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Proof of a conjecture involving the second largest D-eigenvalue and the number of triangles
Linear Algebra and its Applications 472 (2015) 48–53
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
Proof of a conjecture involving the second largest D-eigenvalue and the number of triangles Huiqiu Lin Department of Mathematics, East China University of Science and Technology, Shanghai, 200237, China
a r t i c l e
i n f o
Article history: Received 31 July 2014 Accepted 23 January 2015 Available online xxxx Submitted by J.y. Shao MSC: 05C50
a b s t r a c t Let G be a connected graph of order n with tr triangles and D be the distance matrix of G. Let λ1 (D) ≥ λ2 (D) ≥ · · · ≥ λn (D) be the D-eigenvalue of the graph G. Fajtlowicz (1998) [4] conjectured that λ2 (D) ≤ tr when the independent number α(G) ≤ 2. In this paper, the conjecture is confirmed and the extremal graph when the equality holds is characterized. © 2015 Elsevier Inc. All rights reserved.
Keywords: D-eigenvalue The number of triangles λ2 (D)
1. Introduction In this paper, a graph means a simple connected undirected graph. A graph is denoted by G = (V, E), where V is its vertex set and E is its edge set. The order of G is the number n = |V | of its vertices and its size is the number m = |E| of its edges. For S ⊂ V (G), let G[S] denote the subgraph induced by S. Let Gc denote the complement of the graph G, A and Ac be the adjacency matrix of G and Gc , respectively. Let N (v) denote the neighbor set of v and tr be the number of triangles in G. The independence E-mail address: [email protected]. http://dx.doi.org/10.1016/j.laa.2015.01.034 0024-3795/© 2015 Elsevier Inc. All rights reserved.
H. Lin / Linear Algebra and its Applications 472 (2015) 48–53
49
number (also the stability number) of G, denoted by α(G), is the cardinality of the maximal independent set of G. Denote V (G) = {v1 , . . . , vn } the vertex set of G. Let D(G) = (dij ) be the distance matrix of a graph G, where dij = dG (vi , vj ) is the length (i.e. the number of edges) of the shortest path from vi to vj . Let λ1 (D) ≥ λ2 (D) ≥ · · · ≥ λn (D) denote the D-eigenvalue of G. The inertia of D is given by (n+ (D), n0 (D), n− (D)), where n+ (D), n0 (D) and n+ (D) denote the number of positive, 0, and negative eigenvalues of D, respectively. The research for distance matrix can be dated back to the papers [3,5], which present an interesting result that the determinant of the distance matrix of trees with order n is always (−1)n−1 (n − 1)2n−2 , independent of the structure of the tree. Recently, the distance matrix of a graph has received increasing attention, see [6,7,9]. For more results on the distance matrix, you can refer to the excellent survey [1]. Fajtlowicz [4] proposed several conjectures involving the D-eigenvalue and the graph invariants. The following conjecture is one of them. Conjecture 1.1. (See [4].) Let G be a connected graph with independence number α(G) ≤ 2. Then λ2 (D(G)) ≤ tr, where tr denotes the number of triangles in G. In this paper, we confirm the conjecture as follows. Theorem 1.2. Let G be a connected graph with independence number α(G) ≤ 2. Then λ2 (D(G)) ≤ tr with equality holding if and only if G ∼ = K2,2 . Remark 1.3. Note that the number of triangles of K n2 , n2 (n ≥ 6) is zero, α(K n2 , n2 ) = n2 and λ2 (D(K n2 , n2 )) = n2 − 2 > 0. Therefore, the condition α(G) ≤ 2 cannot be deleted. 2. Proof Before giving the proof of the conjecture, the following lemmas are needed. Lemma 2.1. (See [8].) Let Tr (n) be the r-partite Turán graph with order n. If G is a Kr+1 -free graph of order n, then λ1 (A(G)) < λ1 (A(Tr (n))) unless G = Tr (n). Hermitian matrices have real eigenvalues. The Cauchy interlace theorem states that the eigenvalues of a Hermitian matrix A of order n are interlaced with those of any principal submatrix. Lemma 2.2 (Cauchy interlacing theorem). Let A be a Hermitian matrix with order n, and let B be a principal submatrix of A with order m. If λ1 (A) ≥ λ2 (A) ≥ · · · ≥ λn (A) lists the eigenvalues of A and μ1 (B) ≥ μ2 (B) ≥ · · · ≥ μm (B) the eigenvalues of B, then λn−m+i (A) ≥ μi (B) ≥ λi (A) for i = 1, 2, . . . , m. The following inequalities are well-known Courant–Weyl inequalities.
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H. Lin / Linear Algebra and its Applications 472 (2015) 48–53
Fig. 1. The graphs G1 –G5 .
Lemma 2.3 (Courant–Weyl inequalities). Let A and B be n × n Hermitian matrices and C = A + B. Then λi (C) ≤ λj (A) + λi−j+1 (B)
(n ≥ i ≥ j ≥ 1),
λi (C) ≥ λj (A) + λi−j+n (B)
(1 ≤ i ≤ j ≤ n).
Lemma 2.4. Let G be a connected graph with diameter 2 and D be the distance matrix of G. Then λ2 (D) ≤ −1 + λ1 (Ac ). Proof. Note that diam(G) = 2. Then D = J − I + Ac . Then by Lemma 2.3, the result follows. 2 Lemma 2.5. (See [5].) Let G be a tree with order n and D be the distance matrix of G. Then the inertia of D is given by (n+ (D), n0 (D), n− (D)) = (1, 0, n − 1). Lemma 2.6. (See [2].) Let G be a unicyclic graph with order n and D be the distance matrix of G. Then we have the following statements. (i) If the length of the cycle is 2k + 1, then the inertia of D is given by (n+ (D), n0 (D), n− (D)) = (1, 0, n − 1). (ii) If the length of the cycle is 2k, then the inertia of D is given by (n+ (D), n0 (D), n− (D)) = (1, k − 1, n − k). Now, we are ready to present the proof of Theorem 1.2 (see Fig. 1). Proof of Theorem 1.2. Let G be a connected graph with independence number α(G) ≤ 2 and D be the distance matrix of G. By Lemma 2.5, if n = 2, then the result holds. So we may assume that n ≥ 3. If α(G) = 1, then G ∼ = Kn and tr = n3 ≥ 1 > −1 = λ2 (D). So in the following, suppose that α(G) = 2, then diam(G) ≤ 3. We first suppose that diam(G) = 3 and let d(u, v) = 3, then we can partition V (G) = V0 ∪ V1 ∪ V2 ∪ V3 where V0 = {u}, Vi = {w|d(u, w) = i} for i = 1, 2, 3. Suppose that |Vi | = ni for i = 0, 1, 2, 3. Since α(G) = 2, G[Vi ] is a complete graph for i = 1, 2, 3, c st ∈ E[V2 , V3 ] for s ∈ V2 , t ∈ V3 and |E(V1 , V2 )| ≥ n2 . Then G is a bipartite graph with partitions V0 ∪ V1 , V2 ∪ V3 . Since n2 ≥ 1, we have λ1 (Ac ) < (n1 + 1)(n2 + n3 ). Note that D = J − I + Ac + B where B = A(K1,n3 ), then by Lemma 2.3, √ λ2 (D) ≤ −1 + λ1 Ac + λ1 (B) < −1 + (n1 + 1)(n2 + n3 ) + n3 . So we shall consider the following three cases.
H. Lin / Linear Algebra and its Applications 472 (2015) 48–53
51
Case 1. n3 ≥ 3. Then n1 + 1 n2 + n3 + 3 3 (n2 + n3 )(n2 + n3 − 1)(n2 + n3 − 2) n1 + 1 + = 6 3 n1 + 1 + n2 + n3 ≥ 3 √ ≥ −1 + (n1 + 1)(n2 + n3 ) + n3
tr ≥
> λ2 (D). Case 2. n3 = 2. Then λ2 (D) < −1 +
√ (n1 + 1)(n2 + 2) + 2 and if n2 ≥ 2, then
n1 + 1 n2 + 2 + 3 3 (n2 + 2)(n2 + 1)n2 n1 + 1 + = 6 3 n1 + 1 + n2 + 2 ≥ 3
tr ≥
> λ2 (D). If n2 = n1 = 1, then λ2 (D) < 0 by Lemma 2.6(i) and the result follows. If n2 = 1 and n1 = 2, then either G ∼ = G1 or G ∼ = G2 . Note that λ2 (D(G1 )) = −0.5505 and λ2 (D(G2 )) = −0.4384. Then the result follows. If n2 = 1 and n1 ≥ 3, then tr ≥ n1 + 2 >
3(n1 + 1) +
√
2 − 1 > λ2 (D).
Case 3. n3 = 1. Then λ2 (D) < (n1 + 1)(n2 + 1) and tr ≥ n13+1 + n23+1 . If either n1 ≥ 3 or n2 ≥ 3, then the result follows. If n1 = n2 = 2, then λ2 (D) < 3. If a vertex of V1 has two neighbors in V2 , then tr ≥ 3 > λ2 (D). Otherwise G ∼ = G3 , and note that λ2 (D) = λ2 (D(G3 )) = 0 < tr. If n1 = 1 and n2 = 2, then G ∼ = G4 and then λ2 (D) = λ2 (D(G4 )) = −0.4521 < tr. If n1 = 2 and n2 = 1, then either G ∼ = G4 or ∼ G ∼ . In both cases, we have λ (D) < 0 ≤ tr. If n = n = 1, then G G = 5 = P4 , by 2 1 2 Lemma 2.5, we have λ2 (D) < 0 = tr. In the following, we assume that diam(G) = 2 and let u, v be two non-adjacent vertices of G. Let V1 = N (u) ∩ N (v) = {w1 , · · · , ws }, V2 = N (u)\V1 = {u1 , · · · , up } and V3 = N (v)\V1 = {v1 , · · · , vq }. Without loss of generality, we may assume that p ≤ q. Obviously, V1 = ∅ and V (G) = V1 ∪ V2 ∪ V3 ∪ {u, v}. Note that Gc is triangle-free and G is connected, then by Lemma 2.1, λ1 (Ac ) < n2 . We shall distinguish the result by the following two cases.
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H. Lin / Linear Algebra and its Applications 472 (2015) 48–53
Case 1. G[V1 ] is a complete graph. Then λ1 (Ac ) < following three subcases.
p+q+s+2 . 2
We shall discuss by the
Subcase 1.1. s ≥ 2. Note that diam(G) = 2 and α(G) = 2. Then each vertex of V3 is either adjacent to a vertex of V1 or adjacent to all vertices of V2 . Then
p+1 q+1 s s tr ≥ + +2 + +q−1 3 3 2 3 p+1 q+1 =s+q−1+ + 3 3 p + q + 2s −1 2 p+q+s+2 −1 ≥ 2 > λ2 (D). ≥
Subcase 1.2. s = 1. Then n = p + q + 3, and then λ1 (Ac ) < p+q+3 . 2 ∼ If p = q = 0, then G = P3 , and then the result follows from Lemma 2.5. If p = 0 and q ≥ 1 and note that diam(G) = 2, then each vertex of V3 is adjacent to w1 . Thus by Lemma 2.4,
q+1 tr = q + 3
≥ −1 +
q+3 > λ2 (D). 2
∼ C5 or |E(G)| ≥ 6 since diam(G) = 2. If G ∼ If p = q = 1, then either G = = C5 , then by Lemma 2.6, tr = 0 > λ2 (D). If |E(G)| ≥ 6, then w1 is adjacent to at least one of {u1 , v1 }. If w1 u1 , w1 v1 ∈ E(G), then tr ≥ 2. Note that each vertex of Gc has degree at most 2, then λ1 (Ac ) ≤ 2. Thus we have tr ≥ 2 > λ1 (Ac ) − 1 = λ2 (D). If w1 u1 ∈ E(G), w1 v1 ∈ / E(G), then u1 v1 ∈ E(G). Then Gc ∼ = P5 , by Lemma 2.5, we have λ2 (D) < 0 < tr. λ2 (D) ≤ −1 + λ1 (Gc ) < 1 = tr. If p ≥ 1 and q ≥ 2, then p+q+1 n −1= 2 2 p+1 q+1 ≤ +q−1+ 3 3
λ2 (D) <
≤ tr. Case 2. G[V1 ] is not a complete graph. Then s ≥ 2 and let w1 , w2 ∈ V1 such that w1 w2 ∈ / E(V1 ). Since α(G) = 2, so each vertex in V2 ∪ V3 is adjacent to either w1 or w2 . Thus we have tr ≥
p+1 q+1 + + 2E G[V1 ] + p + q. 3 3
H. Lin / Linear Algebra and its Applications 472 (2015) 48–53
If s ≥ 4, then 2|E(G[V1 ])| ≥ s(s − 1) − s2 /2 =
s(s−2) 2
53
≥ s. Thus
tr ≥ s + p + q = n − 2 > −1 + λ1 Ac ≥ λ2 (D). If s = 3, then n = p + q + 5, it follows that tr ≥ 2 + p + q ≥
p+q+5 − 1 > λ2 (D). 2
If s = 2, then n = p + q + 4 and λ2 (D) < p+q+2 . If p = q = 0, then G ∼ = K2,2 and thus 2 we have tr = 0 = λ2 (D). If p = 0 and q = 1, then tr ≥ 1 and either |E[V1 , V3 ]| = 1 or |E[V1 , V3 ]| = 2. Thus either Gc ∼ = P3 ∪ P2 or Gc ∼ = P5 . In both cases, we have λ1 (Ac ) < 2 and thus λ2 (D) < 1 ≤ tr. If p, q ≥ 1, then tr ≥ p + q ≥
p+q+2 > λ2 (D). 2
Through the proof, we see that the graph with λ2 (D) = tr is K2,2 . Therefore we complete the proof. 2 Acknowledgement We thank the referee for her/his valuable comments which helped to improve the presentation of the paper. The author was supported by the by the National Natural Science Foundation of China (Nos. 11401211 and 11471211), the China Postdoctoral Science Foundation (No. 2014M560303) and Fundamental Research Funds for the Central Universities (No. 222201414021). References [1] M. Aouchiche, P. Hansen, Distance spectra of graphs: a survey, Linear Algebra Appl. 458 (2014) 301–386. [2] R. Bapat, S.J. Kirkland, M. Neumann, On distance matrices and Laplacians, Linear Algebra Appl. 401 (2005) 193–209. [3] M. Edelberg, M.R. Garey, R.L. Graham, On the distance matrix of a tree, Discrete Math. 14 (1976) 23–29. [4] S. Fajtlowicz, Written on the wall: conjectures derived on the basis of the program Galatea Gabriella Graffiti, Technical report, University of Houston, 1998. [5] R.L. Graham, H.O. Pollack, On the addressing problem for loop switching, Bell Syst. Tech. J. 50 (1971) 2495–2519. [6] Y.-L. Jin, X.-D. Zhang, Complete multipartite graphs are determined by their distance spectra, Linear Algebra Appl. 448 (2014) 285–291. [7] H. Lin, Y. Hong, J. Wang, J. Shu, On the distance spectrum of graphs, Linear Algebra Appl. 439 (2013) 1662–1669. [8] V. Nikiforov, Bounds on graph eigenvalues II, Linear Algebra Appl. 427 (2007) 183–189. [9] X. Zhang, On the distance spectral radius of some graphs, Linear Algebra Appl. 437 (2012) 1930–1941.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
Proof of a conjecture involving the second largest D-eigenvalue and the number of triangles Huiqiu Lin Department of Mathematics, East China University of Science and Technology, Shanghai, 200237, China
a r t i c l e
i n f o
Article history: Received 31 July 2014 Accepted 23 January 2015 Available online xxxx Submitted by J.y. Shao MSC: 05C50
a b s t r a c t Let G be a connected graph of order n with tr triangles and D be the distance matrix of G. Let λ1 (D) ≥ λ2 (D) ≥ · · · ≥ λn (D) be the D-eigenvalue of the graph G. Fajtlowicz (1998) [4] conjectured that λ2 (D) ≤ tr when the independent number α(G) ≤ 2. In this paper, the conjecture is confirmed and the extremal graph when the equality holds is characterized. © 2015 Elsevier Inc. All rights reserved.
Keywords: D-eigenvalue The number of triangles λ2 (D)
1. Introduction In this paper, a graph means a simple connected undirected graph. A graph is denoted by G = (V, E), where V is its vertex set and E is its edge set. The order of G is the number n = |V | of its vertices and its size is the number m = |E| of its edges. For S ⊂ V (G), let G[S] denote the subgraph induced by S. Let Gc denote the complement of the graph G, A and Ac be the adjacency matrix of G and Gc , respectively. Let N (v) denote the neighbor set of v and tr be the number of triangles in G. The independence E-mail address: [email protected]. http://dx.doi.org/10.1016/j.laa.2015.01.034 0024-3795/© 2015 Elsevier Inc. All rights reserved.
H. Lin / Linear Algebra and its Applications 472 (2015) 48–53
49
number (also the stability number) of G, denoted by α(G), is the cardinality of the maximal independent set of G. Denote V (G) = {v1 , . . . , vn } the vertex set of G. Let D(G) = (dij ) be the distance matrix of a graph G, where dij = dG (vi , vj ) is the length (i.e. the number of edges) of the shortest path from vi to vj . Let λ1 (D) ≥ λ2 (D) ≥ · · · ≥ λn (D) denote the D-eigenvalue of G. The inertia of D is given by (n+ (D), n0 (D), n− (D)), where n+ (D), n0 (D) and n+ (D) denote the number of positive, 0, and negative eigenvalues of D, respectively. The research for distance matrix can be dated back to the papers [3,5], which present an interesting result that the determinant of the distance matrix of trees with order n is always (−1)n−1 (n − 1)2n−2 , independent of the structure of the tree. Recently, the distance matrix of a graph has received increasing attention, see [6,7,9]. For more results on the distance matrix, you can refer to the excellent survey [1]. Fajtlowicz [4] proposed several conjectures involving the D-eigenvalue and the graph invariants. The following conjecture is one of them. Conjecture 1.1. (See [4].) Let G be a connected graph with independence number α(G) ≤ 2. Then λ2 (D(G)) ≤ tr, where tr denotes the number of triangles in G. In this paper, we confirm the conjecture as follows. Theorem 1.2. Let G be a connected graph with independence number α(G) ≤ 2. Then λ2 (D(G)) ≤ tr with equality holding if and only if G ∼ = K2,2 . Remark 1.3. Note that the number of triangles of K n2 , n2 (n ≥ 6) is zero, α(K n2 , n2 ) = n2 and λ2 (D(K n2 , n2 )) = n2 − 2 > 0. Therefore, the condition α(G) ≤ 2 cannot be deleted. 2. Proof Before giving the proof of the conjecture, the following lemmas are needed. Lemma 2.1. (See [8].) Let Tr (n) be the r-partite Turán graph with order n. If G is a Kr+1 -free graph of order n, then λ1 (A(G)) < λ1 (A(Tr (n))) unless G = Tr (n). Hermitian matrices have real eigenvalues. The Cauchy interlace theorem states that the eigenvalues of a Hermitian matrix A of order n are interlaced with those of any principal submatrix. Lemma 2.2 (Cauchy interlacing theorem). Let A be a Hermitian matrix with order n, and let B be a principal submatrix of A with order m. If λ1 (A) ≥ λ2 (A) ≥ · · · ≥ λn (A) lists the eigenvalues of A and μ1 (B) ≥ μ2 (B) ≥ · · · ≥ μm (B) the eigenvalues of B, then λn−m+i (A) ≥ μi (B) ≥ λi (A) for i = 1, 2, . . . , m. The following inequalities are well-known Courant–Weyl inequalities.
50
H. Lin / Linear Algebra and its Applications 472 (2015) 48–53
Fig. 1. The graphs G1 –G5 .
Lemma 2.3 (Courant–Weyl inequalities). Let A and B be n × n Hermitian matrices and C = A + B. Then λi (C) ≤ λj (A) + λi−j+1 (B)
(n ≥ i ≥ j ≥ 1),
λi (C) ≥ λj (A) + λi−j+n (B)
(1 ≤ i ≤ j ≤ n).
Lemma 2.4. Let G be a connected graph with diameter 2 and D be the distance matrix of G. Then λ2 (D) ≤ −1 + λ1 (Ac ). Proof. Note that diam(G) = 2. Then D = J − I + Ac . Then by Lemma 2.3, the result follows. 2 Lemma 2.5. (See [5].) Let G be a tree with order n and D be the distance matrix of G. Then the inertia of D is given by (n+ (D), n0 (D), n− (D)) = (1, 0, n − 1). Lemma 2.6. (See [2].) Let G be a unicyclic graph with order n and D be the distance matrix of G. Then we have the following statements. (i) If the length of the cycle is 2k + 1, then the inertia of D is given by (n+ (D), n0 (D), n− (D)) = (1, 0, n − 1). (ii) If the length of the cycle is 2k, then the inertia of D is given by (n+ (D), n0 (D), n− (D)) = (1, k − 1, n − k). Now, we are ready to present the proof of Theorem 1.2 (see Fig. 1). Proof of Theorem 1.2. Let G be a connected graph with independence number α(G) ≤ 2 and D be the distance matrix of G. By Lemma 2.5, if n = 2, then the result holds. So we may assume that n ≥ 3. If α(G) = 1, then G ∼ = Kn and tr = n3 ≥ 1 > −1 = λ2 (D). So in the following, suppose that α(G) = 2, then diam(G) ≤ 3. We first suppose that diam(G) = 3 and let d(u, v) = 3, then we can partition V (G) = V0 ∪ V1 ∪ V2 ∪ V3 where V0 = {u}, Vi = {w|d(u, w) = i} for i = 1, 2, 3. Suppose that |Vi | = ni for i = 0, 1, 2, 3. Since α(G) = 2, G[Vi ] is a complete graph for i = 1, 2, 3, c st ∈ E[V2 , V3 ] for s ∈ V2 , t ∈ V3 and |E(V1 , V2 )| ≥ n2 . Then G is a bipartite graph with partitions V0 ∪ V1 , V2 ∪ V3 . Since n2 ≥ 1, we have λ1 (Ac ) < (n1 + 1)(n2 + n3 ). Note that D = J − I + Ac + B where B = A(K1,n3 ), then by Lemma 2.3, √ λ2 (D) ≤ −1 + λ1 Ac + λ1 (B) < −1 + (n1 + 1)(n2 + n3 ) + n3 . So we shall consider the following three cases.
H. Lin / Linear Algebra and its Applications 472 (2015) 48–53
51
Case 1. n3 ≥ 3. Then n1 + 1 n2 + n3 + 3 3 (n2 + n3 )(n2 + n3 − 1)(n2 + n3 − 2) n1 + 1 + = 6 3 n1 + 1 + n2 + n3 ≥ 3 √ ≥ −1 + (n1 + 1)(n2 + n3 ) + n3
tr ≥
> λ2 (D). Case 2. n3 = 2. Then λ2 (D) < −1 +
√ (n1 + 1)(n2 + 2) + 2 and if n2 ≥ 2, then
n1 + 1 n2 + 2 + 3 3 (n2 + 2)(n2 + 1)n2 n1 + 1 + = 6 3 n1 + 1 + n2 + 2 ≥ 3
tr ≥
> λ2 (D). If n2 = n1 = 1, then λ2 (D) < 0 by Lemma 2.6(i) and the result follows. If n2 = 1 and n1 = 2, then either G ∼ = G1 or G ∼ = G2 . Note that λ2 (D(G1 )) = −0.5505 and λ2 (D(G2 )) = −0.4384. Then the result follows. If n2 = 1 and n1 ≥ 3, then tr ≥ n1 + 2 >
3(n1 + 1) +
√
2 − 1 > λ2 (D).
Case 3. n3 = 1. Then λ2 (D) < (n1 + 1)(n2 + 1) and tr ≥ n13+1 + n23+1 . If either n1 ≥ 3 or n2 ≥ 3, then the result follows. If n1 = n2 = 2, then λ2 (D) < 3. If a vertex of V1 has two neighbors in V2 , then tr ≥ 3 > λ2 (D). Otherwise G ∼ = G3 , and note that λ2 (D) = λ2 (D(G3 )) = 0 < tr. If n1 = 1 and n2 = 2, then G ∼ = G4 and then λ2 (D) = λ2 (D(G4 )) = −0.4521 < tr. If n1 = 2 and n2 = 1, then either G ∼ = G4 or ∼ G ∼ . In both cases, we have λ (D) < 0 ≤ tr. If n = n = 1, then G G = 5 = P4 , by 2 1 2 Lemma 2.5, we have λ2 (D) < 0 = tr. In the following, we assume that diam(G) = 2 and let u, v be two non-adjacent vertices of G. Let V1 = N (u) ∩ N (v) = {w1 , · · · , ws }, V2 = N (u)\V1 = {u1 , · · · , up } and V3 = N (v)\V1 = {v1 , · · · , vq }. Without loss of generality, we may assume that p ≤ q. Obviously, V1 = ∅ and V (G) = V1 ∪ V2 ∪ V3 ∪ {u, v}. Note that Gc is triangle-free and G is connected, then by Lemma 2.1, λ1 (Ac ) < n2 . We shall distinguish the result by the following two cases.
52
H. Lin / Linear Algebra and its Applications 472 (2015) 48–53
Case 1. G[V1 ] is a complete graph. Then λ1 (Ac ) < following three subcases.
p+q+s+2 . 2
We shall discuss by the
Subcase 1.1. s ≥ 2. Note that diam(G) = 2 and α(G) = 2. Then each vertex of V3 is either adjacent to a vertex of V1 or adjacent to all vertices of V2 . Then
p+1 q+1 s s tr ≥ + +2 + +q−1 3 3 2 3 p+1 q+1 =s+q−1+ + 3 3 p + q + 2s −1 2 p+q+s+2 −1 ≥ 2 > λ2 (D). ≥
Subcase 1.2. s = 1. Then n = p + q + 3, and then λ1 (Ac ) < p+q+3 . 2 ∼ If p = q = 0, then G = P3 , and then the result follows from Lemma 2.5. If p = 0 and q ≥ 1 and note that diam(G) = 2, then each vertex of V3 is adjacent to w1 . Thus by Lemma 2.4,
q+1 tr = q + 3
≥ −1 +
q+3 > λ2 (D). 2
∼ C5 or |E(G)| ≥ 6 since diam(G) = 2. If G ∼ If p = q = 1, then either G = = C5 , then by Lemma 2.6, tr = 0 > λ2 (D). If |E(G)| ≥ 6, then w1 is adjacent to at least one of {u1 , v1 }. If w1 u1 , w1 v1 ∈ E(G), then tr ≥ 2. Note that each vertex of Gc has degree at most 2, then λ1 (Ac ) ≤ 2. Thus we have tr ≥ 2 > λ1 (Ac ) − 1 = λ2 (D). If w1 u1 ∈ E(G), w1 v1 ∈ / E(G), then u1 v1 ∈ E(G). Then Gc ∼ = P5 , by Lemma 2.5, we have λ2 (D) < 0 < tr. λ2 (D) ≤ −1 + λ1 (Gc ) < 1 = tr. If p ≥ 1 and q ≥ 2, then p+q+1 n −1= 2 2 p+1 q+1 ≤ +q−1+ 3 3
λ2 (D) <
≤ tr. Case 2. G[V1 ] is not a complete graph. Then s ≥ 2 and let w1 , w2 ∈ V1 such that w1 w2 ∈ / E(V1 ). Since α(G) = 2, so each vertex in V2 ∪ V3 is adjacent to either w1 or w2 . Thus we have tr ≥
p+1 q+1 + + 2E G[V1 ] + p + q. 3 3
H. Lin / Linear Algebra and its Applications 472 (2015) 48–53
If s ≥ 4, then 2|E(G[V1 ])| ≥ s(s − 1) − s2 /2 =
s(s−2) 2
53
≥ s. Thus
tr ≥ s + p + q = n − 2 > −1 + λ1 Ac ≥ λ2 (D). If s = 3, then n = p + q + 5, it follows that tr ≥ 2 + p + q ≥
p+q+5 − 1 > λ2 (D). 2
If s = 2, then n = p + q + 4 and λ2 (D) < p+q+2 . If p = q = 0, then G ∼ = K2,2 and thus 2 we have tr = 0 = λ2 (D). If p = 0 and q = 1, then tr ≥ 1 and either |E[V1 , V3 ]| = 1 or |E[V1 , V3 ]| = 2. Thus either Gc ∼ = P3 ∪ P2 or Gc ∼ = P5 . In both cases, we have λ1 (Ac ) < 2 and thus λ2 (D) < 1 ≤ tr. If p, q ≥ 1, then tr ≥ p + q ≥
p+q+2 > λ2 (D). 2
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