Cospectrality of graphs

Size: px

Start display at page:

Download "Cospectrality of graphs"

Augusta Johnston
5 years ago
Views:

Linear Algebra and its Applications 451 (2014) 169 181 Contents lists available at ScienceDirect Linear Algebra and its Applications www.elsevier.

Institute for Research in Fundamental Sciences (IPM), P.O.

1 Linear Algebra and its Applications 451 (2014) Contents lists available at ScienceDirect Linear Algebra and its Applications Cospectrality of graphs Alireza Abdollahi a,b,, Mohammad Reza Oboudi a,b a Department of Mathematics, University of Isfahan, Isfahan , Iran b School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box , Tehran, Iran article info abstract Article history: Received 17 October 2013 Accepted 15 February 2014 Available online xxxx Submitted by R. Brualdi MSC: 05C50 05C31 Keywords: Spectra of graphs Measures on spectra of graphs Adjacency matrix of a graph Richard Brualdi proposed in Stevanivić (2007) [6] the following problem: (Problem AWGS.4) Let G n and G n be two nonisomorphic graphs on n vertices with spectra λ 1 λ 2 λ n and λ 1 λ 2 λ n, respectively. Define the distance between the spectra of G n and G n as λ ( ( G n,g ) n ( n = λi λ ) 2 i or use i=1 ) n λi λ i. i=1 Define the cospectrality of G n by Let cs(g n )=min { λ ( G n,g n) : G n not isomorphic to G n }. cs n =max { cs(g n ): G n agraphonn vertices }. Problem A. Investigate cs(g n ) for special classes of graphs. Problem B. Find a good upper bound on cs n. In this paper we study Problem A and determine the cospectrality of certain graphs by the Euclidian distance. * Corresponding author. addresses: a.abdollahi@math.ui.ac.ir (A. Abdollahi), mr.oboudi@sci.ui.ac.ir (M.R. Oboudi) / 2014 Published by Elsevier Inc.

2 170 A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) Let K n denote the complete graph on n vertices, nk 1 denote the null graph on n vertices and K 2 +(n 2)K 1 denote the disjoint union of the K 2 with n 2 isolated vertices, where n 2. In this paper we find cs(k n ), cs(nk 1 ), cs(k 2 + (n 2)K 1 )(n 2) and cs(k n,n ) Published by Elsevier Inc. 1. Introduction Throughout the paper all graphs are simple, that is finite and undirected without loops and multiple edges. By the spectrum of a graph G, we mean the multiset of eigenvalues of adjacency matrix of G. Richard Brualdi proposed in [6] the following problem: (Problem AWGS.4) Let G n and G n be two nonisomorphic graphs on n vertices with spectra λ 1 λ 2 λ n and λ 1 λ 2 λ n, respectively. Define the distance between the spectra of G n and G n as λ ( G n,g ) n ( ( n = λi λ ) 2 i or use i=1 Define the cospectrality of G n by ) n λi λ i. i=1 cs(g n )=min { λ ( G n,g n) : G n not isomorphic to G n }. Thus cs(g n ) = 0 if and only if G n has a cospectral mate. Let cs n =max { cs(g n ): G n a graph on n vertices }. This function measures how far apart the spectrum of a graph with n vertices can be from the spectrum of any other graph with n vertices. Problem A. Investigate cs(g n ) for special classes of graphs. Problem B. Find a good upper bound on cs n. In this paper we study Problem A and determine the cospectrality of certain graphs by the Euclidian distance, that is λ ( G n,g ) n ( n = λi λ ) 2. i i=1

3 A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) For a graph G, V (G) ande(g) denote the vertex set and edge set of G, respectively; G denotes the complement of G and A(G) denotes the adjacency matrix of G. Fortwo graphs G and H with disjoint vertex sets, G + H denotes the graph with the vertex set V (G) V (H) and the edge set E(G) E(H), i.e. the disjoint union of two graphs G and H. The complete product (join) G H of graphs G and H is the graph obtained from G + H by joining every vertex of G with every vertex of H. Inparticular,nG denotes G + + G }{{} n and n G denotes G G G }{{}. n We denote by Spec(G) the multiset of the eigenvalues of the graph G. For positive integers n 1,...,n l, K n1,...,n l denotes the complete multipartite graph with l parts of sizes n 1,...,n l.letk n denote the complete graph on n vertices, nk 1 = K n denote the null graph on n vertices and P n denote the path with n vertices. By the previous notation, for any integer n>2, K 2 +(n 2)K 1 denotes the disjoint union of the K 2 with n 2 isolated vertices. In this paper we find cs(k n ), cs(nk 1 ), cs(k 2 +(n 2)K 1 ) (n 2) and cs(k n,n ). In particular, we find that there exists a unique graph G H such that λ(h, G H )=cs(h) ifh {K n,nk 1,K 2 +(n 2)K 1,K n,n }. The main results of our paper are the following: Theorem 1.1. For every integer n 2, cs(nk 1 )=2. In particular, λ(nk 1,G)=cs(nK 1 ) for some graph G if and only if G = K 2 +(n 2)K 1. Theorem 1.2. For every integer n 3, cs(k 2 +(n 2)K 1 )=2( 2 1) 2. Also, cs(k 2 )= λ(k 2, 2K 1 )=2. In particular, λ(k 2 +(n 2)K 1,G)=cs(K 2 +(n 2)K 1 ) for some graph G if and only if G = P 3 +(n 3)K 1. Theorem 1.3. For every integer n 2, cs(k n )=n 2 +n n n 2 +2n 7 2. In particular, λ(k n,g)=cs(k n ) for some graph G if and only if G = K n \ e, wherek n \ e is the graph obtaining from K n by deletion one edge e. Theorem 1.4. Let n 2 be an integer. Then cs(k n,n )=2(n n 2 1) 2. In particular, λ(k n,n,g)=cs(k n,n ) for some graph G if and only if G = K n 1,n Cospectrality of graphs with at most one edge In this section we will determine the cospectrality of graphs with at most one edge. Let G be a simple graph of order n and size m. Letλ 1 λ n be the eigenvalues of G. Itiswellknownthatλ λ n =0andλ λ n 2 =2m. Wenowgivethe proof of Theorem 1.1 in which we determine the cospectrality of graphs with no edge. Proof of Theorem 1.1. Let G n be a simple graph of order n and size m with eigenvalues λ 1 λ n. Since the eigenvalues of nk 1 are λ 1 = = λ n =0,then λ ( nk 1,G n) = λ λ n 2 =2m.

4 172 A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) Since G n is not isomorphic to nk 1, m 1. So the minimum value of λ(nk 1,G n)is2 and it happens for G n = K 2 +(n 2)K 1. Proof of Theorem 1.2. Let n 3 be an integer and let G n be a simple graph of order n and size m with eigenvalues λ 1 λ n. Since the eigenvalues of K 2 +(n 2)K 1 are λ 1 =1,λ 2 = = λ n 1 =0,λ n = 1, λ ( K 2 +(n 2)K 1,G ) ( n = λ 1 1 ) 2 + λ λ 2 ( n 1 + λ n +1 ) 2 =2m +2 2λ 1 +2λ n. Now, we want to find the minimum value of m λ 1 + λ n among all graphs of order n and size m. By Perron Frobenius Theorem (see [2, Theorem 0.13]), λ n λ 1.Thus λ ( K 2 +(n 2)K 1,G n) =2m +2 2λ 1 +2λ n 2m 4λ We claim that for every graph G n K 2 +(n 2)K 1,P 3 +(n 3)K 1, the following holds Since λ ( K 2 +(n 2)K 1,G n) > 2( 2 1) 2. λ ( K 2 +(n 2)K 1,P 3 +(n 3)K 1 ) =2( 2 1) 2, the validity of our claim completes the proof. To prove the claim we consider the following cases: Case 1. Let λ Then (λ 1 1) Thus λ ( K 2 +(n 2)K 1,G ) n = ( λ 1 1 ) 2 + λ λ n ( λ n +1 ) > 2( 2 1) 2. Case 2. Let λ 1 < Then for m 4, 2m 4λ 1.Since λ ( K 2 +(n 2)K 1,G n) 2m 4λ 1 +2, it follows that λ(k 2 +(n 2)K 1,G n) 2 and so we are done. To complete the proof of our claim it remains to compute λ(k 2 +(n 2)K 1,G n) for all graphs G n with at most 3 edges. These graphs are as follows, nk 1, K 2 +(n 2)K 1,2K 2 +(n 4)K 1, P 3 +(n 3)K 1,3K 2 +(n 6)K 1, P 3 + K 2 +(n 5)K 1, K 3 +(n 3)K 1, P 4 +(n 4)K 1 and K 1,3 +(n 4)K 1. One can easily see that for the latter graphs the claim is valid. This completes the proof.

5 A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) Cospectrality of the complete graph In this section we will determine the cospectrality of the complete graphs. Let G be a simple graph of order n and size m. In this section we show that for every integer n 2, cs(k n )=λ(k n,k n \ e), where e is an arbitrary edge of K n. First we prove some lemmas. Theorem 3.1. (Theorem of [3]) LetG be a graph of order n and H be an induced subgraph of G with order m. Suppose that λ 1 (G) λ n (G) and λ 1 (H) λ m (H) are the eigenvalues of G and H, respectively. Then for every i, 1 i m, λ i (G) λ i (H) λ n m+i (G). Let X be a graph. Recall that a partition π of V (X) with cells C 1,...,C r is equitable if the number of neighbors in C j of a vertex u in C i is a constant b ij, independent of u. Theorem 3.2. (Theorem and Exercise 3 in page 213 of [3]) If π is an equitable partition of a graph X, then the characteristic polynomial of A(X/π) divides the characteristic polynomial of A(X). Moreover, the spectral radius of A(X/π) is equal to the spectral radius of A(X). Note that in Theorem 3.2, X/π denotes the directed graph with the r cells of π as its vertices and b ij arcs from the ith to the jth cells of π is called the quotient of X over π, and denoted by X/π. The entries of the adjacency matrix of this quotient are given by A(X/π) ij = b ij. Let K t n be the graph obtained from K n by deleting n t 1 edges from one vertex of K n. By the following lemma one can compute the spectrum of K t n. Lemma 3.3. Let V (K n ) = {v 0,v 1,...,v n 1 }. Let 1 t n 2 and K t n = K n \ {v 0 v t+1,v 0 v t+2,...,v 0 v n 1 }. Suppose that f(λ) :=λ 3 (n 3)λ 2 (n+t 2)λ+t(n t 2). Then Spec ( Kn) t = { 1,..., 1,λ }{{} 1,λ 2,λ 3 }, n 3 where λ 1, λ 2, λ 3 are the roots of the polynomial f(λ) and 1 max{λ 1,λ 2,λ 3 }. Proof. Let A = {v 0 }, B = {v 1,...,v t } and C = {v t+1,...,v n 1 }. It is easy to see that the partition {A, B, C} of vertices of Kn t is an equitable partition with the following matrix 0 t 0 M = 1 t 1 n 1 t, 0 t n t 2

6 174 A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) where the first (second and third) row and column corresponds to A (B and C, respectively). Since Spec(K n 1 )={n 2, 1,..., 1} and K }{{} n 1 is an induced subgraph of Kn, t n 2 by interlacing Theorem 3.1 we conclude that Kn t has at least n 3 eigenvalues 1. By Theorem 3.2, the three remaining eigenvalues of Kn t are the eigenvalues of the matrix M and the largest eigenvalue of Kn t is among the latter three eigenvalues. On the other hand f(λ) =det(λi M) andf( 1) 0. This completes the proof. Corollary 3.4. For every integer n 2 and every arbitrary edge e of K n, { n 3+ n2 +2n 7 Spec(K n \ e) =, 0, 1,..., 1, n 3 } n 2 +2n 7. 2 }{{} 2 n 3 Proof. By the notation of Lemma 3.3, K n \ e = K n 2 n. This implies the result. Remark 3.5. Let G be a graph of order n and size m. Letλ 1 λ n be the eigenvalues of G. Sincen are the eigenvalues of K n, one can obtain that λ(k n,g)=2m 2nλ 1 + n 2 n. This equality shows that to obtain cs(k n )itis sufficient to obtain a graph G for which the parameter m nλ 1 has the minimum value. In sequel we show that among all graphs G of order n except K n, the minimum value of m nλ 1 (G) is attained on the graph K n \ e. Lemma 3.6. For every integer n 3 and every edge e of K n, λ(k n,k n \e) < 2. Ifn =2, then λ(k n,k n \ e) =2. Proof. For n = 2, there is nothing to prove. Let n 3. Using Corollary 3.4 and Remark 3.5 one can obtain the result. Lemma 3.7. Let n 3 be an integer. Let K = K n \{e, e },wheree and e are two adjacent edges in K n.thenλ(k n,k) >λ(k n,k n \ e). Proof. Note that K = Kn n 3.ByLemma 3.3 λ 1 (K) is a root of the polynomial f(λ) = λ 3 (n 3)λ 2 (2n 5)λ + n 3. Since the roots of the derivation f (λ) (with respect to λ) off(λ) are n 3± n 2 6 3, f is an increasing function on the interval [ n 3+ n 2 6 3, ). Using Corollary 3.4 one can see that λ 1 (K n \ e) 1 n > n 2 > n 3+ n Itis not hard to see that f(λ 1 (K n \ e) 1 n ) > 0. On the other hand f is increasing on [ n 3+ n 2 6 3, ). Thus f(λ) > 0 for every λ λ 1 (K n \ e) 1 n.sinceλ 1(K) isaroot of f(λ), we conclude that λ 1 (K) < λ 1 (K n \ e) 1 n(n 1) n. This shows that 2 2 nλ 1 (K) > n(n 1) 2 1 nλ 1 (K n \ e). Equivalently, λ(k n,k) > λ(k n,k n \ e) (see Remark 3.5).

7 A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) We need the following theorems to prove the main result of this section. Theorem 3.8. (See [5], and also Theorem 6.7 of [2].) A graph has exactly one positive eigenvalue if and only if its non-isolated vertices form a complete multipartite graph. Lemma 3.9. Let G be a graph with eigenvalues λ 1 λ 2 λ 3 λ n. The graph G is isomorphic to one the following graphs if and only if λ 1 > 0, λ 2 0 and λ 3 < 0. (1) G = K n, (2) G = K 1 + K n 1, (3) G = K 2,1,...,1 = K n \ e for an edge e of K n. Proof. Suppose that λ 1 > 0, λ 2 0andλ 3 < 0. By Theorem 3.8, there exist some integers t 1, n 1,...,n t 1andr 0such that G = rk 1 + K n1,...,n t.sinceλ 3 < 0, r 1. Thus it is enough to investigate the cases G = K n1,...,n t and G = K 1 + K n1,...,n t. Note that t 2, otherwise G = K n, a contradiction. Suppose that G = K 1 +K n1,...,n t.ifn i 2forsomei, theng has a 3K 1 as an induced subgraph and so it follows from Interlacing Theorem that λ 3 0, a contradiction. Therefore G = K 1 + K n 1 in this case. Now assume that G = K n1,...,n t.ifn 1 = = n t =1,thenG = K n.thuswemay assume that n i 2forsomei. Suppose that n i,n j 2 for some distinct i and j. Thus the cycle C 4 of length 4 is an induced subgraph of G. Since the eigenvalues of C 4 are 2, 0, 0, 2, it follows from Interlacing Theorem that λ 3 0, a contradiction. Thus we can assume that n 1 2andn 2 = = n t =1.Ifn 1 =2,thenG = K 2,1,...,1 = K n \ e for some edge e. Therefore we may assume n 1 3. This shows that the star K 1,3 is an induced subgraph of G. Since the eigenvalues of K 1,3 are 3, 0, 0, 3, it follows from Interlacing Theorem 3.1 that λ 3 0, a contradiction. The converse follows from Corollary 3.4 and the fact that the eigenvalues of the complete graph K n are n 1, 1,..., 1. Theorem (See [4].) Let G be a graph without isolated vertices and let λ 2 (G) be the second largest eigenvalue of G. Then0 <λ 2 (G) 2 1 if and only if one of the following holds: (1) G = ( t (K 1 + K 2 )) K n1,...,n m. (2) G = (K 1 + K r,s ) K q. (3) G = (K 1 + K r,s ) K p,q. Now, we are in a position to prove the main result of this section. Theorem Let G be a graph of order n. IfG K n and G K n \e, thenλ(k n,g) 2 or λ(k n,g) >λ(k n,k n \ e).

8 176 A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) Proof. Since G K n,k n \e, n 3. Let λ 1 λ n be the eigenvalues of G. Therefore λ(k n,g)= ( λ 1 (n 1) ) 2 +(λ2 +1) 2 + +(λ n +1) 2. It is easy to see that if one of the following cases holds, then λ(k n,g) 2. Case 1. λ 1 (n 1) 2. Case 2. λ 2 λ 3 0. Case 3. λ Now, suppose that none of the above cases occurs. Therefore we may assume that λ 1 >n 1 2, λ 2 < 2 1andλ 3 < 0. Suppose that λ 2 0. Thus it follows from Lemma 3.9 and the hypothesis, G = K 1 + K n 1. Therefore λ(k n,g)=2. Now suppose that λ 2 > 0. Thus λ 1 >n 1 2, 0 <λ 2 < 2 1andλ 3 < 0. Hence Theorem 3.10 can be applied. Case a. G = ( t (K 1 + K 2 )) K n1,...,n m.ift =0,thenG = K n1,...,n m, which is not possible by Theorem 3.8. Ift 2, then (K 1 + K 2 ) (K 1 + K 2 ) is an induced subgraph of G. Since Spec ( (K 1 + K 2 ) (K 1 + K 2 ) ) = { ,.41421,.26795, 1, 1, }, by Interlacing Theorem λ 3 0, a contradiction. Now, suppose that t = 1. If there exists i such that n i 3, K 1,3 is an induced subgraph of G. Now Interlacing Theorem implies that λ 3 0, a contradiction. Now, we may assume that n i 2 for all i. Ifm =1andn 1 =1,G = (K 1 + K 2 ) K 1, then Spec(G) ={ ,.31111, 1, 1, } and λ(k 4,G) > 2. If m =1andn 1 =2,G =(K 1 + K 2 ) K 2,then Spec(G) ={ ,.32164, 0, 1, }. Hence λ 3 0, a contradiction. Thus we may assume that m 2. If there exist i and j such that n i,n j 2, then C 4 is an induced subgraph of G, and since Spec(C 4 )= {2, 0, 0, 2}, Interlacing Theorem implies that λ 3 0, a contradiction. Thus we can assume that G = (K 1 + K 2 ) K n 3 or G = (K 1 + K 2 ) K 2,1...,1.WemayalsowriteG as follows: (1) G = (K 1 + K 2 ) K n 3 = K n \{e, e } or (2) G = (K 1 + K 2 ) (K n 3 \ e )=K n \{e, e,e }

9 A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) where e, e,e E(K n ), e and e have a common vertex and e, e and e,e are pairwise non-adjacent in K n. If (1) happens then it follows from Lemma 3.7 that λ(k n,g) > λ(k n,k n \ e). Now suppose that (2) happens. Assume e = {v 1,v 2 }, e = {v 1,v 3 } and e = {v 4,v 5 }, where v 1,...,v 5 V (K n ). The induced subgraph on vertices v 1,...,v 5 is (K 1 + K 2 ) K 2.Since Spec ( (K 1 + K 2 ) K 2 ) = { ,.32164, 0, 1, }, it follows from Interlacing Theorem that λ 3 0, a contradiction. Now it remains to investigate the following cases: Case b. G = (K 1 + K r,s ) K q. Case c. G = (K 1 + K r,s ) K p,q. In both cases b and c, if either of r or s is greater than 1, then an induced subgraph isomorphic to K 1 + K 1,2 in G occurs. It now follows from Interlacing Theorem that λ 3 0asSpec(K 1 + K 1,2 )={ 2, 0, 0, 2}. Now we may assume that r = s =1. In both cases b and c, if either of q or p is greater than 1, then an induced subgraph isomorphic to H =(K 1 + K 2 ) K 2 in G occurs. Since Spec(H) ={ , 1, 0,.32164, } it follows from Interlacing Theorem that λ 3 0, a contradiction. Therefore q = p =1 and so G is isomorphic to one the following graphs: (i) G = G 1 =(K 1 + K 2 ) K 1, (ii) G = G 2 =(K 1 + K 2 ) K 2. Since and Spec(G 1 )={ , 1,.31111, } Spec(G 2 )={ , 1, 1,.35793, }, λ(k 4,G 1 ) > 2andλ(K 5,G 2 ) > 2. This completes the proof. Proof of Theorem 1.3. By Lemmas 3.6 and 3.7 and Theorem 3.11, it follows that cs(k n )=λ(k n,k n \ e) =n 2 + n n n 2 +2n 7 2. This completes the proof. 4. Cospectrality of complete bipartite graphs Let K m,n be the complete bipartite graph with parts of sizes m and n. Itisknownthat if G is a cospectral mate of K m,n,theng is bipartite and it follows from Theorem 3.8 that G = K r,s + tk 1 for some positive integers r and s andanintegert 0, where

10 178 A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) r + s + t = n + m and rs = mn. Therefore if m and n are positive integers such that there exist integers r > 0ands>0 satisfying rs = mn and r + s<m+ n, then cs(k m,n )=0. Proposition 4.1. Let m and n be positive integers. Then cs(k m,n ) > 0 if and only if the minimum of x + y for all positive integers x, y such that xy = mn is attained on {x, y} = {m, n}. If n>0, then cs(k n,n ) > 0byProposition 4.1. We now compute the value of cs(k n,n ). We need the following result in the latter computation. Theorem 4.2. (Theorem 2 of [1]) LetG be a graph of order n without isolated vertices. Then 0 <λ 2 (G) < 1 3 if and only if G = (K 1 + K 2 ) K n 3,whereλ 2 (G) is the second largest eigenvalue of G. Proof of Theorem 1.4. Since for every positive integer m and n Spec(K m,n )={ mn, 0,...,0, mn}, }{{} m+n 2 it follows that λ(k n,n,k n 1,n+1 )=2(n n 2 1) 2.Ifn = 2, then the result follows from Fig. 3, where the cospectrality of all graphs of order 4 is computed. Now assume that n 3. Suppose, for a contradiction, that G is a graph non-isomorphic to K n,n and K n 1,n+1 such that λ(k n,n,g) < 2(n n 2 1) 2. Assume λ 1 λ 2n are the eigenvalues of G. Sinceforn 3, 2(n n 2 1) 2 < 1 9 we obtain λ 2 < 1 3.Sincefor every graph except of the complete graph, the second largest eigenvalue is non-negative (see [1, part 3 of Theorem 1]), the latter inequality shows that 0 λ 2 < 1 3.Wecan distinguish the following cases: Case 1. Suppose that λ 2 =0.ThusbyTheorem 3.8, there exist some positive integers k and n 1,...,n k and an integer t 0 such that G = tk 1 + K n1,...,n k.ifk =1,then G = K 2n and so λ(k n,n,g)=2n 2 18, a contradiction. If k =2,thenG = tk 1 + K r,s for some r and s such that r +s =2n t. Inthiscasewehaveλ(K n,n,g)=2(n rs) 2. It is not hard to see that if {r, s} {n, n} and {r, s} {n 1,n+1}, then2(n rs) 2 > 2(n n 2 1) 2. Therefore k 3. If n 1 = = n k =1,thenG = tk 1 + K 2n t and so λ(k n,n,g) > 2(n n 2 1) 2, a contradiction. Now we may assume that G has K 1,1,2 as an induced subgraph. Since λ 3 (K 1,1,2 )= 1, then it follows from Interlacing Theorem that λ 2 2n 1 1andsoλ(K n,n,g) 1, a contradiction. Case 2. Assume that 0 <λ 2 < 1 3.ByTheorem 4.2, we conclude that there exists an integer t 0 such that G = tk 1 +(K 1 + K 2 ) K 2n t 3.If2n t 3 = 1, then it is easy to λ(k n,n,g) > 2(n n 2 1) 2, a contradiction. If 2n t 3 > 1, then G has K 1,1,2 as an induced subgraph and the rest is similar to previous part. This completes the proof.

11 A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) Fig. 1. Cospectrality of graphs with 2 vertices. Fig. 2. Cospectrality of graphs with 3 vertices. 5. Cospectrality of graphs of order at most 4 In this section (see Figs. 1, 2 and 3), we find the cospectrality of all graphs of order at most 4. Based on cospectrality of graphs with at most 4 vertices, one can observe the following facts: (1) It is not in general true that if cs(g) =λ(g, H) for some graph H, thencs(h) = λ(g, H); for example cs(a 4 )=λ(a 4,A 3 ) but cs(a 3 )=λ(a 3,A 2 )(seefig. 2); also cs(b 3 )=λ(b 3,B 4 ) but cs(b 4 )=λ(b 4,B 9 )(seefig. 3). (2) It is not in general true that if G is a regular graph and cs(g) =λ(g, H) forsome graph H, thenh is also regular; for example cs(a 4 )=λ(a 4,A 3 )(seefig. 2); also cs(b 9 )=λ(b 9,B 4 )(seefig. 3). The following question is natural to ask. Question 5.1. Let G and H be two graphs such that λ(g, H) =cs(g) =cs(h). For which graph theoretical property ρ, ifg has ρ then so does H? It is well-known that if λ(g, H) =cs(g) =cs(h) = 0 (that is Spec(G) =Spec(H)), then the answer of Question 5.1 for graph properties such as being bipartite and regularity is positive. We propose the following question to finish the paper. Question 5.2. Is there a constant c for which if cs(g) = λ(g, H), then E(G) E(H) c?

12 180 A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) Fig. 3. Cospectrality of graphs with 4 vertices. Remark 5.3. Concerning Question 5.2 we have the following remarks. By Figs. 2 and 3, c can be chosen 1 for graphs with at most 4 vertices. However for graphs of order 5, it is easy to see that cs(k 2,3 )=2( 6 2) 2 and λ(k 2,3,G)=2( 6 2) 2 for some graph G if and only if G = K 1,4 or G = K 2,2 + K 1. By similar arguments given in Section 4, it is not hard to see that for all n 3, cs(k n,n+1 )=λ(k n,n+1,k n 1,n+2 )= 2( n 2 + n n 2 + n 2) 2 and the graph K n 1,n+2 is the only graph G satisfying the equality λ(k n,n+1,g) = cs(k n,n+1 ). Therefore for graphs with at least 2n +1 7 vertices if the constant c exists it is at least 2. Acknowledgements We are grateful to referees for their useful comments. The research of the first author was in part supported by a grant (No ) from School of Mathematics, Institute for Research in Fundamental Sciences (IPM). The research of the second author was

13 A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) in part supported by a grant (No ) from School of Mathematics, Institute for Research in Fundamental Sciences (IPM). The research of the authors is financially supported by the Center of Excellence for Mathematics, University of Isfahan. References [1] D. Cao, H. Yuan, Graphs characterized by the second eigenvalue, J. Graph Theory 17 (3) (1993) [2] D.M. Cvetković, M. Doob, H. Sachs, Spectra of Graphs, Theory and Application, Academic Press, Inc., New York, [3] C. Godsil, G. Royle, Algebraic Graph Theory, Springer, New York, [4] M. Petrović, On graphs whose second largest eigenvalue does not exceed 2 1, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. 4 (1993) [5] J.H. Smith, Symmetry and multiple eigenvalues of graphs, Glas. Mat. Ser. III 12 (1) (1977) 3 8. [6] D. Stevanivić, Research problems from the Aveiro Workshop on Graph Spectra, Linear Algebra Appl. 423 (2007)

Linear Algebra and its Applications

Linear Algebra and its Applications xxx (2008) xxx xxx Contents lists available at ScienceDirect Linear Algebra and its Applications journal homepage: wwwelseviercom/locate/laa Graphs with three distinct