 Thank you very much. Thank you for the invitation to this wonderful place. I already spent a week here listening to great lectures and it's a pleasure to give lectures myself. So what I'm going to talk about is an approach to random matrices which is a little bit different than the usual approaches. In the sense you may think about it as somehow making geometry more prominent than analysis. So I'm going to think about random matrices as having some geometric structure and when we try to understand limiting things such as eigenvalue distributions and various things that you usually take limits of in random matrices we'll try to keep this structure as opposed to forget it and just focus on the tiny thing that we care about. Another predecessor of this approach is that of David Aldous who actually had the whole program to understand probability this way which is called the objective method. So when you want to take understand limits of some random objects for example random trees you don't just take a statistic and take a limit of it but try to take a limit of as large of, try to capture as much of the object as you can. You know not just one statistic but perhaps the whole thing if possible. And then you know if you do that it's often actually easier to do and you can deduce theorems that would be harder to prove otherwise. So that's the point of this approach. So we're going to start with something that you have seen before which is a spectral measure. So I was told to define everything so we'll start with that. So let's say that A is a symmetric matrix it's n by n. So it's a very simple thing. You know that it has real eigenvalues and real eigenvectors also. So for a symmetric matrix A you can define a measure sigma. This is the spectral measure at the first entry which is just sum over i equals 1 to n of delta mass at lambda. So far it looks like the eigenvalue counting measure but there is a weight which which captures the fact that we want to understand this matrix at its first entry and the weight is phi 1 i squared. So it's the i-th normalized eigenvector. Let me do it right like phi i1 squared. So this is the i-th normalized eigenvector and you take the one entry the first entry of that. So these are the weights of the spectral measure. So this is the spectral measure at 1. And you can imagine how to define the spectral measure at any other vertex too and so on right. Just replace this one by two. So what do you know about this spectral measure? Well actually one thing that you can prove so let me already give you an exercise. I'll give you all kinds of exercises of things that I don't want to do here. So a simple exercise is the following that if you look at the k-th moment of this spectral measure right so this is a this is a probability measure because these things sum to one right because these are normalized eigenvectors. So then the k-th moment of the second vector is just that you look at the k-th power of a and look at the one one entry of that okay. So a simple example as you have seen before is when you take g a graph and a the adjacency matrix right. So aij is one if i is a neighbor of j and 0 otherwise can decide what to do and their loops so maybe even if they're loops. Okay then then this is just in that case this is just the number of paths of length k right number of paths from number of closed paths from one to one right vertex one to vertex one. So here is an exercise which you will be able to do in the next step. So you need to wait maybe 10 minutes. So this of course works this is obvious when the graph is finite but let's say that a is infinite for g infinite then you know you can still make sense of this formula right and you could think of this that this defines the measure. Okay so so g infinite and let's say that it's bounded degree then this formula defines a unique spectral probability measure. Okay so so so you do this exercise and you already see that that the spectral measure is defined for infinite graphs as well. Okay so so why do we care about the spectral measure? Well um the nice thing about the spectral measure is that it behaves nicely with graph convergence. So so let's define some notions of graph convergence. Okay so this is called the local limit or actually rooted limit rooted limit rooted convergence. Okay so so you have a sequence of graphs. Typically you'd like to them to be bounded degrees so universal bound on their degree but this can be relaxed g and o and every graph has a root so it's just a specific vertex. Okay and we say this converges to some graph which is possibly infinite. This can also be infinite it doesn't matter. Okay the the if for every r the r neighborhood eventually stabilizes okay to that in g so it's very simple right so you use the graphs converges if you look around in the ball around this o what you see eventually is it doesn't change and it's just going to be what you see in g. So let's let's look at I use this one I use this one let's look at some examples so let's look at the n cycle this is very simple with the root somewhere. Well what is the rooted limit of that? Well this is this obvious right it's just the nearest neighbor graph of g okay and and then you can look at the n path and now what is the rooted limit of that? Well actually it could be many things right if you put the root here then it goes to z plus right if you put the root some finite distance away from the the beginning then it goes to z plus with a root at that finite distance away from the beginning and if you put the root you know with the distance growing away from both growing from both endpoints then of course you'll get the side so there there are many options to hear and again there is a nice exercise this is just for those of you too who who have seen the notion of of of deregular graphs of course if you have seen al's talk last time then he has done this exercise for you okay so sure that the random deregular graph so g and d converges to td which is the deregular tree okay and I guess this is the improbability right with respect to rooted convergence so that's that's a little bit harder exercise but it's not that hard just have to make sure that you know it's unlikely that you close the cycle and so this is this is rooted convergence what is it good for well you know we're talking about random matrices here and so on so so eventually eigenvalues have to come up and I've already already discussed eigenvalues so so what what is the statement well of course uh so even this is a remark even that the specter the spectral measure is continuous as a is a is continuous as with respect to rooted convergence right if you have g and o converging to some g o then sigma n just the spectral measure at the root here this converges to the sigma weekly okay and why is that yeah yes because of this formula right number of path of length k is the kth moment so that stabilizes because this is an extremely strong version of convergence right you eventually this number which is the number of path of length k stabilizes to that in to the corresponding thing in j so that's not that it's not only that the moments converge the moments actually just freeze actually scores convergence of moments implies convergence of of probability measures I don't need a sponge yet yeah that's why they hit it from me so okay so let's look at again an example so let's look at the n cycle so let's look at what are the what are the eigenvalues of this and what is the spectral measure well let's let's look at so this is actually a little bit tricky so so let's look at what are the eigenvalues okay so the adjacent symmetrics of the n cycle of course you can write as t plus t inverse where t is just the right shift matrix okay it's just the matrix sphere it's the adjacent symmetrics of this directed graph right and of course these things commute okay and what are the eigenvalues where you just write down you know it's just this matrix three of things of of the diagonal one below this is also the discrete Fourier transform so so so it's diagonalized by the Fourier transform so so the eigenvalues are actually just the roots of unity so so what are the eigenvalues of this the eigenvalues of this are the eigenvalues are the inverses of the eigenvalues of this right so so the eigenvalues of this are the i plus the i inverse okay which are the roots of the roots of units unity and of course this is just twice the real part of the i okay so so here is a here is a picture for this let me let me do this picture here the roots of unity let's let me put this on a let me glue them up by two okay because of these two here and then just take the real part so real part is just project them down to the real line okay so so that's so so this is this is so this is now what we have this we understand this is the mu sub n which is the eigenvalue distribution it's just uniform on this now what is the spectral measure but first of all one thing that you know is that the spectral measure is of course the same at every vertex because this graph looks the same it's a transitive graph okay the other thing you know from this formula okay is is is this is an exercise because this is again very nice and simple if you look at the spectral measure it's it's actually immediate from that formula if you think about it and sum over all the vertices or all the all the indices the same way doesn't matter not only some but you take the average okay now what do you get well you get the eigenvalue distribution okay so the average of the spectral measure with respect to the choice of root gives you the eigenvalue distribution just because when you sum this overall entries you get one because these are normalized eigenvector that's all that's the only thing you use right that these are the only thing you use is that these have length one you don't even use that they're orthogonal excuse me it's not an exercise yeah it's too easy i told you the solution okay okay so so we got we got to this point um okay so so so all of these all of these guys see in this case are equal and and their average is mu okay so in this case this is just sigma n so so what does this tell you well sigma n converges to where well we know where it converges on the one hand this is just the projection of this equally distributed lines on the points in the circle when the number of points grow then of course it will just be the projection of the circle to the real line right so you take uniform measure on the circle and project it to the real line okay so this measure its projection is called the arc sign distribution uh on minus two two and of course this converges to the sigma of z because the graph converges to z so we have just concluded that the that the spectral measure of the integers is is is arc sign it's just uh in this in this in this example okay so so let me say this so sigma z is equal to arc sign so nice and simple let's um let's go a little bit further and and see you see now we have have understood somehow you think of this as in an understanding limits of spectral measures in terms of limits of graphs right or limits of structures but how can you understand the limit of eigenvalue distributions in terms of limits of graphs okay so this thing gives you a hint this over here so if you take the average of the spectral measures you get an eigenvalue distribution okay so what this shows you is that if you somehow take a limit of graphs in an average sense then then then you should get the limit of the eigenvalue distribution and this is the content of well this is why it's useful to use the Benjaminian Schramm convergence okay so so so what is that now you have a sequence of say finite unrooted graphs okay and you make this into random rooted graphs by picking a root at random uniformly at random okay and you say g o which is now a random rooted graph so this let's let's let's say that what you'd like is that g and o should convert to some g o which is a limiting random rooted graph okay and with what sense well it's obvious what sense right it's just just locally and weekly right so we know what local convergence is so this is just weak convergence with respect to the local convergence notion okay we know what it means for random rooted graphs to converge for rooted graph to converge to rooted graphs so we know what it means for random rooted graph to converge to random rooted graphs okay and what this really means is if you want it more simply and not so abstractly all this means is that if you look at the statistics of the neighborhoods that you see around your roots what is the probability of seeing this particular thing that converges for each particular thing and for each r you see in each neighborhood you have a probability and those probabilities converge that's all it doesn't mean anything more offensive than that and we again have a corollary right so well corollary of the definition I don't know corollary of what we wrote right so if g n converges to some g o in the Benjamin Nisham sense then mu n which is the spectral measure well sorry the eigenvalue distribution and the empirical eigenvalue distribution converges to to what well this here uh the root is actually the expectation of sigma right is the average of the spectral measures what does it mean to take an expectation of a probability measure well it just means that the measure of a set is just the expected measure of the set here so so this is very simple what it means so you can easily this is you know as long as you have real linearity you can take expectations and and you know everything is fine here because these are probability measures all the numbers you care about are bounded so you don't have to worry about any kind of technicalities okay so so so so what do we know so mu n actually mu n actually converges to the expectation of sigma of the g o why because this is the expectation of sigma n you know that sigma n converges to sigma weekly right because here is weak convergence and that's a continuous function so just by a continuity theorem okay and and then by taking expectations which you can so bounded convergence this this this also converges to that so it's it's extremely extremely simple okay so what do we know let me see and now I need a sponge okay so so let's just do an example again we I'm going to do very simple things now so see if you if you look at the end path right then this converges in Benjamin Shum sense to Z we have seen seen it because most of the points are going to be big order I mean growing distance away from the boundary so when you pick a random route it will be far away from the boundary and all those sequences actually do converge to Z and so and so this tells you that mu of the end path converges to mu n here converges to arc sign it's a it's a proof we notice that we did we did no no computation so far um okay uh nice so um so we have this nice notion of spectral measure and and it works very well with graph convergence so let me tell you you know so let me give you two exercises because I go for one of them is kind of obvious so if you look at the end by end box you know you can do this more precisely if you like it it takes a little bit of work to make it actually precise and by end this converges to Z D Z 2 and for general D so this is one did it make this precisely and the second exercise which I really like show that if you take finite trees okay T n and these never converge in Benjamin Shum sense to the D regular tree if D is greater than or equal to 3 and that's that's uh that's actually surprising if you first see it so really like to approximate this tree it's try hard and doesn't work okay you could you could do it a lot way also that's a that's the worst way but now I tell you that you can't you can prove that you can't of trees okay that's the right thing to do here have you read that in some of the Texas schools they they have reintroduced the corporal punishment is nothing friends why is that an important piece of information hope not yeah actually they need the agreement of the parents but let's still so have we been okay so so so we have seen that this uh spectral measure sigma is something you know something important about matrices so it would be nice to introduce an equivalence relation so let's say that a is equivalent to b okay if sigma a equal to sigma b and now we're talking about n by n symmetric matrices okay so so right this is somehow something that comes up here because you want to understand the spectral measure how does you know what part of the matrix is responsible for the spectral measure can we extract that part no no this is this is at one one yes so it's at the first entry um so so well maybe this question maybe you've seen this question or if not you have maybe seen there's like a different question you know what if what if you do the same thing with me right the eigenvalue distribution well that question you know the answer to right so when are two when the two matrices have the same eigenvalues well there is a structural understanding of that to just say well they can all be conjugated by an orthogonal matrix to some diagonal matrix which say has increasing values on the diagonal right for non-decreasing values on the diagonal so on the one hand you can say well there is a way to transform them to one another so in fact in fact more simply they are equal they are they are this happens if and only if the two matrices are conjugated by an orthogonal matrix uh so so there is a notion of transformation if you have that and then there is also a notion of class actually what class representative right there's a unique class representative which is just the diagonal matrix with non-decreasing entries okay so so what was the so question is what is the the corresponding thing here um so so actually um there are there is an infecting many ways it is nicer um and and here is the here is the here is the the theorem okay so first of all a and b is conjugate to b in this sense if and only if there exists there exists an orthogonal matrix o which is n minus one by n minus one such that a is you know you can conjugate one to the other so q q b q inverse where q is just you know this blog diagonal thing which is very simple I just have a one here and an o here okay this is o right then zero is here so this is one thing um well you have to be this is almost completely true okay um you have to have to actually assume one thing for this which is um at least for the next theorem next part of the theorem so we'll we'll make an assumption okay so the assumption is that the vector e one so the first coordinate vector uh is cyclic for a okay so so so what does that mean it means that if you take the vector e maybe just call it e vector e and a e and a squared e and so on all the way to a to the n minus one e then these things are linearly independent okay so so so for example for example um you know you cannot just have this cannot be an eigenvector it's it's one of the things that it says and of course if they're linearly independent then they form a basis so it's like it's a quivalent to asking that to form a basis if they form a basis you can gram schmidt them and so on the various nice things you can do i'll get to that um okay and the second thing is that that each each equivalence class so so again we're looking at matrices like this okay we forget all others we throw them out of course if you look at random matrices this will happen with probability one so this won't be an issue okay so so each equivalence class contains a unique Jacobi matrix okay and what's a Jacobi matrix okay i should be somewhat consistent with my board use here okay so a Jacobi matrix it's just a matrix which have ai okay so try diagonal matrix okay where the ai's are are real and the bi's are positive that's that's what the Jacobi matrix is um so so this is this is the world that that we live in right so so if you want to understand the spectral measure it's very natural to try to understand Jacobi matrices for that so so how do you uh so how does how does that go well actually i i want i want to i want to do the proof i want to do the proof of this at least one step okay so here's how the proof goes we'll show that we'll show that that's direct first of all step one is there exists the Jacobi matrix with the which is which is equivalent to ai okay that's the first step and the second step is that if two Jacobi matrices have the same spectral measure than they're equal okay okay so i leave this one as an exercise it's actually a nice exercise and if you put these two together then it implies both of these okay just think about it um all right so i forgot to tell you this is really i should have told you so everything i'm talking to you about today and maybe in the next next lecture is you can find lecture notes on my website just just google my name and uh it's it's right there um okay so let's see the first one so the first one actually you know historically the way this came up is is it came up in the Lanza Shah algorithm okay so what's the Lanza Shah algorithm this was this was a way to you know let's say that you you you want to store the eigenvalues of a huge matrix okay um but you're only given the matrix it's an n by n matrix and you like to keep them for your grand children but only the eigenvalues so don't care about the matrix so of course we could one way to do it is to is to compute all the eigenvalues write them write them on a piece of paper you know or upload them in the cloud or something that's that's one way but but but but the problem is that when you want to compute the eigenvalues you have to solve algebraic equations right so that may not be that nice to do right so so also you lose precision and various things it's you have to do some hard things computationally and algebraically so so so the Lanza Shah algorithm is a way to go around this it's one of one of the goal which is that you you want to reduce your your n squared pieces of data to just order n pieces of data right and and so what do you do we just try to find a jacobi matrix okay which has the same eigenvalues as your original matrix okay so so it's essentially essentially this problem right except you only care about the eigenvalues you'll get the you'll get the weights you know just as a bonus from from that so how does it work so you you take your matrix um let's let's see so so you take your matrix a and which which you're gonna write like this and you conjugate by q okay so so let's look at q let's let me write a also in this form that the form that q is so you take a a is gonna be written like this so so this matrix a i'm gonna write it like this so there is a vector b there is then b transpose and then here is the vectors here's a small matrix c which is n minus n minus one by n minus one and you now conjugate but a q that is of the form that we wanted so you want to put one zero zero and an o here and you want to put one zero zero and an o inverse here so so what happens when you do that um well it's not hard to compute right what you get is just something like what you had before so this a actually stays fixed okay the b gets multiplied by o well this is still gonna stay a symmetric matrix so you get your o b transpose and the c just gets conjugated by o so so so what is this good for you well it's good for now you have a choice of this o okay and you choose it so that it will rotate your your um this is just a general orthogonal matrix so you can rotate your vector b to any direction you like okay so choose your o so that b is gonna be this this this vector which is just the length of b well the length has to change has to stay the same but apart from it like that it's gonna be zeros otherwise right okay so so this is gonna be now like this okay and again the length of b and and you have already done a step towards your your Jacobi matrix you have you have put zeros here where they're supposed to be zeros and then you just repeat so you do the same thing with with a matrix where the top corner orthogonal matrix where the top two by two corner is the identity okay and and you see that it will work okay so keep going okay if you do this you get the Jacobi matrix and and of course it's clear that I didn't tell you but of course it's clear that conjugation by matrices of this form don't change the spectral measure that's uh that follows simply from the from the from many things for example from the way the eigen vectors change right so the eigenvectors will get multiplied by this matrix when when you conjugate by o and so the first entry is one change because this matrix doesn't change the front first entries we said that that's one way of seeing it but also by the by expanding the path the o's will cancel that's another way of doing so okay so so so so first so so so that gave you basically a proof of that and and this is called I mean you can actually have a choice of b's which are certain reflections they're called household reflections it's the standard choice but it doesn't matter which o's you use now here is something funny when you do this thing here you see you have the feeling that they're like a huge amount of choice in the you know okay because the only thing about you care about it is where you take this vector where it takes this vector b so there are many many o's that do that and so it looks like there could be many Jacobi matrices that you get to get this way by choosing the o's but but in fact there is just one okay so this choice is an illusion okay I'm not talking about life I'm talking about this matrix all right so what do we do I think we can go to GOE now okay so so you know GOE the Gaussian orthogonal ensemble so it's just m plus m transpose over root 2 where m is the matrix with iid normal 0 1 okay so that's the GOE root t root 2 um and you can check that om o inverse has the same distribution as sorry oh so let's call this a so oa o inverse has the same distribution uh as is a for any orthogonal matrix so that's why it's called the GOE you wish an orthogonal ensemble contained orthogonal matrices but it's not true it's it's orthogonal invariant um so so so what we this idea right uh which is due to Trotter 82 okay he'll Trotter he was at Princeton I actually met him in the 2000s and he said the thing that he did he didn't think it's going to be any importance but he liked it so he wrote it down uh he said that you can apply the same same same algorithm okay to to go so so what do you do well okay so now that's a GOE so what's happening to a well a is a normal okay and b is just a vector of normals independent normals so when you do this first step uh this is going to be a normal and this is going to be what a chi right which is just the length of a normal vector chi n minus one because it's a normal vector with n minus one entries distribution of that is called chi um now what is this going to mean this is a bit more tricky um well if you look at the sub matrix of a GOE from this definition you see that's also a GOE just this lower dimensions so the C was a GOE of lower dimensions and if you conjugate a GOE by an orthogonal you get a GOE but you have to make sure that the or that the orthogonal you conjugate by is independent of the GOE otherwise you could get a diagonal matrix which is not a GOE right so so so it's important well it should be either a constant or independent is also okay but you can see that it was independent because c was independent of a and b c from the definition completely and or the choice of all just dependent on b nothing else okay so it's a still dependent this is still a GOE and the same is true for the the next steps so what does this tell you you can imagine okay so so so basically trotter theorem tells you that the GOE is equivalent in the sense that we we showed before right to this matrix which is n1 and 2 and sub n the n are normal 0 2 because of this the diagonals are just just added so we're going to be square root of two terms are normal and and then you have the chi n minus 1 chi n minus 2 chi 1 okay when you do the second step you just get the normal vector of length n minus 2 and and the chi's are just I'm just right the distributions and these chi eyes are chi's and they're all or independent so there's a matrix with independent entries there's another matrix with the independent entries and their equivalent or you can couple them in an equivalent way so that's that's trotter's theorem and guess what trotter used this theorem for who can guess this is obvious guess he used it to prove the Wigner semicircle law okay and at 82 that was a news but but but but it was a new proof so basically what I'm going to show you is is a modern version of trotter's proof it's it's it's it's said in a different language but essentially it's what he did um and so so so here is uh again what we need is an exercise but this is the following so if you look at chi n then this is a sympathetic to square root of n okay so that's that's important so chi n over square root of n converges to to 1 in probability but in fact you can say something more precise that that this is s in total to square root of 1 plus plus this okay so it's a normal of variance one half so you can reduce it from the clt and some trickery because chi squared is a sum of independent things so so chi squared has a clt but then you have to push this square root the clt through the square root which is not hard so this means that if you subtract the chi n minus root n then it converges to normal in distribution okay and so so let's think of this matrix here as a graph okay so you have this Jacobi matrix j and corresponding to ge what does it look like as a graph well it's a graph it loops right the all right the all Jacobi matrices look like this the weights in here are positive right so this is the chi n minus one okay and the weights here are just you know these normals so it's a weighted graph a graph with edge weights um and the edge weights go to infinity so let's normalize so let's look one over root n time this so this means i'm waiting i'm waiting the i'm waiting the edges uh so i want to take a benium in a shum limit now there is a slight issue that this is a weighted graph and i didn't tell you what the benium in a shum limit of a weighted graph is but it's kind of obvious right a local limit of the weighted graph it just means the structure stabilizes and the weights also converge so so then the benium in a shum limit is the same and it's not so hard to check that all the statements about the spectrum and so on will go through uh so where does it go well you divide by n so all these labels they disappear okay the these normals they go away and not only that but the randomness in the chi's will go away too because that's also order one normal okay so so you could say essentially this thing just looks like for this kind of convergence right and then you have here root k let's see n minus k over root n is this kind of convergence you may as well think of this as a fixed fixed random graph sorry a fixed graph not the random graph the randomness is not important for this kind of convergence the randomness is not important for the ring nurse semicircle law anymore so so where does it converge well this is just the path so first of all it has to converge to z the graph structure has to converge to z okay now you have to understand the labels how do the labels converge well you know what you do is pick a random place and you zoom in there to see what what's around that if you pick a random space you zoom in there the labels don't change very much okay so in the limit all the labels are going to be the same but they're not going to be deterministic they're gonna be random because of you because of the choice of location that was random all right and what are these if you look at this this n minus k over n right where you where you choose k uniformly at random right this converges just to a uniform zero one random variable it's called the u you choose k uniformly at red one between zero between one and n or zero and then this number just converges to that right so so you have a have a have a graph of z with labels that are square root of u okay and these are the same u they're not not independent but the same u so it's really non-ergodic if you like it's this is the opposite very far from regarding so you could just write it as square root of u times z okay that's your graph okay so that's that's we have proved that basically and what is the limit here because yeah Benjamin is sharp of course and improbability right so that's the proper because this is random this is something deterministic you have convergence in probability in distribution also if you like but it's the same so so so of course we also know that the eigenvalue distribution u n converges again in probability to the expected spectral measure of this and the expectation is only over u because everything else is deterministic so what was this what was the spectrum so basically you just take the spectral measure of z and scale it by square root of u so what so let's let's put this geometrically right so you have a circle of radius two okay you scale this circle by a by a square root of a uniform okay you pick a random point and project it down to the line that's the eigenvalue distribution of that's the eigenvalue distribution on this okay but when you scale a circle by a square root of a uniform and pick a random point it's just the same as if I took the disk here and pick the random point and projected it down to the line because the distribution of the radius of a random point in the circle is just it's just the square root of a uniform the square the square is uniform so so this thing is the projection of the circle to the to the line or the disk to the line the uniform measure on the disk to the line so this is the Wigner semicircle okay so so here is here is the proof uh no computation yes no it doesn't matter there's no no difference I mean you know just think about it or the moment things go through as long as things are bounded and so on okay so that's a proof of the Wigner semicircle law and and and it is a kind of proof that I promise you it's a proof of a simple thing but but it's still it's a proof that you know takes geometry out of this matrices and then tries to take a limit with the geometry there and they do things from say the eigenvalue distribution of the matrix using this geometry so so I have 30 more minutes I want to tell you two things one of them is another proof of the Wigner semicircle law which is also very quick even quicker than this and the other one is um is we're going to talk about the top eigenvalue so we can use this approach to show the commutative result and I'll tell you that in a second what it is okay so let's let's try to quickly do this other proof of the semicircle one okay and the reason I'm telling you is because it's a proof that takes a different operator limit and actually it's surprising that you get the same answer it's not it's it's something for contemplation okay so so here is what here is the simplest another way of taking this the Wigner semicircle law so let's look at sigma n okay so the spectral measure uh at at at at one of g e okay so so so you can take this and the Gekomi matrix right here and now you take a rooted limit to understand sigma n so you take the limit at this vertex so what do you see well you have this chi n minus one chi n minus two again the randomness doesn't matter the same same idea and what do you see you see that this measure this graph this rooted graph which is rooted graph converges now in the rooted limit in probability again because this is random and you get a deterministic thing well what to just z plus so right so this z plus with this root so this so this tells you the sigma n which is continuous it converges to sigma of z plus i know okay and this is true for a fixed choice of root but it's true for every choice of root no matter what because the g o e was of course invariant okay so it's also true in average actually this very simple way of concluding from here is just talked to me by a student in my last lecture so it's very very happy to see that so this is true for all choice oh so therefore mu n also the sigma that plus no okay so so so then i know you know now you can do various things you could say well this is an independent proof of the Wigner semicircle law as long as you know that the the the spectral measure of z plus here is the semicircle law you could actually do that prove that by simply computing the spectral measure of of of an n path at the endpoint you can compute that and just check that that converges to the semicircle law by hand so that's one way or you can use the previous proof and just say well the conclusion is that the conclusion is that z plus has the semicircle law here has a spectral measure okay you can do that that's probably the fastest thing it's no computation and what does this imply well it implies that the moments of the semicircle law are actually the dick paths right so just the paths in z plus that return to to to the root the case moment right so so so integral x to the k the sigma for the semicircle or maybe semicircle is equal to write it like this okay so these are just the tick paths or maybe zero zero number of paths in z plus they start from o and return to o okay so the next thing i want to show you and hopefully i will be able to a completely lost track of these bars i'm sorry it's maybe i'm able to continue get to a good point where you can stop and continue next time is is the komlos freddy result or freddy komlos okay so what is freddy komlos it just shows that if you if you look at the top the first eigenvalue of lambda one lambda one of ge it's more general but so works and you divide by root n then well what does the semicircle law tell you right right the eigenvalue distribution converges to the semicircle if you divide by root n the semicircle is supported from minus two to two so it tells you that this eigenvalue this this limit you know of course this is random but but anything that it can converge to it has to be at least two right because otherwise that would contradict the semicircle there has to be all this okay so say so so the freddy komlos tells you that it is actually two it's a simple thing you may learn and so the only thing we have to show is that is that so so so so yeah so the ge tells you the semicircle law tells you this that this is what flesher on the semicircle law and then you have to get the other direction okay so let's now let's do a lemma for this and this is a lemma that's true for general Jacobi matrices okay so for any Jacobi matrix you have an upper bound on the on the top eigenvalue which is even by the max over i of ai plus bi plus bi minus one okay so remember a is where the diagonals and b is where the off diagonals okay so so let's prove this okay so so so here here I write I just recall that this was j right so so let's take this matrix m which is the following so take zero root b1 minus root b1 root b2 minus root b2 root b3 minus root b3 I think you get the point and I'm going to write this so so j is equal to minus m m transpose okay plus a diagonal matrix with entries over here this okay those are the entries of the diagonal matrix well you know that's not surprising right to take mm transpose you get this b squared that's how we made it and then you're going to get something something on the diagonal you have to fix that so that's where the diagonal matrix it's it so the ij entry of this is just the inner product of the corresponding rows or columns so that's how you get it so this is obvious okay and then we're done with the proof actually because look at that right this is positive definite this mm m m m transpose it's every such matrix just has non-negative eigenvalues okay so minus mm transpose just has non-positive eigenvalues okay and so the max eigenvalue of this is less than or equal to the max eigenvalue of this plus the max eigenvalue of that why is that actually what do you prove that it's a norm almost it's not quite the norm because the norm has negative eigenvalues but yeah so yeah essentially the norm so you use the regular characterization right the max eigenvalue is a super overall vectors of length one of the bilinear form and then then you just just drops out of that okay so so it end up proof this is going way too fast actually um and so what does it tell you right so the lambda one of GOE is less than or equal to the max over i what so you have a chi n minus i plus chi n minus i minus one this is bn minus one plus plus a norm plus a normal i i think normal i so what are so what do these look like right so these are a lot for roughly square root of n plus a normal this is just a normal okay so so you can easily say that this is less than or equal to square root of n twice square root of n for these two things plus some constant times square root of log n because this is the maximum of n normals the chi's it's easy to check that they also have Gaussian tails and when you take maximum van Gaussians you get this kind of thing right this is a random constant but tight okay so so you get a pretty good bound on the top eigenvalue right so so it's not even not even terrible um so the truth is in fact this is two square root of n plus n to the minus one six times the tracividum distribution which is some probability distribution we'll talk about later I guess so so this is the actual truth so we're off by still we're still off by a little factor a little power but but it's not terrible it's much more than what we need okay all right so maybe I can tell you about another thing we can get here sorry yes can you recall what you mean by t w u oh so this is just this this is uh okay so so this is here sorry and and and i'm not i didn't tell you what this is okay what i'm saying this is a this is a theorem what i'm stating here but i'm not proving the theorem okay so the theorem is that there exists a probability distribution which is defined by this theorem even such that this holds okay so if you want to make it precise you know you take this thing lambda one g o e you subtract two root n you multiply by n to the one six and that converges in law to something that's what the statement is so here is another thing i want to tell you which you can understand through this operator limit and it's it's very timely now because jared bernard bennaroos just had his 60th birthday and and his most famous result which is not especially proud of because it's not one of his strongest results but but it's a nice result is is what's called the bike bernard specie transition okay so bike did i did i do it wrong no t terrible i'm sorry thanks all right so so what's the vibe when you're especially especially transitions transition well you know so but anyway bennaroos well you know it's terrible that you do all these strong things and then you then you look at famous for the one that is not this strong and then i told him that you're still better off than fubini so um anyway so so oh yeah he was a pretty strong geometry actually no yes um that'd be that was some more of this some okay so so where does it come from so it comes from where the entire random matrix theory is coming from right we do random matrix theory and we know it doesn't really come from the greeks like almost all other parts of mathematics or the egyptians but it comes from the 1920s from scotland right from this guy with this heart which heart who who was studying correlation matrices right so covariance matrices so the the point is you know you want to do principal component analysis you take some piece of data you normalize the entries so that's you know that that you don't care by variances and so on and so you think of this as now some some sample you know various sample from so samples from a population and there are various statistics like like i don't know height and weight and my strength and i don't know all kinds of things so you read their numbers and you want to understand if these things have have a have some kind of non-trivial correlation right so what you do is you normalize these matrices you form the sample covariance matrix right and you look at the top eigenvalue and if the top eigenvalue is large then you say okay there is a correlation that comes from this top eigenvalue so but of course you want to understand what does it mean large right and the way you do it is you you compare to the null case so you create fake data okay and and the fake data is going to be this matrix m okay just an n by n iid matrix for our case it doesn't have to be symmetric this doesn't have to be square this can be a rectangular matrix but just for now let's make it square and and you look at the you look at the covariances in here okay so look at the empirical covariance matrix of the columns of this okay and and see what is the top eigenvalue so you know whatever is smaller than that in in in an actual dataset is is just noise right obviously so so that's that's that's what that's what we sure did essentially i mean they did sort of finite dimensional versions of this and so so it's kind of classical and it's kind of not not not surprising right that that that bbp become an important thing so so let me give you the theorem which we will not prove but we'll prove versions some analogs of this and so you look at one over n times the top eigenvalue of the following so you'll take your matrix m you remember this is iid Gaussian matrix and you look at the diagonal matrix and it looks like the first entry is one plus a squared and then you put just put ones and then you look at m transpose okay so so again so you take this this this is so what is it okay so so you can think of this matrix as the population covariance matrix okay and then what you get is you take some data and you get and you get an empirical covariance matrix the sample covariance matrix this is the top eigenvalue of the sample covariance matrix now you may be telling me i was cheating because i told you to normalize everything to a roughly one variance but here i made this variance one plus a squared so and actually it's still diagonal we should i should make put here correlations right and and and not make this matrix diagonal but because of the variance of the model the only thing that matters of the of the of the of the population covariance matrix is its eigenvalues nothing else matters they're non-negative eigenvalues but the only that's the only thing that shows up if i change it to another different matrix here with different eigenvalues i still have this exactly the same distribution because of the variance of m so so actual covariance is actual correlations in the data you can just encode as one large eigenvalue here okay and you can think of that things were normalized and then i found some covariate correlations and then i conjugated that matrix and i got something like this so so so what is the vibe this is the bbp theorem okay so so actually they proved it for g u e but it's and then they prove much more much much more than this it's just the the simplest case that this thing converges to some phi of a squared okay and and what is this phi of a the bbp bbp bbp transition function okay so phi of a is equal to 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 what does this mean okay so this is how the singular value the the top singular value behaves so what is this thing so here is here is a and here is phi of a let's see so a is one and here is two right so we started two and here is and this thing actually is differentiable here but not twice differentiable i think well i have to check certainly not three times differentiable and this is a plus one over a so it sort of converges it's asymptotic to this diagonal this is not a great picture but so what you see is that that for a while if this a is small if my if my correlations are small then you don't see an effect and if they're because they're the same as if they're none so for a while this is just case two and then suddenly you know some effect is going to stick its head out after a is greater than one and you can see in the data the correlation and the interesting thing is that this transition happens not by scale but actually by just changing a constant that's why it's an interesting that's why it's a phase transition so if you think about it it's not that surprising because you have you know you have you have the all these eigenvectors and you sort of do a rank one a small perturbation of your matrix okay then for a while the perturbation you don't see because it's washed away by the other actuations but suddenly it will overtake everybody else and then and then and then you'll see it but and this is this is the exact way that it happens okay um i should stop at 30 is that the plan or 20 what what's usual 30 okay okay okay so i'm going to give you some version of this theorem uh because we so first of all this model also has a tridiagonal version okay so this is a exercise which is that try diagonalize this and you can prove this theorem using that but i don't want to go through this way rather i'm gonna stick with our g oe because we've seen that very well and just change the theorem to the g oe okay so so this is a theorem which is the g oe with mean okay the mean g oe so you take the g oe and add to it some non-trivial expectation to all the entries and you know look at this is a natural question right what's going to happen to that so you look at one over n of lambda one times g oe n plus uh you look at this all one matrix right that's uh that's uh this is the matrix with all ones this is the all one vector and if you do this you get all one matrix it's it's column vectors and so this is gonna be actually a rank one matrix and you add to add to the g oe this non-trivial mean if i if i put here just this then it's gonna say that the all the entries will have mean one okay and i want to see what is the top eigenvalue of that and actually if you make this interesting uh this is way too much mean to add so this will actually go to infinity should be root n here but you have to add here is a over root n and then this converges to five away in probability okay and that's the theorem we're gonna prove it's gonna have a perfect proof of this but but let me tell you what is the motivation like why is this five way no that's actually one over n yeah it's because it's yeah it's square so that's a covariance thing and this little scaling is slightly off but they all have a reason okay so instead of giving the proof to you now i give you another theorem it's an exercise actually okay so give me a break okay because it's an exercise it's not actually it's it's helpful to do it for the proof it doesn't you can't really follow at least i don't know how to just deduce it from this exercise the proof but it certainly is helpful to understand what's happening uh and it does give you where this comes from okay so if you look at lambda one of z plus okay so the z plus plus a loop of weight a at zero so the top eigenvalue which you can define by the Rayleigh quotient formula again of this graph uh where you just put a and all of these are one it's one okay and this is five way okay so this is the bike benderu speciale transition for z plus and the Rayleigh quotient formula okay you'd like to know that okay so so in general the top eigenvalue right of some operator where this works so of course you look at the adjacency matrix here you define it by the soup overall uh function f in l2 in fact of length one of the inner product f af right and and so when a is a matrix for example then if you plug in here the top eigenvector then you get exactly the top eigenvalue so it's actually a max in that case but in general therefore z plus this may also makes perfect sense so you can take any function in little l2 you can apply the adjacency matrix to it this way right this is just again and and and we can take this soup so that's that's how you get this lambda one okay so we'll continue from here next time thank you