Lecture 09: Spectral Graph Theory Lecture 09: Spectral Graph Theory Sparsest Cuts Let G = (V ; E) be an undirected graph Definition (Sparsity of a Cut) [j1S (u) − 1S (v)j] σ(S) := E(u;v)2E E(u;v)2V 2 [j1S (u) − 1S (v)j] Definition (Sparsity of a Graph) σ(G) := min σ(S) S⊆V : S6=;;S6=V E(S;V −S) For a d-regular graph G, σ(S) = djSjjV −Sj=jV j Lecture 09: Spectral Graph Theory Edge Expansion Let G be a d-regular undirected graph Definition (Edge Expansion of a Set) E(S; V − S) φ(S) := d jSj Definition (Edge Expansion of a Graph) φ(G) := min φ(S) S : jSj6jV j=2 Lecture 09: Spectral Graph Theory Relation between Sparsity and Edge Expansion Lemma For a regular graph G, 1 σ(G) φ(G) σ(G) 2 6 6 Proof is trivial Lecture 09: Spectral Graph Theory Expander Graph Family Definition (Family of Expander Graphs) Let fGngn>d be a family of d-regular graphs such that φ(Gn) > φ. This family is called a family of expander graphs. Lecture 09: Spectral Graph Theory Connectivity of Expanders Lemma Let φ(G) > φ > 0. Consider any 0 < " < φ. On removal of any " jEj edges from G, there exists a connected component of G of size at least (1 − "=2φ) jV j. Let E 0 be any set of at most " jEj edges in G 0 Let C1;:::; Ct be the connected components of G (the graph obtained from G by removal of edges E 0) in non-decreasing order If jC1j 6 jV j =2: 0 1 X 1 X E E(Ci ; Cj ) = E(Ci ; V − Ci ) > 2 2 i6=j i 1 X d jV j φ jC j φ = > 2 i 2 i This is impossible Lecture 09: Spectral Graph Theory Proof Continued... If jC1j > jV j =2: 0 E > E(C1; V − C1) > dφ jV − C1j "d jV j =) jV − C j 1 6 2dφ Hence, jC1j > (1 − "=2φ) jV j Lecture 09: Spectral Graph Theory Hermitian Matrices ∗ P hx; yi := x y = i xi yi hx; xi = kxk 2 n×n ∗ A Hermitian matrix M 2 C satisfies M = M If Mx = λx then λ is the eigenvalue and x is the corresponding eigenvector Lecture 09: Spectral Graph Theory Eigenvalues of Hermitian Matrices Lemma All eigenvalues of a Hermitian M are real Suppose λ is an eigenvalue and x is its corresponding eigenvector Consider hMx; xi = hx; M∗xi = hx; Mxi Note that hMx; xi = λhx; xi and hx; Mxi = λhx; xi Hence, λ = λ Lecture 09: Spectral Graph Theory Eigenvectors of Hermitian Matrices are Orthogonal Lemma Let x and y be eigenvectors of a Hermitian M corresponding to two different eigenvalues. Then, hx; yi = 0. Let λ and λ0 be eigenvalues corresponding to x and y respectively Note that hMx; yi = λhx; yi Note that hx; Myi = λ0hx; yi Since λ 6= λ0, we have hx; yi = 0 Lecture 09: Spectral Graph Theory Variational Characterization of Eigenvalues Theorem (Courant-Fischer Theorem) n×n Let M 2 R be a symmetric matrix. Let λ1 6··· 6 λn be a sequence of non-decreasing eigenvalues with multiplicities. Let x1;:::; xi be the eigenvectors corresponding to λ1; : : : ; λi . Then hx; Mxi λk+1 = min n x2R −{0g: x?hx1;:::;xk i hx; xi Lecture 09: Spectral Graph Theory Proof Using Spectral Theorem: The eigenvectors form an orthonormal basis If x ? hx ;:::; x i then x = P a x 1 k n>i>k i i Then we have: P 2 hx; Mxi i>k ai λi = P 2 > λk+1 hx; xi i>k ai Corollary If λ1 6··· 6 λn then hx; Mxi λk = min max dim(V )=k x2V −{0g hx; xi Lecture 09: Spectral Graph Theory Basics of Spectral Graph Theory Definition (Laplacian) Let G be a d-regular undirected graph with adjacency matrix A. The normalized Laplacian is defined to be: 1 L := I − · A d 1 P 2 Note that hx; Lxi = d (u;v)2E (xu − xv ) Lecture 09: Spectral Graph Theory Graph Properties from Laplacian Theorem Suppose the eigenvalues of L are λ1 6··· 6 λn. Then: λ1 = 0 and λn 6 2 λk = 0 if and only if G has > k connected components λn = 2 if and only if a connected component of G is bipartite Lecture 09: Spectral Graph Theory Proof n Note that hx; Lxi > 0, for all x 2 R − 0 Note that a constant vector is a eigenvector with eigenvalue 0 Therefore, λ1 = 0 If λk = 0 then there exists a vector space V such that for any P 2 x 2 V we have (u;v)2E (xu − xv ) = 0 So, x is constant within each component k = dim(V ) 6 number of connected components in G Converse if easy to see using constant functions over each connected component Lecture 09: Spectral Graph Theory Proof Continued 1 P 2 Note that 2hx; xi − hx; Lxi = d (u;v)2E (xu + xv ) So, λn 6 2 Suppose λn = 2 Consider x as its corresponding eigenvector There exists an edge (u; v) such that xu = a and xv = −a Let A be the set fv : xv = ag Let B be the set fv : xv = −ag Note that no edge connects two vertices within A or two vertices within B Note that no edge connects any vertex in A with a vertex outside B Note that no edge connects any vertex in B with a vertex outside A (A; B) form a connected component and is bi-partite Lecture 09: Spectral Graph Theory.