of all, alpha plus beta is equal to zero. Here we have alpha plus beta over two, which is zero plus alpha minus beta times square root of 5/2 which is equal to one. Which implies that alpha minus beta is equal to two divided by square root of five. But since alpha is equal to minus beta, we have this and this, which means that alpha is equal to one over square root of five and beta is equal to minus one over square root of five. Finally, we get this formula minus I will write it like this. Okay? Does everybody agree? You see you get a very nice formula. And it's so surprising because these are integers, or at least non zero integers. The first 01 is, but this is written as some of n's powers of actually irrational numbers. This one is well known as a golden ratio. You have the golden ratio to the n plus. It's a conjugate version with minus instead of p to the n divided by one square of five. First of all, for every n is going to be an integer. But also, for instance, you notice that this number grows much faster than this number. And in fact, this absolute value is less than one. As, as grows, these numbers grow as N power of the golden ratio. This is something that people oftentimes notice. It's like, wow, this is something very important. It's like sacred geometry or something like this. Here it follows just from the fact that there are two Eigen values. One of the absolute value is less than 11, greater than one, and therefore, the solution is given by a combination of powers. Therefore, symptotically behaves as the N's power of the larger Eigen value just happens to be the golden ratio. This is my example of diagonalizable operator. Okay, now let's go back to the general theory. Let's find out under what conditions an operator is diagnizable for that, we need one more definition. Actually, let me do it here. So we have the notion of an eigenspace. If lambda is an element of, and you have an operator from V to V, we will define the following subspace of v, call of lambda, which is a vectors in V such that minus, let's just call it maybe to simplify, is an all space space of minus lambda I. That is to say, all vectors in V such that V is equal to lambda v subspace, smaller subspace that contains all the eigen vectors of, with eigenvalue lambda. And this is called the eigenspace corresponding to lambda and okay, now, uh there. So first of all, suppose that lambda one, et cetera, lambda M are distinct, then the sum lambda one plus et cetera, lambda is a direct sum and it dimension or the sum of the dimensions of lambda, IT, I from one to, is going to be less than equal to the dimension of V. Okay? So let me remind you what the direct sum means. Direct sum means that it means that if you have an expression u one plus, et cetera, um, where each UI is in E of lambda T, this is zero and all UIs are equal to zero. In other words, you cannot obtain zero element from non zero elements of the subspaces. Remember the sums we discussed earlier on example. A typical example of direct sum is when you have two lines which intersect just at the origin. Or you have a plane like the stage and a vertical line. In this case, if you take the sum, it's a subspace. Well, in this case, it's an entire space. But you cannot express vector from one as a vector from another, which is equivalent to this statement. We need to show this. Now let's suppose that such a formal exists. Let's call it star. Suppose suppose holds for some non zero elements. In that case, we always say that there must be a smallest possible number of non zero vectors of this form which, whose sum is equal to zero. Let K be the smallest number, smallest number of non zero, smallest number of non zero terms. Without loss of generality, then we can relabel the indices so that they will be from one up to UK. Because I don't want to overload my nation without loss of generality. I assume that these are one up to UK'woup UK plus one. Just relabel them to make them into one UK. Then you have one, et cetera. Uk is equal to zero. All of them are non zero. Okay? All of them are non zero. We know for sure that such an equation cannot be satisfied. For a smaller number of non zero elements, there has to be some smallest possible number, that's k. Now what I'm going to do, I'm going to apply T minus lambda k to this formula, to this equation. On the right hand side, I'll still get zero, but on the left, since one is lambda 113 times one will be lambda one and lambda K will give you lambda k. So this will be lambda one minus lamb de K times one, right? Plus, for two applied to you. Two is lambda 22. I subtract lamb de k. It will be lambda two minus lambda k times two plus et cetera, plus the k minus one. It will be lambda k minus one minus lambda k K. Because the same to formula will be true if I put k here, minus one, minus one. This guy will be annihilated by this because it's an eigenvector with eigenvalue plus zero equals zero. Now by our assumption, all of these terms are non zero. We got ourselves another equation where each element is non zero. Each term is non zero because each of the US is non zero. Each of these numbers is non zero. But there are fewer terms. Now in is one terms, which is a contradiction contradiction. That means that this is a direct sum. If it is a direct sum, then the sum of the dimensions, this sum is actually the dimension of this direct sum. But that's a subspace. It's a subspace of V, so its dimension is less than equal to the dimension of V for sure. Okay. So that's the first result. The next result is, is a statement which gives an equivalent expression of the condition that linear operator is diagnoizable. The following are equivalent. Equivalent is diagonalizable. I remind you, T is from V to V. V has a basis of eigenvectors of T. V is actually equal to the direct sum of lambda. This direct sum that we just discussed, where lambda is are distinct, are the distinct eigenvalues. Okay? Finally, the dimension of V is equal to the sum of dimensions to me. Here, lambda one, lambda m is actually just the list of all eigenvalues, all distinct eigenvalues, lamb one lambda. This is fairly straightforward, running out of time. I will leave it for you to read. It's one of the theorems in the book. The only non statement is to link the last formula to the rest of ittlogical. Now we are in position to finally to prove the statement that I promised. Before I do that, I'll give a very quick corollary of this theory. Suppose you have an operator acting on a vector space of dimension n has exactly n distinct tagenvalues in the field. This is oversight then is diagonalizable. You see I mentioned that last week at the end of my lecture exactly one week ago, and I didn't have time to prove it. But now we can prove it easily from this theorem. That's what is the situation. Situation is that you have n distinct tagonvalues. I have given a number of examples where you've seen that there are fewer than distinct tagonvaluesrI' given such examples on Tuesday and also last Thursday. In those cases, we've seen that it may or may not be diagonalizable. But if the operator has exactly n distinct tagonvalues, where n is the dimension of the vector space, then it is diagonalizable. The reason is very simple that if so, then there exists a non zero vector such that I is lambda I V I. By the previous there, we know that first of all, there's V one up to N of them, because there are distinct egenvalues, they are linearly independent. Let's call this, I guess this would be the two, and that is the three by theorem two linear independent, but because the n of them linear independent subset is a basis, it's a basis which consists of eigenvectors, then by this there three, it has a base of eigenvectors, therefore is diagonoizable. Okay? This is the most basic case and it's a generic case. However, there are many other interesting cases which are not covered by this. Then we ask, is the criterion in terms of minimal polynomial, that's the last theorem, is diagonalizable. If and only if the minimum polynomial of t is equal to the product of Z minus lambda I, I from one to m, where these guys are distinct no factors of the form z minus lambda to some power, power. I don't know where r is greater than one. No squares. My square looks like r. I guess they don't need to, I don't need to change out of all polynomials. There's a special class Palomas, which can be written as a product of linear factors. For some fields that's automatic, we take it for granted. For the field of complex numbers, this is always so for the field of real numbers, not always. Now, suppose you are in that case, then there are still two subcases. One is where in this product there is no repetition of factors. Each factor appears exactly once. That's what it means to say that the lambdas are distinct. Equivalently we say no factors of the form z minus lambda squared squared, or Q, or force power, and so on. No such factors appear on the first degree. On first power, the polynomals are called multiplicity free. Multiplicity free. Now, let's compare this result to the very first. There, very first theorem was about the case when the minimal polynomial is a product of linear factors. We saw that this is equivalent to the operator having a triangular form. Now we impose a much stronger condition. Namely, we say that we ask that the minalplynomial be a product of linear factors without repetition, without multiplicities. It turns out that this is exactly what is equivalent to being diagonalizable. Okay, I don't want to rush things, I want to maybe give you one more time to give you an example to illustrate what's going on. So, even in the two by two case, we can distinguish between these. Actually, I am returning to the example in which, which I gave last time on Tuesday this week. Right? Let's suppose that is represented by this matrix. What is the minimal polynomial if lambda one is not equal to lambda two? This is not a diagonal matrix. Therefore, according to what I said earlier, the minimal polynomial cannot have degree one. It has degree two. In fact, the principal polynomial is minus lambda one times Z minus lambda two. We are in the situation of this theorem four. Indeed, the operator is diagonalizable. It's actually presented to us as a diagonal matrix and we see that the minimal Paloma indeed is a product of distinct linear factors. Next example, suppose they are equal, lambda one and lamb two are equal, I call it lambda. In this case it's a multiple of the identity matrix. That's exactly the case where the minimal polynomial has a degree one and z minus lambda. Again, it is of the form that we want is a product of linear factors. Guys, there is a lecture going on. If you don't mind, please wait outside. I know you have exam coming up. We'll finish on time but please do not create commotion here. Please close the door. Please close the door and do not interrupt my class, please. May I ask you to close the door, please? Yeah. Thank you. Okay. So in this case, the minimal panoma is z minus lambda. Okay? And it is diagnoizable indeed, it's a diagnometri. Guys, please close the door and do not interrupt the class. It's not over yet. Please have some respect, what are you doing, Jesus? The third case is when you can have a triangular form like this, okay? In this case it's not proportional to diagnometrix scalar matrix, so it is a degree two point. And guess what it is, it is actually z minus lambda squared. In this case, it does have a triangular form, but it does not have a diagonal form. It does not have an gen basis. That's how you can appreciate that the appearance of the square shows that you cannot diagnoize in the minimal polynomial, not in the redasic polynomial, the characteristic polynomial, which we'll discuss next week. And for this is the same but minimal polynom knows how to distinguish between the two cases. Okay? How to prove this? First of all, if it is diagonalizable, this way is easy. Okay? You have an operator which is diagnoizable, will have the form, let's say even lambda one appears several times, then lambda two appears several times, and so on. You can see easily that you don't need to take a square or higher power. Okay. Again, Jesus, could you please close the door? The class is not finished yet. You started today, your studies come on. Okay? In this case, see for example here, you don't need to overkill it. You don't need to take a square for this matrix. The first power will annihilate. Likewise, even if it has length 34, and so on, the first power will annihilate that block. Every block will be annihilated by degree one minus B1b2 and so on. Therefore, you don't need to take squares, squares of cubes, and so on. This way is very easy. This way is a little harder. The argument is very similar to the argument we used when we were proving the statement for the upper triangular matrices. In this case is induction by m by m. Again, the idea is to take the range for the inductive step. Take the range of minus sum m, we get a subspace of v, which will have smaller dimension. It's a very similar argument to what we already did. I ationally running out of time. I want to tell you about this week and what happens will be next week. There's one more section for this week. Before I get to that, we have three sections. This one is about upper triangular matrices, one is about diagonal diagonalization, one about commuting apparatus. In the diagonization, you are not responsible for Gerzgoren Disc Theorem which comes on the last two pages of section five D, but you are responsible for the commuting apparatus. And I didn't have time to talk about this, but it's a very straightforward material and we will come back to it later in discourse. So this will complete chapter five. And then since we talked about, I mentioned Jordan canonical form and so on, so I decided to switch the order. And while all of these materials fresh in our memory to talk about Jordan canonical form, all the preliminary material for it, which is chapter eight next week and the week after the break. Okay, And then we'll come back. So don't worry that we'll skip the chapter on inner product spaces. We will come back to them after we deal with this more refined information on the inner iterators. Okay, five. It's the material of the previous week. Just five? Yes. Okay, this is a caveat. You cannot say that. It's just about that it's focused on, that's going to use previously it might use something from five, won't be specifically focused on five. Okay. You're welcome.