Math 778S Spectral Graph Theory Hadout #3: Eigevalues of Adjacecy Matrix The Cartesia product (deoted by G H) of two simple graphs G ad H has the vertex-set V (G) V (H). For ay u, v V (G) ad x, y V (H), (u, x) is adjacet to (v, y) if either u = v ad xy E(H) or uv E(G) ad x = y. Lemma 1 Suppose λ 1,..., λ are eigevalues of the adjacecy matrix of a graph G ad µ 1,..., µ m are eigevalues of the adjacecy matrix of a graph H. The the eigevalues of the adjacecy matrix of the Cartesia product G H are λ i + µ j for 1 i ad 1 j m. Proof: Let A (or B) be the adjacecy matrix of G (or H) respectively. For ay eigevalue λ of A ad ay eigevalue µ of B, we would like to show λ + µ is a eigevalue of G H. Let α be the eigevector of A correspodig to λ ad β be the eigevector of B correspodig to µ. We have Equivaletly, for ay u V (G), α v = λα u ; for ay x V (H), Aα = λα (1) Bβ = µβ. (2) β y = µβ x. Let α β be the m colum vector defied by etries (α β) u,x = α u β x. Let C be the adjacecy matrix of G H. We would like to show α β is a eigevector of C. We have, for ay (u, x) V (G H), (α β) v,y = (v,y) (u,x) = α v β y (v,y) (u,x) α u β y + α v β x (u,y) (u,x) (v,x) (u,x) = α u β y + α v β x = α u β y + β x = α u µβ x + β x λα u = (λ + µ)(α β) u,x. α v 1
This is equivalet to C(α β) = (λ + µ)(α β). Thus, λ + µ is a eigevalue of G H. For 1 i ad 1 j m, λ i + µ j are eigevalues of G H. Sice G H has m vertices, these eigevalues (with multiplicity) are all eigevalues of G H. Remark: The adjacecy matrix of G H ca be writte as A I m +I B. Here is tesor product of matrices. Hypercube Q : The vertices of Q are poits i -dimesioal space over the field of two elemets F 2 = {0, 1}. Two poits are adjacet i Q if ad oly if they differ by exactly oe coordiate. We have Q 1 = P 2, Q 2 = C 4, ad Q 3 is the cube i 3-dimesioal space. We have Q +1 = Q 1 Q. The eigevalues of Q ca be determied from the eigevalues of Q 1 ad the above lemma. Q 1 = P 2 has eigevalues ±1. Q has eigevalues 2i with multiplicity ( ) i for 0 i. Regular graphs: The degree of a vertex v i G is the umber of edges icidet to v. If all degrees are equal to d, the G is called a d-regular graph. Let 1 be the colum vector of all etries equal to 1. If G is a regular graph, the A1 = d1. Hece, 1 is a eigevector for the eigevalue d. Eigevalues of K : Let J = 1 1 be the -matrix with all etries 1. Sice J is a rak 1 matrix, J has eigevalues 0 with multiplicity 1. It is easy to see that the ozero eigevalue of J is. The complete graph K has the adjacecy matrix J I. Thus, K has a eigevalue 1 of multiplicity 1 ad 1 of multiplicity 1. 0 1 0 0... 0 0 0 1 0... 0 Eigevalues of C : Let Q = 0 0 0 1... 0......... 1 0 0 0... 0 (Q ca be viewed as the adjacecy matrix of the directed cycle.) We have A = Q + Q. Note that Q = I. Let λ be the eigevalue of Q. We have λ = 1. The eigevalues of Q are precisely -th root of 1: ρ k = cos( 2kπ ) + 1 si( 2kπ ), for 0 k 1. Note Q = Q 1. Thus, A = Q + Q has eigevalues for k = 0, 1, 2,..., 1. ρ k + ρ k( 1 = 2R(ρ k ) = 2 cos( 2kπ ) 2
Let µ 1 µ 2... µ be the eigevalues of the adjacecy matrix of a graph G. We refer µ 1 = µ max ad µ = µ mi. We have µ max = sup x Ax µ mi = if x Ax Suppose f(x) = x Ax reaches the maximum at α o the uit sphere. The all coordiates of α are o-egative. Lemma 2 If H is a subgraph of G, the we have µ max (G) µ max (H). Proof: Without loss of geerality, we assume V (H) = V (G). (Otherwise, we add some isolated vertices to H. It does t chage the maximum eigevalue of H.) Let α be the eigevector A H correspodig to µ max (H). We have µ max (H) = α A H α = 2 ij E(H) 2 ij E(G) = α A G α α i α j α i α j sup x A G x = µ max (G). Let δ be the miimum degree ad be the maximum degree of G. We have the followig boud o µ max. Lemma 3 For every graph G, we have δ(g) µ max (G) (G). Proof: Let α be a eigevector for eigevalue µ = µ max (G). Sice α 0, we ca assume α has at least oe positive coordiate. (If all coordiates are oe-positive, we cosider α istead.) Let α k = max i α i be the largest coordiate of α. Sice Aα = µα, we have µα k = (Aα) k = i k α i α k. Thus, µ. 3
Now we show µ max (G) δ(g). µ max = sup x A G x 1 1 1 A G 1 = 1 2 i j a ij = 2 E(G) δ(g). A k-colorig of a graph G is a map c: V (G) [k] = {1, 2,..., k}. A k- colorig is said to be proper if the ed vertices of ay edge i G receive differet colors. I.e., c(u) c(v) for ay u v. I this case, we say G is k-colorable. The chromatic umber deoted by χ(g) is the miimum iteger k such that G is k-colorable. For example, χ(k ) =. χ(g) = 2 if ad oly if G is a oempty bipartite graph. There is a simple boud o χ(g). Theorem 1 For every G, χ(g) 1 + (G). Proof: Give ay order v 1, v 2,..., v, we color vertices oe by oe usig + 1 colors. At time i, we assume v 1,..., v i 1 has bee colored properly. Note that v i has at most eighbors i v 1,..., v i 1. We ca pickup a distict color for v i other tha those eighbors received. The resulted colorig is a proper colorig.. Theorem 2 (Wilf 1967) For every G, χ(g) 1 + λ max (G). Proof: I the proof of the previous lemma, the graph G is k-colorable if v i has at most k 1 eighbors i the iduced subgraph o v 1, v 2,..., v i for all i = 1, 2,...,. Sice the order of the vertices ca be arbitrary, we choose v to be the vertex havig the miimum degree. For i =, 1,..., 1, let v i be the vertex havig miimum degree i the iduced subgraph G i o v 1, v 2,..., v i. Note δ(g i ) µ max (G i ) µ max (G). Thus, uder this order, the previous greedy algorithm results a proper k-colorig for ay k 1 + µ max (G). 4
Remark: Brook s theorem states that if G is a simple coected graph other tha the complete graph ad odd cycles the χ(g) (G). It is ukow whether similar result ca be proved usig µ max (G) istead. Assume µ 1 > µ 2 >... > mu k are distict eigevalues of A. The φ(x) = k i=1 (x µ k) is called the miimal polyomial of A. We have φ(a) = 0. Ay polyomial f(x) with f(a) = 0 is divisible by φ(x). For ay pair of vertices u, v, the distace d(u, v) is the shortest legth of ay uv-path. The diameter of graph G is the maximum distace amog all pairs of vertices which belogs to the same coected compoet. Theorem 3 The diameter of a graph is less tha its umber of distict eigevalues. Proof: Without loss of geerality, we ca assume G is coected. Let k be the umber of distict eigevalues. The miimum polyomial φ(x) has degree k. Sice φ(a) = 0, A k ca be expressed as a liear combiatio of I, A,..., A k 1. Suppose the diameter of G is greater tha or equal to k. There exists a pair of vertices u ad v satisfyig d(u, v) = k. We have (A k ) uv 1 ad (A i ) uv = 0 for i = 0, 1, 2,..., A k 1. This is a cotradictio to the fact A k is a liear combiatio of I, A,..., A k 1. This result is tight for the hypercube Q. 5