 Thanks for all of you who came last time and went to the problem-solving session, so Diane told me that there are unexpectedly many problems have been solved, so that's great. Let me recall what we did last time. We defined these tri-diagonal models, tri-diagonal models for GOE. It was a Jacobi matrix of the form A1, A2, and so on, B1, B2, and zeros here, and the AIs were IID normal 0, 2, and the BI were chi of n minus i, and all of these things were independent. What we proved is that this matrix has the same eigenvalues as GOE, and moreover not just the same eigenvalues with also the same spectral measure at the first entry. That's what we did last time. We also used this representation to prove the Wigner semicircle law, namely that if you looked at the eigenvalue distribution of this matrix, let's call this J, so maybe J over root n is the right scaling, and this converges to the Wigner semicircle law, it's a weakly end in probability. That's roughly what we did. Today, I want to start with doing some refinements of this, just proving some classical things using this representation. Then we're going to go on and talk more about how you can write general beta ensembles this way. The first result that we can prove here is about the top eigenvalue. If you have this GOE, you know that the eigenvalue distribution converges to Wigner semicircle law, and that tells you that the top eigenvalue has to be on the top edge of the Wigner semicircle law, or it could also be above, because this is a convergence in distribution, and there's only a convergence of measures, and there's only a measure of 1 over n on the top eigenvalues, so it may disappear in the limit. It could still converge to the Wigner semicircle law, even if the top eigenvalue would be much higher. It couldn't be lower, of course, because that would be a contradiction, but it possibly could be higher. So how can you ensure that this doesn't happen? This is first was proved by Komloš and Feredi. This is an old classical hazard, and we actually want to show that if you look at the top eigenvalue, so we call lambda 1, and you divide this by square root of n, and actually this converges to 2 in probability. So, in fact, this is going to be fairly short, and here is the lemma that we need for it. And the lemma is extremely simple, so the lemma just says that if you're looking at lambda 1 of a Jacobi matrix, this is the top eigenvalue of the Jacobi matrix, and this is less than or equal to the max over i of ai plus bi plus bi minus 1, with the convention that B0, which doesn't exist, and Bn, which doesn't exist, you just set them to be 0. Okay, so this is very simple. Here is how you prove it. So actually, right, so the simplest truth is that I know is actually maybe two lines. So let's write it like this. So I'm going to write it as J as minus AA transpose, okay, where A is going to be some matrix, plus the diagonal matrix with these entries. Okay, so if I can do that, and actually let me write what A is, so it's just going to be the matrix. I think you have 0 here. It doesn't matter. You could, and you have B1 minus B1 and B2 minus B2 B3 minus B3. You have to put square root in here. Okay, and then you just put 0s everywhere else. So I hope I did this right. Let me see. Maybe I have to do AA transpose. Yeah, I think I did it right. So right if you take AA transpose, then the entries of this matrix are the inner products of the various rows of A. Okay, and as you can see, what's going to happen is when you take this inner product, then on the off-diagonal, if you further away from the diagonal than 1, you just get 0 because there is no, because these vectors only have to, because the way these vectors work, it's, if you're on the diagonal, you just get the sum of the squares of these entries. So the first diagonal, we'll just have B1. The second one will have B1 plus B2 and so on. And then on the off-diagonals, you just have the Bs. So, okay, so I guess that's the end of the proof, basically, because AA transpose only have non-positive eigenvalues, right? It's a non-negative definite matrix. And the top eigenvalue of the sum of matrices is less than or equal to the sum of the top eigenvalues. So the top eigenvalue of this is that, okay? So that's the end of proof. And so here we just have to, in this, in our particular case, right, so the lambda 1 of GOE is just less than or equal to the max over i of this normal i plus chi, i plus chi i minus 1. And so you remember that these chi i's, they looked like i plus roughly a normal 0, 1, half, like that. So most of this max will come from the large chi, as you can see that right away. And there are, you know, there's some power of n of those that will contribute. And they have Gaussian tails. This normal circle actually, in the chi's will actually do have Gaussian tails. So this you can give an upper bound of square root of n plus some c times root log n, which comes from the Gaussian tails. So in fact, we get a reasonably strong bound. I mean, it's not optimal. Sorry, there's a 2 square root of n. Yeah, because there's a square root of n here, there's a square root of n there. Okay, yeah. Yeah, thank you. This is n minus i and this should be n minus i plus 1. But it doesn't change anything with what I said. So that's simple enough, I hope. And right, this is only for GOE, that's true. And in general, you know, this kind of things can only be used for GOE and some invariant ensembles, I'll talk about that a little bit later. At least they haven't been progressed on how to use it for any other models. Right, so this gives you, this gives you that top eigenvalue cannot go further than an extra log n away. And in fact, you know, lambda 1 of GOE, you know this from tracewidam is equal to 2 root n plus some tracewidam distribution, tracewidam 1 distribution times n to the minus 1, 6 plus smaller order terms. Okay, so it's not precise, but it's fairly nice. Okay, so that's comm-lotion-furity. I want to show you another thing which is, again, as I mentioned, last time is perhaps the most famous thing now from random matrix theory. It's taught everywhere in engineering, which is the Bayek-Benerou-Spachet transition. Okay, so what is the BBP transition? So it actually goes back to, goes back all the way to the origins of random matrix theory, right? As you know, random matrix theory doesn't quite start with the Romans and the Egyptians like everything else in math, but maybe with the Scots, with Wishart in the 1920s. And basically, you know, he was using it to study thick, thick correlations, right? So when you, when you do principal component analysis, then, right, you look at the top eigenvalues of a sample covariance matrix and that indicates a correlation, but you have to make sure that that's not just coming from noise. So that's why you use the Wishart ensemble to put in just random, random data values and see what kind of correlations you would get if you just had noise and nothing else. If, so, so, you know, the simplest model for Wishart is I'm just gonna take g, which is a square. You can, of course, make this rectangular, but I don't want to do that. So this is IID normal matrix. Okay. And, and then you can just look at GG, GG transpose and look at the top eigenvalue of that. So the simplest case of Bisoner's Pachet is when you look at the non-null case, right? So when there is actually some structure in your data in your population, what will, what will the sample covariance matrix look like? What will the top eigenvalue of that matrix look like? So in this Gaussian model, this, the population covariance matrix only matters up to its eigenvalues. That's because of the invariance. So you don't really care what the, what the exact structure of that matrix is, or you care about what is the, what are the eigenvalues. The data will have the same distribution as long as the eigenvalues are the same. So, so in the simple, simplest setup, you just, you know, put here some diagonal matrix and you look at the top eigenvalues of this. Okay. So some, some diagonal matrix. And the even simplest thing, even more simple thing, of course, if it's the identity then that's what you had before. But let's just say that there is one strong correlation. So there's one serious eigenvalue. Let's call it, let's call it strength a squared. Okay. So let's look at this particular matrix. What's going to happen in this case? And so the eigenvalue, I think there is a scaling of n here. And the BB2 transmission tells you that this first eigenvalue converges to phi of a squared. Okay, where phi of a is this interesting function. So let's see. So when a is one, well, phi of a, I just write it here. Okay, I write it here. So this is the BB2 transition function. This is just two if a is less than or equal to one. And it's equal to a plus one over a. If a is greater than or equal to one. So it looks like this for a while is two. And then here is the x equals y diagonal. And then it looks like that. Okay, and here it's continuously differentiable at that point, but the second derivative is not continuous. So what is the interpretation, right? If there is some correlation in your data, then for a while you don't see any of it in the top eigenvalue, in the limiting distribution of the top eigenvalue, right? As long as a is between zero and one, you don't detect anything. Which is of course a serious problem for statistics because you'd like to detect something. And then when a goes above one, then you suddenly start to detect the top eigenvalue. And when a is very large, then you sort of have a linear relationship between the sample and the population and the sample top eigenvalue. Okay? So this is actually a theorem. And I think in this form, I think by Bernoulli's Pache, in the original paper they did it for GUE, but this can be done for GUE as well. And it has been done since then in various ways. And let me say that, of course, the results in this paper are much more refined. So first of all, you can take more than just one as long as there are finitely many perturbations here. That's one thing. And you get something similar. And the other thing is that what they really studied, although this thing has proved to be sort of less used later on, is what happens exactly at this point. So if A is actually below one, then the top eigenvalue still have a tracy-vidom distribution limit. If it's strictly above one, it will have Gaussian fluctuations. And then in between here, there is a small window in which there is some kind of deformed tracy-vidom. And actually in the GUE case, that deformed tracy-vidom was first defined using these tri-diagonal models and their limits. So I'm going to talk about that next time. And it was defined for the GUE case, which is the unitary matrix case in this BBP paper. Any questions? Okay. So my goal is to give you an idea of how this transition arises. Okay, so. And actually, we'll prove a version of it. Now, this particular one is very nice, and you can prove it using tri-diagonal models, but I leave it for the problem session because you can actually do it. And this is the real thing. So I'll do a version of it for GUE, because we have been working with GUE. So let me just stick with that. Okay. So what's the version for GUE? Again, it's a very natural question for GUE as well. And this is about the non-zero mean GUE. So you can look at what is the top eigenvalue, lambda one, of just the GUE and my n. And what you change the mean. Okay. So you just add some constant times and all one matrix. So I'm going to have to write here what the right constant is. So of course, if the constant is small, then this lambda one is not going to change. And if it's large, then it's going to be huge. So we have to decide what I should write here. And the theorem says, and this one we're actually going to prove, is that what you should put here is some a over root n. Okay. So in some sense it's, you know, this, if you first look at it, it's surprising because you, you know, you take a GUE and you increase the mean of those normals by a tiny amount, just one over root n. And already you see change. And the answer is, right, is that this lambda one, if you divide by root n, because that's the typical thing, it's going to be two if a is less than or equal to one and a plus one over a, if a is greater than or equal to one. Okay. So it's just the same function over there. Fire away. So, um, so why, where does this fire away come from? Well, it comes from that. But let me give you an even simpler setup where it shows up just so that you have some feeling. And generally it shows up in these rank one perturbations of matrices. But, but here is, here's an even simpler thing. So if you look at lambda one of the following operator, so you just have, okay, so this is also a theorem, although I'm not going to prove it. It sort of helps you think about this GUE case. So if you look at z plus, okay, this, this, this, this infinite operator acting on L two and you put a loop or weight a at, at, at zero. I guess, I guess that plus is going down. And look at the top, the top lambda one of this. So what does lambda one mean is just mean the top of the spectrum. It's this spectrum of this is has a continuous part always. And depending on a, you will actually have an eigenvector. If a is large enough, it will have an eigenvector and the rest of it is continuous. But lambda one is still defined by, you know, really quotient formula. So it's just the, the, the, the maximum of the really quotients you can get overall vectors. So just the top of the spectrum. So this is actually, again, as you, as you expect, it's just two, if a is less than or equal to one and a plus one over a is greater than equal to one. Okay. And, and so this now is so simple, right? It's so simple that if anywhere from this, you should be able to understand this, this rank one perturbation story, this back when there was special history. Again, right, you have the same thing, the, the spectrum of this plus the top of the spectrum doesn't change if you put the loop, which is small. It changes if you put the loop, which is large. And, and, and why is this change? Well, actually the change is simple. There is an eigenvector, which just looks like one a inverse, a minus two, a to the minus three, and so on. Okay. So this thing actually is always an eigenvector in the sense that it satisfies the eigenvalue equation for every a, but it's only now two when a is greater than one, right? Then you can, it's, it's simple. You just plug it in. You see it's an eigenvector. In fact, it's an eigenvector with eigenvalue a plus one over a. And, and if there is an eigenvector, then you have, you get this. And if this is in L two, then of course it will show up in the spectrum. And there'll be some atom at a plus one over a in the spectrum measure, for example. So, so this phenomenon of, of, you know, doing some rank one perturbation and, and it doesn't change for a while is very common. As you can see in these three cases. And so let's, let's, this is the one we're going to prove. We're going to use this as a, as a guidance. The proof is actually very simple. It's mean one G O E. Okay. So, so where does the proof come from? Well, it just comes from the tridiagonalism argument. So the first observation is that because this is an invariant ensemble, it doesn't matter what the vector I write here. It's only the length of the vector that matters, right? Because you could just rotate, if I put another vector of the same length, I could just rotate the whole thing back to this thing without changing the eigenvalues. So instead of putting this vector one, one, t, I could just put E one, E one, t. But then, you know, the straight off is that the length of this is one, the length of that is, is root n. So, so I have to put square root of n a. Okay. So if I replace this term by this term, I get exactly the same distribution for the lambda one. So that's the first observation. The second observation is that when I do the tridiagonalization, remember, I didn't actually change this. So what does this look like, right? So this looks like a GOE and to this corner, you added this term, square root of n a. So you had whatever you had here, you had some normal and now you added this square root of n a to that corner. So when you did the tridiagonalization, this corner didn't actually change. Okay. So you remember, that was, that was actually just untouched because when you conjugate by those orthogonal matrices, which have ones in the top corner and zeros, first row and first column, that, that, that, that corner doesn't change. So, so we get, right, we get the lambda one of this, of this Jacobi matrix where you have square root of n a because there is an a and then there is an a one. So sorry about that. And then you have this chi or maybe put here a normal. Okay. This is chi n minus one, chi n minus two at the dot, at the dot. Okay. Here you just have a normal again, chi minus one here. Sorry. So since it's one over root n, lambda one of this over root n, right? So you need to show that this converges to phi of a in probability. Well, so let's look at two cases. Case one is when a is less than or equal to one. Well, you've already seen that wherever, you know, any limit point has to be at least two. That's just because of the Wigner semicircle law. Right? That's true even if I don't put the say. When I put in the say, that's a positive perturbation. I want a to be greater than zero. So all the eigenvalues move up a little bit. So you still, any limit point has to be at least two. So now you just have to show that it's at most two. But actually when a is less than or equal to one, you can just use the, you can still use the top eigenvalue bound that we had above with the commotion 3d. Right? Because you, a one has grown by, the first entry has grown by scrote of n. But remember the first entry, we had something like max of ai plus bi plus bi minus one. Okay. That's what it was. So when i equals zero, where i equals one, this guy was zero. And this could have been of the size square root of n because everywhere else it was square root of n. So if you replace, if you increase ai by a one by square root of n, this max is still not really changing. So just the original bound works in this case. So, so again, so, so for the upper bound, you have, it works just as well as before. So now we have to go to a is greater than one. And here's what you do. So