Ma/CS 6b Class 3: Eigenvalues in Regular Graphs By Adam Sheffer Recall: The Spectrum of a Graph Consider a graph G = V, E and let A be the adjacency matrix of G. The eigenvalues of G are the eigenvalues of A. The characteristic polynomial φ G; λ is the characteristic polynomial of A. The spectrum of G is spec G = λ 1,, λ t m 1,, m t, where λ 1,, λ t are the eigenvalues of A and m i is the multiplicity of λ i. 1
Example: Spectrum A = det λi A 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 = det λ 1 0 1 1 λ 1 0 0 1 λ 1 1 0 1 λ = λ λ λ +. v 1 v v 4 v 3 spec C 4 = 0 1 1 Slight Change of Notation Instead of multiplicities, let λ 1,, λ n be the not necessarily distinct eigenvalues of n. 1 For example, if the spectrum is, we write λ 1 = λ = and λ 3 = λ 4 = 1 (instead of λ 1 =, m 1 =, λ = 1, m = ).
Recall: The Spectral Theorem Theorem. Any real symmetric n n matrix has real eigenvalues and n orthonormal eigenvectors. By definition, any adjacency matrix A is symmetric and real. The algebraic and geometric multiplicities are the same in this case. n We have φ A; λ = i=1 λ λ i. The multiplicity of an eigenvalue λ is n rank λi A. More Examples We already derived the following: 1 n 1 spec K n = n 1 1 spec K n,m = 0 mn mn m + n 1 1 Our next goal is to study regular graphs. Can you come up with an eigenvector of any regular graph with n vertices? 3
Eigenvalues of Regular graphs If A is the adjacency matrix of a d-regular graph, then any row of A contains exactly d 1 s. Thus, the vector 1 n = 1,1,, 1 is an eigenvector of A with eigenvalue d. Theorem. Let G be a connected graph. The eigenvalue of G of largest absolute value is the maximum degree if and only if G is regular. Proof A n n adjacency matrix of a graph G. Δ G the maximum degree of G. x = x 1,, x n eigenvector of eigenvalue λ of largest absolute value. x j = max x i. i λ x j = Ax j = x i So λ Δ G. v i N v j deg v j x j Δ G x j, 4
Δ G the maximum degree of G. We proved that the absolute value of any eigenvalue of A is at most Δ G, using λ x j = Ax j = x i v i N v j deg v j x j Δ G x j. For equality to hold, we need deg v j = Δ G. x i = x j for each v i N v j. That is, v j and all of its neighbors are of degree Δ G. Repeating the same argument for a neighbor v i implies that v i s neighbors are also of degree Δ G. We continue to repeat the argument to obtain that the graph is regular. Multiplicity The previous proof also shows that any eigenvector x 1,, x n of the eigenvalue d satisfies x 1 = x = = x n. Thus, the space of eigenvectors of the eigenvalue d is of dimension 1 (that is, d has multiplicity 1). 5
The Spectrum of the Petersen Graph The Petersen graph G = V, E is a 3- regular graph with 10 vertices. We know that it has eigenvalue 3 with eigenvector 1 10. To find the other eigenvalues, we notice some useful properties: If u, v E then u and v have no common neighbors. If u, v E then u and v have exactly one common neighbor. The Adjacency Matrix The number of neighbors shared by v i and v j is A ij. That is 3, if i = j, A ij = 0, if i j and v i, v j E, 1, if i j and v i, v j E. That is, A + A I = 1 n n. 6
The Additional Eigenvalues We have A + A I = 1 n n. Since 3 is an eigenvalue with eigenvector 1 n, the other eigenvectors are orthogonal to 1 n. Thus, for an eigenvector v of eigenvalue λ 3: 1 n n v = 0 n. That is, A + A I v = 0 n. If v is an eigenvector of λ then we have λ v + λv v = 0 n. Thus, the additional eigenvalues of A are 1,. spec G = 3 1 1 m m 3. The Multiplicities spec G = 3 1 1 m m 3. Recall that 10 i=1 λ i = trace A = 0. That is, 3 + m m 3 = 0. Combining this with m + m 3 = 9 and m, m 3 0, we obtain the unique solution m = 5, m 3 = 4. spec G = 3 1 1 5 4. 7
Petersen Graph and K 10 Problem. Can we partition the edges of K 10 into three disjoint sets, such that each set forms a Petersen graph? 10 K 10 has = 45 edges, and the Petersen graph has 15. In K 10 every vertex is of degree 9, and in the Petersen graph 3. Disproof Assume, for contradiction, that the partition exists, and let A, B, C be the adjacency matrices of the three copies of the Petersen graph. The adjacency matrix of K 10 is 1 10 10 I. That is, A + B + C = 1 10 10 I. V A, V B the vector subspaces of eigenvectors corresponding to the eigenvalue 1 in A and B. We know that dim V A = dim V B = 5. Since both V A and V B are orthogonal to 1 10, they are not disjoint (otherwise we would have a set of 11 orthogonal vectors in R 10 ). 8
Disproof (cont.) A, B, C the adjacency matrices of the three copies of the Petersen graph in K 10. A + B + C = 1 n n I. V A, V B the vector subspaces of eigenvectors corresponding to the eigenvalue 1 in A and B. V A and V B are not disjoint. Let z V A V B. Since every vector in V A and V B is orthogonal to 1 n, so is z. We have Cz = 1 n n I A B z = 0 z Az Bz = 3z. Contradiction since -3 is not an eigenvalue of C. Four Things You Did Not Know About The Petersen Graph It has 15 edges and 000 spanning trees. It is the smallest 3-regular graph of girth 5 (this is called a 3,5 -cage graph). It likes gardening, ballet, and building airplane models. It has gotten divorced three times. 9
Moore Graphs A Moore graph is a graph that is dregular, of diameter k, and whose number of vertices is k 1 1 + d d 1 i. i=0 As can be easily checked, this is the minimum possible number of vertices of any graph of diameter k and minimum degree d. Examples of Moore Graphs A Moore graph is a graph that is d-regular, of diameter k, and whose number of vertices is k 1 1 + d d 1 i. i=0 What Moore graphs do we know? The Petersen graph is 3-regular, of diameter, and contains 1 + 3 1 i=0 3 1 i = 10 vertices. 10
Examples of Moore Graphs A Moore graph is a graph that is d-regular, of diameter k, and whose number of vertices is k 1 1 + d d 1 i. i=0 The Petersen graph is 3-regular, of diameter, and contains 1 + 3 1 i=0 3 1 i = 10 vertices. Is there a Moore graph that is -regular and of diameter? C 5 Moore Graphs of Diameter and Girth 5 Recall that the girth of a graph is the length of the shortest cycle in it. Theorem. There exist d-regular Moore graphs with diameter and girth 5 only for d =,3,7, and possibly 57. The case of d = 57 is an open problem. If it exists, it has 350 vertices, and 9,65 edges. The case of d = 7 11
G = V, E a d-regular graph of diameter, girth 5, and with 1 V = 1 + d d 1 i = 1 + d. i=0 Since the girth is five, if v i, v j E then v i and v j have no common neighbors. Since the diameter is two, if v i, v j E then v i and v j have exactly one common neighbor. Thus, the adjacency matrix A satisfies d, if i = j, A ij = 0, if i j and v i, v j E, 1, if i j and v i, v j E. That is, we have A + A d 1 I = 1 n n. Proof (cont.) G = V, E a d-regular graph of diameter, girth 5, and with V = 1 + d. A + A d 1 I = 1 n n. λ d an eigenvalue of A with eigenvector v. Since λ d, v is orthogonal to 1 n. Thus A + A d 1 I v = 0 n, or λ v + λv d 1 v = 0. This implies that λ + λ d + 1 = 0, so λ,3 = 1 ± 1 + 4 d 1 = 1 ± 4d 3. We thus have spec G = d λ λ 3 1 m m 3. 1
Finding the Multiplicities We have λ = 1+ 4d 3, λ 3 = spec G = d λ λ 3 1 m m 3. 1 4d 3 0 = trace A = d + λ m + λ 3 m 3 = d m + m 3 + m 3 m 4d 3. Since m + m 3 = n 1 = d, we have d d = m 3 m 4d 3. This can happen if either m = m 3 or 4d 3 = s for some integer s. If m = m 3 than d d = 0, implying d =., The Case of 4d 3 = s d d = m 3 m 4d 3. Assume that 4d 3 = s for some integer s. That is, d = s + 3 /4. Substituting into the above equation: s + 3 s + 6 = m 16 4 3 m s. Setting m 3 m = m 3 d, we have s 4 + 6s + 9 8s 4 = 3m 3 s s 5 6s 3 9s. s 5 + s 4 + 6s 3 s + 9 3m 3 s = 15. So s must divide 15, and we get s ± 1,3,5,15, which implies d 1,3,7,57. The case d = 1 leads to K. Not a Moore graph. 13
Recall: Independent Sets Consider a graph G = V, E. An independent set in G is a subset V V such that there is no edge between any two vertices of V. Finding a maximum independent set in a graph is a major problem in theoretical computer science. No polynomial-time algorithm is known. Past Bounds Let G = (V, E) be a graph. Already proved: G has an independent set of size at least 1. 1 + deg v v V If E = V d, then G has an independent set of size at least V /d. 14
An Upper Bound Theorem. For a d-regular graph G = V, E with smallest (most negative) eigenvalue λ n, the size of the largest independent set of G is at most Example. spec C 4 = 0 1 1. 4 So at most =. 1 n 1 d/λ n. v 1 v v 4 v 3 Recall: The Rayleigh Quotient The Rayleigh quotient is R A, x = R(x) = xt Ax for n n matrix A and x x T x Rn. Lemma. Let A be a real symmetric n n matrix. Then R x attains its maximum and minimum at eigenvectors of A. (We do not prove the lemma.) Question. What is R x when x is an eigenvector of eigenvalue λ? xt Ax x T x = xt λx x T x = λ. Thus, the min and max values of R x are the min and max eigenvalues of A. 15
λ n the most negative eigenvalue of G. S a largest independent set of G. 1 S = x 1,, x n a vector with x i = 1 if v i S (otherwise x i = 0). y = n1 S 1 n S. y T Ay = n 1 S T A1 S S n 1 S T A1 n + S 1 n T A1 n. Since S is an independent set, we have 1 S T A1 S = A ij i,j S = 0. Since G is d-regular, 1 S T A1 n = 1 S T d1 n = d S, and also 1 n T A1 n = 1 n T d1 n = dn. Combining the above, we have y T Ay = 0 S n d S + S dn = S dn. y T y = n 1 S T 1 S S n1 S T 1 n + S 1 n T 1 n = n S S n + S n = S n n S. Completing the Proof By the lemma, we have y T Ay y T y λ n. We have Thus λ n y T Ay = S dn. y T y = S n n S. S dn S n n S d S = n S. n λ n n S d S S 1 d/λ n 16
The End 17