On the second Laplacian eigenvalues of trees of odd order

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1 Linear Algebra and its Applications ) On the second Laplacian eigenvalues of trees of odd order Jia-yu Shao, Li Zhang, Xi-ying Yuan Department of Applied Mathematics, Tongji University, Shanghai , China Received 28 December 2005; accepted 18 May 2006 Available online 28 July 2006 Submitted by R.A. Brualdi Abstract We study the largest value and the second largest value of the second Laplacian eigenvalue λ 2 T ) among all trees T of a given odd order 2t + 1 the case of the even order 2t was already settled by Zhang and Li in [Xiao-dong Zhang, Jiong-sheng Li, The two largest Laplacian eigenvalues of Laplacian matrices of trees, J. Univ. Sci. Technol. China 28 5) 1998) ]). We show that the largest value of λ 2 T ) among all trees T of order 2t + 1 t 4) is 2 1 t ) t 2 + 2t 3, while the second largest value of λ 2 T ) is the second largest root of the cubic equation x 3 2t + 3)x 2 + t 2 + 3t + 3)x 2t + 1) = 0. We also determine the unique tree T 1 of order 2t + 1 whose λ 2 T 1 ) reaches this largest value, and determine the unique tree T 2 of order 2t + 1 whose λ 2 T 2 ) reaches this second largest value Elsevier Inc. All rights reserved. AMS classification: 05C50 Keywords: Tree; Laplacian eigenvalue; Subgraph 1. Introduction Unless stated otherwise, all graphs in this paper are finite, undirected and simple. Let G = V, E) be a graph with vertex set V ={v 1,v 2,...,v n } and edge set E ={e 1,e 2,...,e m }. Denote the order of G by G, and the maximum degree of G by ΔG). Pendant vertices are the vertices of Research supported by the National Natural Science Foundation of China Corresponding author. address: jyshao@sh163.net J.-y. Shao) /$ - see front matter 2006 Elsevier Inc. All rights reserved. doi: /j.laa

2 476 J.-y. Shao et al. / Linear Algebra and its Applications ) degree one. A pendant edge is an edge incident with a pendant vertex. We will abuse the language by writing v G and uv G, rather than v V and uv E, to indicate that v is a vertex of G and uv is an edge of G, respectively. Denote the degree of a vertex v i by dv i ). The Laplacian matrix LG) = DG) AG) is the difference of DG) = diagdv 1 ), dv 2 ),...,dv n )) and the adjacency matrix AG) of G. It is well known that LG) is positive semidefinite symmetric and singular. Denote its eigenvalues by ρg) = λ 1 G) λ 2 G) λ n G) = 0, ρg) is called the Laplacian spectral radius of G, and λ k G) is called the kth Laplacian eigenvalue of the graph G. Let G 1 = V 1,E 1 ) and G 2 = V 2,E 2 ) be graphs with V 1 V 2 =. A connected sum of G 1 and G 2 is any graph G = V, E) with V = V 1 V2, and E differing from E 1 E2 by the addition to E 1 E2 of a single edge joining some arbitrary) vertex of V 1 to some arbitrary) vertex of V 2. Throughout this paper, we shall denote by ΦB) = ΦB, x) = detxi B) the characteristic polynomial of a square matrix B. In particular, if B = LG), we sometimes use ΦG) to denote ΦLG)). For the second Laplacian eigenvalues λ 2 T ) of the trees T of order n, Zhang and Li [4] obtained an upper bound λ 2 T ) n 2 where x denotes the smallest integer no less than x), and they also showed that this bound is attained if and only if n is even and T is one of the three trees of order n listed in [4]. So for the case when n = 2t + 1 is odd, their upper bound λ 2 T ) n 2 =t + 1 is not a sharp bound namely, the strict inequality λ 2 T ) < n 2 holds when n is odd). In this paper, we study the largest value and the second largest value of λ 2 T ) among all trees T of a given odd order n = 2t + 1. We show that the largest value of λ 2 T ) among all trees T of order 2t + 1 t 4) is 2 1 t t 2 + 2t 3 ), together with the determination of the unique tree T 1 of order 2t + 1 whose λ 2 T 1 ) reaches this value. We also show that the second largest value of λ 2 T ) among all trees T of order 2t + 1 t 4) is the second largest root of the cubic equation x 3 2t + 3)x 2 + t 2 + 3t + 3)x 2t + 1) = 0, and we also determine the unique tree T 2 of order 2t + 1 whose λ 2 T 2 ) reaches this second largest value. 2. Lemmas and results Let G be a graph and let G = G + e be the graph obtained from G by adding a new edge e to G. It follows by the well-known Courant Weyl inequalities that the following is true. Lemma 2.1 [6]. The Laplacian eigenvalues of G and G interlace, that is, λ 1 G ) λ 1 G) λ 2 G ) λ 2 G) λ n G ) λ n G) = 0. The following inequalities are known as Cauchy s inequalities and the whole theorem is also known as the interlacing theorem. Lemma 2.2 [6]. Let A be a Hermitian matrix of order n with eigenvalues λ 1 λ 2 λ n and B is a principal submatrix of order m. Let B have eigenvalues µ 1 µ 2 µ m. Then the inequalities λ n m+i µ i λ i i = 1, 2,...,m)hold.

3 J.-y. Shao et al. / Linear Algebra and its Applications ) Lemma 2.3 [1]. Let T beatreeofordern. Then for any positive integer a, there exists a vertex v VT),such that there is one component of T v with order not exceeding max{n 1 a,a}, and all the other components of T v have orders not exceeding a. Lemma 2.4 [5]. Let G be a graph with degree sequence d 1 d 2 d n, and let λ 1 G) be the largest eigenvalue of the Laplacian matrix LG) of G. Then λ 1 G) 2 + d 1 + d 2 2)d 1 + d 3 2). Let a and b be positive integers and let Sa,b) denote a tree of diameter 3, having exactly two nonpendant vertices and they are adjacent, one of the two nonpendant vertices is connected to a pendant vertices and the other one is connected to b pendant vertices. In particular, the order of Sa,b) is a + b + 2. Lemma 2.5 [2]. ΦLSa, b))) = xx 1) n 4 x 3 n + 2)x 2 + 2n + ab + 1)x n), where n is the order of the tree Sa,b). Lemma 2.6. Let T be a tree with order n = t + 1, and T does not contain K 1,t 1 as a subgraph namely ΔT ) t 2), then λ 1 T)<t. Proof. It is obvious from the hypothesis that t 4. Case 1. t = 4. ΔT ) 2, T is a path with five vertices. λ 1 T ) = λ 1 P 5 ) = cos π 5 < 4 = t. Case 2. t 5. By [4], we know that the Laplacian spectral radius of St 3, 2) is the biggest among all the trees of order t + 1 with ΔT ) t 2. The degree sequence of St 3, 2) is t 2, 3, 1,...,1. By Lemma 2.4, wehaveλ 1 T ) λ 1 St 3, 2)) 2 + d 1 + d 2 2)d 1 + d 3 2) = 2 + t )t ) = 2 + t 1)t 3) <t. If v G, let L v G) be the principal submatrix of LG) obtained by deleting the row and column corresponding to the vertex v. Lemma 2.7. Let v be a vertex of the tree G = K 1,t 1, then λ 1 L v G)) < t. Proof. Let M denote the entrywise) absolute value of the matrix M, and ρm)denote the spectral radius of a square matrix M. Let BG) = DG) + AG), then LG) = DG) AG) = DG) + AG) = BG) and also L v G) =B v G).Sowehave λ 1 L v G)) ρ L v G) ) = ρb v G)). Since G is bipartite, BG) and LG) are similar, so we have λ 1 BG)) = λ 1 LG)) = t. Also BG) is a nonnegative irreducible symmetric matrix since G is connected, and B v G) is a proper principal submatrix of BG), so by the well-known Perron Frobenius theorem we have ρb v G)) < ρbg)) = λ 1 BG)). Combining the above relations we have λ 1 L v G)) < t.

4 478 J.-y. Shao et al. / Linear Algebra and its Applications ) Lemma 2.8. Let v be a vertex of a tree T.Let T 1,T 2,...,T k be all the connected components of T v. Let v j be the unique vertex in T j with vv j ET ) for j = 1,...,k. Let T j be the tree obtained from T j by adding a vertex v and an edge vv j to T j j = 1,...,k).Then under suitable ordering of the vertices of Tlet v be the first one, and any vertex of T i precedes any vertex of T j for 1 i<j k), we have L v T ) = L v T 1 ) L v T 2 ) L v T k ), where means the direct sum of matrices. Proof. By directly looking at the matrix LT ). Lemma 2.9. Let T beatreeofordern t + 1, where t 4. Let v be a vertex of T with T v/= K 1,t 1, then λ 1 L v T )) < t. Proof. Case 1. n<t. By Lemma 2.2, wehave λ 1 L v T )) λ 1 LT )) n<t. Case 2. n = t. Subcase 2.1. T = K 1,t 1. By Lemma 2.7, wehave λ 1 L v T )) < t. Subcase 2.2. T /= K 1,t 1. By Lemma 2.2, wehave λ 1 L v T )) λ 1 LT )) < t. Case 3. n = t + 1. Subcase 3.1. ΔT ) t 2. By Lemmas 2.2 and 2.6,wehave λ 1 L v T )) λ 1 LT )) < t. Subcase 3.2. ΔT ) = t 1. Let u be the vertex of T with maximum degree. Then v is adjacent to u or equal to u since T v/= K 1,t 1. Subcase v is a vertex of degree one. Then we have x t x detxi L v T ))= x x x 1 =x 1) t 4 x 4 t + 3)x 3 + 3t + 2)x 2 t + 4)x + 1).

5 J.-y. Shao et al. / Linear Algebra and its Applications ) Let fx)= x 4 t + 3)x 3 + 3t + 2)x 2 t + 4)x + 1, then λ 1 L v T )) is the largest root of the equation fx)= 0. Now we can see that fx)= x 2 x t)x 3) + xx t) + xx 4) + 1 > 0 for x t 4, so we have λ 1 L v T )) < t. Subcase v = u. By Lemma 2.8, wehave L v T ) = L v P 3 ) L v P 2 ) L v P 2 ). So by Lemma 2.7, wehave λ 1 L v T )) = max{λ 1 L v P 3 )), λ 1 L v P 2 ))} < 3 <t. Subcase v is the vertex of degree 2. By Lemma 2.8, wehave L v T ) = L v P 2 ) L v K 1,t 1 ). So by Lemma 2.7, wehave λ 1 L v T )) = max{λ 1 L v P 2 )), λ 1 L v K 1,t 1 ))} <t. Subcase 3.3. ΔT ) = t. Then we have T = K 1,t. Since T v/= K 1,t 1, then v is the center of K 1,t. By Lemma 2.8, wehave so L v T ) = L v P 2 ) L v P 2 ) L v P 2 ), λ 1 L v T )) = λ 1 L v P 2 )) 2 <t. Lemma Let T = St 2, 1) with t 4, then t<λ 1 T)<t Proof. From Lemma 2.5, we can easily get ΦLT )) = xx 1) t 3 x 3 t + 3)x 2 + 3t + 1)x t 1). Let hx) = x 3 t + 3)x 2 + 3t + 1)x t 1, then λ 1 T ) is the largest root of hx) = 0. Now h0) = t 1 < 0, h1) = t 2 > 0, ht) = 1 < 0,

6 480 J.-y. Shao et al. / Linear Algebra and its Applications ) and also hx) = x t)x 2 3x + 1) 1. Let a = t + 1 3, then we have h t + 1 ) = a2 3a + 1) 1 and 3h t + 1 ) = a 2 3a 2 = a 4)a + 1) + 2 > 0, 3 so we have h t + 1 ) > 0. 3 It follows from the above inequalities that t<λ 1 T)<t Lemma Let T beatreeofordern = 2t + 1, where t 4. If T does not contain 2K 1,t 1 as a subgraph, then λ 2 T ) t. Proof. Take a = t in Lemma 2.3, then n 1 a = t. There exists a vertex v VT), such that there is one component of T v, say T 0, with order T 0 max{n 1 a,a} =t, and all the other components of T v, say T j j = 1, 2,...,s), have orders not exceeding t. Suppose v 0,v 1,...,v s are vertices of T, and v j VT j ), vv j ET ) j = 0, 1,...,s). For each j0 j s), let T j be the tree obtained from T j by attaching the vertex v to its vertex v j by an edge. Then T j = T j +1 t + 1 j = 0, 1,...,s). Case 1. One of T j j = 0, 1, 2,...,s), say T 0,isK 1,t 1. Let T vv 0 = T 0 G. It is obvious that λ1 T 0 ) = t. And G =t + 1. By hypothesis the tree G does not contain K 1,t 1 as a subgraph, so λ 1 G)<tby Lemma 2.6. Thus by Lemma 2.1 we have λ 2 T ) λ 1 T vv 0 ) = max{λ 1 T 0 ), λ 1 G)} =t. Case 2. None of T j j = 0, 1, 2,...,s)is K 1,t 1. By Lemma 2.8, L v T ) = L v T 0 ) L v T 1 ) L v T s ). Since T j = T j +1 t + 1 and T j v = T j /= K 1,t 1, by Lemmas 2.2 and 2.9, wehave λ 2 T ) λ 1 L v T )) = max{λ 1 L v T 0 )), λ 1L v T 1 )),...,λ 1L v T s ))} <t. Lemma 2.12 [3]. Let G = G 1 u : vg 2 be a connected sum of G 1 and G 2 obtained by joining a vertex u of the graph G 1 to a vertex v of the graph G 2 by an edge where the vertex sets of G 1 and G 2 are disjoint). Then ΦLG)) = ΦLG 1 ))ΦLG 2 )) ΦLG 1 ))ΦL v G 2 )) ΦLG 2 ))ΦL u G 1 )). 2.1)

7 J.-y. Shao et al. / Linear Algebra and its Applications ) Fig The tree T 1. Fig The tree T 2. Lemma Let G 1 = K 1,t and G 2 = K 1,t 1 with t 3. 1) If T 1 = G 1 u : vg 2 as defined in Lemma 2.12), where u is a pendant vertex of K 1,t and v is the center of K 1,t 1 see Fig. 2.1), then we have ΦLT 1 )) = xx 1) 2t 4 x 2 t + 1)x + 1)x 2 t + 3)x + 2t + 1) 2.2) and thus λ 2 T 1 ) = 1 t t t 3 ). 2) If T 2 = G 1 u : vg 2, where u is the center of K 1,t and v is the center of K 1,t 1 see Fig. 2.2), then we have ΦLT 2 )) = xx 1) 2t 3 x 3 2t + 3)x 2 + t 2 + 3t + 3)x 2t + 1)) 2.3) and t <λ 2T 2 )<t ) If T 3 = G 1 u : vg 2, where u is the center of K 1,t and v is a pendant vertex of K 1,t 1, then we have and ΦLT 3 )) = xx 1) 2t 4 x 4 2t + 4)x 3 + t 2 + 6t + 4)x 2 t<λ 2 T 3 )<t t 2 + 4t + 2)x + 2t + 1)) 2.4)

8 482 J.-y. Shao et al. / Linear Algebra and its Applications ) Proof. It is well known that ΦLG 1 )) = ΦLK 1,t )) = xx t 1)x 1) t 1, 2.5) ΦLG 2 )) = ΦLK 1,t 1 )) = xx t)x 1) t ) 1) In this case we have ΦL u G 1 )) = x 1) t 2 x 2 t + 1)x + 1), 2.7) ΦL v G 2 )) = x 1) t ) Using Lemma 2.12, and 2.5) 2.8) we obtain 2.2). Furthermore, from 2.2) we can easily obtain λ 2 T 1 ) = 1 t t t 3 ). 2) In this case we have ΦL u G 1 )) = x 1) t, ΦL v G 2 )) = x 1) t ) Using Lemma 2.12 and 2.5), 2.6), 2.8) and 2.9) we obtain 2.3). Furthermore, let fx)= x 3 2t + 3)x 2 + t 2 + 3t + 3)x 2t + 1). Then we can easily verify that f0) = 2t 1 < 0. Now fx)= x t) 2 x 3) + t 1)1 3x t)), so Also, f f Thus we have t + 1 ) > 0. 3 t + 1 ) = t 1 8 < 0. t <λ 2T 2 ) = λ 2 f)<t+ 1 2, where λ 2 f ) is the second largest root of the equation fx)= 0. 3) In this case we have ΦL u G 1 )) = x 1) t, ΦL v G 2 )) = x 1) t 3 x 2 tx + 1). 2.10) Using Lemma 2.12 and 2.5), 2.6), 2.9) and 2.10) we obtain 2.4).

9 J.-y. Shao et al. / Linear Algebra and its Applications ) Furthermore, let gx) = x 4 2t + 4)x 3 + t 2 + 6t + 4)x 2 2t 2 + 4t + 2)x + 2t + 1. Then we can easily check that g0) = 2t + 1 > 0, g1) = t2 t) < 0, gt) = 1 > 0. Let a = t + 1 3, then we have ga) = a t)[a t)a 2 4a + 4 2t) 2t 1) 2 ]+1, using a t = 1 3,wehave 9ga) = a 2) 2 2t 6t 1) =[a 2) 2 t 1) 2 ] 5t + 2)t 2) <0, namely, g t + 1 ) < 0. 3 From the above relations we can obtain t<λ 2 T 3 ) = λ 2 g)<t+ 1 3, where λ 2 g) is the second largest root of the equation gx) = 0. Now we can obtain our main result of this paper. Theorem 2.1. Let T beatreeofordern = 2t + 1 with t 4, then we have a) λ 2 T ) 1 t t t 3 ) 2.11) with equality if and only if T is the tree T 1 in 1) of Lemma 2.13 see Fig. 2.1). b) If T /= T 1, then λ 2 T ) λ 2 f ), where λ 2 f ) is the second largest root of the following equation: fx)= x 3 2t + 3)x 2 + t 2 + 3t + 3)x 2t + 1) = 0 with equality if and only if T is the tree T 2 in 2) of Lemma 2.13 see Fig. 2.2). 2.12) Proof. First, from 2) of Lemma 2.13 it can be easily seen that 1 t t t 3 ) >t+ 1 2 >λ 2f)>t+ 1 3 and we already know in 1), 2) of Lemma 2.13 that equality holds in 2.11) when T = T 1, and equality holds in 2.12) when T = T 2. So we are now going to show that the strict inequality holds in 2.12) for all T /= T 1 and T /= T 2. Now we suppose that T /= T 1 and T /= T 2. Case 1. T does not contain 2K 1,t 1 as a subgraph.

10 484 J.-y. Shao et al. / Linear Algebra and its Applications ) Then from Lemma 2.11, we have Fig The tree T. λ 2 T ) t<t+ 1 3 <λ 2f ). Case 2. T contains 2K 1,t 1 as a subgraph. Let H = 2K 1,t 1 be such a subgraph of T and let x be the unique vertex of T not in the subgraph H. Subcase 2.1. There is an edge between x and some pendant vertex of H. Then T is a connected sum of K 1,t 1 and St 2, 1). So there exists an edge e such that T e = St 2, 1) K 1,t 1, thus from Lemma 2.10 we have λ 2 T ) λ 1 T e) = max{λ 1 St 2, 1)), λ 1 K 1,t 1 )} <t+ 1 3 <λ 2f ). Subcase 2.2. There is no edge between x and some pendant vertex of H. Then there is an edge between x and one of the centers of the two K 1,t 1 of H. Thus T is a connected sum of K 1,t and K 1,t 1, namely, T = G 1 u : vg 2 where G 1 = K 1,t and G 2 = K 1,t 1 ). i) If u is a pendant vertex of G 1 and v is a pendant vertex of G 2. Then T can also be viewed as a connected sum of K 1,t 1 and St 2, 1). The result follows by similar reasoning as subcase 2.1. ii) In all other cases i.e., not in the case i)), T is either T 1, T 2 or T 3 in Lemma But we have assumed that T /= T 1 and T /= T 2,soT = T 3. Thus from 3) of Lemma 2.13, we have λ 2 T)<t+ 1 3 <λ 2f ). At the end of this paper, we propose the following conjecture. Conjecture. Let t, k be positive integers with k 3 and t 2, let T be the tree of order kt + 1 as in Fig Then λ k T ) = max{λ k T ) T isatreeoforderkt + 1}. References [1] Jia-yu Shao, Bounds on the kth eigenvalues of trees and forests, Linear Algebra Appl ) [2] R. Grone, R. Merris, Ordering trees by algebraic connectivity, Graphs Combin ) [3] Ji-ming Guo, On the second largest Laplacian eigenvalue of trees, Linear Algebra Appl )

11 J.-y. Shao et al. / Linear Algebra and its Applications ) [4] Xiao-dong Zhang, Jiong-sheng Li, The two largest Laplacian eigenvalues of Laplacian matrices of trees, J. Univ. Sci. Technol. China 28 5) 1998) [5] Jiong-sheng Li, Xiao-dong Zhang, A new upper bound for eigenvalues of the Laplacian matrix of a graph, Linear Algebra Appl ) [6] D.M. Cvetković, M. Doob, H. Sachs, Spectra of Graphs Theory and Applications, VEB Deutscher Verlag d. Wiss./Academic Press, Berlin/New York, 1979.

A lower bound for the Laplacian eigenvalues of a graph proof of a conjecture by Guo

A lower bound for the Laplacian eigenvalues of a graph proof of a conjecture by Guo A. E. Brouwer & W. H. Haemers 2008-02-28 Abstract We show that if µ j is the j-th largest Laplacian eigenvalue, and d