 So a lecture by Valin Virag on Beethoven's books. Thank you very much. Thanks for the invitation. It's hard to give a lecture in this spotlight. It's hard to be an actor. So I have the fortune to give the first lecture after lunch in the dark in the third week. So if you want to fall asleep, just feel free to do it. You know, if I should speak more quietly, let me know. The topic today is I'm going to talk about beta ensembles. Actually, that's the main title of the lecture series. But today I'm just going to talk about tri-diagonal matrices and operators. So what I'm going to talk about is basically today is basically an idea that appeared in a paper of Hale Trotter in 1982. He published it and he said, probably there's not going to be much of this idea, but I'll publish it anyway. I had it, so. And since then there's been a lot of work that's been built on it. Very serious results, like recent results about maximum of characteristic polynomials and random unitary matrices and many others. So my goal is to introduce you today very slowly to this world. And that's what we'll do. So the first object that I want to talk about is the spectral measure. Okay, so let's say that you have a matrix A. It's symmetric, okay? And we can define its spectral measure. It's going to be a spectral measure at the first coordinate vector, but we'll suppress that according to the following. So if you take the spectral measure sigma and look at its k-th moment, okay, then this actually should equal the x. So it's a probability measure which satisfies the following. If you look at its k-th moment, it is going to be equal to the 1-1 entry of the k-th power away. Okay? So you can check, and it's a nice exercise, that this actually defines this measure well, and you can also write it explicitly. So sigma is the following. So it's just the measure that puts weights on the eigenvalues. So there is delta lambda i of your matrix A. These are the eigenvalues of A. But it's not the eigenvalue counting measure or eigenvalue distribution, what we will call. It's not the same because the weights are not all the same. They're not one over n, but the weights are actually, they sort of consider the fact that we're looking at this coordinate 1-1. Okay? So it's kind of a local property around this point 1-1. So the weights are actually going to be phi i 1 squared. Okay? So this is the i-th normalized eigenvector. Look at its first coordinate and you square it. So it's an exercise to check that these two are actually the same. Now, let me give you an example. So for example, if sigma, if A is the adjacent symmetric, A is an adjacent symmetric of a graph. Okay? So what is the adjacent symmetric? The graph is just indexed by the vertices of the graph and two vertices are connected with an edge then you put a 1 to the corresponding place in this matrix A. Okay? So if A is an adjacent symmetric of a graph, then A to the k 1-1, and 1 is, you know, say a labeled vertex here in the graph, then A to the k 1-1 is just the number of paths of length k in this graph that return to 1. Right? That's what it is. That's how you do matrix powers. And so in fact, this definition even makes sense when the graph is infinite as long as the degrees are bounded. Okay? So at least this thing makes sense. You can say what it is. So another exercise, because it's a course so you'll get lots of exercises, is to show that if you have an infinite and bounded degree graph, then the spectral measure actually exists as a measure. Okay? So I told you what its moments are, but why is there a measure of those moments? Okay, well. Okay, so that's the next exercise. Okay, so why is the spectral measure interesting? Well, it's interesting for taking graph limits. Of course, we are studying large matrices, right? That's what this whole, or at least large part of this whole school is about, large matrices. So you want to take limits of matrices. That's kind of a complicated thing. Taking limits of graphs is more hands-on. So let's look at a simple graph limit, which is called the local convergence, or rooted convergence. Maybe I'll use that word. Okay? So you have some graphs, GN, and they're bounded degree. Let's say they're at degree at most five or something. And they have a root. So one vertex is distinguished, and we call that a root. Okay? And we say the sequence of graphs converges. The sequence of graphs converges if for every R, the R neighborhood of the root stabilizes eventually. Okay? So eventually if you look at a ball around U at R, it will be something fixed after some time. Okay? So this converges to some graph G. So let's look at some examples. For example, if we take the N cycle and put the roots wherever you want, right? And this converges in this rooted convergence to the graph Z, the graph of the integers. Okay? That's one example. If you take a large box in ZD, right, and you pick something, some vertex in the middle, then again, and that will be the root, then it will converge to the entire ZD. So what's the relationship between the two? Well, you see right away that in this, if you look at rooted convergence, then spectral measure at the root is continuous, is a continuous functional, right? In terms of weak convergence. So if graphs converge rooted to some limiting graph, then the spectral measures at the root converge to the spectral measure of the root of the limiting graph. That's just because the moments, the number of paths converge, right, because this neighbor has to stabilize, and because of that, the moments of the measure converge. So the moments, converges of moments imply convergence of measures. Let's rebound the degree graphs. Okay, so it's a simple, very simple fact. Okay, so here is another exercise. I'll give you two exercises. One of them is let's look at the N path and look at spectral measure of this root. Okay? Well, this is different than if you look at the vertex in the middle, because now this converges to Z plus. Okay? And of course then the spectral measure here converges to the spectral measure there. Now what's the spectral measure there in Z plus? You should probably have seen this before, right? So the case moment of this measure is the number of paths that are of length k and stay in this graph. So there are dick paths, right? Dick paths, I don't know. And so the spectral measure, and the measure whose moments are the dick paths are the Wigner semicircle law. Okay. So that's an exercise. Another exercise I can show is that maybe it's another exercise in random matrix theory, which is nice and connected to this, show that if you take a random irregular graph, so you know what it is, right? So if you take, say, a uniform measure, or you take some slightly, take the configuration model on N vertices. Okay, let's say D3. Then this converges to the deregular tree. Okay, so this is improbability. Okay, with respect to rooted convergence. Okay, so this is TD. Okay, so that's not hard to prove. You just check that the chance of creating, in these models, creating cycles is small. If you pick a root, it's very unlikely that you will create cycles there. And so this also implies that in this random matrix model, which is the adjacency matrix of the random matrix, you have convergence to the spectral measure of this deregular tree, which is something well known. It's called the Kasten-McKay measure. Okay. So I want to talk about another notion of convergence, which is called the Benjamin-Nisham convergence. Okay. And here it's very similar to local convergence. But here you have graphs, that don't have roots. Okay. And what you do is pick a root uniformly at random. Okay, so you're given a sequence of deterministic graphs, and you have the following notion of convergence. You pick a root uniformly at random, and you say the sequence converges if this random rooted graph that you have converges to some random rooted graph locally in distribution. Okay. So what does it mean? It just means that if you pick a random place and the statistics of the graphs in an arbol that you see around yourself converges, the probability of seeing a certain specific graph converges for every specific graph. That's all it means. Okay, but this is slightly more abstract. So what you see is GO. It's going to be a random rooted graph. Okay. And the randomness just comes from the fact that we picked a root at random here. Okay, so that's Benjamin-Nisham convergence. So there are two nice things here. One of them, well, several nice things. One of them is what could be the limit here? Okay, so what are the possible limits that you can get this way? Well, there is something that, you know, you pick the root uniformly at random. So there is something that in this model, which is just a random rooted graph, that has to remember that somehow. And there is a concept called unimogularity. There, which basically just means if the degrees are fixed, it just means that if you make a random walk step in this limiting graph, then this graph together with the random walk step has the same distribution as if you reverse this random walk step. Okay, that's unimogularity. And the big open question here, it's one of the main open questions in this entire area is whether all unimogular graphs can be realized as limits of finite graphs. This is called Soficity Conjecturing. So I won't give you that as an exercise. But you could use an exercise, you could show as an exercise the statement. So the fact that every limit of, I mean if Benjamin Nishram, limit of graphs is actually unimogular in the sense that I told you. So what does Benjamin Nishram convergence has to do with anything? Well, here is a nice fact. Let's say that you have this GM. Can you see this board like that? No? Okay, so this GM, it has a spectral measure of sigma at the root. The spectral measure of the adjacent symmetrics. But then you pick the root at random. So it's a random measure. So you look at its expectation of the random measure. So the measure of a stat under this expectation is going to be just the expectation of the measure of the set. So we call this mu. Okay, so what is mu? That's my question. Who can tell me what is mu? It's something that you know. Excuse me? Exactly. So mu is the empirical, is the eigenvalue distribution. Okay, so why is that? Well, so it's just the one where this weight is replaced by 1 over n. Right? So why is that? Well, look at that. If you pick the average of this, if you sum this over all eigenfunction, then you have to get 1. Okay? That's because these guys form an orthonormal basis. So if you average, you'll get 1 over n. Okay, so this is a nice way to represent kind of probabilistically the eigenvalue distribution. Okay? And so, again, I'll give you the most boring example. For example, if you take this cycle, okay, then it converges in banyan minishom to z. Okay? It's not random, it's deterministic. If you take a path of length n, okay, then the root typically tends to be, if you pick it at random, far away from the boundary, okay? And it will just converge to z again in banyan minishom. If you take some random tree, or some fixed tree, sequence of trees, you may want to use a sequence of trees to approximate the infinite tree, the infinite tree regular tree. Okay, so that may be another example, but it's a nice exercise to show that that example doesn't exist. You cannot approximate the infinite tree regular tree by trees. You can approximate it by the random graph. So, it's another effect here. Okay. So, again, well, what does banyan minishom convergence has to do with eigenvalues? Well, the expected spectral measure here is just the eigenvalue distribution. Okay, so the limit of the expected spectral measure, which is the expected spectral measure of this limiting graph, is the limit of the eigenvalue distribution. Okay? So, if you have a banyan minishom convergent sequence, then the spectral, the eigenvalue distribution also converges. That's a continuous functional. Okay? So, that implies, right, that implies this fact that I told you, the random deregular graph will, the eigenvalue distribution of that converges to the deregular tree, that of the deregular tree. The expected spectral measure of the deregular tree, but the deregular tree is not random, so it's just the spectral measure of the deregular tree. All right. So, I'll come back to this in a second. Just one thing that, if you have graphs, it's weights on them. So, you could put on each edge a weight. So, that corresponds to general matrices A. Okay, general symmetric matrices A. The whole story still goes through. Okay? You just have to make sure, in the convergence notions, you have to put in that the weights also have to converge. So, there's nothing special when you put in weights. The moments, you have to be a bit careful that the weights have all the moments that you need. All right. So, it turns out the spectral measure is a very nice object to analyze random matrices and many things. And as you see, it has a little bit more information than the eigenvalue distribution. So, the eigenvalue distribution, you know what it is, just what the eigenvalue matrix has, but it also has some information about locally what this graph or matrix looks like. So, you... So, me ask the following question. You know, what does it mean that two matrices have the same spectral measure? Okay. Can you give some characterization of this? I think in terms of other properties. For example, let me give you an analogy. You know what it means for two symmetric matrices to have the same eigenvalues, right? It's just equivalent to saying that they're conjugate to each other with some orthogonal matrix, right? And there's also a diagonal matrix with the same eigenvalues, just put the eigenvalues in the diagonal. So, if you think of this as an equivalence class, two matrices are equivalent if they have the same spectrum, or an equivalence relation, then there is a unique representative in each class, which is, say, a diagonal matrix using eigenvalues in the diagonal. And... In each... In an equivalence class, it's characterized by conjugation. So, two things are conjugated in that, right? So, you can ask the same question about spectral measure. What does it mean for two matrices to have the same spectral measure? So, let me tell you. So, there is... Okay, let's maybe say this is a theorem. So, it's... This theorem works in not all the cases, but in the generic case. And the generic case is when E1, so it's the first coordinate vector, is cyclic. And cyclic with respect to your matrix. So, what does that mean? It means that if you take E1 and apply to it A, and then apply A to that, and so on, and you do it N-1 times, then this gives you a basis. Okay? Or equivalently, they're linearly independent. So, that's when we call these things... That's when we call this E1 cyclic. Typical case, if A is diagonal, it doesn't work. And if A is some blocks, it doesn't work, but generically, it's easy to check that this works. Okay, so this will be our assumption. E1 is a cyclic vector, and... So, what does it mean that to... So, first of all, you get that A and sigma A is equal to sigma B. Okay, so the spectral measure of A is equal to the spectral measure of B. If and only if you can conjugate A to B. There exists an orthogonal matrix, such that A conjugated by O. So, this is... Which one is it? O inverse, right? Equal to B. And it's a special O. Of course, this is just isospectral, so far. And you have to say that O is of the form that you have a 1 here and 0s there, so it doesn't move the first coordinate. Okay? There are too many O's, but... Okay, so this is 0, there's a 1, and it's of that form. Okay? So that's... And this is orthogonal. This is 1. And then, how about representatives? Right? So, representatives are Jacobi matrices. Okay, so what's a Jacobi matrix? So, a Jacobi matrix is just of this form. You have A's on the diagonal, B's on the off-diagonal. It's symmetric, of course. The B, I are positive, and the A, I are 0s elsewhere. The A, I are arbitrary. Okay, so that's a Jacobi matrix. And the second statement is that there exists a unique Jacobi matrix so that in each equivalence class, so let's say that sigma of J is equal to sigma of A. Okay? So in this world, in the world of spectral measures, the Jacobi matrices are the ones that play the role of the diagonal matrices in some sense. So, uniqueness is an exercise, and maybe Diana will do it with you guys if you want. So, I'll let you prove uniqueness. So, what does uniqueness mean? It means that if you have two Jacobi matrices that have the same spectral measure, then they're the same matrix. So, I encourage you to think about it. It's not hard. And let me show you existence. Okay? So, existence is shown by this so-called reflections or Lanzuch algorithm. The original goal of this algorithm was the following. Let's say that you have a matrix. You're given a matrix and you would like to store its eigenvalues for future generations who are interested in them. But you don't actually want to compute them because it's hard to compute eigenvalues, right? At least if you want to sort them somewhat precisely, it's hard because you have to solve algebraic equations. But let's say that you want to reduce the dimension of your data, which is n squared, something around n, without doing anything fancy like that, just linear transformations. So, that's what the Lanzuch algorithm starts with. And the way it works is that... So, let's say you have your matrix. This is your general matrix A, and you write it like this. So, this is gonna be A, this is gonna be B. This is gonna be B transpose and something C here. And you conjugate it by an orthogonal matrix of the kind that I wrote up there, okay? Which was... I write it backwards now, so O prime is of the form 1O and then zeros here. I think O's and sigma's are zeros that are clashing. Okay, so when you do that conjugation, you check what happens. The C gets conjugated by your O. The B gets multiplied by your O. Let's say from here, I'm not sure, but for now. Okay, so this is O B transpose. It still stays symmetric. And this is your new matrix. So, what you can... So, this is... And notice that since this fixes the first coordinate vector, it also fixes the spectral measure. So, this is not so hard to check. If you fix the first coordinate vector, you fix the spectral measure there. You're using the spectral measure, but you are doing this transformation to the matrix. So, what you'll do is you'll try to ensure that this O B is of the form 1, 0, 0, 0, 0, 0. Or not 1, but actually... Well, you're rotating your B, so what we'll get here is actually the length of B. The Euclidean length. And of course, you can do that. You just pick your O to rotate it to that coordinate. Okay, and once you have done that, then you have a matrix here which has... You've managed to get zeros over here. This is a big zero here. There is the length of B. This A is unchanged, and you get those big X here are zeros. And then, of course, you just go on one dimension and keep going like that. And then you'll get to a tri-diagonal matrix. Okay? So this way we proved existence. There exists a Jacobi matrix such that sigma A is sigma j, sigma A. Okay? And just to encourage you to do this exercise, notice that you have a huge amount of choice in your O here. The only thing you require from this orthogonal matrix is to take one vector to its right coordinate. There is a lot of freedom. So this freedom makes you think that actually by choosing O differently, you may get a different Jacobi matrix as you go on. And the answer is that you don't. No, there is uniqueness. So your freedom will be canceled out. You may do something differently, but that will influence your later steps to do the same. Okay? And there is another nice fact. Again, I leave that as an exercise. Which is that if you take E and AE and A squared E and so on, and you apply Gram-Schmidt to that, Gram-Schmidt dad. So you get an orthonormal basis. You write your matrix A in this basis. Okay? Then you get your matrix J. It's going to be tri-diagonal. So that's what you're doing. Some sense you're doing orthogonal polynomials here. We're not going into that. Okay? So what was Trother's idea? Well, Trother's idea was to apply this to GOE. Okay? So what will happen when you apply this to GOE? Well, this A is the normal. Okay? Right? In the GOE. This C itself is a smaller GOE. Okay? The B is just a bunch of Gaussians. So the length of B is just going to be a chi, random variable with n minus 1 degrees of theta. Right? Because it's a length of an n-dimensional Gaussian vector. And then C becomes a GOE conjugated by an O. Well, that's why it's called GOE. Okay? Because when you conjugate it by an O, it will still be a GOE. It's invariant under orthogonal conjugation. That's the reason it's called GOE. But the problem is that your orthogonal conjugation was dependent on your matrix, right? It forced your B to go to its... So that may be a problem. But you realize that B was independent of C before. So it depended on your matrix, but only on independent things. So then it's fine. Then you can conjugate by O, and you still have the same distribution. Okay? So if you conjugate by an orthogonal matrix, or you conjugate by an independent orthogonal matrix, you will still get the same. Okay? So you keep doing this operation, and what will you get? Well, you get, you know, a matrix like this, like this. I think maybe I write it... So here you're gonna have some normal, I think in the GOE, it's 0, 2, if I remember. Or 0, 1, maybe. Or 0, 2. Okay? So they're independent. On the diagonal. And the off-diagonal, you have this chi. So it's chi in minus 1. Chi in minus 1. Chi in minus 1. Yes. Chi in minus 2. Yeah, sorry, thank you.