1 Replacement Product and Zig-Zag Product

Transcription

1 U.C. Berkeley CS9: Spectral Methods and Expanders Handout 7 Luca Trevisan March 8, 06 Lecture 7 In which we analyze the zig-zag graph product. In the previous lecture, we claimed it is possible to combine a d-regular graph on D vertices and a D-regular graph on N vertices to obtain a d -regular graph on ND vertices which is a good expander if the two starting graphs are. Let the two starting graphs be denoted by H and respectively. Then, the resulting graph, called the zig-zag product of the two graphs is denoted by z H. We will use λ() to denote the eigenvalue with the second-largest absolute value of the normalized adjacency matrix d A of a d-regular graph. If 0 = λ λ λ n are the eigenvalues of the normalized Laplacian of, then λ() = max{ λ, λ n }. We claimed that if λ(h) b and λ() a, then λ( z H) a + b + b. In this lecture we shall recall the construction for the zig-zag product and prove this claim. Replacement Product and Zig-Zag Product We first describe a simpler product for a small d-regular graph on D vertices (denoted by H) and a large D-regular graph on N vertices (denoted by ). Assume that for each vertex of, there is some ordering on its D neighbors. Then we construct the replacement product (see figure) r H as follows: Replace each vertex of with a copy of H (henceforth called a cloud). For v V (), i V (H), let (v, i) denote the i th vertex in the v th cloud. Let (u, v) E() be such that v is the i-th neighbor of u and u is the j-th neighbor of v. Then ((u, i), (v, j)) E( r H). Also, if (i, j) E(H), then u V () ((u, i), (u, j)) E( r H). Note that the replacement product constructed as above has ND vertices and is (d + )-regular.

2 B B B C A E B C D F r H A E A A A E E E B C D D D D F F F C F C r H Zig-zag product of two graphs iven two graphs and H as above, the zig-zag product z H is constructed as follows (see figure): The vertex set V ( z H) is the same as in the case of the replacement product. ((u, i), (v, j)) E( z H) if there exist l and k such that ((u, i)(u, l), ((u, l), (v, k)) and ((v, k), (v, j)) are in E( r H) i.e. (v, j) can be reached from (u, i) by taking a step in the first cloud, then a step between the clouds and then a step in the second cloud (hence the name!).

3 B B B C A E B C D F z H A E A A A E E E B C D D D D F F F C F C z H It is easy to see that the zig-zag product is a d -regular graph on ND vertices. Let M R ([N] [D]) ([N] [D]) be the normalized adjacency matrix of z H. Using the fact that each edge in r H is made up of three steps in r H, we can write M as BAB, where { 0 if u v B[(u, i), (v, j)] = d if u = v and {i, j} H And A[(u, i), (v, j)] = if u is the j-th neighbor of v and v is the i-th neighbor of u, and A[(u, i), (v, j)] = 0 otherwise. Note that A is the adjacency matrix for a matching and is hence a permutation matrix. A Technical Preliminary We will use the following fact. Suppose that M = d A is the normalized adjacency matrix of a graph. Thus the largest eigenvalue of M is, with eigenvector ; we have λ() = max x x T Mx = max x M x ()

4 which is a corollary of the following more general result. Recall that a vector space S R n is an invariant subspace for a matrix M R n n if Mx S for every x S. Lemma Let M be a symmetric matrix, and S be a k-dimensional invariant subspace for M. Thus, (from the proof of the spectral theorem) we have that S has an orthonormal basis of eigenvectors; let λ λ k be the corresponding eigenvalues with multiplicities; we have max λ x T Mx i = max i=,...k = max M x Proof: If the largest eigenvalue in absolute value is λ k, then max λ x T Mx i = λ k = max i=,...k and if it is λ (because λ is negative, and λ > λ n ) so we have From Cauchy-Schwarz, we have and so max λ x T Mx i = λ = min i=,...k = max xt Mx max λ x T Mx i max () i=,...k max x T Mx Mx x T Mx max M x Finally, if x,..., x k is the basis of orthonormal eigenvectors in S such that Mx i = λ i, then, for every x S, we can write x = i a ix i and () Mx = i λ i a i x i = i λ i a i max λ i i=,...,k i a i = max i=,...,k λ i so max Mx and the Lemma follows by combining (), () and (). max i=,...,k λ i ()

5 Analysis of the zig-zag Product Theorem Let be a D-regular graph with n nodes, H be a d-regular graph with D nodes, and let a := λ(), b := λ(h), and let the normalized adjacency matrix of z H be M = BAB where A and B are as defined in Section. Then λ( z H) a + b + b Proof: Let xr n D be such that x. We refer to a set of coordinates of x corresponding to a copy of H as a block of coordinate. We write x = x + x, where x is constant within each block, and x sums to zero within each block. Note both x and x are orthogonal to, and that they are orthogonal to each other. We want to prove We have (using the fact that M is symmetric) x T Mx a + b + b (5) x T Mx = x T Mx + x T Mx + x T Mx And it remains to bound the three terms.. x T Mx a x Because, after writing M = BAB, we see that Bx = x, because B is the same as I n ( A d H), the tensor product of the identity and of the normalized adjacency matrix of H. The normalized adjacency matrix of H leaves a vector parallel to all-ones unchanged, and so B leaves every vector that is constant in each block unchanged. Thus x T Mx = x T Ax Let y be the vector such that y v is equal to the value that x has in the block of v. Then x T Ax = y v y w = y T A y = ad y a x {(v,i),(w,j)} E z H because y and y = D x 5

6 . x T Mx b x Because, from Cauchy-Schwarz and the fact that permutation matrices preserve length, we have x T BABx Bx ABx = Bx Now let us call x v the restriction of x to coordinates of the form (v, i) for i =,..., D. Then each x v is orthogonal to the all-one vector and A Hx v db x v, so Bx = v d A H x v v b x v = b x. x T Mx b Because, from Cauchy-Schwarz, the fact that Bx = x and the fact that permutation matrices preserve length, we have and we proved above that so x T BABx Bx ABx = x Bx Bx b x x T BABx b x x b ( x + x ) = b 6