 Okay, so I'm going to continue my series of lectures on eigenvector problems in random matrix theory and estimation problems using eigenvectors. So, today I'll be using quite a bit of, of free probability theory and and transform that I've written there. And so we'll be using them today so it's a bit abstract like that it's just a bunch of formulas. Well, first, I'll start with reviewing those that I've done for it for for addition so, and then I'll re raise it because today I'll be mostly doing multiple matrix multiplication. But in the case of addition, this is much more. Suppose, whoever does free probability or, or, or, or use a free policy in random matrices knows about the additive problem, but just, let me just review it. Suppose that I have a matrix C that's built out of the sum of two free matrices, which here I've expressed the fact that a and B are free by explicitly rotating be the matrix B by a random rotation. If there's a random rotation between a and B, then, then you can use this, but you don't necessarily need this random rotation. If, if either matrix is, is by self rotation invariance like a Wigner matrix or white wishart. Okay, but to make sure. And so, when you have such a, such a matrix addition a free addition, then there's a transform that's extremely useful called the R transform, and the R transform is additive, additive. It behaves a little bit like the generating function of the cumulants in, like the log of the log of the characteristic function in the standard probability theory. So the log of the characteristic function in standard probability theory, give you the cumulants, and the cumulants are additive under normal convolution, normal addition of independent variables. So the key for free non-commutative variables is this R transform that additive, and how is the R transform build. Well, first you build the resolvent which is a random matrix built from the random matrix A. You take the normalized trace for me tau here is always trace divided by the size of the matrix. So it's okay, but, but in informal, informal algebraic setting, you just have an operator. So what I'm going to introduce I think can me forgot what's your notation is fire. It's fire. Yeah, so some people use five some people use tau, I prefer tau because it's closer to the trace. But so my towel is chemist fine. Yeah. And so this is a normalized trace operator, and it converges for Z outside of where the eigenvalues are this actually converges to a deterministic function. Whereas, you should really think as this, this object here is a is an n by n matrix it doesn't converge to anything. It's a, it's a random object. It's a random object with poles at the eigenvalues and and projectors. And the residue is a projector on the, on the eigenvector, but this is a this is a function that converges typically has a branch cut where the eigenvalues sometimes as directs, direct masses show up as poles. So this is a function G, and it's, and since I'm looking at bounded objects, it's always nice and analytic away from the where the eigenvalues are so near near infinity, if things if things are bounded. It never extend all the way to infinity. So there's always a region where it where G is well defined it's monotonic. And so if it's well defined and monotonic then it can be inverted. So kind of a abusive notation but I find very useful. I have G, G of Z, and then it's functional inverse I write Z of G. So Z of G is just a functional inverse of this guy. And it's as a Taylor series at infinity, it always starts with one over G, which I subtract to build the R transform, and then then this becomes regular. It becomes regular at zero. Okay, so this function, the R transform is regular at zero. It's the functional inverse of the steel just transform minus one over G, and another, and this, these relationship. The fact that it's additive, and from the definition you can work it backwards. It's a subordination relation, which tells you that the steel just transform of C, at some point. Z is related to the steel just transform of one of the two matrices. In this case I chose a evaluated the point, which depends on the R transform of B, and the steel just transform itself I'm trying to compute this so this is a this is a nasty equation because it's a implicit equation I have GFC here and GFC there, but it's actually quite useful. And what's very strong is that this is true for these are still just transform so these are functions, this is a function of a complex variable into the complex plane. But this, these relationship extend to these big matrices here that are random. Well, but instead of getting, but, but of course, this is not an equality between random matrices it's an equality between the expectation value of a random matrix. And so here for instance I assume that the matrix a is deterministic. So if the matrix a the deterministic, this is a this is a deterministic function of its argument. And so, if I average, I should say here, if I average over the noise of five over B, or five over the rotations of B, then I have this in relationship that this big, this big matrix G of a, or GFC in this case, so which would be z identity minus C inverse. So if the expectation with respect to the noise and be, then I get a relationship with with G of a which is the identity minus a, but where is the, the argument is the same argument as this guy here. And that's extremely useful to compute, because in, in, in this big matrix, I don't I have eigenvalue information but also have eigenvector information because I have projectors on all the eigenvectors. So I have, and so I have a relationship about average eigenvectors and when I'm going to sandwich this relationship with, with certain vectors that I'm interested in, I can compute overlap. So, how the offer for me that well, when I say overlap I really mean square overlap which is the dot product square of two vectors. Okay, so, but today I'm not going to use this I'm going to use something something slightly more complicated which is a multiplicative case. So there's a whole zoo of, of transform for the multiplicative case. I want to explain right now the multiplicative case, I'll, I'll, I'll explain it when I need it. Okay, so when when I, it's just the one thing that the for the multiplicative case you could you could use the, the still just transform, but the equations are a bit more clumsy. And so there's a transformer that's more natural here is the, what I call the T transform. So, and so, well, the object I'll be looking at is not no longer z identity minus a inverse is the matrix, the product of this times the matrix a. Okay. It turns out that this object of the equation are much simpler. So, I'll try to build. If I try if I can build this combination matrix a, I use data, because I use a different Z because I want to when I take the inverse. I can say that this is the inverse of this function and not the inverse of that function. This is the, the drawback of this notation saying that function. But the point is that I want to build a function which is a times zeta identity minus a inverse. And if I can build this function, then I can say a lot of things about this because of all these powerful relationship that I explained better when when I need it. So the other thing I want to say is that when I talked about the, the problem of phase retrieval. Remember, I define a function H of Z, which was really the normalized trace of H H transpose. And I didn't compute this object and I still won't do it, but I'll show just give you a hint that the technology I'll develop today for this estimation problem can be used to compute this object. Okay, but it will be to take too much time as just a computation. But once you know how to use, if you know how to use this technology, then it's relatively straightforward to compute something like this. So G to two was, was of Z was an object of this type Z one. The identity minus M to two inverse and M to two is really a color was really a color wish hard was really age F H transpose where F is a diagonal matrix of the objects F of Y. And the important thing is that it was F of Y only depends on the element one, which are no longer in the matrix to two. So F is really the diagonal element of F were independent of this this wish are these these white matrices. So basically what I have here is some sort of colored wish art. So a color which are with with the color in the middle rather than the outside, but it's still some sort of colored wish art matrix. And I needed to compute, and I have it's resolvent, and I wanted to compute the normalized trace of a white wish art with the colored version. And if I can use and and the and the and the this G here is will be related to the T transform and have a subordination relation for a T transform, so I can compute all this, and it will depend. In the end it will depend on the S transform I haven't really told you what the S transform of this F. Okay, so, so basically if I know the statistics of F, I can compute its S transform, and I can compute all this. So I won't do this but we will do something similar in this problem of, but I just want to say that problems like that are solvable. That's that's all. And then, well, just advertisement from my book. It's in there. Okay. But now, my point was to go back to to covariance matrices so I talked about covariance matrices on Monday. So today the two sessions will be on this on on covariance matrices. And so really there's, there's a direct problem and there's the inverse problem. So the direct problem is I have a matrix C that I know. And I measure some data age, and I convinced you that you can model the data square root of C H zero. Okay, so H is some rectangular matrix, where the columns have covariance C. Okay, and a zero is a white rectangular matrix. So, so a zero is just a n by T ID and zero one. And then I color it by multiplying by square root of C. And if you if you actually do the random number generation, this is the way to generate correlated data. Okay, you multiply by a matrix on the left, and then all the columns are correlated within the columns and then each column is independent. Okay, so you have independent columns and correlation within the columns in this setting. Okay, so the direct problem and then. I measure a sample covariance matrix as one over T H H transpose, which can be modeled as one over T square root of C, H, H, H not. And this combination, H not H not transpose over T is a white wishart. So I can really write this as square root of C. And I put a Q here, because a wishart depends on the parameter Q and Q is the ratio of N over T. Okay, and remember, standard statistics is when T goes to infinity at M six, so standard statistics is when Q goes to zero. But we're interested in a limit where Q is finite. And I told you that in the direct problem. In the direct problem what happens is that if Q is greater than one, then he has some zeros, I get zero again values, and actually quite a lot of them. Okay, and but also if Q is greater than zero. It's variance. So trace of E squared minus trace of E squared is is equal to the variance of C plus Q. Okay, so this this this this variance here that I computing, you can think of it as a variance of the eigenvalue distribution. The eigenvalue distribution is wider for the sample covariance matrix and from the from the true covariance matrix. It's wider by a factor of Q. So again if Q is zero, then the has the same or if Q is zero, really E converges to C. If it's one zero then he doesn't converge to see you can already see that it's he has the same mean, which I will always choose to be one. So I'll normalize things always that. So, so I'll choose tower C to be one. Okay, so tower C equals one means that the, the, the typical eigenvalue of C is one, the average eigenvalue of C is one. So this is just a normalization I can always scale my problem to have this condition. And so, so see some variance. except, well, see, see could be the identity sees the identity then see has no variance. And then this term is zero. So when sees the identity, the data, the idea that the eigenvalue distribution of the identity. So this is row of lambda. It's just a direct one. Okay, so this is for C equals the identity. And in this case, the variance is zero, but the variance of e is non zero, it's q. Okay, and actually in that case, we know e, e is called the martian capacitor distribution. So, in that case you would have that e as the shape from lambda minus two lambda plus with lambda plus or minus equals one plus or minus q squared. And again, if q goes to zero, then lambda plus equals lambda minus equals one, and the martian capacitor conversion Iraq. Okay, but in general, for q non zero, but this is true for q between. Well, this is, I won't discuss much the q that the case q greater than one, but in the case q greater than one, you also have a dirac at zero. Okay, but but the the martian capacitor. At least the field just transformed the martian capacitor has a pole at zero precisely when this happened but anyway, I won't have time so let's forget about the case q equals one let's just assume now that q is between zero and one. But the case q greater than one can be solved is just so they're not too, too complex if I too much and they're going to talk about. Okay, so. And I found, and the other thing is that I, I showed the equation for a slightly very different version of this. I showed that that there's a there's a there's an integral equation. If you if you write E, not like this one, but a similar problem. If you write this problem. Okay. Then by considering E as a sum of projectors. You can write an integral equation related to the eigenvalues of C to this, the eigenvalues, the spectrum of E, and this was actually the work of martian coin pastore and 67. But by by by moving around matrices and considering matrices that have similar eigenvalues you can actually find a spectrum the same method for this problem is just a bit more cumbersome. There's a more direct method to compute this, and it's using these transforms. Okay, so this is again, still in the direct problem. And now can erase. So actually today we won't be doing the additive case at all. I guess can read but the additive. I mean, if you, if you need to know one transform should know the are the s, the s you should only know you can forget about it. And, and relearn about it when you need to do multiplicative problems, but the are you should know by heart mean the R is super useful. It's everywhere. Okay, and there are even relationship between the R and the S. But anyway, but today we're going to talk about the S. So maybe it's a good time now to. So you see. Okay, the notation is bad because it's on this. So my sample covariance matrix. So maybe you just write this is here. This is important. So my sample covariance matrix is the square root of a matrix. You multiply and the, well, this is also just it happens naturally but you want it's important that the matrix that is symmetrical. So, or if you can do all this with with with her mission matrices, and in the complex case, and it's this case is would be an Hermitian matrix. So it has real eigenvalues, and it can be diagonalized. So it's nice to work with the symmetric product. So if I go back here, I really want to do the product of the matrix also doesn't. So here, really morally what this is the product of the matrix a and the matrix B. So I'm assuming that a and B are positive definite. Okay, I mean I can relax these assumptions but for the will be considering they could have zero eigenvalues and one of them could be non positive definite but let's just assume it's a simpler to assume that both matrices are positive. They are positive definite. So I can take the square root, the square root of the matrix I recall you, you can, you can there are many ways of seeing it but that's the one that's that that's symmetrical is you, you diagonalize the matrix. You take the square root of the eigenvalues which you can do because it's positive definite, and then you, you go back in the physical basis. Okay, so, so by square root of the matrix. So if I have a which is equal to all D transpose square root of a is really defined as all square root of D transpose, and square root of a diagonal matrix is really square root of the elements on the diagonal. So this, and, and I want to put a square root of a on both sides, sort of a symmetric matrix. Okay. This is kind of object and I'm, and I mentioned this matrix Oh here is only really needed to ensure that the problem is rotation invariant here in the middle of a white wishart. And the white wishart is invariant by, you know, in law, a rotated white wishart is just as likely as the original white wishart. For a white if B is a white wishart I don't need these matrices, I have I do have a free product. Okay, so what I say is that in this problem there exists a transform S, which is multiplicative so the transform as the S transform of this matrix C will be equal to the S transform matrix a and the S transform of the matrix B. Okay, and Of course, of course, all this is a synthetic. Yeah, yeah, yeah, yeah, it's only true. Yeah, so basically, at finite and they're one over and corrections all this. Okay, so, so yeah, of course, yeah, so, yeah, so this notion of freeness doesn't apply. So basically you. So, large, rotationally invariant matrices are free up to factor one over N. So, if you look at the equation that you need to satisfy to actually be mathematically free. You see that there's a one their terms in one over and, and so it's only that's why we say that large that random matrices are asymptotically free, they're free. They, when the size goes to infinity, but for us we're going to be talking about matrices of size 1000 in real applications and one over 1000 small enough given all the other modeling error that we make that this is not the main error. Okay. So, so, Yeah, so the, I have these are these S transformer multiply and the definition of the S transform is similar to the R which I've, so you need you have a you have a transform on the matrix itself in this case you're just kind of the equivalent of a we call maybe a T resolvent or in a T matrix. And then the normalized rate of this T matrix is some function. So it converges in a complex plane outside of where the eigenvalues are, it's actually related to the steel just transform. So it's just, it's just zeta g of zeta minus one. So it's, it's, it's very nicely analytic outside of where it is very similar has it has a branch cut where the eigenvalues it can have poles. So the eigenvalues it's, it's analytical, and, and it's a regular at infinity. This is just that the behavior at infinity slightly different, is that the, the exponent of one over Z are shifted by one, and you don't have the one over Z pole at infinity. Okay. It's actually new overseas but it doesn't matter. Anyway, so from this, you can compute the S transform, which has this for some reason. It's T plus one over T in an inverse of this T transform. And more, more beautifully, you have these, these subordination relation, so the T transform evaluated at one point of the product matrix is the T transform of one of the matrices evaluated at the different point that depends on the S transform of the other. And what, what I'll be using for to compute eigenvectors is this subordination relation. So if you think of this T matrix T matrix again, like the like the resolvent it's a random matrix but I can take its expectation value with respect to be. So typically I'll take the expectation value with respect to the, to the white wishart, which we can think of it as a random matrix. So if I have if, if a is a fixed matrix, and B is a white wishart for instance, then I can average over the white wishart. And then I get the relationship that the average of, of this object for for the product matrix is given by this object for the determinants a evaluate at some point to the same point as here. And this point depends on the S transform. And the last thing I want to say is that why this is also useful is because white wishart have a very simple S transform. So if I have a white wishart parameter q s q of T is very simple is one over one plus QT. Okay, so, so if I'm averaging over a white wishart this term is relatively simple here. Okay, in general, it can be quite nasty. This S transform. Yeah, the subordination relation on the on the T matrices here that this generalized the resolvent, does it become true without the average in the large size limit. No, no, because this is a, this is a matrix with the eigenvectors are all over the place. It's an n square object. It doesn't. Yeah, but I mean the resolvent for example, can this one yes. Yeah, this is this is a scalar equation. No, no, but the resolvent seen as a matrix that there are all these notions of local laws and they, and in certain scales I mean you have concentrations at the level of the element, but I'm not too close to the real axis now. Because this this is a this is an object with n square entries. And I don't see how all of them know that the eigenvectors fluctuate. So, for instance, I think this way if, if a is diagonal, if a is diagonal T of a is also diagonal. And then without the average, it would mean that T of C is diagonal. And it just cannot be it's a random matrix, it will have off diagonal entries, is it that on average the off diagonal entries are zero, but I think they may tend to zero in the large size limit. And for the resolvent of a being there, I think that's the case. We can discuss some I really don't think so. I think that that you would get absurdity to actually use that in some of your papers. Yeah, you can get generalized elements that you can do it for any vector and so you can take a unit vector with just one component what you want. And this shows that each element essentially concentrates if you're not too close to the. Okay, yeah. Okay, yeah, but it depends how you use it, because yeah. Okay, so I see what you mean. But it with this kind of argument, you say that a bigger matrix is a zero matrix. Because, because all the elements are small in a bigger matrix, all the elements are small. They all, if you look at one element of a bigger matrix, it converges to zero. And but the whole matrix itself when you multiply it, you multiply and and such random matrices, such a random elements. And the fact that you have any of them. They're all small, but it becomes big. So, so indeed, okay, this is quite subtle, and thank you for pointing that out. Indeed. But I think this is the best way to see it a bigger matrix is essentially element wise, the zero matrix, but collectively, it's definitely not the zero matrix. Yeah, so you can, yeah, so it's okay that this. So, indeed, a point wise if you only look zoom on one element, and you're not going to use n elements, then, then you're right. Yes. Okay, now I see what you're trying to say. But you have to be careful with these things, you can't treat them collectively anymore. If you say one element converges to zero. And then such elements don't, it's very dangerous to treat n elements, because because they convert to zero as one of our square root of n, or was one of ran. And so when you have any of them, then, then you then you get a spectrum of eigenvalue, this is the bigger matrix definitely not the zero matrix it has eigenvalues that are between minus two and two. The zero matrix has all eigenvalues equal to zero. So, so, and of course, the, the eigenvalues as a function of all the elements. So function, so you can each element convert to zero, but all the elements can treat it together, they're definitely not non zero. So, okay, so if to be careful. Okay, I mean, being a physicist you have to, to, to be careful what you do where we're going to addition you make precise statements, and, and, and vague statements need to be, you need to be careful with vague statements. Okay, anyway, so, I'm going to rewrite. So basically, if I have E equals square root of C, we should q square root of C, then the subordination relation tells me that T E of zeta equals T C of zeta. And I know also that the S of a white wishart of parameter q is one over one plus qt. So I have a subordination relation which I can write as zeta one plus q T C of theta. And this is equivalent to the ugly integral equation that Martin point best or wrote. Okay, so if you know C. You can compute values of C, you can compute this it's T transform, and this is an this is an implicit equation because T E here with T E. Yeah. Okay, so, so it's an implicit equation because I have T E here, but I also have T here. In general, it's pretty ugly. This one simple case where C is the identity. So the T matrix is identity, zeta identity minus identity inverse. And this is just essentially the matrix one over zeta minus one. It's really matrix. Okay, maybe I'll write it this way. It's one of zeta minus one times the identity. Okay, so the T matrix of the identity is super easy to compute and therefore the T transform. So the T of the identity is just one over zeta minus one. And then I won't do it. But if I say that see here as this very simple T transform. I can develop all this, I get a quadratic equation in T. And if I solve this quadratic equation in T, I get the T transform of the marching purpose to her law. And with the relationship between the steel just and the, and the T transform, I can recover this this there. I can I can also recover the density. I can also you can also there, you can also recover directly the density. I mean, the from the T transform there's a branch cut, and the branch cut is instead of just being the density is lambda times the density. So, so the T, so basically here what I'm saying, if I do this in this, this very simple case, the white wish art. Then I get a quadratic equation, and there's a branch cut offer for T that I get. And on the branch cut I can recover the the Martian Cooper store law. Okay, so this is a very powerful way of recovering Martian Cooper store, but it is what it's very powerful because I can also do any colored one. There's one case also that's solvable. That you know, you should know about. I'll talk more about it. Is what it's called the inverse which art. It's one of the few. Well, they're basically only two, two random matrix that are very simple. They're there, not maybe, but they're, they're as transforms are typically very hard to compute, but they're few simple case one of them is the wish art, and the inverse which art, which would, which is the inverse of a wish art matrix where I rescaled it to have mean one and variance P, it would have S of P equals to one plus PT. So in another case of a very simple. So, the inverse we started to take a wish art matrix, you invert it. Well, the inverse doesn't have quite mean one so you change the mean, it has some some some variance which depends on q, which you rename P. So this is a, this is a matrix with mean one and variance P, but it's really built from from the inverse of a wish art matrix. And this object has a very simple S transform, which means that if I injected here. I also get a quadratic equation. That's a bit more nasty because it's a quadratic equation that will depend on P and q, but I can do and I can have an explicit quadratic formula and basically the only thing we can do. In life or quadratic functions anything cubic or Fort order is too messy to, to, to, to work with, but for this case you get a quadratic equation. And, and this is going to be very important, because so the Jacobi ensemble of random matrices is essentially the product of a wish art and inverse which art. That's one one application. The other one is in Bayesian inference. If you if you're trying to, if you this was this this problem that I'm going to talk about inferring the matrix C was done by statisticians. I think in around 1980 or late 70s. And they say well the only thing you could do is an inverse wish art, because that's the conjugate prior. So if you know about Bayesian inference. So if you look at the prior that's the easiest to work with. It turns out that inverse wish art are is a, and you can do that in finite dimension. I mean there's a Pierre Palo showed you the, the wish or distribution for finite and the inverse wish art is also known for finite and and for finite and their conjugate in the in the Bayesian sense. So if you have a Bayesian theory of random matrices, then the easiest prior you can work with is inverse wish art, and you can do this inference problem and if you assume an inverse wish art, you get very simple answers. Okay. If, if, if your matrix C is either white or inverse wish art, you get simple answers. Okay, in other cases, you have to do things numerically. Okay. Yeah, let me just start. Let me state before we break the inference problem that I'm going to talk about. Okay, so So basically, I have a matrix C that I don't know, I measure, but, but I measure. E, which is, I model a square root of C white wish art square root of C, but I can model with any multiple, I could actually So in theory, I'm going to develop is a bit more general than that. I can say, I measure, I have a matrix C that I don't know, and I have multiple some some multiplicative noise. Just going to call this w, but w is not necessarily wish art matrix is some noisy matrix that is one on average, but could be and in some from financial application. It's just that that you have all sorts of correlation here we're talking about the, you can have fluctuating variants for instance the fact that volatility fluctuates. There are all sorts of noise that can happen that most of the kids doing that noise that could happen here. And it's just as easy to do theory, not just for a white wish art but for any multiplicative noise. Okay, so my problem. So the two things that I want is that. So, on average, my noise is one, and that might that w is rotation invariant. So so I can be so I can use. So this is really the free product. So really, he is in a sense that is a is a is C is a symmetric free product with with with C on the with C on the outside. Okay. So, the question is how I want to build an estimator. All right, Kai. Kaiser. So we can be a. So I observe E so it has to be a function of E. And I want Kai of E to be an estimator of C. So I don't know C. I don't know what it was generated by C, but I don't know it and I observe E so I want to build an estimator from C. Okay, and the problem is rotation invariant the noise is rotation invariant. And I don't know what the noise is rotation invariant and the matrix C itself. I don't know what it is. So I could do it sort of a Bayesian approach. I would need to make to build a prior, and I'm saying is that I have no prior for the eigenvectors. Okay. Of of of C of the truth see another way of saying that is that since the problem is completely rotation invariant I'm going to look in the class of estimators that are rotation invariant what's a rotation invariant. This is an important concept as a rotation E invariant estimator is a. So if I, if I'm given a rotated matrix E, then I'm going to have the same estimate as a non rotated E up to a rotation. So, an estimator that has this property. Okay, and what I argue that since if you have no prior on C, then, then necessarily your estimator should be in this class. Another way, and then the consequence of this. Okay. And maybe you'll have more intuition on the consequence is that your estimator E commutes with E. In another way, another way, kind of E is in the same is in the same basis. Okay, so if you have an estimator in this class, or you can think directly as maybe it's easier to build the intuition about about this statement is that you don't know anything my eigenvectors, you're trying to estimate the matrix see but you really have no prior knowledge of its eigenvectors, and the only eigenvectors you have in the problems are those that you observe those of E. So, so, so, so, in a sense you're stuck with those eigenvectors, there's no, no, nobody else will give you a direction in space. Other than the eigenvectors of E. So which means that your estimator should stay in the eigenvectors of E, and this is a very strong. basis. And of course, like, to example where this fails. If you think that he might be the C might be sparse. So if you have a canonical basis. So, so if, well, first of all, if you have a canonical basis if you truly believe that there exists a basis in which C is natural. And then C might be diagonal on that basis and this is this breaks rotation variance. You might have a collective modes is your, your, your, like in finance for instance, there's a, there's a, there's a canonical basis which is every stock is stock one stock two stock three that that that's a basis. There's there's actually an eigen mode that's very easy to call the market mode where all the stocks go up or all the stocks go down together, and that breaks rotation variance. Okay, and the other type of thing that can break rotation variance is the notion of sparsity. So if you say that that the eigenvectors of C are sparse, that also breaks rotation variance. So, so this part is very strong. And, and in some cases not justified. But, but, but on the other hand it's very powerful, because essentially, and I'm going to stop here. You go from an n square problem you're trying to estimate and the n square elements of the matrix C. Okay, but you really only need to me just write finally that Kai of E, you can write it as sum over K. Psi k, vk, vk transpose, and just to set the notation, E I diagonalize as sum over K is lambda k, vk, vk transpose. Okay, so if I diagonalize the matrix E, and the V, the V case or its eigenvector, I'm saying that my estimator must be in the same basis. But, but the fact that it's in the same basis doesn't say anything about the coefficient. Okay, so basically I'm I reduce my problem from an square problem to estimating and coefficient and estimating and coefficient is much easier than estimating n square matrix entries. So this rotation variance hypothesis is extremely strong. It's strong in two way it reduces that radically the national table problem, and it allows you to use very powerful tool of, of rotation invariant matrices which, which, when they're large become asymptotically free. And then you can use all the work of the school on on on freedom. Okay, so I'm going to break here and reconvene unless there are questions about this. Is it so clear that this class estimator in, I mean, apart from the natural argument that, yeah, you don't have prior and so the only basis you know is the one of the data. Is there like stronger argument that shows that it is like the Bayesian optimal that belongs to that class. Yes, if you if you write a Bayesian estimation, and you say that your prior on C is rotation invariant. Then you can show, you can show this property. Yeah. So, so yeah, that's what that's to me that from, I like Bayesian estimation so I am comfortable with it. So in that sense, to me that's the best proof of this is that you write P of C given E as P of E given C times a prior P of C, you say the prior P of C is rotation invariant you look at that little terms in equation, and you can show that you can show first that that the estimator satisfies this property. And, and, and you can easily convince yourself that it will commute with the matrix E and it says actually with, I won't do this because it's almost it's too abstract that it's super quick to do it really is kind of kind in a, in a, in a free. If you forget about random matrices and you just think of these as non commuting objects, and you can actually write that this estimator is just going to be the conditional expectation of, of C given E. If you write this abstractly, you get very quickly the little apache equations, but I think this is a bit too abstract. I have a little section in the book where I do that but, but you don't get any, any insight on where it comes from. And then then in free. Well, I think, yeah. Again, I'm not an expert on free probability but but for for people in free policies it's pretty obvious that the conditional expectation of E of C given E must commute with E, but that's a bit abstract to me so yeah. Yes, can can do. So this can be the hypothesis about rotation variance can be relaxed to permutation and variance. I don't know how to solve the permutation there and probably been indeed permutation variance is much more natural. But, but, but much harder to. But can you do can you do the inference from Okay, unfortunately, unfortunately, okay. Okay, so because I'd be really curious of the permutation and variance prior. Okay, so that. Okay. Yeah, so I don't know but I don't know if in the end you get, you can get a practical answer that you can build a portfolio and what do you tomorrow send me the answer that I won't be there unfortunately but okay so let's break maybe.