Lectures 2 3 : Wigner s semicircle law

Transcription

1 Fall 009 MATH 833 Random Matrices B. Való Lectures 3 : Wigner s semicircle law Notes prepared by: M. Koyama As we set up last wee, let M n = [X ij ] n i,j=1 be a symmetric n n matrix with Random entries such that X i.j = X j,i X i,j s are iid for all i < j, and X jj are iid for all j with E[X ij] = 1, E[X i j] = 0 All moments exists for each entries. We considered the eigenvector of this random matrix; λ 1 λ λ n which turns out to be random elements depending continuously on M n ; Lemma 1. If H n is a topological space of n n matrix with topology derived from the usual metric on product Lebesgue measurable space, then λ i (H) is a continuous function on H n. Proof. Let H = [h ij ] n i,j=1 be an element in H n. We now that H = T r(h )) = λ i So for example, H = i λ i Note that therefore H max(λ n, λ 1 ). Our goal is to obtain λ in terms of H. So it is good if we can say lim H λ n because λ n dominates all the other eigen vectors, maybe except λ 1. Clearly, this logic might not wor because of the presence of negative eigen values including λ 1. To fix this problem we may just shift the matrix by H. In particular, we can claim lim T r((h + H I) ) λ n + H 1

2 To be more precise, λ n (H) + H T r((h + H I) ) (1) n(λ n (H) + H ) () λ n (H) + H. (3) Having obtained λ n...λ, we can inductively obtain the λ 1 by simply taing the limit of T r((h + H I) ) (λ n (H) + H ) i=1 This allows us to induce the random measure ν n = 1 n δ λi n. The Wigner s semicircle law claims that this ν n has a nice distributional limit. Theorem. where ν dx = 1 π 4 x 1 ( x ). 1 δ λi n n ν We will use Borrel Cantelli lemma and Carleman s condition for moment problem to show this fact. Consider the random variable X n, = x dν n. We will show 1. EX n, c = x dν (4). Var(X n, ) c n (5) How do they help? Suppose these two statements are true. Then we can use Borrel Cantelli lemma to show that P ( X n, EX n, > 1 4 n ) E(X n, EX n, ) n (6) = V ar(x n, ) n (7) = O(1/n 3/ ) (8)

3 Thus P ( X n, EX n, > 1 4 n i.o) = 0 and X n, EX n < 1 4 n for some large n almost surely. If this is the case, then ν n can be shown to be tight because this means X n, is bounded by some constant C and hence by Chebyshev ν n ({x : x > m}) < C m. We can therefore choose a converging subsequence ν n(j) of measures that converge to ν. We would now lie to show that any of these subsequencial limits ν of converging subsequences equals to ν. In this way, we can establish that any subsequence ν n() has further subsequence that converges to ν. This can be done if we can characterize ν by its moments, because we now that ν s moments for all subsequence agrees by the claim (1). This can be done using the following useful criterion. Theorem 3. (Carleman s condition: ) Suppose =1 1 µ 1/ =. Then there is at most one measure F such that x df (x) = µ for all positive integer. This criterion can be made stronger: in fact, the conclusion above holds if lim sup µ1/ = r <. The logic behind the proof of this claim follows from the fact that the characteristic function E[exp(iXt)] charaterizes the distribution of X. We can consider the Taylor polynomial of exp(ixt). If E[exp(iXt)] = (it) E[X ]!, then the moment indeed determines the characteristic function. Let s hence chec if the Carleman s condition applies to our case. Put c = 1 π x 4 x dx. If is odd, then c = 0. Therefore put = n. Then c = 1 x n 4 x π dx (9) [0,] = 1 sin n (t) cos (t) n+ dt (10) π [0,π/] = 1 n+ ( sin n (t) sin n+ t ) dt (11) π [0,π/] = 1 ( ) (n)! π (n + )(n + 1) n+ π n! n 1 4(n + 1) (1) We used the fact = ( n n ) 1 n + 1 < 4n (13) [0,π/] sin l (t)dt = 3 (l)! π (l!) l

4 Therefore µ1/ < (4/ ) 1/ = 4 4 and the claim follows. Therefore, it remains to show (1) and () in (0.4) and (0.5). Proof of (1) Let us begin with (1). We will achieve this by a way of controlled brute force. Note that E x dν n = E 1 n ( λ i n ) (14) To organize this, whenever we have -tuple (i 1, i,... i ) = I, put = n 1 E(T rm n ) (15) = n 1 E(Xi1,i X i,i 3 X i3,i 4... X i,i 1 ) (16) E(I) = E(X i1,i X i,i 3 X i3,i 4... X i,i 1 ). First, observe that E(I) is bounded by some constant B. This can be seen by applying Cauchy Shwartz inequality inductively. Let us represent each I by a directed closed path with vertices {1,, 3,... n} = V (I) and edges ξ(i) = {(i a, i a+1 ); a = 1,..., i +1 = i 1 } For example, if I = (, 3, 1,,, 1) then this will correspond to the directed adjacency matrix (17) Now, seleton of a directed graph is a undirected graph induced by the directed graph by replacing all the multiedges by edges. For example, the seleton of the graph above is given by the adjacency matrix entries in the a ij represents the number of edges from vertex i to vertex j. entries in the a ij represents the number of edges between vertex i to vertex j.. (18) 4

5 Here, remar that that E[I] = 0 unless every edge in the seleton is used at least twice. If, for example, an edge (i, j) happens only once, then E[I] = E[X i,j ]E = 0 e ξ(i)\{(i,j)} X e This implies that if E(I) 0 then ξ(i). This bound let us also put a bound on V (I) ; Lemma 4. Given any graph G, denote the vertex set by V (G) and edge set by E(G). V (G) E(G) + 1. Then Proof. To see this, first assume that G is a tree. Note that removing a leaf from the graph removes one edge and one vertex. We may continue removing leaves from the Graph until K (complete graph of vertices) remains. Removing a leaf from K results in K 1. Thus V (G) = E(G) + 1 in this case. For a generic graph G, we may remove edges from the graph until we obtain its spanning tree G. If we removed m edges in this process, then V (G) = E(G ) m and the claim follows. Thus, we have E(I) 0 then V(I) + 1. We are now in position to bound the expectation of X n,. Lemma 5. E [ x dν n ] c n Proof. E x dν n = 1 n /+1 E(I) (19) I = E(I) (0) V (I) +1 B n 1+/ {I; V (I) + 1} (1) Temporarily, consider V (I) = l for a fixed l. How many ways can we choose I? Most naive bound on this number is indeed n l l. It turns out that this naive bound suffices. From the inequality that we obtained above, we see that if l < + 1 then the terms with V (I) = l will vanish in limit. Thus we can ignore the odd all together in the limit. Let us therefore consider the case of even 5

6 . When is even, we see that V (I) + 1. If the inequality is strict, again E x dν n 0 in the limit. Therefore, asymptotically, we can restrict our case to when V(I) = + 1 and ξ(i). Because V (I) ξ(i)+1, we have ξ(i) = necessarily. We are thus considering directed graphs for which the seletons are trees, and there are exactly two edges between two adjacent vertices. This ind of directed graph is called a double tree. Below is a directed adjacency matrix for an example of a double tree; () Now, if I is a double tree, clearly E(I) = E e ξ(i) X e = Thus, all together, we obtain the following statement; Proposition 6. lim E ( x dν n ) = lim 1 n 1+ e ξ(i) E(X e ) = 1 ( Number of double trees with n vertices ) Our proof of (1) is therefore simplified to the counting of the number of double trees with n vertices. To answer this, first let us answer the following question; If I fix a shape of a tree, just how many double trees of that shape exist? We may achieve this by maing a bijection between the shape of a double tree and a random wal on N beginning from 0 and returning in exactly -step. For example, suppose that a double tree is given by the directed adjacency matrix above; then fixing the vertex 1 as the starting point of the wal, the shape of this double tree corresponds to the random wal (0, 1,, 1,, 1, 0) (the th entry is the distance of the waler from the vertex 1 at th step). Counting this way, we will show next wee that the shape of a double tree with edges are given by ( ) Now, given a fixed shape, the number of double trees of that shape is given by ( ) ( ) n }{{ ! }}{{} choosing the vertices permutation Thus at last, we obtain that ( ) lim x 1 1 dν n = lim n / ( ) ( ) 1 n 1 = n n + 1 and the claim follows. + 1 = 1 1) + 1n(n ( n ) (3) (4) 6

7 Fall 009 MATH 833 Random Matrices B. Való Lectures 4 5 : Wigner s semicircle law Notes prepared by: H. Lin Following from last wee, we let is an even number, say, = l. we first briefly show how to derive the number of paths from (0, 0) to (, 0) while not allowed to go below the x axis. For any path from (0, 0) to (l, 0) which intersects with y = 1, we can let (a, 1) be the last intersection, and reflect the part after (a, 1) with respect to y = 1, and get a path from (0, 0) to (l, ). On the other hand, given a path from (0, 0) to (l, ), we can reflect similarly and obtain a path (0, 0) to (l, 0). Since a path from (0, 0) to (l, ) taes l 1 steps upward, and l + 1 steps downward, we have ( ) ( l l 1 such paths. Therefore we have l l 1) paths from (0, 0) to (l, 0) which intersects with y = 1. Now we can claim that there are ( ) ( ) l l = l l 1 ( ) l 1 l l + 1 paths from (0, 0) to (, 0) without going below the x-axis. Proof of () We follow the notation used in the proof of (1). Let I = (i 1, i,..., i n ) [n] be a tuple, and write X I for the product of the entries X i1 i X i i 3...X i i 1. From (16), we have x dν n = n 1 X I. I [n] So V ar( x dν n ) = n I,J [n] cov(x I, X J ). (5) Now we again represent I and J as closed directed path as in the proof of (1), and give the following facts: 1. If there are no common edges in I and J, then X I and X J are independent and hence cov(x I, X J ) = 0. 7

8 . If there is an edge which only appears once in I or J, say, the edge i 1 i, then X i1 i is independent of rest of the terms in X I and X I X J. Since EX i1 i = 0, we now EX I = EX I X J = 0, so cov(x I, X J ) = EX I X J EX I EX J = 0. Let m = V (I J) be the size of the set corresponding to the union of I and J. For a given m, we at most have Cn m ways to select the tuples I and J, where is C is a constant independent of n. If m, then we see the contribution of these term in (5) is of order 1 n. Now we consider the terms with m + 1. From the two facts we just mentioned, I and J give a connected graph, with each edge used at least twice, and hence in the seleton of the graph of I J we only have at most edges. However, m + 1, in this situation we now we must have m = + 1 and I J is actually a double tree. By erasing vertices not belonging to I, we can see that I is also a double tree, and so is J. We now loo at a common edge in both I and J, it appears twice in I and twice in J, and thus four times in I J, which contradicts with the observation that I J is a double tree. To sum up, all terms in (5) satisfy m and are O( 1 n ), which proves (). Before proceeding to the next theorem, we prove the following lemma: Lemma 7. (Hoffman Wielandt) Let A and B be two n n symmetric (or Hermitian) matrices with eigenvalues λ 1 (A) λ (A)... λ n (A), and λ 1 (B) λ (B)... λ n (B), then n λ i (A) λ i (B) T r[(a B) ] (6) i=1 Proof. Since i λ i (A) = T r(a ), we only need to prove T r(ab) n λ i (A)λ i (B). (7) i=1 Because A and B are symmetric, we can write A = UD A U T and B = V D B V T for some diagonal matrices D A, D B and orthogonal matrices U, V. 8

9 Then let W = [w ij ] n i,j=1 = U T V, we get T r(ab) = T r(ud A U T V D B V T ) = T r(d A U T V D B V T U) = T r(d A W D B W T ) = λ i (A)λ j (B)wij 1 i,j n (8) So now we try to maximize i,j λ i(a)λ j (B)v ij with the constraints that v ij 0, n i=1 v ij = 1 for j = 1,..., n, and n j=1 v ij = 1 for i = 1,..., n. Suppose v 11 < 1, then there must exist i and j such that v i1 > 0 and v 1j > 0. Let v = min{v i1, v j1 }. Then define v 11 = v 11 + v, v 1j = v 1j v, v i1 = v i1 v, and v ij = v ij + v. Since λ 1 (A)λ 1 (B)(v 11 v 11 ) + λ 1 (A)λ j (B)(v 1j v 1j ) + λ i (A)λ 1 (B)(v i1 v i1 ) + λ i (A)λ j (B)(v ij v ij ) =v(λ 1 (A) λ i (A))(λ 1 (B) λ j (B)) 0, (9) we see that if we repeat the same argument, we maximize i,j λ i(a)λ j (B)v ij when all v ij = 0 for i j and v ii = 1 for i = 1,,..., n. Therefore (7) is proved and we conclude the proof of the lemma. Now we loo at a more generalized version of Wigner s theorem without assuming finiteness of higher moments: Theorem 8. Let M n = [X ij ] n i,j=1 be a symmetric n n matrix with Random entries such that X ij are i.i.d., with EX ij = 0 and EXij = 1 for all i < j. X ii are i.i.d., with EX ii = 0 and EX ii is finite for 1 i n. Let ν n and ν be defined as before, then we have ν n ν (30) 9

10 Proof. Fix C > 0, for i j define σ (C) = V ar ( ) X ij 1 ( Xij C) and for all i and j define X C ij = X ij1 ( Xij C) EX ij 1 ( Xij C) σ(c) Let Mn = [Xij C]n i,j=1, and define the corresponding λ i and ν n as before, then we see all entries have bounded support and thus M n satisfy all conditions of theorem (we actually didn t use the condition EXii = 1 in the proof of theorem, we only need finiteness), so ν n ν a.s. (31) From the definition of Xij C, we have X ij Xij C = 1 ( ( 1 ) Xij 1 σ(c) ( Xij C ) EX ij 1 ( Xij C )) + 1 Xij σ(c) By lemma (7) and Cauchy-Schwarz inequality, we have 1 n n i=1 λ i λ i n n 1 n T r[(m n M n ) ] = 1 n (X ij Xij C ) i,j n 1 σ(c) i,j ( Xij 1 ( Xij C ) EX ij 1 ( Xij C )) + = (1 σ(c) ) σ(c) O(1) + ( 1 1 ) O(1) as n σ(c) ( 1 ) 1 n σ(c) i,j X ij (3) The last step comes from the observation that as n, 1 ( ) ( Xij n 1 ( Xij C ) EX ij 1 ( Xij C ) V ar Xij 1 ( Xij C ) EX ij 1 ( Xij C )) = 1 σ(c) i,j and 1 n Xij V arx ij = 1 i,j 10

11 Now we loo at any bounded and Lipschitz continuous function f(x). K > 0 such that for all x and y, There exists a constant f(x) f(y) K x y. Hence for some constant K. fd ν n fdν n 1 f( λ i ) f( λ i ) n n n i K 1 ( λ i λ i ) n n n i K (1 σ(c) ) σ(c) + ( 1 1 ) σ(c) (33) So lim sup lim sup fdν n fd ν n fdν fdν + lim sup K (1 σ(c) ) σ(c) + ( 1 1 σ(c) fdν n fd ν n (34) ). This holds for any C > 0, so we let C go to infinity, and obtain lim fdν n fdν ) = 0 a.s. (35) Tae an arbitrary function that is Lipschitz continuous with the following conditions 1. f(x) = 1 for x and x = 0 for x f(x) 1 for all real number x. Immediately from (35) we have lim fdν n = fdν 11

12 Recall that ν is a measure with support [, ], and we can also get fdν = 1. Since we let 0 f(x) 1 have support [ 3, 3], it follows that ν n ([ 3, 3]) 1 as n approaches infinity. Now define a new measure ν n (A) = ν n (A [ 3, 3]). Now we claim ν n ν a.s.. It suffices to show that x d ν n x dν for all Z +. This is clear from previous argument because x d ν n = x 1 ( x 3) dν n, and x 1 ( x 3) is bounded and Lipschitz continuous. So for any bounded and continuous function f(x), lim lim fdν n fdν n fdν fd ν n + lim fdν fd ν n (36) f lim ν n(r\[ 3, 3]) + 0 =0 This concludes the proof of the theorem. Remar 9. When we proved (35) for Lipschitz functions, we could get that for each t, the corresponding characteristic function c n (t) = e itx dν n converges to c(t) = e itx dν almost surely. Another fact is that there is a countable selection of bounded Lipschitz funcitons on R which determines the convergence in distribution. So in this way we can prove the theorem. We can relax the conditions even further and give the following two theorems: Theorem 10. For each n Z +, let M n = [X (n) ij ]n i,j=1 entries such that be a symmetric n n matrix with Random X (n) ij are independent with mean zero and variance 1. sup i,j,n E X (n) ij 4 < C for some constant C. If we define ν n and ν as before, then ν n ν a.s. The second condition could also be replaced by some sort of uniform integrability of variance. 1

13 Theorem 11. For each n Z +, let M n = [X (n) ij ]n i,j=1 be a symmetric n n matrix. Assume the r(n) matrix EM n has ran r(n), with lim n = 0. If also assuming V arx(n) ij = 1 and sup E X (n) ij EX (n) ij 4 <, i,j,n then for any bounded and continuous function f(x), fdν n = fdν With some tightness on ν n the above conditions imply ν n ν. Note that when all entries have the same mean, then r(n) = 1, which gives a special case of the theorem. Another note is that if A is symmetric matrix, then the eigenvalues of A and A+λee T are interlaced for λ R, e R n. For matrices with complex entries, we have the following theorem: Theorem 1. M n = [X ij ] n i,j=1 be an n n matrix with Random entries such that X ij = X ji X ij s are i.i.d for all i < j, and X ii are i.i.d for all i. For all 1 i, j n, E X ij = 1, E[X i j] = 0 All moments exists for each entry. Define ν n and ν as before, then ν n ν. This is analogue of theorem, and the proof is very similar, with the only difference X ij = X ji, so we will have X ij X ji = X ij in our computation. 13