Unicyclic graphs with exactly two main eigenvalues

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Applied Mathematics Letters 19 (006) 1143 1147 www.elsevier.

1 Applied Mathematics Letters 19 (006) Unicyclic graphs with exactly two main eigenvalues Yaoping Hou a,,fengtian b a Department of Mathematics, Hunan Normal University Changsha, Hunan , China b Institute of System Sciences, Academy of Math. and System Sci., Beijing , China Received 1 November 005; accepted 1 November 005 Abstract An eigenvalue of a graph G is called a main eigenvalue if it has an eigenvector the sum of whose entries is not equal to zero, and it is well known that a graph has exactly one main eigenvalue if and only if it is regular. In this work, all connected unicyclic graphs with exactly two main eigenvalues are determined. c 006 Elsevier Ltd. All rights reserved. Keywords: Spectra of a graph; Main eigenvalues; Unicyclic graphs; Harmonic graphs; -walk linear graphs 1. Introduction Throughout this work, G is a simple graph with vertex set V ={v 1,v,...,v n } and edge set E and adjacency matrix A = A(G) with eigenvalues λ 1,λ,...,λ n. We call λ 1,λ,...,λ n the eigenvalues of G. Aneigenvalue of a graph G is called a main eigenvalue if it has an eigenvector the sum of whose entries is not equal to zero. We denote by m(g) the numberof main eigenvaluesof G.We may indexthe eigenvaluesof G so that the first m(g) are main. It is well known that a graph is regular if and only if it has exactly one main eigenvalue. Along-standing problem posed by Cvetković (see[1]) is that of how to characterize graphs with exactly k (k ) main eigenvalues. Recently, Hagos []gaveanalternative characterization of graphs with exactly two main eigenvalues. The aim of this work is to describe exactly all unicyclic graphs with precisely two main eigenvalues. The organization of this work as follows: in Section we define -walk linear graphs and obtain some basic facts about -walk linear graphs and graphs with exactly two main eigenvalues; and in Section 3, all connected -walk linear unicyclic graphs (i.e., connected unicyclic graphs with exactly two main eigenvalues) are determined.. -walk linear graphs and main eigenvalues For a graph G,thenumber of walks of length k of G starting at v is denoted by d k (v) and the number of all walks in G of length k is denoted by W k = W k (G). Clearly, one has d 0 (v) = 1, d 1 (v) = d(v), thedegree of vertex v, W 0 = n = V, W 1 = E,andd k+1 (v) = u N G (v) d k(u), foreveryintegerk 0, where N G (v) ={u V there is an edge between u and v in G} is theset of all neighbors of v in graph G. Corresponding author. addresses: yphou@hunnu.edu.cn (Y. Hou), ftian@mail.iss.ac.cn (F. Tian) /$ - see front matter c 006 Elsevier Ltd. All rights reserved. doi: /j.aml

2 1144 Y. Hou, F. Tian / Applied Mathematics Letters 19 (006) A graph G is called -walk (a, b)-linear if there exist uniquerationalnumbersa, b such that d (v) = ad(v) + b holds for every vertex v V (G). (1) Remark. In view of the above definition, a -walk linear graph must be irregular, since for an r-regular graph, one has d (v) = rd(v) + 0andalsod (v) = 0d(v) + r. If b = 0, then -walk (a, b)-linear graphs are a-harmonic irregular graphs in [3], and harmonic graphs have been investigated recently [3 5]. Let G be a graph, and let j denote the all-one vector. We call the matrix W(G) = ( j, Aj,...,A n j) the walk-matrix of G. Recently, Hagos [] showedthat Theorem 1 ([,Theorem.1]). The rank of the walk-matrix W(G) is equal to the number of main eigenvalues. Lemma ([, Corollary.3]). If G has exactly m main eigenvalues λ 1,...,λ m,then m p i (A λ p I) j is an eigenvector corresponding to λ i, i = 1,,...,m. In particular, m p=1 (A λ p I) j = 0. Put m := m(g) and M(x) := (x λ 1 )(x λ ) (x λ m ).ItwasshownthatM(x) is arational polynomial [6]. In fact, M(x) is an integral polynomial by the following result. Proposition 3. Let λ 1,λ,...,λ m be all the main eigenvalues of G. Then M(x) = (x λ 1 )(x λ ) (x λ m ) is an integral polynomial. Proof. Since G has exactly m main eigenvalues, the rank of the walk-matrix W(G) is m and j, Aj,...,A m 1 j are linear independent and j, Aj,...,A m 1 j, A m j are dependent. Thus there exist unique rational numbers a 0, a 1,...,a m 1 such that A m j = a 0 j + a 1 Aj + +a m 1 A m 1 j. By Lemma, alla i must be the elementary symmetric functions of λ 1,λ,...,λ m, i = 0, 1,...,m 1. Since λ 1,λ,...,λ m are algebraic integers and the sums and products of algebraic integers are also algebraic integers, a 0, a 1,...,a m 1 are also algebraic integers. Hence, all a i are integers by the rationality of a i for i = 0, 1,..., m 1. In [],Hagos showedthat a graph G has exactlytwo maineigenvaluesif and only if G is -walk linear. Moreover, if G is a -walk (a, b)-linear connected graph, then the two main eigenvalues λ 1,λ of G are λ 1, = a± a +4b,i.e., one has λ 1 + λ = a,λ 1 λ = b.hence, in order to find all graphs with exactly two main eigenvalues, it is sufficient to find all -walk linear graphs. For -walk (a, b)-linear graphs, since (x λ 1 )(x λ ) is an integral polynomial, a and b must be integers. Thus, we have Corollary 4. Let G be a -walk (a, b)-linear graph. Then both a and b are integers. If a -walk linear graph G is connected, then λ 1 λ and a 0. It was shown that a -walk (0, b)-linear graph is bipartite semi-regular [3] and it is easy to see that, if a -walk (0, b)-linear graph has at least one pendent vertex, then it must be a star graph. In [4], all finite and infinite harmonic trees were determined, and it was shown in [5] thatnoconnected irregular unicyclic or bicyclic graphs can be harmonic, and there exist exactly 4 and 16 connected irregular tricyclic and tetracyclic harmonic graphs, respectively. In the next section, we determine all -walk linear connected unicyclic graphs, and hence all connected unicyclic graphs with exactly two main eigenvalues. 3. -walk linear unicyclic graphs For a, the tree T a has been defined in [4] tobethetreewith one vertex v of degree a a + 1while every neighbor of v has degree a and all remaining vertices are pendent. The star graph K 1,n is the tree with n + 1vertices

3 Y. Hou, F. Tian / Applied Mathematics Letters 19 (006) where one central vertex is adjacent to all of the remaining n pendent vertices; and the double star graph S n+1,n+1 is the tree with n + vertices, where two central vertices u,v are adjacent and they are adjacent to exactly n pendent vertices, respectively. It is easy to check that T a is -walk (a, 0)-linear, K 1,n (n ) is -walk (0, n)-linear and S n+1,n+1 is -walk (1, n)-linear. Theorem 5 ([7]). T a,k 1,n (n ), and S n+1,n+1 are the only -walk linear trees. Proof. This follows immediately from the results in [7]. Lemma 6. Let G be a connected and -walk (a, b)-linear graph (a > 0). If a a + b + 1 > 0 and (a a + b + 1) + 4(a 1)b > 0 holds, then d(u) a a + ab + 1 holds for every vertex u V (G), and G has a vertex u with d(u) = a a + ab + 1 if and only if a = 1 and G = S b+1,b+1,ora, b = 0, and G = T a. Proof. Let v 1,v,...,v d(u) be the vertices adjacent to u and let w ij, j = 1,,...,d(v i ) 1beall vertices adjacent to v i other than u. Then, applying Eq. (1) to vertex v i,wehave d(v i ) 1 ad(v i ) + b = d(u) + d(w ij ) d(u) + d(v i ) 1. () j=1 Thus (a 1)d(v i ) d(u) b 1. On the other hand, applying Eq. (1) to vertex u,wehavead(u) + b = d(u) i=1 d(v i). Hence d(u) a(a 1)d(u) + (a 1)b = (a 1) d(v i ) d(u)(d(u) b 1), i=1 that is, d(u) (a a + b + 1)d(u) (a 1)b 0. Since (a a + b + 1) + 4(a 1)b > 0anda a + b + 1 > 0, we have d(u) a a + b (a a + b + 1) + 4(a 1)b a a + b a a + b + 1 +(a 1) b a a + b (a 1) b a a + ab + 1, The above inequalities become equalities if and only if a = 1orb = 0andd(w ij ) = 1forallvertices w ij.if b = 0thenG must be the tree T a by Lemma. of [4]. If a = 1thend(u) = b + 1. By Eq. (),wehaved(v) = 1or d(v) = b + 1for every vertex adjacent to u. Byad(u) + b = v N G (u) d(v), there exists exactly one neighbor of u with degree b + 1. Hence G = S b+1,b+1. Lemma 7. Let G be a connected and -walk (a, b)-linear graph. If G has a vertex u with exactly a + b 1 > 0 pendent vertices, then G = T a in the case a, and G = S b+1,b+1 in the case a = 1. Proof. Since a + b 1 1, a + b. Since d(u) = a + b and u has a + b 1pendent vertices, u has only one more neighbor, say v, thatisnot a pendent vertex. Applying Eq. (1) to the vertex u we have a(a + b) + b = a + b 1 + d(v), that is, d(v) = a a + ab + 1. Since a + b, we have a a + b + 1 (a 1) + > 0and(a a + b + 1) + 4(a 1)b ((a 1) + ) + 4(a 1)( a) = (a 1) (a 1) >0. Therefore Lemma 7 follows from Lemma 6.

4 1146 Y. Hou, F. Tian / Applied Mathematics Letters 19 (006) Lemma 8. Let G be a connected -walk (a, b)-linear graph possessing at least one cycle. Suppose there exists an edge such that G e has two components G 1 and G with G 1 acyclic. Then G 1 has exactly one vertex. Proof. Let w be the vertex of G 1 incident to the edge e; Lemma 8 claims that w is the only vertex of G 1.Assume the opposite, namely that G 1 has vertices other than w. Because G 1 is acyclic, some of thesemust be pendent vertices; choose one of the pendent vertices u that is at the greatest distance from w. Letu be adjacent to vertex x (it is immaterial whether x does or does not coincide with w). Then d(x) = a + b. Among the neighbors of x, ifone belongs to the path connecting w and u and its degree is greater than 1, then all remaining a + b 1vertices must be pendent vertices because otherwise u would not be the vertex at the greatest distance from w. Thus exactly a + b 1 pendent vertices are attached to vertex x. ByLemma 7, G must be the harmonic tree T a or the double star graph S b+1,b+1,whichcontradicts the assumption that G has at least a cycle. The following terminology comes from [8,9]. We call a vertex v in a graph G a bud if v is not a pendent vertex, yet all neighbors of v except at most one are pendent vertices, and a knob if v has exactly two neighbors that are not pendent vertices, and describe as the trunk sk(g) of a graph G the graph obtained by removing all pendent vertices and edges of G. Lemma 7 says that a connected -walk linear graph with at least one bud must be T a or S b+1,b+1. Lemma 8 says that the edge connectivity of the trunk of a connected -walk linear graph with at least a cycle must be more than 1. Lemma 9. Let v 0,v 1,v,v 3,v 4 be a path or a cycle (if v 0 = v 4 )ofa-walklinear graph Gsuchthat v 1,v,v 3 are knobs and v 0,v 4 are not pendent vertices, and let p i = d(v i ), i = 1,, 3,denote the number of pendent vertices adjacent to v i. If p > 0,thend(v 0 ) = d(v 1 ) = d(v ) = d(v 3 ) = d(v 4 ). Proof. Let G be a -walk (a, b)-linear graph. Since p > 0, we have d(v ) = a + b and v has exactly a + b > 0 pendent vertices. Applying Eq. (1) to vertex v we have a + b + d(v 1 ) + d(v 3 ) = a(a + b) + b.sincea 1and a +b 1, 4 a(a +b 1)+ = d(v 1 )+d(v 3 ),where the equality holds if and only if d(v 1 ) = d(v 3 ) =, a = 1 and b =. If d(v 1 ) = d(v 3 ) =, then a = 1, b =. Applying Eq. (1) to v 3 we have a + b + d(v 4 ) = ad(v 3 ) + b. Thus d(v 4 ) = 1, a contradiction to d(v 4 )>1. If d(v 1 ) = a + b and d(v 3 ) =, applying Eq. (1) to vertex v,wehavea + b + a + b + = a(a + b) + b, and therefore (a + b)( a) = b.sincea, b are integers, we have a =, b = 0ora = 0, b = 0, a contradiction. Similarly, the case d(v 3 ) = a + b and d(v 1 ) = wouldyieldacontradiction. Thus, we must have d(v 1 ) = d(v ) = d(v 3 ) = a + b > andhence d(v 0 ) = d(v 1 ) = d(v ) = d(v 3 ) = d(v 4 ) by Eq. (1) Let C r be a cycle of length r; denote by Cr k the graph attained from C r by attaching k 0pendent vertices to every vertex of C r.itiseasy to check that Cr k is -walk (, k)-linear for every integer k > 0. Theorem 10. Thegraphs C k r (k > 0) are the only connected -walk linear unicyclic graphs Proof. If G is a -walk (a, b)-linear graph, then, by Lemma 8, sk(g) contains no pendent vertices, that is, sk(g) = C r for some r.sinceg is not a cycle, G must possess pendent vertices and hence a > 0. Suppose the vertex v possesses at least one pendent vertex, then d(v) = a + b and v has exactly a + b > 0pendent vertices. Case 1: Thelength r of a cycle is at least 4. By Lemma 9, weknow that G must be the graph Cr a+b. Case : The length r of the cycle C r is 3. Let the vertices of the 3-cycle be v, u,w and d(v) = a + b >. If d(u) =, applying Eq. (1) to v and u, wehavea + b + + d(w) = a(a + b) + b and a + b = a + b + d(w), respectively. Hence a(a + b) = a. Sincea > 0, this implies a + b =, a contradiction. Hence d(u) = a + b. Similarly, d(w) = a + b.hence G = C3 a+b. Corollary 11. Thegraphs Cr k (k > 0) are all unicyclic connected graphs with exactly two main eigenvalues. Acknowledgments The authors thank Andreas Dress for his help, Dragan Stevanovićforsending preprints and the referee for careful reading and helpful suggestions. The work was done while Hou was visiting the Academy of Math. and Sys. Sci.,

5 Y. Hou, F. Tian / Applied Mathematics Letters 19 (006) Chinese Academy of Sci., and was supported by NSFC( ) and SRP(03B019) from the Education Committee of Hunan Province. References [1] D. Cvetković, P. Rowlinson, S. Simić, Eigenspaces of Graphs, Cambridge University Press, Cambridge, [] E.M. Hagos, Some results on graph spectra, Linear Algebra Appl. 356 (00) [3] A. Dress, I. Gutman, On the number of walks in a graph, Appl. Math. Lett. 16 (003) [4] S. Grünewald, Harmonic trees, Appl. Math. Lett. 15 (8) (00) [5] B. Borovićanin, S. Grünewald, I. Gutman, M. Petrović, Harmonic graphs with small number of cycles, Discrete Math. 65 (003) [6] D. Cvetković, M. Peterić, A table of connected graphs on six vertices, Discrete Math. 50 (1984) [7] Y. Hou, H. Zhou, Trees with exactly two main eigenvalues, Acta of Hunan Normal University 8 () (005) 1 3 (in Chinese). [8] A. Dress, S. Grünewald, D. Stevanović, Semiharmonic graphs with fixed cyclomatic number, Appl. Math. Lett. 17 (6) (004) [9] A. Dress, S. Grünewald, D. Stevanović, Semiharmonic bicyclic graphs, Appl. Math. Lett. 18 (11) (005)

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