 It's 45. So let's resume the lecture. So in this last part, we will turn this combinatorial result in something which is more analytical and allow you, as I promised, with your computer to compute the first spectrum. And then, if time allows, I will talk about the odd tires. Okay, so let's start with the first problem of estimating the eigenvalue density. Here is a three-step receipt. First, as I announced, made the replacement of the complex number by diagonal matrices. And I will show in action how to do that on the normalized trace by the diagonal operator. Then you write your favorite fixed point equation, the one presented by Mark Potter yesterday, like the fixed point equation. You take your fixed point equations that have been presented by Mark Potter's with this replacement. And then you enjoy it because don't forget this. So let me show in action what we do. Let's consider, I will restrict to this case, the question of the sum of two matrices. Actually, in free probability, if you want to compute the spectrum of a polynomial in matrices, there is a linearization trick which allows you to reduce the problem for a sum of matrices. This is extremely powerful. In classical probability, if you want to compute the density of a function of independent random variable, there is not such a trick. If you have a function or a polynomial in independent variable, and you want to understand what is the density of this function and compute it with a computer, it will be difficult. But for random matrices or free variables, there is this magic. I'm not presenting this aspect, but it is actually the same receipt of another replacement and amalgamation. For our rational function, it works as well. And then what is another non-commutative function? I don't know. But yeah, you can do that if you have a, if your expression involves inversions, don't worry if you work as well. And if you look at Mingo's special books, you have all the receipts. Again, it's amalgamation over a good algebra. So let's just focus on this aspect, but don't think that it is just focusing on a single very tiny problem. It will solve all the problems you want. Okay. So first, let me remind what happened when we are in the case of application of freeness. That's it. We assume that xn, yn are independent. One of the matrices is unitary invariant. Or one of the matrices is a Vignard matrix. It's not unitary invariant, but it is asymptotically free. It is unitary invariant. Then assuming that the matrices are bounded, there is an asymptotic subordination property is that denoting the still-gest transform g of, let's say, an arbitrary matrix an is the expectation of the normalized trace of the resolvent lambda identity minus an as you know, we also need the air transform, which has an implicit definition. And this is like this. The still-gest transform of an equals lambda minus the air transform in the still-gest transform minus one. So the air transform is a unique analytic map defined on some domain. I'm not going to specify, which makes this fixed point equation characterizing the still-gest transform. So if you are able to know what is the air transform of an, you can solve this fixed point equation and you will be able to compute the still-gest transform. Then, and this is a theorem of recoulescu, the subordination property with this definition, which are for generic matrices. It should be for the limit for numberables, but let's say we define it like this. The still-gest transform of the sum of these two matrices, xn plus yn in a parameter lambda, in the large n limit, it will be very close to the still-gest transform of yn applied in lambda minus, there's no identity, minus the air transform of x applied to the still-gest transform of the sum. Is it correct? So you have a fixed point equation for the quantity of interest involving two ingredients in a non-symmetric way. That is the still-gest transform of this guy and the air transform of this guy. I have chosen to write this equation. There is another characterization like the air transform of the sum is the sum of air transform, but I like this one because usually there is one model for which we know the air transform, but the other one is an arbitrary data, so we treat it like this. Okay, let's imagine we do that, and if you know no air transforms, there is also a fixed point equation just involving a cushy transform. Okay, so let's apply the recipe and see what happened for not three elements, but three other diagonals. So we take independent elements, we assume that one matrix is permutation invariant, possibly with the variance profiling in the sense that I defined, so is variance profiled permutation invariant matrix. So I refer to the second part of the section for this definition, which would just take a permutation invariant matrix. So what I will do is that I will define new objects using this rule. G of an now will be a diagonal matrix because we make this replacement. Do you think it was a complex number? No, it's a diagonal matrix. So parameter lambda, I will change the notation by putting a big lambda, since now lambda is a diagonal matrix, and we assume that the imaginary part of the diagonal element is positive to ensure that there is no singularity. So we are not taking the expectation of the normalized trace, but the delta, and the delta without expectation, this is important. Here, there is a complex parameter we will replace it by lambda, and this is done. So the still just transform over the diagonal, the name of this guy is just the diagonal of the generalized resolvent, generalized in the sense that here it's not a scalar matrix, but that's it. Then you define the air transform over the diagonal, I will keep the same symbol because it's already an uppercase symbol, but it's solution of the same equation with replacement big G and big lambda. And then it's done. We have the subordination property over the diagonal, which is true in the large end limit in some sense that I'm not going to specify because it's a bit technical to talk about this. But if you have written your algorithm on your computer for the classical case, it won't be painful to just adapt, just be painful for your computer, because now the fixed point equation is about diagonal matrices. So you have much more computation to do, but you know that you can do that. So the diagonal of the linear high resolvent of the sum is the diagonal resolvent of Yn apply to this function. And that's it. And if you do that for matrix models, so I will try to show you a picture. So here it is. So I don't know if online you can see it, so I should do something. Here it is. So can I zoom? So in this picture, there are several matrix models. So there's a name of the matrix model. I refer to the article if you want to see that. In the left-hand side, we have here a GRI matrices with variance profile. On the right-hand side, here we have a Bernoulli matrix or a permutation matrix, which is symmetrized. There are three information in each graph. The true histogram of a random matrix that we sample. This is a light blue histogram. We have a blue curve that fits very well this density. And this is what is done with this algorithm. In the blue line, it is a spectrum of an operator which will be very the sum of two, three elements over the diagonal. For finite matrices, not exactly three over the diagonal, it's just asymptotic. But as you can see, I don't remember what is the size of matrices on this simulation, but it was made with a simpler computer, so it's not huge. And it fits quite very well. Especially in cases where the spectrum is very singular, we see that the approximation is not bad. Here we have some singularities and they are really considered by the blue line. What is the red line? It would be the solution of the algorithm if we assume that the variables are free. So for the red line, we just run the algorithm but with the complex equation, the one I have written in the one part. And as we see, the matrices are not free in the cases we have chosen there. We have cooked matrices to to not be free. So we see that in red, we have something which is smoother because being free turns to smooth the spectrum. And when we go over the diagonal, we can look at something which is much singular. And as you see, you have a recipe, you have an algorithm which is the same as usual, but with these different objects. Is there any question about this? What? Now it's not free. You see it is quite similar on this part, but here you're not right. But there is a phenomenon in this example that could be explained is why those two curves only differ after this point. I don't know. And it's tricky to consider this. I just consider a model where I know that they are not unitary invariant and I cook them so that it is a bad situation. It's not a GUI. It's a GUI with a variance profile in this case. And in the articles, the models are specified. If you take a GUI, it will be free with everything. So when it is written GUI, there is a parameter GUI, which is actually a variance profile that makes a percolation, if I remember well. I removed a lot of entries. So it's a GUI with a variance profile, actually. And here we have Bernoulli matrices. We have matrices cook from the FFT matrix, the fast Fourier transform matrix. We have different examples. We have a Bronian motion from a permutation matrix to a unitary matrix. So just look at different examples where it is clearly not free, but free over the diagonal. I don't know. It's quite mysterious. Why does the diagonal play a specific role for permutation invariant matrices? I would be happy to hear an realistic text from this, but I don't know it yet. Yes? Yeah. Don't break a second mic. So you need to first compute the equation for the implicit equation for the R-transform. It's a vectorial equation because both matrices are diagonal, but it's a large n equation. So when you do the fixed point, I think even finding the R-transform for each lambda must be difficult, no? It's time consuming for your computer because if you have matrices of size 1,000, it is a vector equation between vectors of size 1,000. And you need to do it for every lambda? Yeah. So the reason another question is how to make this algorithm efficient. Clearly, I'm not the guy who will be able to answer this question, but because of course you have to, then what you want to get is the R-transform, the stigest-transform in the classical sense. Because you know that if you have this object, you draw the density by taking minus imaginary part divided by pi. So here you have a receipt to have this big object, but don't forget that the R-transform of the matrix, sorry, the stigest-transform can be recovered thanks to this guy by taking the expectation of the normalized trace of this generalized stigest-transform applied in the matrix lambda identity. And actually what you want is x plus a small parameter for any x on your line. And for any x, you must run the algorithm. So it will be a lot of computation, but at least if you're patient or if you have a big computer, you will get the solution. Yes? The mic. So you need to solve it for, let's say, pick n-thousand, right? But if you solve for n-thousand and n-thousand and one, there should be no much difference in the solution in the curve, right? Like is there some limiting, like, I mean, can you solve it just for n-them, for example, or the error will be very big? Oh, but in general, you don't choose the size of your matrices if you have one-thousand stocks in your market and you want to consider one-thousand stocks and you want to reduce it. You will probably have data which is not a random matrix from a model, but a true matrix given by your, your modeling problem. But I thought in pre-probability, it only holds in the limit that for very large, absolutely, and very large for a free limit in the nice sense, can be 20. For matrices of size 20, you will see that they are almost free. Actually, here, you will need to reach higher bound, of course, because it's need more, there are less randomness. So the convergence towards the free object of the diagonal is much slower. Estimating here, I'm not talking about how can we estimate the error between the two elements. And that's an open problem in this general context. So I cannot say you, you must take matrices of this size to have such an error because I don't have this estimate. What I have is just an evidence that if I take one single matrix and one single matrix, I have nice pictures. And actually, this fact is not the one that we predict because we predict things in average. And it turns out to be true point-wise, almost surely. I don't know if it's true, but we have the evidence that in this context, it is efficient. Okay, I don't understand that you said for small n, like 1000, you, so in free-probability, you don't compute it for some fixed energy, still like stages transform, you only work in the end. So let me introduce a concept. I think it will be useful to pursue this discussion. So here I have an approximate symbol. What does that mean? It means that in the limits, we have the free object. But usually in application, you don't have a sequence of matrices with size goes to infinity. You have a fixed matrix. So what you introduce is what is called a deterministic equivalent. Imagine Xn is a GUI matrix with a variance profile. You will have an approximation of this air matrix. And if I have time, I will give the expression. Okay, what you introduce is the guy who is the exact solution of this guy, of this equation. For your matrices, you know that it is only approximate. But so this is not exactly the still just transform over the diagonal of your son. It is what is called a deterministic equivalent. A deterministic equivalent means that it is actually the still just transform of the sum of two operators, not matrices, but operators, where this guy has the same distribution as the finite size matrix as well. This guy has the distribution of the finite size matrix, but they belong in a space of infinite dimension where they are exactly three over the diagonal. For your matrices, they have this exact distribution, but they are just approximately three over the diagonal. And then what you want to do, if you want to have competitive estimate, is to estimate the difference, let's say in norm between the deterministic equivalent and your still just transform over the diagonal. You want to estimate the true xn plus yn minus g tilde. In the definition g tilde, this is just a notation. It doesn't depend on the sum, it depends on xn, it depends on yn. But here, you want to have an estimate. And if you can do that, it will be much more powerful than just what I have presented. But if you don't care about proving estimate and just want an algorithm, you have all the ingredients. Okay, does it clarify what you need? You know, you can use the free probability setting of large and limited just to have the tool, but at the end, you have finite size matrices, and the recipe is to, on the language, is to use this deterministic equivalent. Okay, so I have seven minutes to talk about otlias, so maybe it's a bit... So we'll ask my questions after. I can ask my questions at the end then. Oh no, we can just keep this part and answer questions now. I think the answer is no, but for example, in the examples that you cooked here, is it not the case that you can find free matrices such that there are some would construct this weird spectra there? Again, you can cook... Like here, you selected specific matrices which are not free, so that you get this weird spectra, right? Which are described by... Actually, it's quite complicated because often you expect they are not free, but as they turn out to be free. Is it possible to show that you cannot find a pair of free matrices, whose sum gives you the same spectra? You cannot find a pair of free matrices. A pair of matrices such that if you sum them, the resulting matrix has this spectra in blue, which would then be described by free probability. But if you have free matrices, the blue and the red will coincide. Yes, it's not obvious, but... Nope, so then my question is, can you find a pair of free matrices such that you would get this blue line and the red line matching? Yes, if you take a GUI on an arbitrary matrices, there will be both free and free over the day. No, I understand, but for these specific cases... Oh, they are not free. Yes, but could you find a pair of free matrices, a new pair of matrices? By picking like this and another one... In each case, I have a pair of matrices, so when you... Yeah, but let's say you have... So these pair of matrices are not free, so that's why you have a mismatch between the blue and the... Now, take... Can you... Could you construct a pair of free matrices? Another picture? Yes, but whose eigenvalue density would end up being the blue curves there? Or it's clear that... Okay, okay, so here the blue line is the spectrum of the sum of the free over the diagonal guys. Exactly. Can I discover exactly the same spectrum? Yes. But I have... Now, I will give you a trivial answer. You take the first matrix with spectrum is this one and you add the zero. They are free. No, no, I don't know. I'm not sure that I can answer positively your question. Okay. That's a good... Yeah, they're very nice. They're very good answer. Freeness, regularizing a lot, so you will probably not get something which is weird, like... Yes, I discovered yesterday that the free convolution gets you always the same kind of edge decay in many, many... But you see that on this wave cure, so there's not some... It's not obvious because there are some singularities. Yeah, yeah, so probably not the case. In any case, it's much more so. And the second comment which is probably nonsense, but... So we went from a theory of freeness whose amalgamation relations were in term of scalar quantities and we can describe these three matrices. So here we are at the next level, you can describe a broader set of matrices and the amalgamation is in term of vectors whose size is now the size of the diagonal of these matrices. So if you extrapolate, you may think is there a kind of even larger theory where amalgamation would be between matrices, but then at the end it would not probably get anything because it would be as hard as solving the direct problem. No, there is an angle where you simplify things still, but you will have a matrix equation. So if you look at the work of Erdos, Erdos is actually with his group in Vienna working with his vector on the matrix Dyson equation, which is just another name for the subordination property. It is equivalent just that it is not considering matrix parameters only complex parameter here, but still taking the diagonal or taking the full matrix and still it's giving you something. There is a subordination property of the level of matrices which is useful. And actually yesterday Max Potters talked about this to talk about these eigenvectors, problems, or outliers localization. And this is what I would have explained if I had time. Let me just show the idea or remind what Max said, but at the level of freeness over the diagonal. So let's say that this Xn is a variance profile GUI matrix. If you take the expectation of the result, not the diagonal, not the trace. I'm taking expectation, but I'm in the world of the GUI, so I have a concentration. So this day I take expectation whereas I told you don't take the expectation in the general case. In the GUI you can do that. Then this guy is actually close to the resultant of Y applied in another parameter that I call omega, which depends on the model where the omega lambda is this lambda minus the transform of Xn applying this guy, no, not in this guy, in the diagonal of the resultant in this guy G of Xn plus 1. So this is not exactly your question. This is the recipe to have the outliers but it shows you that at some point this subordination property it's very similar. It's not true only for the stigest transform, but it is true at some point for the resolvent itself. The only thing that we have mentioned yesterday is that this approximation is not true uniformly in the operator norm. It is true entrywise for the matrices, which means that if you take the deterministic equivalent defining the same way, you don't want to control the norm of this generalized resolvent minus the norm of the deterministic equivalent, you will not converge to zero. But if you take this guy and take the scalar product between two vectors, you will have a good approximation. Is it what people, is this related to this literature on local laws? So local laws is more about, so it's different. Let me explain this at the level of the stigest transform of the diagonal. What is a local law? Let's say we have a parameter lambda or lambda identity. A local law is an estimate which holds when lambda can be small when n goes to infinity. And a strong local law is when you get the good estimate where you have the best rate of convergence for lambda. Why is it important? Is it because when you want to write the density, to get the density from the stigest transform, you must take for a point x some x plus a little complex parameter. So if you make this parameter as small as possible, you get something much more precise. Or in another way, when you're doing the stigest transform, it is a convolution with a Cauchy distribution. So if you're too small, you have this, which is not a good approximation. If you're too large, you have something which is too uniform. And local law means I find the good rate of convergence to have something which is as close as possible to the real density. So proving the local law is like the next step. First step is proving that in some way we have this approximation or in the limits, we can describe the limiting object for which we have the spectrum which coincide with the limit. But local law is making a precise estimate. And this is quite difficult. This is what we do with Jeremie Bigo for this outlier's problem. And we don't write a strong local law because it's not optimal because it's a bit difficult and technical for us. But still we have a method which in practice, I'll show you a very nice estimation of a target. But correctly, if I'm wrong, at least in vignamatrices and quite generic type of vignamatrices, the local law holds even at the scale one over n. Yeah, one over n plus something, plus a little eta for any eta positive. You mean it's not exactly one over n. If I remember, it's true for one over n plus eta for any eta positive. Yeah, okay. But you are at the scale of the eigenvalues essentially. Yes, you just have to avoid to be in this situation and just slightly slower than one over n. Okay, so it's over. Thank you very much for your attention. It was a pleasure to present this work. I hope you enjoy it. Thank you very much. Hello, I have a question by Monday idea. Please, the second method you talk about can it still arrive at the input you explain? I'm a bit confused about the question. So if you can specify a little bit the question, I will be happy to answer it. But I apologize. I'm not really understanding. So do we have something more today? Okay.