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aa362, April 2nd, 2007 Lecture 28 – Spectral Analysis of Simple Graphs A matrix A can be written as: a11 = M a m1 A = VT D V K a1n e1 0 a11 K am1 O M O M O M L amn 0 en a1n L amn = Left EigenVector * EigenValues * Right Eigenvector This decomposition can be used when say you have a matrix of documents to word occurrences. Such a matrix will have huge dimensions. However, you can simplify the problem by decomposing that matrix, using Singular Value Decomposition, into “concepts” represented by eigenvalues. Lemma 1 Let G be a connected regular graph (where a regular matrix is one in which all vertices have same degree) of degree d. Let A be the adjacency matrix of this graph. Then the eigenvalues of A satisfy: d = λ 1 > λ 2 ≥ Kλ n ≥ − d Where λ1 > λ2 Because the graph is connected While λn ≥ −d Equality happens in this expression only if G is a bipartite graph. Proof Let U = (1, 1, …., 1) be an eigenvector of A with eigenvalue d. Therefore: AU = dU Let X be another eigenvector of A. By the way any constant times an eigenvector is also an eigenvector. So, by another we mean , an eigenvector that is no proportional to U. aa362, April 2nd, 2007 Let Xmax be the maximum component of X. Let S = { I | Xi = Xmax} In other words S is a subset of vertices of G where the value assigned to that vertex by X is Xmax. There is at least one vertex outside S that will be connected to a vertex j inside S because the graph is connected. The value assigned by X to such a vertex(ices) is < Xmax. Now multiply X with jth row of A. The jth row of A contains d 1’s. (A)(X) = λ (X) = λ Xj Since there is at least 1 ‘d’ in Aj that was outside of S, therefore: A Xj < d Xmax λ Xj < d Xmax Since Xj = Xmax λ<d We have proven one half of the lemma. The other half of the proof can be proven similarly using bipartite graphs. Lemma 2 If G is not connected and has exactly k components then: d = λ 1 = λ 2 = Kλ k > λ k + 1 K Proof by Example Lets say we have two components: B A= 1 0 0 B2 There are no edges between B1 and B2. Suppose I have an eigenvector for B1 then we can get an eigenvector for A. How? Say the eigenvector for B1 is simply (x1, x2, x3), then we can extend it with zeroes and multiply it with A to get an eigenvector for A. aa362, April 2nd, 2007 B1 0 x1 x1 x2 x2 x3 0 x3 = λ B2 0 0 M M 0 0 x1 x2 x3 Then is the eigenvector for A. 0 M 0 Eigenvalues of a graph that contains k, instead of two, components can be computed by taking the union as a multiset of eigenvalues of the individual components. From our earlier lemma, for each of these blocks, you are gonna have one of the eigenvalues be d. For k components, you are going to have k copies of this eigenvalue. Hence, the proof. Lemma 3 Let G1 and G2 be graphs where G1 ⊆ G2 which means that G2 is obtained by adding edges to G1. The max eigenvalue of G2 is at least as large as the max eigenvalue of G1. Proof Let A1 and A2 be adjacency matrices. Let λi(Ai) be the max eigenvalue of Ai. Let v1 be the eigenvector associated with λ1(A1). v1 has non-negative coordinated because A1 only has 1’s and 0’s in it. Now we know that Av = λ v v Av = λ vTv vTAv = λ T T λ1(A1) = v Av ≤ v A2 v (Since A2 has a few more 1’s than A1) T T λ1(A2) = max| x|= 1 x A2 x ≥ v A2 v ≥ λ 1 ( A1 ) T aa362, April 2nd, 2007 Star Graphs Graphs in which a single vertex has edges to all other vertices each one of which only have degree 1. The corresponding adjacency matrix A for such a graph would be: 0 1 1L1 1 0 0L 0 A= 1 0 0L 0 M An eigenvalue and eigenvector can be found for this matrix by making the following guess: n − 1 n − 1 n − 1 0 1 1L1 n − 1 1 1 = n − 1 = n − 1 1 1 0 0L 0 1 1 0 0L 0 M M M M 1 n − 1 1 n − 1 1 is the eigenvector and Therefore, 1 M 1 n − 1 is the eigenvalue for this matrix. Since the rank of A is 2, because all rows except the first one are identical, therefore, there are n – 2 eigenvalues that are zero (trivial) and 2 non-trivial eigenvalues one of which is n − 1 . To calculate the other we can do the following calculation: n − 1 − (n − 1) n − 1 0 1 1L1 n− 1 −1 −1 = n − 1 = − n − 1 − 1 1 0 0L 0 − 1 1 0 0L 0 M M M M −1 n − 1 −1 Therefore, − n − 1 is the other eigenvalue. Lemma 4 Let G be a connected, undirected, irregular graph. Let dmin and dmax be its min and max degrees. Then aa362, April 2nd, 2007 max{d min , d max } ≤ λ 1 ≤ min{d max , 2 | E |} Let A be the adjacency matrix and U = (1, 1, …1) AU = degree vector ≥ dmin U AU = degree vector ≥ min degree vector UTAU ≥ dmin UTU xT Ax u T Au λ1 = max T ≥ T ≥ d min x x u u dmin ≤ λ1 To prove: d max ≤ λ1 Let Gstar be star formed by highest degree vertex in G. Then from discussion on star graphs: λ1(Gstar) = d max What is the relationship between λ1(Gstar) and λ1(G)? Since one can only add edges to Gstar to get G, therefore, λ1 can only increase. Therefore: λ1(G) ≥ λ1(Gstar) = d max