- Email: [email protected]
Cospectrality of graphs
Linear Algebra and its Applications 451 (2014) 169–181
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
Cospectrality of graphs Alireza Abdollahi a,b,∗ , Mohammad Reza Oboudi a,b a
Department of Mathematics, University of Isfahan, Isfahan 81746-73441, Iran School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box 19395-5746, Tehran, Iran
b
a r t i c l e
i n f o
Article history: Received 17 October 2013 Accepted 15 February 2014 Available online 2 April 2014 Submitted by R. Brualdi MSC: 05C50 05C31 Keywords: Spectra of graphs Measures on spectra of graphs Adjacency matrix of a graph
a b s t r a c t Richard Brualdi proposed in Stevanivić (2007) [6] the following problem: (Problem AWGS.4) Let Gn and Gn be two nonisomorphic graphs on n vertices with spectra λ1 λ2 · · · λn
λ1 λ2 · · · λn ,
and
respectively. Define the distance between the spectra of Gn and Gn as n 2 λ Gn , Gn = λi − λi
i=1
n or use λi − λi . i=1
Define the cospectrality of Gn by cs(Gn ) = min λ Gn , Gn : Gn not isomorphic to Gn . Let csn = max cs(Gn ): Gn a graph on n vertices . Problem A. Investigate cs(Gn ) for special classes of graphs. Problem B. Find a good upper bound on csn . In this paper we study Problem A and determine the cospectrality of certain graphs by the Euclidian distance. * Corresponding author. E-mail addresses: [email protected] (A. Abdollahi), [email protected] (M.R. Oboudi). http://dx.doi.org/10.1016/j.laa.2014.02.052 0024-3795/© 2014 Elsevier Inc. All rights reserved.
170
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
Let Kn denote the complete graph on n vertices, nK1 denote the null graph on n vertices and K2 + (n − 2)K1 denote the disjoint union of the K2 with n − 2 isolated vertices, where n 2. In this paper we find cs(Kn ), cs(nK1 ), cs(K2 + (n − 2)K1 ) (n 2) and cs(Kn,n ). © 2014 Elsevier Inc. All rights reserved.
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 Gn and Gn be two nonisomorphic graphs on n vertices with spectra and λ1 λ2 · · · λn ,
λ1 λ2 · · · λn
respectively. Define the distance between the spectra of Gn and Gn as
λ
Gn , Gn
=
n
λi −
2 λi
n λi − λi . or use
i=1
i=1
Define the cospectrality of Gn by cs(Gn ) = min λ Gn , Gn : Gn not isomorphic to Gn . Thus cs(Gn ) = 0 if and only if Gn has a cospectral mate. Let csn = max cs(Gn ): Gn 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(Gn ) for special classes of graphs. Problem B. Find a good upper bound on csn . In this paper we study Problem A and determine the cospectrality of certain graphs by the Euclidian distance, that is λ
Gn , Gn
=
n i=1
2 λi − λi .
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
171
For a graph G, V (G) and E(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. For two 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. In particular, nG denotes G + · · · + G and ∇n G denotes G∇G∇ · · · ∇G .
n
n
We denote by Spec(G) the multiset of the eigenvalues of the graph G. For positive integers n1 , . . . , n , Kn1 ,...,n denotes the complete multipartite graph with parts of sizes n1 , . . . , n . Let Kn denote the complete graph on n vertices, nK1 = Kn denote the null graph on n vertices and Pn denote the path with n vertices. By the previous notation, for any integer n > 2, K2 +(n−2)K1 denotes the disjoint union of the K2 with n−2 isolated vertices. In this paper we find cs(Kn ), cs(nK1 ), cs(K2 +(n−2)K1 ) (n 2) and cs(Kn,n ). In particular, we find that there exists a unique graph GH such that λ(H, GH ) = cs(H) if H ∈ {Kn , nK1 , K2 + (n − 2)K1 , Kn,n }. The main results of our paper are the following: Theorem 1.1. For every integer n 2, cs(nK1 ) = 2. In particular, λ(nK1 , G) = cs(nK1 ) for some graph G if and only if G ∼ = K2 + (n − 2)K1 . √ Theorem 1.2. For every integer n 3, cs(K2 + (n − 2)K1 ) = 2( 2 − 1)2 . Also, cs(K2 ) = λ(K2 , 2K1 ) = 2. In particular, λ(K2 + (n − 2)K1 , G) = cs(K2 + (n − 2)K1 ) for some graph G if and only if G ∼ = P3 + (n − 3)K1 . √ Theorem 1.3. For every integer n 2, cs(Kn ) = n2 +n−n n2 + 2n − 7−2. In particular, λ(Kn , G) = cs(Kn ) for some graph G if and only if G ∼ = Kn \ e, where Kn \ e is the graph obtaining from Kn by deletion one edge e. √ Theorem 1.4. Let n 2 be an integer. Then cs(Kn,n ) = 2(n − n2 − 1)2 . In particular, λ(Kn,n , G) = cs(Kn,n ) for some graph G if and only if G ∼ = Kn−1,n+1 . 2. 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. Let λ1 · · · λn be the eigenvalues of G. It is well known that λ1 + · · · + λn = 0 and λ1 2 + · · · + λn 2 = 2m. We now give the proof of Theorem 1.1 in which we determine the cospectrality of graphs with no edge. Proof of Theorem 1.1. Let Gn be a simple graph of order n and size m with eigenvalues λ1 · · · λn . Since the eigenvalues of nK1 are λ1 = · · · = λn = 0, then 2 2 λ nK1 , Gn = λ1 + · · · + λn = 2m .
172
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
Since Gn is not isomorphic to nK1 , m 1. So the minimum value of λ(nK1 , Gn ) is 2 and it happens for Gn = K2 + (n − 2)K1 . 2 Proof of Theorem 1.2. Let n 3 be an integer and let Gn be a simple graph of order n and size m with eigenvalues λ1 · · · λn . Since the eigenvalues of K2 + (n − 2)K1 are λ1 = 1, λ2 = · · · = λn−1 = 0, λn = −1, 2 2 2 2 λ K2 + (n − 2)K1 , Gn = λ1 − 1 + λ2 + · · · + λn−1 + λn + 1 = 2m + 2 − 2λ1 + 2λn . Now, we want to find the minimum value of m − λ1 + λn among all graphs of order n and size m . By Perron–Frobenius Theorem (see [2, Theorem 0.13]), λn −λ1 . Thus λ K2 + (n − 2)K1 , Gn = 2m + 2 − 2λ1 + 2λn 2m − 4λ1 + 2. We claim that for every graph Gn K2 + (n − 2)K1 , P3 + (n − 3)K1 , the following holds √ λ K2 + (n − 2)K1 , Gn > 2( 2 − 1)2 . Since √ λ K2 + (n − 2)K1 , P3 + (n − 3)K1 = 2( 2 − 1)2 , the validity of our claim completes the proof. To prove the claim we consider the following cases: √ Case 1. Let λ1 1 + 0.5. Then (λ1 − 1)2 0.5. Thus λ K2 + (n − 2)K1 , Gn √ 2 2 2 2 = λ1 − 1 + λ2 + · · · + λn−1 + λn + 1 0.5 > 2( 2 − 1)2 . Case 2. Let λ1 < 1 +
√ 0.5. Then for m 4, 2m 4λ1 . Since
λ K2 + (n − 2)K1 , Gn 2m − 4λ1 + 2, it follows that λ(K2 + (n − 2)K1 , Gn ) 2 and so we are done. To complete the proof of our claim it remains to compute λ(K2 + (n − 2)K1 , Gn ) for all graphs Gn with at most 3 edges. These graphs are as follows, nK1 , K2 + (n − 2)K1 , 2K2 + (n − 4)K1 , P3 + (n − 3)K1 , 3K2 + (n − 6)K1 , P3 + K2 + (n − 5)K1 , K3 + (n − 3)K1 , P4 + (n − 4)K1 and K1,3 + (n − 4)K1 . One can easily see that for the latter graphs the claim is valid. This completes the proof. 2
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
173
3. 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(Kn ) = λ(Kn , Kn \ e), where e is an arbitrary edge of Kn . First we prove some lemmas. Theorem 3.1. (Theorem 9.1.1 of [3]) Let G be a graph of order n and H be an induced subgraph of G with order m. Suppose that λ1 (G) · · · λn (G) and λ1 (H) · · · λm (H) are the eigenvalues of G and H, respectively. Then for every i, 1 i m, λi (G) λi (H) λn−m+i (G). Let X be a graph. Recall that a partition π of V (X) with cells C1 , . . . , Cr is equitable if the number of neighbors in Cj of a vertex u in Ci is a constant bij , independent of u. Theorem 3.2. (Theorem 9.3.3 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 bij 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 = bij . Let Knt be the graph obtained from Kn by deleting n − t − 1 edges from one vertex of Kn . By the following lemma one can compute the spectrum of Knt . Lemma 3.3. Let V (Kn ) = {v0 , v1 , . . . , vn−1 }. Let 1 t n − 2 and Knt = Kn \ {v0 vt+1 , v0 vt+2 , . . . , v0 vn−1 }. Suppose that f (λ) := λ3 −(n−3)λ2 −(n+t−2)λ+t(n−t−2). Then Spec Knt = {−1, . . . , −1, λ1 , λ2 , λ3 },
n−3
where λ1 , λ2 , λ3 are the roots of the polynomial f (λ) and −1 max{λ1 , λ2 , λ3 }. Proof. Let A = {v0 }, B = {v1 , . . . , vt } and C = {vt+1 , . . . , vn−1 }. It is easy to see that the partition {A, B, C} of vertices of Knt is an equitable partition with the following matrix ⎡ ⎤ 0 t 0 ⎢ ⎥ M = ⎣1 t − 1 n − 1 − t⎦, 0 t n−t−2
174
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
where the first (second and third) row and column corresponds to A (B and C, respectively). Since Spec(Kn−1 ) = {n−2, −1, . . . , −1} and Kn−1 is an induced subgraph of Knt ,
n−2
by interlacing Theorem 3.1 we conclude that Knt has at least n − 3 eigenvalues −1. By Theorem 3.2, the three remaining eigenvalues of Knt are the eigenvalues of the matrix M and the largest eigenvalue of Knt is among the latter three eigenvalues. On the other hand f (λ) = det(λI − M ) and f (−1) = 0. This completes the proof. 2 Corollary 3.4. For every integer n 2 and every arbitrary edge e of Kn , Spec(Kn \ e) =
n−3+
√ √ n − 3 − n2 + 2n − 7 n2 + 2n − 7 , 0, −1, . . . , −1, .
2 2 n−3
Proof. By the notation of Lemma 3.3, Kn \ e = Knn−2 . This implies the result. 2 Remark 3.5. Let G be a graph of order n and size m. Let λ1 · · · λn be the eigenvalues of G. Since n − 1 −1 · · · −1 are the eigenvalues of Kn , one can obtain that λ(Kn , G) = 2m − 2nλ1 + n2 − n. This equality shows that to obtain cs(Kn ) it is sufficient to obtain a graph G for which the parameter m − nλ1 has the minimum value. In sequel we show that among all graphs G of order n except Kn , the minimum value of m − nλ1 (G) is attained on the graph Kn \ e. Lemma 3.6. For every integer n 3 and every edge e of Kn , λ(Kn , Kn \ e) < 2. If n = 2, then λ(Kn , Kn \ 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. 2 Lemma 3.7. Let n 3 be an integer. Let K = Kn \ {e, e }, where e and e are two adjacent edges in Kn . Then λ(Kn , K) > λ(Kn , Kn \ e). Proof. Note that K = Knn−3 . By Lemma 3.3 λ1 (K) is a root of the polynomial f (λ) = λ3 − (n − 3)λ2 − (2n −√5)λ + n − 3. Since the roots of the derivation f (λ) (with respect √ n−3± n2 −6 n−3+ n2 −6 to λ) of f (λ) are , f is an increasing function on the interval [ , ∞). 3 √3 2 Using Corollary 3.4 one can see that λ1 (Kn \ e) − n1 > n − 2 > n−3+3 n −6 . It is not hard to see that f (λ1 (Kn \ e) − n1 ) > 0. On the other hand f is increasing on √ 2 [ n−3+3 n −6 , ∞). Thus f (λ) > 0 for every λ λ1 (Kn \ e) − n1 . Since λ1 (K) is a root of f (λ), we conclude that λ1 (K) < λ1 (Kn \ e) − n1 . This shows that n(n−1) −2− 2 n(n−1) nλ1 (K) > − 1 − nλ1 (Kn \ e). Equivalently, λ(Kn , K) > λ(Kn , Kn \ e) (see 2 Remark 3.5). 2
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
175
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 λ1 λ2 λ3 · · · λn . The graph G is isomorphic to one the following graphs if and only if λ1 > 0, λ2 0 and λ3 < 0. (1) G ∼ = Kn , ∼ K1 + Kn−1 , (2) G = (3) G ∼ = K2,1,...,1 = Kn \ e for an edge e of Kn . Proof. Suppose that λ1 > 0, λ2 0 and λ3 < 0. By Theorem 3.8, there exist some integers t 1, n1 , . . . , nt 1 and r 0 such that G ∼ = rK1 + Kn1 ,...,nt . Since λ3 < 0, r 1. Thus it is enough to investigate the cases G ∼ = Kn1 ,...,nt and G ∼ = K1 + Kn1 ,...,nt . ∼ Note that t 2, otherwise G = Kn , a contradiction. Suppose that G ∼ = K1 +Kn1 ,...,nt . If ni 2 for some i, then G has a 3K1 as an induced subgraph and so it follows from Interlacing Theorem that λ3 0, a contradiction. Therefore G ∼ = K1 + Kn−1 in this case. Now assume that G ∼ = Kn1 ,...,nt . If n1 = · · · = nt = 1, then G ∼ = Kn . Thus we may assume that ni 2 for some i. Suppose that ni , nj 2 for some distinct i and j. Thus the cycle C4 of length 4 is an induced subgraph of G. Since the eigenvalues of C4 are 2, 0, 0, −2, it follows from Interlacing Theorem that λ3 0, a contradiction. Thus we can assume that n1 2 and n2 = · · · = nt = 1. If n1 = 2, then G ∼ = K2,1,...,1 = Kn \ e for some edge e. Therefore we may assume n1 3. This shows that the star K1,3 is an √ √ induced subgraph of G. Since the eigenvalues of K1,3 are 3, 0, 0, − 3, it follows from Interlacing Theorem 3.1 that λ3 0, a contradiction. The converse follows from Corollary 3.4 and the fact that the eigenvalues of the complete graph Kn are n − 1, −1, . . . , −1. 2 Theorem 3.10. (See [4].) Let G be a graph without isolated vertices and let λ2 (G) be √ the second largest eigenvalue of G. Then 0 < λ2 (G) 2 − 1 if and only if one of the following holds: (1) G ∼ = (∇t (K1 + K2 ))∇Kn1 ,...,nm . (2) G ∼ = (K1 + Kr,s )∇Kq . (3) G ∼ = (K1 + Kr,s )∇Kp,q . Now, we are in a position to prove the main result of this section. Theorem 3.11. Let G be a graph of order n. If G Kn and G Kn \e, then λ(Kn , G) 2 or λ(Kn , G) > λ(Kn , Kn \ e).
176
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
Proof. Since G Kn , Kn \e, n 3. Let λ1 · · · λn be the eigenvalues of G. Therefore 2 λ(Kn , G) = λ1 − (n − 1) + (λ2 + 1)2 + · · · + (λn + 1)2 . It is easy to see that if one of the following cases holds, then λ(Kn , G) 2. √ Case 1. λ1 − (n − 1) − 2. Case 2. λ2 λ3 0. √ Case 3. λ2 + 1 2. Now, suppose that none of the above cases occurs. Therefore we may assume that √ √ λ1 > n − 1 − 2, λ2 < 2 − 1 and λ3 < 0. Suppose that λ2 0. Thus it follows from Lemma 3.9 and the hypothesis, G ∼ = K1 + Kn−1 . Therefore λ(Kn , G) = 2. √ √ Now suppose that λ2 > 0. Thus λ1 > n − 1 − 2, 0 < λ2 < 2 − 1 and λ3 < 0. Hence Theorem 3.10 can be applied. Case a. G ∼ = (∇t (K1 + K2 ))∇Kn1 ,...,nm . If t = 0, then G ∼ = Kn1 ,...,nm , which is not possible by Theorem 3.8. If t 2, then (K1 + K2 )∇(K1 + K2 ) is an induced subgraph of G. Since Spec (K1 + K2 )∇(K1 + K2 ) = {3.73205, .41421, .26795, −1, −1, −2.41421}, by Interlacing Theorem λ3 0, a contradiction. Now, suppose that t = 1. If there exists i such that ni 3, K1,3 is an induced subgraph of G. Now Interlacing Theorem implies that λ3 0, a contradiction. Now, we may assume that ni 2 for all i. If m = 1 and n1 = 1, G ∼ = (K1 + K2 )∇K1 , then Spec(G) = {2.17009, .31111, −1, −1, 48 119} and λ(K4 , G) > 2. If m = 1 and n1 = 2, G = (K1 + K2 )∇K2 , then Spec(G) = {2.85577, .32164, 0, −1, −2.17741}. Hence λ3 0, a contradiction. Thus we may assume that m 2. If there exist i and j such that ni , nj 2, then C4 is an induced subgraph of G, and since Spec(C4 ) = {2, 0, 0, −2}, Interlacing Theorem implies that λ3 0, a contradiction. Thus we can assume that G ∼ = (K1 + K2 )∇Kn−3 or G ∼ = (K1 + K2 )∇K2,1...,1 . We may also write G as follows: (1) G ∼ = (K1 + K2 )∇Kn−3 = Kn \ {e, e } or (2) G ∼ = (K1 + K2 )∇(Kn−3 \ e ) = Kn \ {e, e , e }
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
177
where e, e , e ∈ E(Kn ), e and e have a common vertex and e, e and e , e are pairwise non-adjacent in Kn . If (1) happens then it follows from Lemma 3.7 that λ(Kn , G) > λ(Kn , Kn \ e). Now suppose that (2) happens. Assume e = {v1 , v2 }, e = {v1 , v3 } and e = {v4 , v5 }, where v1 , . . . , v5 ∈ V (Kn ). The induced subgraph on vertices v1 , . . . , v5 is (K1 + K2 )∇K2 . Since Spec (K1 + K2 )∇K2 = {2.85577, .32164, 0, −1, −2.17741}, it follows from Interlacing Theorem that λ3 0, a contradiction. Now it remains to investigate the following cases: Case b. G ∼ = (K1 + Kr,s )∇Kq . Case c. G ∼ = (K1 + Kr,s )∇Kp,q . In both cases b and c, if either of r or s is greater than 1, then an induced subgraph isomorphic to K1 + K1,2 in G occurs. It now follows from Interlacing Theorem that √ √ λ3 0 as Spec(K1 + K1,2 ) = { 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 = (K1 + K2 )∇K2 in G occurs. Since Spec(H) = {−2.17741, −1, 0, .32164, 2.85577} it follows from Interlacing Theorem that λ3 0, a contradiction. Therefore q = p = 1 and so G is isomorphic to one the following graphs: (i) G ∼ = G1 = (K1 + K2 )∇K1 , ∼ G2 = (K1 + K2 )∇K2 . (ii) G = Since Spec(G1 ) = {−1.48119, −1, .31111, 2.17009} and Spec(G2 ) = {−1.68133, −1, −1, .35793, 3.32340}, λ(K4 , G1 ) > 2 and λ(K5 , G2 ) > 2. This completes the proof. 2 Proof of Theorem 1.3. By Lemmas 3.6 and 3.7 and Theorem 3.11, it follows that √ cs(Kn ) = λ(Kn , Kn \ e) = n2 + n − n n2 + 2n − 7 − 2. This completes the proof. 2 4. Cospectrality of complete bipartite graphs Let Km,n be the complete bipartite graph with parts of sizes m and n. It is known that if G is a cospectral mate of Km,n , then G is bipartite and it follows from Theorem 3.8 that G ∼ = Kr,s + tK1 for some positive integers r and s and an integer t 0, where
178
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
r + s + t = n + m and rs = mn. Therefore if m and n are positive integers such that there exist integers r > 0 and s > 0 satisfying rs = mn and r + s < m + n, then cs(Km,n ) = 0. Proposition 4.1. Let m and n be positive integers. Then cs(Km,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(Kn,n ) > 0 by Proposition 4.1. We now compute the value of cs(Kn,n ). We need the following result in the latter computation. Theorem 4.2. (Theorem 2 of [1]) Let G be a graph of order n without isolated vertices. Then 0 < λ2 (G) < 13 if and only if G ∼ = (K1 + K2 )∇Kn−3 , where λ2 (G) is the second largest eigenvalue of G. Proof of Theorem 1.4. Since for every positive integer m and n √ √ Spec(Km,n ) = {− mn, 0, . . . , 0, mn},
m+n−2
√ it follows that λ(Kn,n , Kn−1,n+1 ) = 2(n − n2 − 1)2 . If n = 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 Kn,n √ and Kn−1,n+1 such that λ(Kn,n , G) < 2(n − n2 − 1)2 . Assume λ1 · · · λ2n are the √ eigenvalues of G. Since for n 3, 2(n − n2 − 1)2 < 19 we obtain |λ2 | < 13 . Since for 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 λ2 < 13 . We can distinguish the following cases: Case 1. Suppose that λ2 = 0. Thus by Theorem 3.8, there exist some positive integers k and n1 , . . . , nk and an integer t 0 such that G ∼ = tK1 + Kn1 ,...,nk . If k = 1, then 2 G∼ K and so λ(K , G) = 2n 18, a contradiction. If k = 2, then G ∼ = 2n = tK1 + Kr,s n,n √ for some r and s such that r + s = 2n − t. In this case we have λ(Kn,n , G) = 2(n − rs)2 . √ 2 It is not hard to see that if {r, s} = {n, n} and {r, s} = {n−1, n+1}, then 2(n− rs) > √ 2(n − n2 − 1)2 . Therefore k 3. If n1 = · · · = nk = 1, then G ∼ = tK1 + K2n−t and so √ 2 2 λ(Kn,n , G) > 2(n− n − 1) , a contradiction. Now we may assume that G has K1,1,2 as an induced subgraph. Since λ3 (K1,1,2 ) = −1, then it follows from Interlacing Theorem that λ22n−1 1 and so λ(Kn,n , G) 1, a contradiction. Case 2. Assume that 0 < λ2 < 13 . By Theorem 4.2, we conclude that there exists an integer t 0 such that G ∼ tK + (K1 + K2 )∇K2n−t−3 . If 2n − t − 3 = 1, then it is easy √= 12 2 to λ(Kn,n , G) > 2(n − n − 1) , a contradiction. If 2n − t − 3 > 1, then G has K1,1,2 as an induced subgraph and the rest is similar to previous part. This completes the proof. 2
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
179
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, then cs(H) = λ(G, H); for example cs(A4 ) = λ(A4 , A3 ) but cs(A3 ) = λ(A3 , A2 ) (see Fig. 2); also cs(B3 ) = λ(B3 , B4 ) but cs(B4 ) = λ(B4 , B9 ) (see Fig. 3). (2) It is not in general true that if G is a regular graph and cs(G) = λ(G, H) for some graph H, then H is also regular; for example cs(A4 ) = λ(A4 , A3 ) (see Fig. 2); also cs(B9 ) = λ(B9 , B4 ) (see Fig. 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, if G 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?
180
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
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(K2,3 ) = 2( 6 − 2)2 and λ(K2,3 , G) = 2( 6 − 2)2 for some graph G if and only if G ∼ = K2,2 + K1 . By similar arguments given in = K1,4 or G ∼ Section 4, it is not hard to see that for all n 3, cs(Kn,n+1 ) = λ(Kn,n+1 , Kn−1,n+2 ) = √ √ 2( n2 + n − n2 + n − 2)2 and the graph Kn−1,n+2 is the only graph G satisfying the equality λ(Kn,n+1 , G) = cs(Kn,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. 92050219) from School of Mathematics, Institute for Research in Fundamental Sciences (IPM). The research of the second author was
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
181
in part supported by a grant (No. 92050012) 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) 325–331. [2] D.M. Cvetković, M. Doob, H. Sachs, Spectra of Graphs, Theory and Application, Academic Press, Inc., New York, 1979. [3] C. Godsil, G. Royle, Algebraic Graph Theory, Springer, New York, 2001. √ [4] M. Petrović, On graphs whose second largest eigenvalue does not exceed 2 − 1, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. 4 (1993) 70–75. [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) 172–181.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
Cospectrality of graphs Alireza Abdollahi a,b,∗ , Mohammad Reza Oboudi a,b a
Department of Mathematics, University of Isfahan, Isfahan 81746-73441, Iran School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box 19395-5746, Tehran, Iran
b
a r t i c l e
i n f o
Article history: Received 17 October 2013 Accepted 15 February 2014 Available online 2 April 2014 Submitted by R. Brualdi MSC: 05C50 05C31 Keywords: Spectra of graphs Measures on spectra of graphs Adjacency matrix of a graph
a b s t r a c t Richard Brualdi proposed in Stevanivić (2007) [6] the following problem: (Problem AWGS.4) Let Gn and Gn be two nonisomorphic graphs on n vertices with spectra λ1 λ2 · · · λn
λ1 λ2 · · · λn ,
and
respectively. Define the distance between the spectra of Gn and Gn as n 2 λ Gn , Gn = λi − λi
i=1
n or use λi − λi . i=1
Define the cospectrality of Gn by cs(Gn ) = min λ Gn , Gn : Gn not isomorphic to Gn . Let csn = max cs(Gn ): Gn a graph on n vertices . Problem A. Investigate cs(Gn ) for special classes of graphs. Problem B. Find a good upper bound on csn . In this paper we study Problem A and determine the cospectrality of certain graphs by the Euclidian distance. * Corresponding author. E-mail addresses: [email protected] (A. Abdollahi), [email protected] (M.R. Oboudi). http://dx.doi.org/10.1016/j.laa.2014.02.052 0024-3795/© 2014 Elsevier Inc. All rights reserved.
170
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
Let Kn denote the complete graph on n vertices, nK1 denote the null graph on n vertices and K2 + (n − 2)K1 denote the disjoint union of the K2 with n − 2 isolated vertices, where n 2. In this paper we find cs(Kn ), cs(nK1 ), cs(K2 + (n − 2)K1 ) (n 2) and cs(Kn,n ). © 2014 Elsevier Inc. All rights reserved.
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 Gn and Gn be two nonisomorphic graphs on n vertices with spectra and λ1 λ2 · · · λn ,
λ1 λ2 · · · λn
respectively. Define the distance between the spectra of Gn and Gn as
λ
Gn , Gn
=
n
λi −
2 λi
n λi − λi . or use
i=1
i=1
Define the cospectrality of Gn by cs(Gn ) = min λ Gn , Gn : Gn not isomorphic to Gn . Thus cs(Gn ) = 0 if and only if Gn has a cospectral mate. Let csn = max cs(Gn ): Gn 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(Gn ) for special classes of graphs. Problem B. Find a good upper bound on csn . In this paper we study Problem A and determine the cospectrality of certain graphs by the Euclidian distance, that is λ
Gn , Gn
=
n i=1
2 λi − λi .
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
171
For a graph G, V (G) and E(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. For two 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. In particular, nG denotes G + · · · + G and ∇n G denotes G∇G∇ · · · ∇G .
n
n
We denote by Spec(G) the multiset of the eigenvalues of the graph G. For positive integers n1 , . . . , n , Kn1 ,...,n denotes the complete multipartite graph with parts of sizes n1 , . . . , n . Let Kn denote the complete graph on n vertices, nK1 = Kn denote the null graph on n vertices and Pn denote the path with n vertices. By the previous notation, for any integer n > 2, K2 +(n−2)K1 denotes the disjoint union of the K2 with n−2 isolated vertices. In this paper we find cs(Kn ), cs(nK1 ), cs(K2 +(n−2)K1 ) (n 2) and cs(Kn,n ). In particular, we find that there exists a unique graph GH such that λ(H, GH ) = cs(H) if H ∈ {Kn , nK1 , K2 + (n − 2)K1 , Kn,n }. The main results of our paper are the following: Theorem 1.1. For every integer n 2, cs(nK1 ) = 2. In particular, λ(nK1 , G) = cs(nK1 ) for some graph G if and only if G ∼ = K2 + (n − 2)K1 . √ Theorem 1.2. For every integer n 3, cs(K2 + (n − 2)K1 ) = 2( 2 − 1)2 . Also, cs(K2 ) = λ(K2 , 2K1 ) = 2. In particular, λ(K2 + (n − 2)K1 , G) = cs(K2 + (n − 2)K1 ) for some graph G if and only if G ∼ = P3 + (n − 3)K1 . √ Theorem 1.3. For every integer n 2, cs(Kn ) = n2 +n−n n2 + 2n − 7−2. In particular, λ(Kn , G) = cs(Kn ) for some graph G if and only if G ∼ = Kn \ e, where Kn \ e is the graph obtaining from Kn by deletion one edge e. √ Theorem 1.4. Let n 2 be an integer. Then cs(Kn,n ) = 2(n − n2 − 1)2 . In particular, λ(Kn,n , G) = cs(Kn,n ) for some graph G if and only if G ∼ = Kn−1,n+1 . 2. 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. Let λ1 · · · λn be the eigenvalues of G. It is well known that λ1 + · · · + λn = 0 and λ1 2 + · · · + λn 2 = 2m. We now give the proof of Theorem 1.1 in which we determine the cospectrality of graphs with no edge. Proof of Theorem 1.1. Let Gn be a simple graph of order n and size m with eigenvalues λ1 · · · λn . Since the eigenvalues of nK1 are λ1 = · · · = λn = 0, then 2 2 λ nK1 , Gn = λ1 + · · · + λn = 2m .
172
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
Since Gn is not isomorphic to nK1 , m 1. So the minimum value of λ(nK1 , Gn ) is 2 and it happens for Gn = K2 + (n − 2)K1 . 2 Proof of Theorem 1.2. Let n 3 be an integer and let Gn be a simple graph of order n and size m with eigenvalues λ1 · · · λn . Since the eigenvalues of K2 + (n − 2)K1 are λ1 = 1, λ2 = · · · = λn−1 = 0, λn = −1, 2 2 2 2 λ K2 + (n − 2)K1 , Gn = λ1 − 1 + λ2 + · · · + λn−1 + λn + 1 = 2m + 2 − 2λ1 + 2λn . Now, we want to find the minimum value of m − λ1 + λn among all graphs of order n and size m . By Perron–Frobenius Theorem (see [2, Theorem 0.13]), λn −λ1 . Thus λ K2 + (n − 2)K1 , Gn = 2m + 2 − 2λ1 + 2λn 2m − 4λ1 + 2. We claim that for every graph Gn K2 + (n − 2)K1 , P3 + (n − 3)K1 , the following holds √ λ K2 + (n − 2)K1 , Gn > 2( 2 − 1)2 . Since √ λ K2 + (n − 2)K1 , P3 + (n − 3)K1 = 2( 2 − 1)2 , the validity of our claim completes the proof. To prove the claim we consider the following cases: √ Case 1. Let λ1 1 + 0.5. Then (λ1 − 1)2 0.5. Thus λ K2 + (n − 2)K1 , Gn √ 2 2 2 2 = λ1 − 1 + λ2 + · · · + λn−1 + λn + 1 0.5 > 2( 2 − 1)2 . Case 2. Let λ1 < 1 +
√ 0.5. Then for m 4, 2m 4λ1 . Since
λ K2 + (n − 2)K1 , Gn 2m − 4λ1 + 2, it follows that λ(K2 + (n − 2)K1 , Gn ) 2 and so we are done. To complete the proof of our claim it remains to compute λ(K2 + (n − 2)K1 , Gn ) for all graphs Gn with at most 3 edges. These graphs are as follows, nK1 , K2 + (n − 2)K1 , 2K2 + (n − 4)K1 , P3 + (n − 3)K1 , 3K2 + (n − 6)K1 , P3 + K2 + (n − 5)K1 , K3 + (n − 3)K1 , P4 + (n − 4)K1 and K1,3 + (n − 4)K1 . One can easily see that for the latter graphs the claim is valid. This completes the proof. 2
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
173
3. 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(Kn ) = λ(Kn , Kn \ e), where e is an arbitrary edge of Kn . First we prove some lemmas. Theorem 3.1. (Theorem 9.1.1 of [3]) Let G be a graph of order n and H be an induced subgraph of G with order m. Suppose that λ1 (G) · · · λn (G) and λ1 (H) · · · λm (H) are the eigenvalues of G and H, respectively. Then for every i, 1 i m, λi (G) λi (H) λn−m+i (G). Let X be a graph. Recall that a partition π of V (X) with cells C1 , . . . , Cr is equitable if the number of neighbors in Cj of a vertex u in Ci is a constant bij , independent of u. Theorem 3.2. (Theorem 9.3.3 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 bij 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 = bij . Let Knt be the graph obtained from Kn by deleting n − t − 1 edges from one vertex of Kn . By the following lemma one can compute the spectrum of Knt . Lemma 3.3. Let V (Kn ) = {v0 , v1 , . . . , vn−1 }. Let 1 t n − 2 and Knt = Kn \ {v0 vt+1 , v0 vt+2 , . . . , v0 vn−1 }. Suppose that f (λ) := λ3 −(n−3)λ2 −(n+t−2)λ+t(n−t−2). Then Spec Knt = {−1, . . . , −1, λ1 , λ2 , λ3 },
n−3
where λ1 , λ2 , λ3 are the roots of the polynomial f (λ) and −1 max{λ1 , λ2 , λ3 }. Proof. Let A = {v0 }, B = {v1 , . . . , vt } and C = {vt+1 , . . . , vn−1 }. It is easy to see that the partition {A, B, C} of vertices of Knt is an equitable partition with the following matrix ⎡ ⎤ 0 t 0 ⎢ ⎥ M = ⎣1 t − 1 n − 1 − t⎦, 0 t n−t−2
174
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
where the first (second and third) row and column corresponds to A (B and C, respectively). Since Spec(Kn−1 ) = {n−2, −1, . . . , −1} and Kn−1 is an induced subgraph of Knt ,
n−2
by interlacing Theorem 3.1 we conclude that Knt has at least n − 3 eigenvalues −1. By Theorem 3.2, the three remaining eigenvalues of Knt are the eigenvalues of the matrix M and the largest eigenvalue of Knt is among the latter three eigenvalues. On the other hand f (λ) = det(λI − M ) and f (−1) = 0. This completes the proof. 2 Corollary 3.4. For every integer n 2 and every arbitrary edge e of Kn , Spec(Kn \ e) =
n−3+
√ √ n − 3 − n2 + 2n − 7 n2 + 2n − 7 , 0, −1, . . . , −1, .
2 2 n−3
Proof. By the notation of Lemma 3.3, Kn \ e = Knn−2 . This implies the result. 2 Remark 3.5. Let G be a graph of order n and size m. Let λ1 · · · λn be the eigenvalues of G. Since n − 1 −1 · · · −1 are the eigenvalues of Kn , one can obtain that λ(Kn , G) = 2m − 2nλ1 + n2 − n. This equality shows that to obtain cs(Kn ) it is sufficient to obtain a graph G for which the parameter m − nλ1 has the minimum value. In sequel we show that among all graphs G of order n except Kn , the minimum value of m − nλ1 (G) is attained on the graph Kn \ e. Lemma 3.6. For every integer n 3 and every edge e of Kn , λ(Kn , Kn \ e) < 2. If n = 2, then λ(Kn , Kn \ 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. 2 Lemma 3.7. Let n 3 be an integer. Let K = Kn \ {e, e }, where e and e are two adjacent edges in Kn . Then λ(Kn , K) > λ(Kn , Kn \ e). Proof. Note that K = Knn−3 . By Lemma 3.3 λ1 (K) is a root of the polynomial f (λ) = λ3 − (n − 3)λ2 − (2n −√5)λ + n − 3. Since the roots of the derivation f (λ) (with respect √ n−3± n2 −6 n−3+ n2 −6 to λ) of f (λ) are , f is an increasing function on the interval [ , ∞). 3 √3 2 Using Corollary 3.4 one can see that λ1 (Kn \ e) − n1 > n − 2 > n−3+3 n −6 . It is not hard to see that f (λ1 (Kn \ e) − n1 ) > 0. On the other hand f is increasing on √ 2 [ n−3+3 n −6 , ∞). Thus f (λ) > 0 for every λ λ1 (Kn \ e) − n1 . Since λ1 (K) is a root of f (λ), we conclude that λ1 (K) < λ1 (Kn \ e) − n1 . This shows that n(n−1) −2− 2 n(n−1) nλ1 (K) > − 1 − nλ1 (Kn \ e). Equivalently, λ(Kn , K) > λ(Kn , Kn \ e) (see 2 Remark 3.5). 2
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
175
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 λ1 λ2 λ3 · · · λn . The graph G is isomorphic to one the following graphs if and only if λ1 > 0, λ2 0 and λ3 < 0. (1) G ∼ = Kn , ∼ K1 + Kn−1 , (2) G = (3) G ∼ = K2,1,...,1 = Kn \ e for an edge e of Kn . Proof. Suppose that λ1 > 0, λ2 0 and λ3 < 0. By Theorem 3.8, there exist some integers t 1, n1 , . . . , nt 1 and r 0 such that G ∼ = rK1 + Kn1 ,...,nt . Since λ3 < 0, r 1. Thus it is enough to investigate the cases G ∼ = Kn1 ,...,nt and G ∼ = K1 + Kn1 ,...,nt . ∼ Note that t 2, otherwise G = Kn , a contradiction. Suppose that G ∼ = K1 +Kn1 ,...,nt . If ni 2 for some i, then G has a 3K1 as an induced subgraph and so it follows from Interlacing Theorem that λ3 0, a contradiction. Therefore G ∼ = K1 + Kn−1 in this case. Now assume that G ∼ = Kn1 ,...,nt . If n1 = · · · = nt = 1, then G ∼ = Kn . Thus we may assume that ni 2 for some i. Suppose that ni , nj 2 for some distinct i and j. Thus the cycle C4 of length 4 is an induced subgraph of G. Since the eigenvalues of C4 are 2, 0, 0, −2, it follows from Interlacing Theorem that λ3 0, a contradiction. Thus we can assume that n1 2 and n2 = · · · = nt = 1. If n1 = 2, then G ∼ = K2,1,...,1 = Kn \ e for some edge e. Therefore we may assume n1 3. This shows that the star K1,3 is an √ √ induced subgraph of G. Since the eigenvalues of K1,3 are 3, 0, 0, − 3, it follows from Interlacing Theorem 3.1 that λ3 0, a contradiction. The converse follows from Corollary 3.4 and the fact that the eigenvalues of the complete graph Kn are n − 1, −1, . . . , −1. 2 Theorem 3.10. (See [4].) Let G be a graph without isolated vertices and let λ2 (G) be √ the second largest eigenvalue of G. Then 0 < λ2 (G) 2 − 1 if and only if one of the following holds: (1) G ∼ = (∇t (K1 + K2 ))∇Kn1 ,...,nm . (2) G ∼ = (K1 + Kr,s )∇Kq . (3) G ∼ = (K1 + Kr,s )∇Kp,q . Now, we are in a position to prove the main result of this section. Theorem 3.11. Let G be a graph of order n. If G Kn and G Kn \e, then λ(Kn , G) 2 or λ(Kn , G) > λ(Kn , Kn \ e).
176
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
Proof. Since G Kn , Kn \e, n 3. Let λ1 · · · λn be the eigenvalues of G. Therefore 2 λ(Kn , G) = λ1 − (n − 1) + (λ2 + 1)2 + · · · + (λn + 1)2 . It is easy to see that if one of the following cases holds, then λ(Kn , G) 2. √ Case 1. λ1 − (n − 1) − 2. Case 2. λ2 λ3 0. √ Case 3. λ2 + 1 2. Now, suppose that none of the above cases occurs. Therefore we may assume that √ √ λ1 > n − 1 − 2, λ2 < 2 − 1 and λ3 < 0. Suppose that λ2 0. Thus it follows from Lemma 3.9 and the hypothesis, G ∼ = K1 + Kn−1 . Therefore λ(Kn , G) = 2. √ √ Now suppose that λ2 > 0. Thus λ1 > n − 1 − 2, 0 < λ2 < 2 − 1 and λ3 < 0. Hence Theorem 3.10 can be applied. Case a. G ∼ = (∇t (K1 + K2 ))∇Kn1 ,...,nm . If t = 0, then G ∼ = Kn1 ,...,nm , which is not possible by Theorem 3.8. If t 2, then (K1 + K2 )∇(K1 + K2 ) is an induced subgraph of G. Since Spec (K1 + K2 )∇(K1 + K2 ) = {3.73205, .41421, .26795, −1, −1, −2.41421}, by Interlacing Theorem λ3 0, a contradiction. Now, suppose that t = 1. If there exists i such that ni 3, K1,3 is an induced subgraph of G. Now Interlacing Theorem implies that λ3 0, a contradiction. Now, we may assume that ni 2 for all i. If m = 1 and n1 = 1, G ∼ = (K1 + K2 )∇K1 , then Spec(G) = {2.17009, .31111, −1, −1, 48 119} and λ(K4 , G) > 2. If m = 1 and n1 = 2, G = (K1 + K2 )∇K2 , then Spec(G) = {2.85577, .32164, 0, −1, −2.17741}. Hence λ3 0, a contradiction. Thus we may assume that m 2. If there exist i and j such that ni , nj 2, then C4 is an induced subgraph of G, and since Spec(C4 ) = {2, 0, 0, −2}, Interlacing Theorem implies that λ3 0, a contradiction. Thus we can assume that G ∼ = (K1 + K2 )∇Kn−3 or G ∼ = (K1 + K2 )∇K2,1...,1 . We may also write G as follows: (1) G ∼ = (K1 + K2 )∇Kn−3 = Kn \ {e, e } or (2) G ∼ = (K1 + K2 )∇(Kn−3 \ e ) = Kn \ {e, e , e }
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
177
where e, e , e ∈ E(Kn ), e and e have a common vertex and e, e and e , e are pairwise non-adjacent in Kn . If (1) happens then it follows from Lemma 3.7 that λ(Kn , G) > λ(Kn , Kn \ e). Now suppose that (2) happens. Assume e = {v1 , v2 }, e = {v1 , v3 } and e = {v4 , v5 }, where v1 , . . . , v5 ∈ V (Kn ). The induced subgraph on vertices v1 , . . . , v5 is (K1 + K2 )∇K2 . Since Spec (K1 + K2 )∇K2 = {2.85577, .32164, 0, −1, −2.17741}, it follows from Interlacing Theorem that λ3 0, a contradiction. Now it remains to investigate the following cases: Case b. G ∼ = (K1 + Kr,s )∇Kq . Case c. G ∼ = (K1 + Kr,s )∇Kp,q . In both cases b and c, if either of r or s is greater than 1, then an induced subgraph isomorphic to K1 + K1,2 in G occurs. It now follows from Interlacing Theorem that √ √ λ3 0 as Spec(K1 + K1,2 ) = { 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 = (K1 + K2 )∇K2 in G occurs. Since Spec(H) = {−2.17741, −1, 0, .32164, 2.85577} it follows from Interlacing Theorem that λ3 0, a contradiction. Therefore q = p = 1 and so G is isomorphic to one the following graphs: (i) G ∼ = G1 = (K1 + K2 )∇K1 , ∼ G2 = (K1 + K2 )∇K2 . (ii) G = Since Spec(G1 ) = {−1.48119, −1, .31111, 2.17009} and Spec(G2 ) = {−1.68133, −1, −1, .35793, 3.32340}, λ(K4 , G1 ) > 2 and λ(K5 , G2 ) > 2. This completes the proof. 2 Proof of Theorem 1.3. By Lemmas 3.6 and 3.7 and Theorem 3.11, it follows that √ cs(Kn ) = λ(Kn , Kn \ e) = n2 + n − n n2 + 2n − 7 − 2. This completes the proof. 2 4. Cospectrality of complete bipartite graphs Let Km,n be the complete bipartite graph with parts of sizes m and n. It is known that if G is a cospectral mate of Km,n , then G is bipartite and it follows from Theorem 3.8 that G ∼ = Kr,s + tK1 for some positive integers r and s and an integer t 0, where
178
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
r + s + t = n + m and rs = mn. Therefore if m and n are positive integers such that there exist integers r > 0 and s > 0 satisfying rs = mn and r + s < m + n, then cs(Km,n ) = 0. Proposition 4.1. Let m and n be positive integers. Then cs(Km,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(Kn,n ) > 0 by Proposition 4.1. We now compute the value of cs(Kn,n ). We need the following result in the latter computation. Theorem 4.2. (Theorem 2 of [1]) Let G be a graph of order n without isolated vertices. Then 0 < λ2 (G) < 13 if and only if G ∼ = (K1 + K2 )∇Kn−3 , where λ2 (G) is the second largest eigenvalue of G. Proof of Theorem 1.4. Since for every positive integer m and n √ √ Spec(Km,n ) = {− mn, 0, . . . , 0, mn},
m+n−2
√ it follows that λ(Kn,n , Kn−1,n+1 ) = 2(n − n2 − 1)2 . If n = 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 Kn,n √ and Kn−1,n+1 such that λ(Kn,n , G) < 2(n − n2 − 1)2 . Assume λ1 · · · λ2n are the √ eigenvalues of G. Since for n 3, 2(n − n2 − 1)2 < 19 we obtain |λ2 | < 13 . Since for 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 λ2 < 13 . We can distinguish the following cases: Case 1. Suppose that λ2 = 0. Thus by Theorem 3.8, there exist some positive integers k and n1 , . . . , nk and an integer t 0 such that G ∼ = tK1 + Kn1 ,...,nk . If k = 1, then 2 G∼ K and so λ(K , G) = 2n 18, a contradiction. If k = 2, then G ∼ = 2n = tK1 + Kr,s n,n √ for some r and s such that r + s = 2n − t. In this case we have λ(Kn,n , G) = 2(n − rs)2 . √ 2 It is not hard to see that if {r, s} = {n, n} and {r, s} = {n−1, n+1}, then 2(n− rs) > √ 2(n − n2 − 1)2 . Therefore k 3. If n1 = · · · = nk = 1, then G ∼ = tK1 + K2n−t and so √ 2 2 λ(Kn,n , G) > 2(n− n − 1) , a contradiction. Now we may assume that G has K1,1,2 as an induced subgraph. Since λ3 (K1,1,2 ) = −1, then it follows from Interlacing Theorem that λ22n−1 1 and so λ(Kn,n , G) 1, a contradiction. Case 2. Assume that 0 < λ2 < 13 . By Theorem 4.2, we conclude that there exists an integer t 0 such that G ∼ tK + (K1 + K2 )∇K2n−t−3 . If 2n − t − 3 = 1, then it is easy √= 12 2 to λ(Kn,n , G) > 2(n − n − 1) , a contradiction. If 2n − t − 3 > 1, then G has K1,1,2 as an induced subgraph and the rest is similar to previous part. This completes the proof. 2
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
179
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, then cs(H) = λ(G, H); for example cs(A4 ) = λ(A4 , A3 ) but cs(A3 ) = λ(A3 , A2 ) (see Fig. 2); also cs(B3 ) = λ(B3 , B4 ) but cs(B4 ) = λ(B4 , B9 ) (see Fig. 3). (2) It is not in general true that if G is a regular graph and cs(G) = λ(G, H) for some graph H, then H is also regular; for example cs(A4 ) = λ(A4 , A3 ) (see Fig. 2); also cs(B9 ) = λ(B9 , B4 ) (see Fig. 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, if G 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?
180
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
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(K2,3 ) = 2( 6 − 2)2 and λ(K2,3 , G) = 2( 6 − 2)2 for some graph G if and only if G ∼ = K2,2 + K1 . By similar arguments given in = K1,4 or G ∼ Section 4, it is not hard to see that for all n 3, cs(Kn,n+1 ) = λ(Kn,n+1 , Kn−1,n+2 ) = √ √ 2( n2 + n − n2 + n − 2)2 and the graph Kn−1,n+2 is the only graph G satisfying the equality λ(Kn,n+1 , G) = cs(Kn,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. 92050219) from School of Mathematics, Institute for Research in Fundamental Sciences (IPM). The research of the second author was
A. Abdollahi, M.R. Oboudi / Linear Algebra and its Applications 451 (2014) 169–181
181
in part supported by a grant (No. 92050012) 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) 325–331. [2] D.M. Cvetković, M. Doob, H. Sachs, Spectra of Graphs, Theory and Application, Academic Press, Inc., New York, 1979. [3] C. Godsil, G. Royle, Algebraic Graph Theory, Springer, New York, 2001. √ [4] M. Petrović, On graphs whose second largest eigenvalue does not exceed 2 − 1, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. 4 (1993) 70–75. [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) 172–181.