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Which Q-cospectral graphs have same degree sequences
Accepted Manuscript Which Q-cospectral graphs have same degree sequences
Guangliang Zhang, Muhuo Liu, Haiying Shan
PII: DOI: Reference:
S0024-3795(16)30607-3 http://dx.doi.org/10.1016/j.laa.2016.12.018 LAA 13976
To appear in:
Linear Algebra and its Applications
Received date: Accepted date:
29 January 2016 13 December 2016
Please cite this article in press as: G. Zhang et al., Which Q-cospectral graphs have same degree sequences, Linear Algebra Appl. (2017), http://dx.doi.org/10.1016/j.laa.2016.12.018
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Which Q-cospectral graphs have same degree sequences∗ Guangliang Zhang1 , Muhuo Liu2 , Haiying Shan3 1
School of Mathematical Science, South China Normal University, Guangzhou, 510631, P.R. China
2
Department of Mathematics, South China Agricultural University, Guangzhou, 510642, P.R. China
3
Department of Mathematics, Tongji University, Shanghai, 200092, China
Abstract Let Cn denote the cycle with n vertices, and let τ (G) be the number of triangles of G. If G is a graph with n vertices, then μ1 (G) ≥ μ2 (G) ≥ · · · ≥ μn (G) denote the signless Laplacian spectrum of G. Two graphs are said to be Q-cospectral if they have the same signless Laplacian spectrum. A graph G is said to be Q − DS if there is no other non-isomorphic graph H such that G and H are Q-cospectral. In this paper, we prove that “Let G be a graph with n ≥ 12 vertices and 1 ≤ μn (G) ≤ μ2 (G) ≤ 5 < n < μ1 (G) or let G be a graph with n ≥ 10 vertices and 0 < μn (G) ≤ μ2 (G) ≤ 4 < n < μ1 (G). If G and H are Q-cospectral, then G and H share the same degree sequences and τ (G) = τ (H)”. As a consequence of our results, we show that all multi-wheel graphs K1 ∨ (Cq1 ∪ Cq2 ∪ · · · ∪ Cqt ) are Q − DS. ∗
Partially supported by NSFC project No. 11571123, No. 11271288, the Training Program for Out-
standing Young Teachers in University of Guangdong Province (No. YQ2015027) and China Scholarship Council. Email: [email protected] (M. Liu), shan [email protected] (H. Shan, Corresponding author)
1
Key words and phrases: Signless Laplacian spectrum, Join graph, Determined by spectrum; Multi-wheel graph AMS 2000 Subject Classifications: 05C50; 15A18; 15A36
1
Introduction
Throughout this paper, G = (V, E) is an undirected simple graph with n vertices, and G denotes the complement graph of G. Let dG (u) and NG (u) be the degree and neighbor set of vertex u in G, respectively. If dG (u) = 1, then we call u a pendent vertex. In the sequel, we enumerate the degrees of G in non-increasing order, i.e., d1 (G) ≥ d2 (G) ≥ · · · ≥ dn (G), where dG (vi ) = di (G) for i ∈ {1, 2, ..., n}. As usual, K1,n−1 , Kn and Cn denote the star, complete graph and cycle with n vertices, respectively, and G1 ∨G2 denotes the join graph of two vertex disjoint graphs G1 and G2 . In other words, G1 ∨ G2 is the graph having vertex set V (G1 ∨ G2 ) = V (G1 ) ∪ V (G2 ) and edge set E(G1 ∨ G2 ) = E(G1 ) ∪ E(G2 ) ∪ {uv : u ∈ V (G1 ), v ∈ V (G2 )}. Let A(G) and D(G) be the adjacency matrix and the diagonal degree matrix of G, respectively. The Laplacian matrix of G is L(G) = D(G) − A(G), and the signless Laplacian matrix of G is Q(G) = D(G) + A(G). Let SQ (G) and SL (G) denote the set of eigenvalues of Q(G) and L(G), respectively. It is easy to see that Q(G) is positive semidefinite and hence its eigenvalues can be arranged as [1]: μ1 (G) ≥ μ2 (G) ≥ · · · ≥ μn (G) ≥ 0. If there is no risk of confusion, we simplify dG (u), di (G), NG (u) and μi (G) as d(u), di , N (u) and μi , respectively. Two graphs are said to be Q-cospectral (respectively, L-cospectral) if they have the same signless Laplacian (respectively, Laplacian) spectrum. Similarly, two graphs are said to be co-degree if they have the same degree sequences. A graph G is said to be determined by its signless Laplacian spectrum, simplified as Q − DS (respectively, Laplacian spectrum, simplified as L − DS) if there does not exist other non-isomorphic graph H such that H and G are Q-cospectral (respectively, L-cospectral). Which graphs are determined by their spectra? This problem was posed by G¨ unthard 2
Figure 1.1: The trees T1 and T2 . and Primas [6] more than 50 years ago in the context of H¨ uckel’s theory in chemistry. Only in the most recent years have mathematicians devoted their attention to this problem and more and more papers focus on this item are now appearing. For detail, we refer the readers to [2, 3, 5, 8, 15] and the references therein. Among these literatures on this item, a great ratios of results are on L − DS, the main reason is that SL (G) = SL (H) if and only if SL (G) = SL (H) (See [2, 11]). Thus, the L − DS problem of sparse graphs can be transferred to the L − DS problem of dense graphs, and vise versa. Different from L − DS problem, there exist graphs G and H such that SQ (G) = SQ (H), but SQ (G) = SQ (H). For instance, let T1 and T2 be two trees as shown in Figure 1.1. It is easy to check that SQ (T1 ) = SQ (T2 ), but SQ (T1 ) = SQ (T2 ). In view of this, as the experience to the authors, when we want to prove that G is Q − DS, we always first prove that H and G are co-degree under the supposition of G and H being Q-cospectral, and then prove that G and H are isomorphic, but the first step is not an easy task, and we always have to spend a large amount of work to solve it [5, 8, 10, 11, 16]. Thus, it is natural for us to consider the following problem. Problem 1.1 Which graphs satisfy the property “every Q-cospectral graph of G is also co-degree with G”? In this paper, we will give partial answer to Problem 1.1 by proving Theorem 1.1 Let G be a graph with n ≥ 12 vertices and 1 ≤ μn (G) ≤ μ2 (G) ≤ 5 < n < μ1 (G). If SQ (G) = SQ (H), then G and H are co-degree. Theorem 1.2 Let G be a graph with n ≥ 10 vertices and 0 < μn (G) ≤ μ2 (G) ≤ 4 < n < μ1 (G). If SQ (G) = SQ (H), then G and H are co-degree. If G is a graph with n vertices, then we use Φ(G, x)=det(xIn − Q(G)) to define the signless Laplacian characteristic polynomial of G, where In is the identity matrix of order 3
Figure 1.2: The graphs F1 (n) and F2 (n) on n vertices. n. It is not difficult for the readers to find that Theorems 1.1 and 1.2 are only suitable for small graph classes. But the following example show that we cannot decrease the condition “μn (G) ≥ 1” from Theorem 1.1 or the condition “μ2 (G) ≤ 4” from Theorem 1.2, respectively. Example 1.1 Let F1 (n) and F2 (n) be the graphs on n ≥ 8 vertices as shown in Figure 1.2. By an elementary calculation (or see [9]), we have Φ(F1 (n), x) = Φ(F2 (n), x) = (x − 1)n−5 (x − 2)2 f (x),
(1.1)
where f (x) = x3 − (n + 5)x2 + 5nx − 12. When n ≥ 8, since f (0) = −12 < 0, f (1) = 4(n − 4) > 0, f (4) = 4(n − 7) > 0, f (5) = f (n) = −12 < 0 and f (n + 1) = n(n − 3) − 16 > 0, by (1.1) we can conclude that SQ (F1 (n)) = SQ (F2 (n)) with 0 < μn (F1 (n)) = μn (F2 (n)) < 1, 4 < μ2 (F1 (n)) = μ2 (F2 (n)) < 5 and n < μ1 (F1 (n)) = μ1 (F2 (n)) < n + 1. It is easy to see that F1 (n) and F2 (n) are not co-degree. Let G be a graph such that every Q-cospectral graph of G is also co-degree with G. If the degree sequence of G is π and the graph with π as its degree sequence is unique, then G and H are isomorphic, in other word, G is Q − DS. From this fact and Theorem 1.2, we can easily deduce the following conclusion, which is the main result of [10]. Corollary 1.1 [10] If n ≥ 10, then S(n, q) is Q − DS for any n ≥ 2q + 1 and q ≥ 0, where S(n, q) ∼ = K1 ∨ (qK2 ∪ (n − 2q − 1)K1 ). The wheel graph is denoted by Wn = K1 ∨ Cn−1 and the multi-wheel graph is denoted by K1 ∨ (Cq1 ∪ Cq2 ∪ · · · ∪ Cqt ). It was proved that Wn is L − DS except for W7 [15], and all multi-wheel graphs with no W7 as its induced subgraphs are also L − DS [2]. In [8], it was proved that Wn is Q − DS. As a consequence of Theorem 1.1, we will show that 4
Theorem 1.3 For any t ≥ 1, K1 ∨ (Cq1 ∪ Cq2 ∪ · · · ∪ Cqt ) is Q − DS.
2
The proof of Theorem 1.1
Let diag{b1 , b2 , . . . , bn } be a diagonal matrix with the i-th diagonal entry being bi , where i = 1, 2, . . . , n. Let θ1 (M ) ≥ θ2 (M ) ≥ · · · ≥ θn (M ) be the eigenvalues of M , where M is a symmetric matrix of order n. Lemma 2.1 [7] Let M be a symmetric matrix of order n. If N is a principal submatrix of order s of M , then θi (M ) ≥ θi (N ) ≥ θn−s+i (M ) holds for any i = 1, 2, . . . , s. From Lemma 2.1, Liu et al. [12] proved the following result, which is crucial to our proofs. Lemma 2.2 [12] Let G be a graph with X ⊆ V (G) and let H ∼ = G[X]. If X = {x1 , x2 , . . . , xs }, then for any i ∈ {1, 2, . . . , s} and 0 ≤ pi ≤ dG (xi ), we have μi (G) ≥ θi (A(H) + diag{p1 , p2 , . . . , ps }) . The following two lemmas present lower or upper bounds for μ1 (G) and μ2 (G). Lemma 2.3 [4] For any graph G with at least two vertices, we have 1 μ2 (G) ≥ d1 (G) + d2 (G) − (d1 (G) − d2 (G))2 + 4 ≥ d2 (G) − 1, 2 where μ2 (G) = d2 (G) − 1 implies that d1 (G) = d2 (G) and v1 is adjacent with v2 . Let uv be an edge of G and v be a vertex of G. Let m(v) denote the average of the d(w)/d(v). Denote by degrees of the vertices being adjacent to v, i.e., m(v) = w∈N (v)
Ψ(uv) =
d(u)(d(u) + m(u)) + d(v)(d(v) + m(v)) . d(u) + d(v)
According to the proof of Theorem 2.10 in [13], we have Lemma 2.4 [13] If G is a connected graph with at least one edge, then μ1 (G) ≤ max{Ψ(uv) : uv ∈ E(G)} ≤ max{d(v) + m(v) : v ∈ V (G)} ≤ d1 (G) + d2 (G), where the first equality holds if and only if G is regular or bipartite semiregular. 5
Throughout this paper, if every component of G contains at least one edge, then we agree that max{d(v) + m(v) : v ∈ V (G)} occurs at the vertex v0 and Ψ(x0 y0 ) = max {Ψ(xy) : xy ∈ E(G)}, where d(x0 ) ≥ d(y0 ). Furthermore, we suppose that d(y0 ) = d in the following. Corollary 2.1 Let G be a graph with n vertices and μ1 (G) > n. If every component of G contains at least one edge, then we have n − d1 (G) + 1 ≤ d(x0 ) ≤ d1 (G), n − d1 (G) + 1 ≤ d(v0 ) ≤ d1 (G) and m(v0 ) ≤
2|E(G)| − dn (G)(n − 1) 2|E(G)| − (n − 1) + dn (G) − 1 ≤ . d(v0 ) d(v0 )
(2.1)
Proof. If d(y0 ) ≤ d(x0 ) ≤ n − d1 (G), by Lemma 2.4 we have μ1 (G) ≤ Ψ(x0 y0 ) ≤ d(x0 ) + d1 (G) ≤ n, a contradiction. Thus, n − d1 (G) + 1 ≤ d(x0 ) ≤ d1 (G). If d(v0 ) ≤ n − d1 (G), by Lemma 2.4, μ1 (G) ≤ d(v0 ) + m(v0 ) ≤ d(v0 ) + d1 (G) ≤ n, a contradiction. Thus, n − d1 (G) + 1 ≤ d(v0 ) ≤ d1 (G). Now, since u∈N (v0 ) d(u) + d(v0 ) + dn (G)(n − d(v0 ) − 1) ≤ u∈V (G) d(u) = 2|E(G)|, we have u∈N (v0 ) d(u) ≤ 2|E(G)| − dn (G)(n − 1) + (dn (G) − 1)d(v0 ), and hence (2.1) holds. Lemma 2.5 [1, 4] If G is a connected graph with n ≥ 2 vertices, then 0 ≤ μn (G) < dn (G) < d1 (G) + 1 ≤ μ1 (G), where μn (G) = 0 if and only if G is bipartite, and μ1 (G) = d1 (G) + 1 if and only if G∼ = K1,n−1 . As usual, G + e denotes the graph obtained from G by adding one edge e ∈ E(G), while G − e denotes the graph obtained from G by deleting one edge e ∈ E(G). Lemma 2.6 Let G be a graph with d(u1 ) ≥ 8 and d(u2 ) ≥ 5. If either u1 u2 ∈ E(G) with |N (u1 ) ∩ N (u2 )| ≥ 2 or u1 u2 ∈ E(G) with |N (u1 ) ∩ N (u2 )| ≥ 1, then μ2 (G) > 5. Proof. Let X = {u1 , u2 , x, y} and let Y = {u1 , u2 , x, z}. Let ⎛ ⎛ ⎛ ⎞ ⎞ 8 1 1 0 8 1 1 0 ⎜ ⎜ ⎜ ⎟ ⎟ ⎜ 1 2 0 1 ⎟ ⎜ 1 3 1 1 ⎟ ⎜ ⎜ ⎜ ⎜ ⎟ ⎟ B1 = ⎜ ⎟ , B2 = ⎜ ⎟ , B3 = ⎜ ⎜ 1 0 2 1 ⎟ ⎜ 1 1 3 1 ⎟ ⎜ ⎝ ⎝ ⎝ ⎠ ⎠ 0 1 1 5 0 1 1 5 6
8 1 1 0
⎞
⎟ 1 5 1 1 ⎟ ⎟ ⎟, 1 1 2 0 ⎟ ⎠ 0 1 0 1
⎛ ⎛
8 1 ⎜ ⎜ 1 5 ⎜ B4 = ⎜ ⎜ 1 1 ⎝ 0 1
1 0
⎞
⎛
8 ⎜ ⎟ ⎜ 1 1 1 ⎟ ⎜ ⎟ ⎟ , B5 = ⎜ ⎜ 1 3 1 ⎟ ⎝ ⎠ 1 2 1
1 5 1 1
⎜ ⎜ 1 1 ⎜ ⎜ ⎟ ⎜ 1 1 ⎟ ⎜ ⎟ ⎟ , B6 = ⎜ ⎜ 3 1 ⎟ ⎜ ⎠ ⎜ ⎜ 1 3 ⎝ ⎞
⎞ 8 1 1 1 1 1 1 5 1 1 1 1 1 1 2 0 0 0 1 1 0 2 0 0 1 1 0 0 2 0
⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎟ ⎟ ⎠
1 1 0 0 0 2 We firstly consider the case of u1 u2 ∈ E(G). Since |N (u1 ) ∩ N (u2 )| ≥ 2, there must exist vertices {x, y} ⊆ N (u1 ) ∩ N (u2 ) such that either G[X] ∼ = C4 or G[X] ∼ = K4 − e. If G[X] ∼ = K4 − e, by Lemma = C4 , by Lemma 2.2 we have μ2 (G) ≥ θ2 (B1 ) > 5.46 . If G[X] ∼ 2.2 we have μ2 (G) ≥ θ2 (B2 ) > 5.75. We secondly consider the case of u1 u2 ∈ E(G). Recall that N (u1 ) ∩ N (u2 ) = Ø. When there exists some vertex z ∈ N (u2 ) \ {N (u1 ) ∪ {u1 }}, either G[Y ] ∼ = K1,3 + e or G[Y ] ∼ = K1,3 + e, by Lemma 2.2 we have = K4 − e, where x ∈ N (u1 ) ∩ N (u2 ). If G[Y ] ∼ μ2 (G) ≥ θ2 (B3 ) > 5.03. If G[Y ] ∼ = K4 − e, by Lemma 2.2 we have μ2 (G) ≥ θ2 (B4 ) > 5.3. When N (u2 ) ∪ {u2 } ⊆ N (u1 ) ∪ {u1 }, we choose {x, y} ⊆ N (u1 ) ∩ N (u2 ). Then, either G[X] ∼ = K4 − e. If G[X] ∼ = K4 , by Lemma 2.2 we have μ2 (G) ≥ θ2 (B5 ) > = K4 or G[X] ∼ 5.12. Otherwise, G[X] ∼ = K4 − e holds for any {x, y} ⊆ N (u1 ) ∩ N (u2 ). By Lemma 2.2, it follows that μ2 (G) ≥ θ2 (B6 ) > 5.05. Lemma 2.7 If G is a graph with n ≥ 12 vertices and 0 < μn (G) ≤ μ2 (G) ≤ 5 < n < μ1 (G), then G is connected. Proof. On conversely, assume that G is disconnected. By Lemma 2.3, d2 (G) ≤ 6. We first suppose that d2 (G) = 6. By Lemma 2.3 it follows that d1 (G) = 6 and μ2 (G) = 5. Furthermore, Lemma 2.4 implies that n < μ1 (G) ≤ d1 (G) + d2 (G) ≤ 12. Thus, we have n ≤ 11, a contradiction. Now, we suppose that d2 (G) ≤ 5. By Lemma 2.4 we have n < μ1 (G) ≤ d1 (G) + d2 (G), which implies that n − 4 ≤ d1 (G) ≤ n − 2. From Lemma 2.5, we can conclude that G∼ = G1 ∪ C3 , where d1 (G1 ) = n − 4 ≥ d2 (G1 ) = d2 (G) = 5. Since |V (G1 )| = n − 3, d1 (G1 ) = n − 4 ≥ 8 and d2 (G1 ) = d2 (G) = 5, we can conclude that v1 v2 ∈ E(G1 ) with |NG1 (v1 ) ∩ NG1 (v2 )| ≥ 1. By Lemma 2.6, μ2 (G) = μ2 (G1 ) > 5, a contradiction. 7
Theorem 2.1 If G is a graph with n ≥ 12 vertices and 0 < μn (G) ≤ μ2 (G) ≤ 5 < n < μ1 (G), then G is connected with d2 (G) ≤ 4 and d1 (G) ≥ n−3. Furthermore, if G contains no pendent vertices, then d1 (G) ≥ n − 2. Proof. In the proof of this result, we simplify μi (G) and di (G) as μi and di , respectively. By Lemmas 2.3 and 2.7, G is connected with d2 ≤ 6. If d2 = 6, then d1 = 6 and μ2 = 5 follow from Lemma 2.3. Furthermore, Lemma 2.4 implies that n < μ1 ≤ d1 + d2 ≤ 12, and hence n ≤ 11, a contradiction. Thus, d2 ≤ 5. To complete the proof, it suffices to prove the following two claims. Claim 1. d2 ≤ 4 and d1 ≥ n − 3. Proof of Claim 1. By Lemma 2.4, it suffices to show that d2 ≤ 4. On conversely, suppose that d2 = 5. Then, Lemma 2.4 implies that d1 ≥ n − 4. Since d1 ≥ n − 4 ≥ 8 and d2 = 5, either v1 v2 ∈ E(G) with |N (v1 )∩N (v2 )| ≥ 2 or v1 v2 ∈ E(G) with |N (v1 )∩N (v2 )| ≥ 1. By Lemma 2.6, μ2 > 5, a contradiction. Claim 2. If dn (G) ≥ 2, then d1 (G) ≥ n − 2. Proof of Claim 2. By Claim 1, we suppose that d1 = n − 3. Thus, Claim 1 and Lemma 2.4 imply that d2 = 4 < n − 3 = d1 . By Corollary 2.1, d(x0 ) ∈ {4, n − 3}. If d(x0 ) = 4 = d2 , since d1 = n − 3 and 2 ≤ d = d(y0 ) ≤ 4, by Lemma 2.4 we have μ1 ≤ Ψ(x0 y0 ) ≤
42 + n − 3 + d + 4 × 2 + d2 + n − 3 + 4 × (d − 1) ≤ n, 4+d
a contradiction. Thus, d(x0 ) = n − 3 = d1 . Now, since 2 ≤ d = d(y0 ) ≤ 4 and n ≥ 12, by Lemma 2.4 we have μ1 ≤ Ψ(x0 y0 ) ≤
(n − 3)2 + 4(n − 4) + d + d2 + n − 3 + 4 × (d − 1) ≤ n, n−3+d
a contradiction. Let τ (G) denote the number of triangles of G, and let nk (G) be the number of vertices with degree k in V (G) \ {v1 }, where k ∈ {1, 2, . . . , d2 (G)}. Lemma 2.8 [1] If G is a graph with n vertices, m edges, and τ triangles, then n i=1
μi =
n i=1
di = 2m,
n i=1
μ2i = 2m +
n i=1
8
d2i , and
n i=1
μ3i = 6τ + 3
n i=1
d2i +
n i=1
d3i .
Remark 2.1 If G and H are Q-cospectral and co-degree, by Lemma 2.8 we have τ (G) = τ (H). Thus, τ (G) = τ (H) holds in Theorems 1.1 and 1.2. Proof of Theorem 1.1. By Lemma 2.5 and Theorem 2.1, we have dn (H) ≥ 2 and dn (G) ≥ 2. Furthermore, Theorem 2.1 implies that d2 (G) ≤ 4 < n − 2 ≤ d1 (G) and d2 (H) ≤ 4 < n − 2 ≤ d1 (H). In the following, we write ni (G) and d1 (G) as ni and d1 , and we write ni (H) and d1 (H) as n∗i and d∗1 , respectively, where i ∈ {2, 3, 4}. From Lemma 2.8, we have ⎧ ∗ ∗ ∗ ⎪ ⎪ ⎨ n2 + n3 + n4 = n2 + n3 + n4 ⎪ ⎪ ⎩
2n2 + 3n3 + 4n4 + d1 = 2n∗2 + 3n∗3 + 4n∗4 + d∗1
(2.2)
4n2 + 9n3 + 16n4 + (d1 )2 = 4n∗2 + 9n∗3 + 16n∗4 + (d∗1 )2
If d1 = d∗1 , then ni = n∗i holds for any i ∈ {2, 3, 4} by (2.2). If d1 = d∗1 , by symmetry we may suppose that d1 = n−2 and d∗1 = n−1. Now, from (2.2) we have n∗3 = n3 +2n−9 > n, a contradiction. Thus, G and H are co-degree.
3
The proofs of Theorem 1.2 and Corollary 1.1
Similarly with Lemma 2.7, we can easily obtain the following result. Lemma 3.1 If G is a graph with n ≥ 10 vertices and 0 < μn (G) ≤ μ2 (G) ≤ 4 < n < μ1 (G), then G is connected. Theorem 3.1 If G is a graph with n ≥ 10 vertices and 0 < μn (G) ≤ μ2 (G) ≤ 4 < n < μ1 (G), then G is connected with d2 (G) ≤ 3 and d1 (G) ≥ n − 2. Proof. In the proof of this result, we simplify μi (G) and di (G) as μi and di , respectively. By Lemmas 2.3 and 3.1, G is connected with d2 ≤ 5. If d2 = 5, then d1 = 5 and μ2 = 4 by Lemma 2.3. Furthermore, Lemma 2.4 implies that n < μ1 ≤ d1 + d2 ≤ 10, a contradiction. Thus, d2 ≤ 4. By Lemma 2.4, we have d1 ≥ n − 3. By Lemma 2.4, to complete the proof, it suffices to prove that d2 ≤ 3. On conversely, suppose that d2 = 4.
9
Let X = {v1 , v2 , x} and let Y = {v1 , v2 , x, y}. ⎛ ⎛ ⎞ 7 1 0 ⎜ 7 1 0 ⎜ 1 4 1 ⎜ ⎟ ⎜ ⎟ D1 = ⎜ = , D ⎜ 2 ⎝ 1 2 1 ⎠ ⎜ 0 1 1 ⎝ 0 1 4 1 1 0 ⎛
7 1 ⎜ ⎜ 1 4 ⎜ D4 = ⎜ ⎜ 1 1 ⎝ 1 1
1 1 3 1
Let ⎞ 1 ⎟ 1 ⎟ ⎟ ⎟, 0 ⎟ ⎠ 2 ⎛ ⎞ 7 ⎜ 1 ⎜ 1 ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎟ , D5 = ⎜ ⎜ 1 1 ⎟ ⎜ ⎠ ⎜ 1 ⎝ 3 1
⎛
7 1 0 1
⎞
⎜ ⎟ ⎜ 1 4 1 1 ⎟ ⎜ ⎟ D3 = ⎜ ⎟, ⎜ 0 1 2 1 ⎟ ⎝ ⎠ 1 1 1 3 ⎞ 1 1 1 1 ⎟ 4 1 1 1 ⎟ ⎟ ⎟ 1 2 0 0 ⎟ ⎟. ⎟ 1 0 2 0 ⎟ ⎠ 1 0 0 2
We first consider the case of v1 v2 ∈ E(G). Since d2 = 4 and d1 ≥ n − 3, there must exist vertex {x} ⊆ N (v1 ) ∩ N (v2 ) such that G[X] ∼ = K1,2 . Since d1 ≥ n − 3 ≥ 7, by Lemma 2.2 we have μ2 ≥ θ2 (D1 ) > 4.36, a contradiction. Now, we consider the case of v1 v2 ∈ E(G). Then, N (v1 ) ∩ N (v2 ) = Ø. We claim that N (v2 ) ∪ {v2 } ⊆ N (v1 ) ∪ {v1 }. Otherwise, assume that there exists some vertex y ∈ N (v2 ) \ {N (v1 ) ∪ {v1 }}. Then, either G[Y ] ∼ = K1,3 + e or G[Y ] ∼ = K4 − e, where x ∈ N (v1 ) ∩ N (v2 ). If G[Y ] ∼ = K1,3 + e, by Lemma 2.2 we have μ2 ≥ θ2 (D2 ) > 4.13, a contradiction. If G[Y ] ∼ = K4 − e, by Lemma 2.2 we have μ2 ≥ θ2 (D3 ) > 4.57, a contradiction. Thus, N (v2 ) ∪ {v2 } ⊆ N (v1 ) ∪ {v1 }. We choose two vertices {x, y} ⊆ N (v1 ) ∩ N (v2 ). If G[Y ] ∼ = K4 , since d1 ≥ n − 3 ≥ 7, by Lemma 2.2 we have μ2 ≥ θ2 (D3 ) > 4.33, a contradiction. Thus, G[Y ] ∼ = K4 − e holds for any Y with {x, y} ⊆ N (v1 ) ∩ N (v2 ). By Lemma 2.2 and d1 ≥ n − 3 ≥ 7, we have μ2 ≥ θ2 (D5 ) > 4.07, a contradiction. Proof of Theorem 1.2. By Theorem 3.1, d2 (G) ≤ 3 < n − 2 ≤ d1 (G) and d2 (H) ≤ 3 < n − 2 ≤ d1 (H). In the following, we simplify ni (G) and d1 (G) as ni and d1 , and we rewrite ni (H) and d1 (H) as n∗i and d∗1 , respectively, where i ∈ {1, 2, 3}. By Lemma 2.8, we have
⎧ ∗ ∗ ∗ ⎪ ⎪ ⎨ n1 + n2 + n3 = n1 + n2 + n3 n1 + 2n2 + 3n3 + d1 = n∗1 + 2n∗2 + 3n∗3 + d∗1 ⎪ ⎪ ⎩ n1 + 4n2 + 9n3 + (d1 )2 = n∗1 + 4n∗2 + 9n∗3 + (d∗1 )2 10
(3.1)
If d1 = d∗1 , then ni = n∗i holds for any i ∈ {1, 2, 3} by (3.1). If d1 = d∗1 , by symmetry we may suppose that d1 = n−2 and d∗1 = n−1. Now, from (3.1) we have n∗2 = n2 +2n−7 > n, a contradiction. Thus, G and H are co-degree. Proof of Corollary 1.1. Suppose that G and S(n, q) are Q-cospectral. If q = 0, then S(n, q) ∼ = K1,n−1 . Since S(n, q) contains In−1 as its principal submatrix, by Lemma 2.1, we have μ2 (G) = μ2 (S(n, q)) ≤ θ1 (In−1 ) = 1, and hence d2 (G) ≤ 2 by Lemma 2.3. If d2 (G) = 2, then d1 (G) = 2 by Lemma 2.3. From Lemmas 2.4 and 2.5, we have μ1 (G) ≤ 4 < n = μ1 (K1,n−1 ), a contradiction. Thus, d2 (G) = 1. By Lemma 2.4 and μ1 (K1,n−1 ) = n, we have d1 (G) = n − 1 and hence G ∼ = K1,n−1 . If q ≥ 1, then S(n, q) is non-bipartite. By Lemma 2.5, μ1 (G) = μ1 (S(n, q)) > d1 (S(n, q)) + 1 = n and μn (G) = μn (S(n, q)) > 0. Let G∗ ∼ = qK2 ∪ (n − 2q − 1)K1 . Note that S(n, q) contains In−1 + Q(G∗ ) as its principal submatrix. Thus, by Lemma 2.1, we have μ2 (G) = μ2 (S(n, q)) ≤ θ1 (In−1 + Q(G∗ )) = 3. Now, from Theorem 1.2, we can conclude that G and S(n, q) are co-degree, which implies that G ∼ = S(n, q).
4
The proof of Theorem 1.3
For any n ≥ 3, it is well-known that (See [2]) 2kπ SQ (Cn ) = 2 + 2 cos : k = 0, 1, . . . , n − 1 . n
(4.1)
Let q (p) define p copies of q, where q and p are two non-negative integers. Lemma 4.1 If t ≥ 1, then SQ (K1 ∨ (Cq1 ∪ Cq2 ∪ · · · ∪ Cqt )) is equal to 2k π 1 i (t−1) n + 4 ± (n − 4)2 + 16 , 5 : 1 ≤ ki ≤ qi − 1, 1 ≤ i ≤ t . , 3 + 2 cos 2 qi Proof. In the proof of this result, we write K1 ∨(Cq1 ∪Cq2 ∪· · ·∪Cqt ) and Cq1 ∪Cq2 ∪· · ·∪Cqt as G and G∗ , respectively. By equation (4.1), θk (In−1 + Q(G∗ )) = 1 + μk (G∗ ) = 5 holds for 1 ≤ k ≤ t. Note that G contains In−1 + Q(G∗ ) as its principal submatrix. Thus, by Lemma 2.1, we have 5 = θt (In−1 + Q(G∗ )) ≤ μt (G) ≤ μt−1 (G) ≤ · · · ≤ μ2 (G) ≤ θ1 (In−1 + Q(G∗ )) = 5. 11
For 1 ≤ i ≤ t, we suppose that V (Cqi ) = {ui1 , ui2 , . . . , uiqi } and θ = 2 is an eigenvalue of A(Cqi ). Since Cqi is regular, we may choose ϕ = (ϕ(ui1 ), ϕ(ui2 ), . . . , ϕ(uiqi ))T as the unit i ϕ(uij ) = 0. Let ψ = (ψ(u1 ), ψ(u2 ), . . . , ψ(un ))T eigenvector corresponding to θ such that qj=1 be the unit eigenvector such that ψ(v) = ϕ(v) holds for v ∈ V (Cqi ) and ψ(v) = 0 holds for v ∈ V (G) \ V (Cqi ). Since Iqi + Q(Cqi ) = 3Iqi + A(Cqi ) is a principal submatrix of Q(G), ψ is an eigenvector of Q(G) corresponding to the eigenvalue 3 + θ. By equation (4.1), {3 + 2 cos( 2kqii π ) : i = 1, 2, . . . , qi − 1} are the eigenvalues of Q(G), where 1 ≤ i ≤ t. Set μ = 12 (n + 4 ± (n − 4)2 + 16 ). It is easy to check that Q(G)ψ1 = μψ1 , where ψ1 = (μ − 5, 1(n−1) ) and μ − 5 is corresponding to the vertex v1 of G. Since μ = 5 and ψ1 is orthogonal with ψ, the result follows. Lemma 4.2 Let G∗ ∼ = K1 ∨ (Cq1 ∪ Cq2 ∪ · · · ∪ Cqt ) with t ≥ 2. If 7 ≤ |V (G∗ )| ≤ 11 and G and G∗ are Q-cospectral, then G and G∗ are co-degree. Proof. We may suppose that |V (G∗ )| = n. In the proof of this result, we write di (G) and μi (G) as di and μi , respectively. By Lemma 2.5 and Lemma 4.1, we have dn ≥ 2 and hence every component of G contains at least one edge. If d1 ≤ n − 3, since 7 ≤ n ≤ 11 and |E(G)| = |E(G∗ )| = 2(n − 1) by Lemma 2.8, by Corollary 2.1 and (2.1) we have 4 ≤ d(v0 ) ≤ n − 3 and hence 2(n − 1) d(v0 ) + m(v0 ) ≤ d(v0 ) + 1 + d(v0 ) 2(n − 1) 2(n − 1) ≤ max 5 + ,n − 2 + 4 n−3 1 ≤ n + 4 + (n − 4)2 + 16 . 2 1 2
(4.2)
Recall that 7 ≤ n ≤ 11. By Lemma 4.1 and (4.2), we have μ1 = d(v0 ) + m(v0 ) = 2 n + 4 + (n − 4) + 16 , which implies that n = 7, μ1 = 8 and μ2 = 5. Furthermore,
μ2 = 5, dn ≥ 2 and Lemmas 2.4 and 4.1 imply that G is a connected regular graph with d1 = 4. It is well known that every Q-cospectral graph with a regular graph is also regular (See Proposition 2 or Lemma 4 of [2]), a contradiction. Thus, d1 ≥ n − 2 and hence G is connected (otherwise, un = 0, a contradiction). Since μ2 = μ2 (G∗ ) = 5 by Lemma 4.1, we have d2 ≤ 6 by Lemma 2.3. 12
If d2 = 6, by Lemma 2.3, we have 6 = d1 ≥ n − 2 and hence 7 ≤ n ≤ 8. When 7 ≤ n ≤ 8, from Lemma 2.8, there is no graph with n vertices such that d1 = d2 = 6 and dn ≥ 2. Thus, d2 ≤ 5. Now, from Lemma 2.8, we have ⎧ ⎪ =n−1 ⎪ ⎨ n2 + n3 + n4 + n5 ⎪ ⎪ ⎩
= 4(n − 1) − d1
2n2 + 3n3 + 4n4 + 5n5
(4.3)
4n2 + 9n3 + 16n4 + 25n5 = 9(n − 1) + (n − 1)2 − d21
If d1 = n − 1, from (4.3), it is easy to check that G and G∗ are co-degree. In the following, we consider the case of d1 = n − 2. We firstly consider the case of 10 ≤ n ≤ 11. If n5 = 0, by (4.3) we have n3 = 8−n < 0, a contradiction. Otherwise, n5 ≥ 1 and hence d2 = 5. Since d1 = n − 2 ≥ 8 and d2 = 5, either v1 v2 ∈ E(G) with |N (v1 ) ∩ N (v2 )| ≥ 2 or v1 v2 ∈ E(G) with |N (v1 ) ∩ N (v2 )| ≥ 1. By Lemma 2.6, μ2 > 5, a contradiction. Now, we consider the case of 7 ≤ n ≤ 9. By solving systems of linear diophantine equations (4.3) for 7 ≤ n ≤ 9, we get the following non-negative integral solutions: (n, d1 ) n 2 n3 n4 n5
(7, 5) 1 2 4 1 , 0 3 1
0
(8, 6) 2 3 3 0 , 1 4 1
(9, 7) 3 2 2 1
0
With the aid of mathematics software “SageMath” [14], there are exactly 306 connected graphs with degree sequences corresponding to these solutions of (4.3), but none of them satisfies the property “μ2 = 5 and μn ≥ 1”, which contradicts Lemma 4.1. Proof of Theorem 1.3. In the proof of this result, we write K1 ∨ (Cq1 ∪ Cq2 ∪ · · · ∪ Cqt ) as G∗ . Suppose that SQ (G) = SQ (G∗ ). Since the case of t = 1 was already proved in [8], we may suppose that t ≥ 2, and hence n ≥ 7 in the following. By Lemma 4.1, we can conclude that 1 ≤ μn (G) ≤ μ2 (G) = 5 <
1 n + 4 + (n − 4)2 + 16 = μ1 (G). 2
If 7 ≤ n ≤ 11, by Lemma 4.2, G and G∗ are co-degree. If n ≥ 12, by Theorem 1.1, G and G∗ are co-degree. Thus, G ∼ = K1 ∨ (Ck ∪ Ck ∪ · · · ∪ Ck ). Now, from Lemma 4.1, we 1
2
can conclude that G ∼ = G∗ . 13
s
Acknowledgement. The authors are much grateful to the anonymous referee for his/her valuable comments on our paper, which have considerably improved the presentation of this paper.
References [1] D. Cvetkovi´c, P. Rowlinson, S.K. Simi´c, Signless Laplacians of finite graphs, Linear Algebra Appl., 423 (2007), 155–171. [2] E.R. van Dam, W.H. Haemers, Which graphs are determined by their spectrum, Linear Algebra Appl., 373 (2003), 241–272. [3] E.R. van Dam, W.H. Haemers, Developments on spectral characterizations of graphs, Discrete Math., 309 (2009), 576–586. [4] K.Ch. Das, On conjectures involving second largest signless Laplacian eigenvalue of graphs, Linear Algebra Appl., 432 (2010), 3018–3029. [5] K.Ch. Das, M. Liu, Complete split graph determined by its (signless) Laplacian spectrum, Discrete Appl. Math., 205 (2016), 45–51. [6] H.H. G¨ unthard, H. Primas, Zusammenhang von Graphtheorie und Mo-Theotie von Molekeln mit Systemen konjugierter Bindungen, Helv. Chim. Acta, 39 (1956), 1645–1653. [7] R.A. Horn, C.R. Johnson, Matrix Analysis, Cambridge University Press, New York, 1990. [8] M. Liu, Some graphs determined by their (signless) Laplacian spectra, Czechoslovak Math. J., 62 (2012), 1117–1134. [9] M. Liu, B. Liu, On the signless Laplacian spectra of bicyclic and tricyclic graphs, Ars Combin., 120 (2015), 169–180. [10] M. Liu, B. Liu, F. Wei, Graphs determined by their (signless) Laplacian spectra, Electron. J. Linear Algebra, 22 (2011), 112–124. 14
[11] M. Liu, H. Shan, K.Ch. Das, Some graphs determined by their (signless) Laplacian spectra, Linear Algebra Appl., 449 (2014), 154–165. [12] M. Liu, Y. Zhu, H. Shan, K. Ch. Das, The spectral characterization of butterfly-like graphs, Linear Algebra Appl., 513 (2017), 55–68. [13] Y.-L. Pan, Sharp upper bounds for the Laplacian graph eigenvalues, Linear Algebra Appl., 355 (2002), 287–295. [14] W.A. Stein, et al., Sage Mathematics Software (Version 6.10), The Sage Development Team, http:// www.sagemath.org, 2015. [15] Y. Zhang, X. Liu, X. Yong, Which wheel graphs are determined by their Laplacian spectra? Comput. Math. Appl., 58 (2009), 1887–1890. [16] Y. Zhang, X. Liu, B. Zhang, X. Yong, The lollipop graph is determined by its Q-spectrum, Discrete Math., 309 (2009), 3364–3369.
15
Guangliang Zhang, Muhuo Liu, Haiying Shan
PII: DOI: Reference:
S0024-3795(16)30607-3 http://dx.doi.org/10.1016/j.laa.2016.12.018 LAA 13976
To appear in:
Linear Algebra and its Applications
Received date: Accepted date:
29 January 2016 13 December 2016
Please cite this article in press as: G. Zhang et al., Which Q-cospectral graphs have same degree sequences, Linear Algebra Appl. (2017), http://dx.doi.org/10.1016/j.laa.2016.12.018
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Which Q-cospectral graphs have same degree sequences∗ Guangliang Zhang1 , Muhuo Liu2 , Haiying Shan3 1
School of Mathematical Science, South China Normal University, Guangzhou, 510631, P.R. China
2
Department of Mathematics, South China Agricultural University, Guangzhou, 510642, P.R. China
3
Department of Mathematics, Tongji University, Shanghai, 200092, China
Abstract Let Cn denote the cycle with n vertices, and let τ (G) be the number of triangles of G. If G is a graph with n vertices, then μ1 (G) ≥ μ2 (G) ≥ · · · ≥ μn (G) denote the signless Laplacian spectrum of G. Two graphs are said to be Q-cospectral if they have the same signless Laplacian spectrum. A graph G is said to be Q − DS if there is no other non-isomorphic graph H such that G and H are Q-cospectral. In this paper, we prove that “Let G be a graph with n ≥ 12 vertices and 1 ≤ μn (G) ≤ μ2 (G) ≤ 5 < n < μ1 (G) or let G be a graph with n ≥ 10 vertices and 0 < μn (G) ≤ μ2 (G) ≤ 4 < n < μ1 (G). If G and H are Q-cospectral, then G and H share the same degree sequences and τ (G) = τ (H)”. As a consequence of our results, we show that all multi-wheel graphs K1 ∨ (Cq1 ∪ Cq2 ∪ · · · ∪ Cqt ) are Q − DS. ∗
Partially supported by NSFC project No. 11571123, No. 11271288, the Training Program for Out-
standing Young Teachers in University of Guangdong Province (No. YQ2015027) and China Scholarship Council. Email: [email protected] (M. Liu), shan [email protected] (H. Shan, Corresponding author)
1
Key words and phrases: Signless Laplacian spectrum, Join graph, Determined by spectrum; Multi-wheel graph AMS 2000 Subject Classifications: 05C50; 15A18; 15A36
1
Introduction
Throughout this paper, G = (V, E) is an undirected simple graph with n vertices, and G denotes the complement graph of G. Let dG (u) and NG (u) be the degree and neighbor set of vertex u in G, respectively. If dG (u) = 1, then we call u a pendent vertex. In the sequel, we enumerate the degrees of G in non-increasing order, i.e., d1 (G) ≥ d2 (G) ≥ · · · ≥ dn (G), where dG (vi ) = di (G) for i ∈ {1, 2, ..., n}. As usual, K1,n−1 , Kn and Cn denote the star, complete graph and cycle with n vertices, respectively, and G1 ∨G2 denotes the join graph of two vertex disjoint graphs G1 and G2 . In other words, G1 ∨ G2 is the graph having vertex set V (G1 ∨ G2 ) = V (G1 ) ∪ V (G2 ) and edge set E(G1 ∨ G2 ) = E(G1 ) ∪ E(G2 ) ∪ {uv : u ∈ V (G1 ), v ∈ V (G2 )}. Let A(G) and D(G) be the adjacency matrix and the diagonal degree matrix of G, respectively. The Laplacian matrix of G is L(G) = D(G) − A(G), and the signless Laplacian matrix of G is Q(G) = D(G) + A(G). Let SQ (G) and SL (G) denote the set of eigenvalues of Q(G) and L(G), respectively. It is easy to see that Q(G) is positive semidefinite and hence its eigenvalues can be arranged as [1]: μ1 (G) ≥ μ2 (G) ≥ · · · ≥ μn (G) ≥ 0. If there is no risk of confusion, we simplify dG (u), di (G), NG (u) and μi (G) as d(u), di , N (u) and μi , respectively. Two graphs are said to be Q-cospectral (respectively, L-cospectral) if they have the same signless Laplacian (respectively, Laplacian) spectrum. Similarly, two graphs are said to be co-degree if they have the same degree sequences. A graph G is said to be determined by its signless Laplacian spectrum, simplified as Q − DS (respectively, Laplacian spectrum, simplified as L − DS) if there does not exist other non-isomorphic graph H such that H and G are Q-cospectral (respectively, L-cospectral). Which graphs are determined by their spectra? This problem was posed by G¨ unthard 2
Figure 1.1: The trees T1 and T2 . and Primas [6] more than 50 years ago in the context of H¨ uckel’s theory in chemistry. Only in the most recent years have mathematicians devoted their attention to this problem and more and more papers focus on this item are now appearing. For detail, we refer the readers to [2, 3, 5, 8, 15] and the references therein. Among these literatures on this item, a great ratios of results are on L − DS, the main reason is that SL (G) = SL (H) if and only if SL (G) = SL (H) (See [2, 11]). Thus, the L − DS problem of sparse graphs can be transferred to the L − DS problem of dense graphs, and vise versa. Different from L − DS problem, there exist graphs G and H such that SQ (G) = SQ (H), but SQ (G) = SQ (H). For instance, let T1 and T2 be two trees as shown in Figure 1.1. It is easy to check that SQ (T1 ) = SQ (T2 ), but SQ (T1 ) = SQ (T2 ). In view of this, as the experience to the authors, when we want to prove that G is Q − DS, we always first prove that H and G are co-degree under the supposition of G and H being Q-cospectral, and then prove that G and H are isomorphic, but the first step is not an easy task, and we always have to spend a large amount of work to solve it [5, 8, 10, 11, 16]. Thus, it is natural for us to consider the following problem. Problem 1.1 Which graphs satisfy the property “every Q-cospectral graph of G is also co-degree with G”? In this paper, we will give partial answer to Problem 1.1 by proving Theorem 1.1 Let G be a graph with n ≥ 12 vertices and 1 ≤ μn (G) ≤ μ2 (G) ≤ 5 < n < μ1 (G). If SQ (G) = SQ (H), then G and H are co-degree. Theorem 1.2 Let G be a graph with n ≥ 10 vertices and 0 < μn (G) ≤ μ2 (G) ≤ 4 < n < μ1 (G). If SQ (G) = SQ (H), then G and H are co-degree. If G is a graph with n vertices, then we use Φ(G, x)=det(xIn − Q(G)) to define the signless Laplacian characteristic polynomial of G, where In is the identity matrix of order 3
Figure 1.2: The graphs F1 (n) and F2 (n) on n vertices. n. It is not difficult for the readers to find that Theorems 1.1 and 1.2 are only suitable for small graph classes. But the following example show that we cannot decrease the condition “μn (G) ≥ 1” from Theorem 1.1 or the condition “μ2 (G) ≤ 4” from Theorem 1.2, respectively. Example 1.1 Let F1 (n) and F2 (n) be the graphs on n ≥ 8 vertices as shown in Figure 1.2. By an elementary calculation (or see [9]), we have Φ(F1 (n), x) = Φ(F2 (n), x) = (x − 1)n−5 (x − 2)2 f (x),
(1.1)
where f (x) = x3 − (n + 5)x2 + 5nx − 12. When n ≥ 8, since f (0) = −12 < 0, f (1) = 4(n − 4) > 0, f (4) = 4(n − 7) > 0, f (5) = f (n) = −12 < 0 and f (n + 1) = n(n − 3) − 16 > 0, by (1.1) we can conclude that SQ (F1 (n)) = SQ (F2 (n)) with 0 < μn (F1 (n)) = μn (F2 (n)) < 1, 4 < μ2 (F1 (n)) = μ2 (F2 (n)) < 5 and n < μ1 (F1 (n)) = μ1 (F2 (n)) < n + 1. It is easy to see that F1 (n) and F2 (n) are not co-degree. Let G be a graph such that every Q-cospectral graph of G is also co-degree with G. If the degree sequence of G is π and the graph with π as its degree sequence is unique, then G and H are isomorphic, in other word, G is Q − DS. From this fact and Theorem 1.2, we can easily deduce the following conclusion, which is the main result of [10]. Corollary 1.1 [10] If n ≥ 10, then S(n, q) is Q − DS for any n ≥ 2q + 1 and q ≥ 0, where S(n, q) ∼ = K1 ∨ (qK2 ∪ (n − 2q − 1)K1 ). The wheel graph is denoted by Wn = K1 ∨ Cn−1 and the multi-wheel graph is denoted by K1 ∨ (Cq1 ∪ Cq2 ∪ · · · ∪ Cqt ). It was proved that Wn is L − DS except for W7 [15], and all multi-wheel graphs with no W7 as its induced subgraphs are also L − DS [2]. In [8], it was proved that Wn is Q − DS. As a consequence of Theorem 1.1, we will show that 4
Theorem 1.3 For any t ≥ 1, K1 ∨ (Cq1 ∪ Cq2 ∪ · · · ∪ Cqt ) is Q − DS.
2
The proof of Theorem 1.1
Let diag{b1 , b2 , . . . , bn } be a diagonal matrix with the i-th diagonal entry being bi , where i = 1, 2, . . . , n. Let θ1 (M ) ≥ θ2 (M ) ≥ · · · ≥ θn (M ) be the eigenvalues of M , where M is a symmetric matrix of order n. Lemma 2.1 [7] Let M be a symmetric matrix of order n. If N is a principal submatrix of order s of M , then θi (M ) ≥ θi (N ) ≥ θn−s+i (M ) holds for any i = 1, 2, . . . , s. From Lemma 2.1, Liu et al. [12] proved the following result, which is crucial to our proofs. Lemma 2.2 [12] Let G be a graph with X ⊆ V (G) and let H ∼ = G[X]. If X = {x1 , x2 , . . . , xs }, then for any i ∈ {1, 2, . . . , s} and 0 ≤ pi ≤ dG (xi ), we have μi (G) ≥ θi (A(H) + diag{p1 , p2 , . . . , ps }) . The following two lemmas present lower or upper bounds for μ1 (G) and μ2 (G). Lemma 2.3 [4] For any graph G with at least two vertices, we have 1 μ2 (G) ≥ d1 (G) + d2 (G) − (d1 (G) − d2 (G))2 + 4 ≥ d2 (G) − 1, 2 where μ2 (G) = d2 (G) − 1 implies that d1 (G) = d2 (G) and v1 is adjacent with v2 . Let uv be an edge of G and v be a vertex of G. Let m(v) denote the average of the d(w)/d(v). Denote by degrees of the vertices being adjacent to v, i.e., m(v) = w∈N (v)
Ψ(uv) =
d(u)(d(u) + m(u)) + d(v)(d(v) + m(v)) . d(u) + d(v)
According to the proof of Theorem 2.10 in [13], we have Lemma 2.4 [13] If G is a connected graph with at least one edge, then μ1 (G) ≤ max{Ψ(uv) : uv ∈ E(G)} ≤ max{d(v) + m(v) : v ∈ V (G)} ≤ d1 (G) + d2 (G), where the first equality holds if and only if G is regular or bipartite semiregular. 5
Throughout this paper, if every component of G contains at least one edge, then we agree that max{d(v) + m(v) : v ∈ V (G)} occurs at the vertex v0 and Ψ(x0 y0 ) = max {Ψ(xy) : xy ∈ E(G)}, where d(x0 ) ≥ d(y0 ). Furthermore, we suppose that d(y0 ) = d in the following. Corollary 2.1 Let G be a graph with n vertices and μ1 (G) > n. If every component of G contains at least one edge, then we have n − d1 (G) + 1 ≤ d(x0 ) ≤ d1 (G), n − d1 (G) + 1 ≤ d(v0 ) ≤ d1 (G) and m(v0 ) ≤
2|E(G)| − dn (G)(n − 1) 2|E(G)| − (n − 1) + dn (G) − 1 ≤ . d(v0 ) d(v0 )
(2.1)
Proof. If d(y0 ) ≤ d(x0 ) ≤ n − d1 (G), by Lemma 2.4 we have μ1 (G) ≤ Ψ(x0 y0 ) ≤ d(x0 ) + d1 (G) ≤ n, a contradiction. Thus, n − d1 (G) + 1 ≤ d(x0 ) ≤ d1 (G). If d(v0 ) ≤ n − d1 (G), by Lemma 2.4, μ1 (G) ≤ d(v0 ) + m(v0 ) ≤ d(v0 ) + d1 (G) ≤ n, a contradiction. Thus, n − d1 (G) + 1 ≤ d(v0 ) ≤ d1 (G). Now, since u∈N (v0 ) d(u) + d(v0 ) + dn (G)(n − d(v0 ) − 1) ≤ u∈V (G) d(u) = 2|E(G)|, we have u∈N (v0 ) d(u) ≤ 2|E(G)| − dn (G)(n − 1) + (dn (G) − 1)d(v0 ), and hence (2.1) holds. Lemma 2.5 [1, 4] If G is a connected graph with n ≥ 2 vertices, then 0 ≤ μn (G) < dn (G) < d1 (G) + 1 ≤ μ1 (G), where μn (G) = 0 if and only if G is bipartite, and μ1 (G) = d1 (G) + 1 if and only if G∼ = K1,n−1 . As usual, G + e denotes the graph obtained from G by adding one edge e ∈ E(G), while G − e denotes the graph obtained from G by deleting one edge e ∈ E(G). Lemma 2.6 Let G be a graph with d(u1 ) ≥ 8 and d(u2 ) ≥ 5. If either u1 u2 ∈ E(G) with |N (u1 ) ∩ N (u2 )| ≥ 2 or u1 u2 ∈ E(G) with |N (u1 ) ∩ N (u2 )| ≥ 1, then μ2 (G) > 5. Proof. Let X = {u1 , u2 , x, y} and let Y = {u1 , u2 , x, z}. Let ⎛ ⎛ ⎛ ⎞ ⎞ 8 1 1 0 8 1 1 0 ⎜ ⎜ ⎜ ⎟ ⎟ ⎜ 1 2 0 1 ⎟ ⎜ 1 3 1 1 ⎟ ⎜ ⎜ ⎜ ⎜ ⎟ ⎟ B1 = ⎜ ⎟ , B2 = ⎜ ⎟ , B3 = ⎜ ⎜ 1 0 2 1 ⎟ ⎜ 1 1 3 1 ⎟ ⎜ ⎝ ⎝ ⎝ ⎠ ⎠ 0 1 1 5 0 1 1 5 6
8 1 1 0
⎞
⎟ 1 5 1 1 ⎟ ⎟ ⎟, 1 1 2 0 ⎟ ⎠ 0 1 0 1
⎛ ⎛
8 1 ⎜ ⎜ 1 5 ⎜ B4 = ⎜ ⎜ 1 1 ⎝ 0 1
1 0
⎞
⎛
8 ⎜ ⎟ ⎜ 1 1 1 ⎟ ⎜ ⎟ ⎟ , B5 = ⎜ ⎜ 1 3 1 ⎟ ⎝ ⎠ 1 2 1
1 5 1 1
⎜ ⎜ 1 1 ⎜ ⎜ ⎟ ⎜ 1 1 ⎟ ⎜ ⎟ ⎟ , B6 = ⎜ ⎜ 3 1 ⎟ ⎜ ⎠ ⎜ ⎜ 1 3 ⎝ ⎞
⎞ 8 1 1 1 1 1 1 5 1 1 1 1 1 1 2 0 0 0 1 1 0 2 0 0 1 1 0 0 2 0
⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎟ ⎟ ⎠
1 1 0 0 0 2 We firstly consider the case of u1 u2 ∈ E(G). Since |N (u1 ) ∩ N (u2 )| ≥ 2, there must exist vertices {x, y} ⊆ N (u1 ) ∩ N (u2 ) such that either G[X] ∼ = C4 or G[X] ∼ = K4 − e. If G[X] ∼ = K4 − e, by Lemma = C4 , by Lemma 2.2 we have μ2 (G) ≥ θ2 (B1 ) > 5.46 . If G[X] ∼ 2.2 we have μ2 (G) ≥ θ2 (B2 ) > 5.75. We secondly consider the case of u1 u2 ∈ E(G). Recall that N (u1 ) ∩ N (u2 ) = Ø. When there exists some vertex z ∈ N (u2 ) \ {N (u1 ) ∪ {u1 }}, either G[Y ] ∼ = K1,3 + e or G[Y ] ∼ = K1,3 + e, by Lemma 2.2 we have = K4 − e, where x ∈ N (u1 ) ∩ N (u2 ). If G[Y ] ∼ μ2 (G) ≥ θ2 (B3 ) > 5.03. If G[Y ] ∼ = K4 − e, by Lemma 2.2 we have μ2 (G) ≥ θ2 (B4 ) > 5.3. When N (u2 ) ∪ {u2 } ⊆ N (u1 ) ∪ {u1 }, we choose {x, y} ⊆ N (u1 ) ∩ N (u2 ). Then, either G[X] ∼ = K4 − e. If G[X] ∼ = K4 , by Lemma 2.2 we have μ2 (G) ≥ θ2 (B5 ) > = K4 or G[X] ∼ 5.12. Otherwise, G[X] ∼ = K4 − e holds for any {x, y} ⊆ N (u1 ) ∩ N (u2 ). By Lemma 2.2, it follows that μ2 (G) ≥ θ2 (B6 ) > 5.05. Lemma 2.7 If G is a graph with n ≥ 12 vertices and 0 < μn (G) ≤ μ2 (G) ≤ 5 < n < μ1 (G), then G is connected. Proof. On conversely, assume that G is disconnected. By Lemma 2.3, d2 (G) ≤ 6. We first suppose that d2 (G) = 6. By Lemma 2.3 it follows that d1 (G) = 6 and μ2 (G) = 5. Furthermore, Lemma 2.4 implies that n < μ1 (G) ≤ d1 (G) + d2 (G) ≤ 12. Thus, we have n ≤ 11, a contradiction. Now, we suppose that d2 (G) ≤ 5. By Lemma 2.4 we have n < μ1 (G) ≤ d1 (G) + d2 (G), which implies that n − 4 ≤ d1 (G) ≤ n − 2. From Lemma 2.5, we can conclude that G∼ = G1 ∪ C3 , where d1 (G1 ) = n − 4 ≥ d2 (G1 ) = d2 (G) = 5. Since |V (G1 )| = n − 3, d1 (G1 ) = n − 4 ≥ 8 and d2 (G1 ) = d2 (G) = 5, we can conclude that v1 v2 ∈ E(G1 ) with |NG1 (v1 ) ∩ NG1 (v2 )| ≥ 1. By Lemma 2.6, μ2 (G) = μ2 (G1 ) > 5, a contradiction. 7
Theorem 2.1 If G is a graph with n ≥ 12 vertices and 0 < μn (G) ≤ μ2 (G) ≤ 5 < n < μ1 (G), then G is connected with d2 (G) ≤ 4 and d1 (G) ≥ n−3. Furthermore, if G contains no pendent vertices, then d1 (G) ≥ n − 2. Proof. In the proof of this result, we simplify μi (G) and di (G) as μi and di , respectively. By Lemmas 2.3 and 2.7, G is connected with d2 ≤ 6. If d2 = 6, then d1 = 6 and μ2 = 5 follow from Lemma 2.3. Furthermore, Lemma 2.4 implies that n < μ1 ≤ d1 + d2 ≤ 12, and hence n ≤ 11, a contradiction. Thus, d2 ≤ 5. To complete the proof, it suffices to prove the following two claims. Claim 1. d2 ≤ 4 and d1 ≥ n − 3. Proof of Claim 1. By Lemma 2.4, it suffices to show that d2 ≤ 4. On conversely, suppose that d2 = 5. Then, Lemma 2.4 implies that d1 ≥ n − 4. Since d1 ≥ n − 4 ≥ 8 and d2 = 5, either v1 v2 ∈ E(G) with |N (v1 )∩N (v2 )| ≥ 2 or v1 v2 ∈ E(G) with |N (v1 )∩N (v2 )| ≥ 1. By Lemma 2.6, μ2 > 5, a contradiction. Claim 2. If dn (G) ≥ 2, then d1 (G) ≥ n − 2. Proof of Claim 2. By Claim 1, we suppose that d1 = n − 3. Thus, Claim 1 and Lemma 2.4 imply that d2 = 4 < n − 3 = d1 . By Corollary 2.1, d(x0 ) ∈ {4, n − 3}. If d(x0 ) = 4 = d2 , since d1 = n − 3 and 2 ≤ d = d(y0 ) ≤ 4, by Lemma 2.4 we have μ1 ≤ Ψ(x0 y0 ) ≤
42 + n − 3 + d + 4 × 2 + d2 + n − 3 + 4 × (d − 1) ≤ n, 4+d
a contradiction. Thus, d(x0 ) = n − 3 = d1 . Now, since 2 ≤ d = d(y0 ) ≤ 4 and n ≥ 12, by Lemma 2.4 we have μ1 ≤ Ψ(x0 y0 ) ≤
(n − 3)2 + 4(n − 4) + d + d2 + n − 3 + 4 × (d − 1) ≤ n, n−3+d
a contradiction. Let τ (G) denote the number of triangles of G, and let nk (G) be the number of vertices with degree k in V (G) \ {v1 }, where k ∈ {1, 2, . . . , d2 (G)}. Lemma 2.8 [1] If G is a graph with n vertices, m edges, and τ triangles, then n i=1
μi =
n i=1
di = 2m,
n i=1
μ2i = 2m +
n i=1
8
d2i , and
n i=1
μ3i = 6τ + 3
n i=1
d2i +
n i=1
d3i .
Remark 2.1 If G and H are Q-cospectral and co-degree, by Lemma 2.8 we have τ (G) = τ (H). Thus, τ (G) = τ (H) holds in Theorems 1.1 and 1.2. Proof of Theorem 1.1. By Lemma 2.5 and Theorem 2.1, we have dn (H) ≥ 2 and dn (G) ≥ 2. Furthermore, Theorem 2.1 implies that d2 (G) ≤ 4 < n − 2 ≤ d1 (G) and d2 (H) ≤ 4 < n − 2 ≤ d1 (H). In the following, we write ni (G) and d1 (G) as ni and d1 , and we write ni (H) and d1 (H) as n∗i and d∗1 , respectively, where i ∈ {2, 3, 4}. From Lemma 2.8, we have ⎧ ∗ ∗ ∗ ⎪ ⎪ ⎨ n2 + n3 + n4 = n2 + n3 + n4 ⎪ ⎪ ⎩
2n2 + 3n3 + 4n4 + d1 = 2n∗2 + 3n∗3 + 4n∗4 + d∗1
(2.2)
4n2 + 9n3 + 16n4 + (d1 )2 = 4n∗2 + 9n∗3 + 16n∗4 + (d∗1 )2
If d1 = d∗1 , then ni = n∗i holds for any i ∈ {2, 3, 4} by (2.2). If d1 = d∗1 , by symmetry we may suppose that d1 = n−2 and d∗1 = n−1. Now, from (2.2) we have n∗3 = n3 +2n−9 > n, a contradiction. Thus, G and H are co-degree.
3
The proofs of Theorem 1.2 and Corollary 1.1
Similarly with Lemma 2.7, we can easily obtain the following result. Lemma 3.1 If G is a graph with n ≥ 10 vertices and 0 < μn (G) ≤ μ2 (G) ≤ 4 < n < μ1 (G), then G is connected. Theorem 3.1 If G is a graph with n ≥ 10 vertices and 0 < μn (G) ≤ μ2 (G) ≤ 4 < n < μ1 (G), then G is connected with d2 (G) ≤ 3 and d1 (G) ≥ n − 2. Proof. In the proof of this result, we simplify μi (G) and di (G) as μi and di , respectively. By Lemmas 2.3 and 3.1, G is connected with d2 ≤ 5. If d2 = 5, then d1 = 5 and μ2 = 4 by Lemma 2.3. Furthermore, Lemma 2.4 implies that n < μ1 ≤ d1 + d2 ≤ 10, a contradiction. Thus, d2 ≤ 4. By Lemma 2.4, we have d1 ≥ n − 3. By Lemma 2.4, to complete the proof, it suffices to prove that d2 ≤ 3. On conversely, suppose that d2 = 4.
9
Let X = {v1 , v2 , x} and let Y = {v1 , v2 , x, y}. ⎛ ⎛ ⎞ 7 1 0 ⎜ 7 1 0 ⎜ 1 4 1 ⎜ ⎟ ⎜ ⎟ D1 = ⎜ = , D ⎜ 2 ⎝ 1 2 1 ⎠ ⎜ 0 1 1 ⎝ 0 1 4 1 1 0 ⎛
7 1 ⎜ ⎜ 1 4 ⎜ D4 = ⎜ ⎜ 1 1 ⎝ 1 1
1 1 3 1
Let ⎞ 1 ⎟ 1 ⎟ ⎟ ⎟, 0 ⎟ ⎠ 2 ⎛ ⎞ 7 ⎜ 1 ⎜ 1 ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎟ , D5 = ⎜ ⎜ 1 1 ⎟ ⎜ ⎠ ⎜ 1 ⎝ 3 1
⎛
7 1 0 1
⎞
⎜ ⎟ ⎜ 1 4 1 1 ⎟ ⎜ ⎟ D3 = ⎜ ⎟, ⎜ 0 1 2 1 ⎟ ⎝ ⎠ 1 1 1 3 ⎞ 1 1 1 1 ⎟ 4 1 1 1 ⎟ ⎟ ⎟ 1 2 0 0 ⎟ ⎟. ⎟ 1 0 2 0 ⎟ ⎠ 1 0 0 2
We first consider the case of v1 v2 ∈ E(G). Since d2 = 4 and d1 ≥ n − 3, there must exist vertex {x} ⊆ N (v1 ) ∩ N (v2 ) such that G[X] ∼ = K1,2 . Since d1 ≥ n − 3 ≥ 7, by Lemma 2.2 we have μ2 ≥ θ2 (D1 ) > 4.36, a contradiction. Now, we consider the case of v1 v2 ∈ E(G). Then, N (v1 ) ∩ N (v2 ) = Ø. We claim that N (v2 ) ∪ {v2 } ⊆ N (v1 ) ∪ {v1 }. Otherwise, assume that there exists some vertex y ∈ N (v2 ) \ {N (v1 ) ∪ {v1 }}. Then, either G[Y ] ∼ = K1,3 + e or G[Y ] ∼ = K4 − e, where x ∈ N (v1 ) ∩ N (v2 ). If G[Y ] ∼ = K1,3 + e, by Lemma 2.2 we have μ2 ≥ θ2 (D2 ) > 4.13, a contradiction. If G[Y ] ∼ = K4 − e, by Lemma 2.2 we have μ2 ≥ θ2 (D3 ) > 4.57, a contradiction. Thus, N (v2 ) ∪ {v2 } ⊆ N (v1 ) ∪ {v1 }. We choose two vertices {x, y} ⊆ N (v1 ) ∩ N (v2 ). If G[Y ] ∼ = K4 , since d1 ≥ n − 3 ≥ 7, by Lemma 2.2 we have μ2 ≥ θ2 (D3 ) > 4.33, a contradiction. Thus, G[Y ] ∼ = K4 − e holds for any Y with {x, y} ⊆ N (v1 ) ∩ N (v2 ). By Lemma 2.2 and d1 ≥ n − 3 ≥ 7, we have μ2 ≥ θ2 (D5 ) > 4.07, a contradiction. Proof of Theorem 1.2. By Theorem 3.1, d2 (G) ≤ 3 < n − 2 ≤ d1 (G) and d2 (H) ≤ 3 < n − 2 ≤ d1 (H). In the following, we simplify ni (G) and d1 (G) as ni and d1 , and we rewrite ni (H) and d1 (H) as n∗i and d∗1 , respectively, where i ∈ {1, 2, 3}. By Lemma 2.8, we have
⎧ ∗ ∗ ∗ ⎪ ⎪ ⎨ n1 + n2 + n3 = n1 + n2 + n3 n1 + 2n2 + 3n3 + d1 = n∗1 + 2n∗2 + 3n∗3 + d∗1 ⎪ ⎪ ⎩ n1 + 4n2 + 9n3 + (d1 )2 = n∗1 + 4n∗2 + 9n∗3 + (d∗1 )2 10
(3.1)
If d1 = d∗1 , then ni = n∗i holds for any i ∈ {1, 2, 3} by (3.1). If d1 = d∗1 , by symmetry we may suppose that d1 = n−2 and d∗1 = n−1. Now, from (3.1) we have n∗2 = n2 +2n−7 > n, a contradiction. Thus, G and H are co-degree. Proof of Corollary 1.1. Suppose that G and S(n, q) are Q-cospectral. If q = 0, then S(n, q) ∼ = K1,n−1 . Since S(n, q) contains In−1 as its principal submatrix, by Lemma 2.1, we have μ2 (G) = μ2 (S(n, q)) ≤ θ1 (In−1 ) = 1, and hence d2 (G) ≤ 2 by Lemma 2.3. If d2 (G) = 2, then d1 (G) = 2 by Lemma 2.3. From Lemmas 2.4 and 2.5, we have μ1 (G) ≤ 4 < n = μ1 (K1,n−1 ), a contradiction. Thus, d2 (G) = 1. By Lemma 2.4 and μ1 (K1,n−1 ) = n, we have d1 (G) = n − 1 and hence G ∼ = K1,n−1 . If q ≥ 1, then S(n, q) is non-bipartite. By Lemma 2.5, μ1 (G) = μ1 (S(n, q)) > d1 (S(n, q)) + 1 = n and μn (G) = μn (S(n, q)) > 0. Let G∗ ∼ = qK2 ∪ (n − 2q − 1)K1 . Note that S(n, q) contains In−1 + Q(G∗ ) as its principal submatrix. Thus, by Lemma 2.1, we have μ2 (G) = μ2 (S(n, q)) ≤ θ1 (In−1 + Q(G∗ )) = 3. Now, from Theorem 1.2, we can conclude that G and S(n, q) are co-degree, which implies that G ∼ = S(n, q).
4
The proof of Theorem 1.3
For any n ≥ 3, it is well-known that (See [2]) 2kπ SQ (Cn ) = 2 + 2 cos : k = 0, 1, . . . , n − 1 . n
(4.1)
Let q (p) define p copies of q, where q and p are two non-negative integers. Lemma 4.1 If t ≥ 1, then SQ (K1 ∨ (Cq1 ∪ Cq2 ∪ · · · ∪ Cqt )) is equal to 2k π 1 i (t−1) n + 4 ± (n − 4)2 + 16 , 5 : 1 ≤ ki ≤ qi − 1, 1 ≤ i ≤ t . , 3 + 2 cos 2 qi Proof. In the proof of this result, we write K1 ∨(Cq1 ∪Cq2 ∪· · ·∪Cqt ) and Cq1 ∪Cq2 ∪· · ·∪Cqt as G and G∗ , respectively. By equation (4.1), θk (In−1 + Q(G∗ )) = 1 + μk (G∗ ) = 5 holds for 1 ≤ k ≤ t. Note that G contains In−1 + Q(G∗ ) as its principal submatrix. Thus, by Lemma 2.1, we have 5 = θt (In−1 + Q(G∗ )) ≤ μt (G) ≤ μt−1 (G) ≤ · · · ≤ μ2 (G) ≤ θ1 (In−1 + Q(G∗ )) = 5. 11
For 1 ≤ i ≤ t, we suppose that V (Cqi ) = {ui1 , ui2 , . . . , uiqi } and θ = 2 is an eigenvalue of A(Cqi ). Since Cqi is regular, we may choose ϕ = (ϕ(ui1 ), ϕ(ui2 ), . . . , ϕ(uiqi ))T as the unit i ϕ(uij ) = 0. Let ψ = (ψ(u1 ), ψ(u2 ), . . . , ψ(un ))T eigenvector corresponding to θ such that qj=1 be the unit eigenvector such that ψ(v) = ϕ(v) holds for v ∈ V (Cqi ) and ψ(v) = 0 holds for v ∈ V (G) \ V (Cqi ). Since Iqi + Q(Cqi ) = 3Iqi + A(Cqi ) is a principal submatrix of Q(G), ψ is an eigenvector of Q(G) corresponding to the eigenvalue 3 + θ. By equation (4.1), {3 + 2 cos( 2kqii π ) : i = 1, 2, . . . , qi − 1} are the eigenvalues of Q(G), where 1 ≤ i ≤ t. Set μ = 12 (n + 4 ± (n − 4)2 + 16 ). It is easy to check that Q(G)ψ1 = μψ1 , where ψ1 = (μ − 5, 1(n−1) ) and μ − 5 is corresponding to the vertex v1 of G. Since μ = 5 and ψ1 is orthogonal with ψ, the result follows. Lemma 4.2 Let G∗ ∼ = K1 ∨ (Cq1 ∪ Cq2 ∪ · · · ∪ Cqt ) with t ≥ 2. If 7 ≤ |V (G∗ )| ≤ 11 and G and G∗ are Q-cospectral, then G and G∗ are co-degree. Proof. We may suppose that |V (G∗ )| = n. In the proof of this result, we write di (G) and μi (G) as di and μi , respectively. By Lemma 2.5 and Lemma 4.1, we have dn ≥ 2 and hence every component of G contains at least one edge. If d1 ≤ n − 3, since 7 ≤ n ≤ 11 and |E(G)| = |E(G∗ )| = 2(n − 1) by Lemma 2.8, by Corollary 2.1 and (2.1) we have 4 ≤ d(v0 ) ≤ n − 3 and hence 2(n − 1) d(v0 ) + m(v0 ) ≤ d(v0 ) + 1 + d(v0 ) 2(n − 1) 2(n − 1) ≤ max 5 + ,n − 2 + 4 n−3 1 ≤ n + 4 + (n − 4)2 + 16 . 2 1 2
(4.2)
Recall that 7 ≤ n ≤ 11. By Lemma 4.1 and (4.2), we have μ1 = d(v0 ) + m(v0 ) = 2 n + 4 + (n − 4) + 16 , which implies that n = 7, μ1 = 8 and μ2 = 5. Furthermore,
μ2 = 5, dn ≥ 2 and Lemmas 2.4 and 4.1 imply that G is a connected regular graph with d1 = 4. It is well known that every Q-cospectral graph with a regular graph is also regular (See Proposition 2 or Lemma 4 of [2]), a contradiction. Thus, d1 ≥ n − 2 and hence G is connected (otherwise, un = 0, a contradiction). Since μ2 = μ2 (G∗ ) = 5 by Lemma 4.1, we have d2 ≤ 6 by Lemma 2.3. 12
If d2 = 6, by Lemma 2.3, we have 6 = d1 ≥ n − 2 and hence 7 ≤ n ≤ 8. When 7 ≤ n ≤ 8, from Lemma 2.8, there is no graph with n vertices such that d1 = d2 = 6 and dn ≥ 2. Thus, d2 ≤ 5. Now, from Lemma 2.8, we have ⎧ ⎪ =n−1 ⎪ ⎨ n2 + n3 + n4 + n5 ⎪ ⎪ ⎩
= 4(n − 1) − d1
2n2 + 3n3 + 4n4 + 5n5
(4.3)
4n2 + 9n3 + 16n4 + 25n5 = 9(n − 1) + (n − 1)2 − d21
If d1 = n − 1, from (4.3), it is easy to check that G and G∗ are co-degree. In the following, we consider the case of d1 = n − 2. We firstly consider the case of 10 ≤ n ≤ 11. If n5 = 0, by (4.3) we have n3 = 8−n < 0, a contradiction. Otherwise, n5 ≥ 1 and hence d2 = 5. Since d1 = n − 2 ≥ 8 and d2 = 5, either v1 v2 ∈ E(G) with |N (v1 ) ∩ N (v2 )| ≥ 2 or v1 v2 ∈ E(G) with |N (v1 ) ∩ N (v2 )| ≥ 1. By Lemma 2.6, μ2 > 5, a contradiction. Now, we consider the case of 7 ≤ n ≤ 9. By solving systems of linear diophantine equations (4.3) for 7 ≤ n ≤ 9, we get the following non-negative integral solutions: (n, d1 ) n 2 n3 n4 n5
(7, 5) 1 2 4 1 , 0 3 1
0
(8, 6) 2 3 3 0 , 1 4 1
(9, 7) 3 2 2 1
0
With the aid of mathematics software “SageMath” [14], there are exactly 306 connected graphs with degree sequences corresponding to these solutions of (4.3), but none of them satisfies the property “μ2 = 5 and μn ≥ 1”, which contradicts Lemma 4.1. Proof of Theorem 1.3. In the proof of this result, we write K1 ∨ (Cq1 ∪ Cq2 ∪ · · · ∪ Cqt ) as G∗ . Suppose that SQ (G) = SQ (G∗ ). Since the case of t = 1 was already proved in [8], we may suppose that t ≥ 2, and hence n ≥ 7 in the following. By Lemma 4.1, we can conclude that 1 ≤ μn (G) ≤ μ2 (G) = 5 <
1 n + 4 + (n − 4)2 + 16 = μ1 (G). 2
If 7 ≤ n ≤ 11, by Lemma 4.2, G and G∗ are co-degree. If n ≥ 12, by Theorem 1.1, G and G∗ are co-degree. Thus, G ∼ = K1 ∨ (Ck ∪ Ck ∪ · · · ∪ Ck ). Now, from Lemma 4.1, we 1
2
can conclude that G ∼ = G∗ . 13
s
Acknowledgement. The authors are much grateful to the anonymous referee for his/her valuable comments on our paper, which have considerably improved the presentation of this paper.
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