On the sum of the two largest Laplacian eigenvalues of unicyclic graphs
- Yirong Zheng^{1, 2}Email author,
- An Chang^{2} and
- Jianxi Li^{2, 3}
https://doi.org/10.1186/s13660-015-0794-1
© Zheng et al. 2015
Received: 31 March 2015
Accepted: 23 August 2015
Published: 17 September 2015
Abstract
Let G be a simple graph and \(S_{2}(G)\) be the sum of the two largest Laplacian eigenvalues of G. Guan et al. (J. Inequal. Appl. 2014:242, 2014) determined the largest value for \(S_{2}(T)\) among all trees of order n. They also conjectured that among all connected graphs of order n with m (\(n \leq m\leq2n -3\)) edges, \(G_{m,n}\) is the unique graph which has maximal value of \(S_{2}(G)\), where \(G_{m,n}\) is a graph of order n with m edges which has \(m-n+1\) triangles with a common edge and \(2n-m-3\) pendent edges incident with one end vertex of the common edge. In this paper, we confirm the conjecture with \(m=n\).
Keywords
MSC
1 Introduction
Let \(G=(V,E)\) be a simple connected graph with vertex set \(V(G)\) and edge set \(E(G)\). Its order is \(|V(G)|\), denoted by \(n(G)\) (or n for short), and its size is \(|E(G)|\), denoted by \(m(G)\) (or m for short). For a vertex \(v \in V(G)\), let \(N(v)\) be the set of all neighbors of v in G and let \(d(v)=|N(v)|\) be the degree of v. Particularly, denote by \(\Delta(G)\) the maximum degree of G. A pendent vertex is a vertex with degree one. The diameter of a connected graph G, denoted by \(d(G)\), is the maximum distance among all pairs of vertices in G. Let \(S_{n}\) and \(P_{n}\) be the star and the path of order n, respectively. Let \(S^{k}_{a,b}\) be the tree of order n obtained from two stars \(S_{a+1}\), \(S_{b+1}\) by joining a path of length k between their central vertices (see Figure 2). For all other notions and definitions, not given here, see, for example, [1] or [2] (for graph spectra).
Let \(A(G)\) and \(D(G)\) be the adjacency matrix and the diagonal matrix of vertex degrees of G, respectively. The matrix \(L(G)=D(G)-A(G)\) is called Laplacian matrix of G. The Laplacian matrix is an important topic in the theory of graph spectra. We use the notation \(I_{n}\) for the identity matrix of order n and denote by \(\phi(G, x) = \det(xI_{n}-L(G))\) the Laplacian characteristic polynomial of G. It is well known that \(L(G)\) is positive semidefinite symmetric and singular. Denote its eigenvalues by \(\mu_{1}(G) \geq\mu_{2}(G)\geq\cdots\geq \mu_{n}(G)\) (or simply \(\mu_{1} \geq\mu_{2} \geq\cdots\geq \mu_{n}\) sometimes for convenience) which are always enumerated in non-increasing order and repeated according to their multiplicity. Note that each row sum of \(L(G)\) is 0 and, therefore, \(\mu_{n}(G)=0\). Fiedler [3] showed that the second smallest eigenvalue \(\mu_{n-1}(G)\) of \(L(G)\) is 0 if and only if G is disconnected. Thus the second smallest eigenvalue of \(L(G)\) is popularly known as the algebraic connectivity of G. The largest eigenvalue \(\mu_{1}(G)\) of \(L(G)\) is usually called the Laplacian spectral radius of the graph G. Recently, some of the research has been focused on \(\mu_{1}\), \(\mu_{2}\) or \(u_{n-1}\) (see [3–9]).
Conjecture 1.1
[11]
Among all connected graphs of order n with m edges (\(n \leq m\leq2n -3\)), \(G_{m,n}\) is the unique graph with maximal value of \(S_{2}(G)\), that is, \(S_{2}(G_{m,n})=m(G_{m,n})+3\).
In this paper, we confirm Conjecture 1.1 with \(m=n\).
2 Preliminaries
In this section, we present some lemmas which will be useful in the subsequent sections. For \(\mu_{1}(G)\), the following results are well known.
Lemma 2.1
Lemma 2.2
[7]
Lemma 2.4
Let T be a tree of order n with \(d(T)\geq4\). Then \(\mu_{1}(T) < n-1\).
Proof
For any tree T of order n with \(d(T)\geq4\), it follows that \(n \geq5\) and \(\Delta(T)\leq n-3\). That is, \(\Delta(T)+2 \leq n-1\) and \(m-\frac{n-1}{2} +2=\frac{n-1}{2} +2 \leq n-1\). Then the result follows from Lemma 2.2. □
The following theorem from matrix theory plays a key role in our proofs. We denote the eigenvalues of a symmetric matrix M of order n by \(\lambda_{1}(M)\geq\lambda_{2}(M)\geq\cdots \geq\lambda_{n}(M)\).
Theorem 2.5
[16]
From Theorem 2.5, the following lemma is immediate.
Lemma 2.6
The following lemma can be found in [2] and is well known as the interlacing theorem of Laplacian eigenvalues.
Lemma 2.7
[2]
Lemma 2.8
[2]
Let A be a symmetric matrix of order n with eigenvalues \(\lambda _{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{n}\) and B be a principal submatrix of A of order m with eigenvalues \(\mu_{1} \geq\mu_{2} \geq\cdots\geq\mu_{m}\). Then the eigenvalues of B interlace the eigenvalues of A, that is, \(\lambda_{i} \geq\mu_{i} \geq\lambda _{n-m+i}\) for \(i=1, \ldots, m\). Specially, for \(v \in V(G)\), let \(L_{v}(G)\) be the principal submatrix of \(L(G)\) formed by deleting the row and column corresponding to vertex v, then the eigenvalues of \(L_{v}(G)\) interlace the eigenvalues of \(L(G)\).
Lemma 2.9
For any tree T of order n, if there exists an edge \(e\in E(T)\) such that \(\min\{e(T_{1}), e(T_{2})\} \geq2\) and \(\min\{d(T_{1}), d(T_{2})\} \geq3\) or \(\max\{d(T_{1}), d(T_{2})\} \geq4\), then \(S_{2}(T) < n(T)+1\), where \(T_{1}\), \(T_{2}\) are the two components of \(T-e\).
Proof
Let \(T-e=T_{1} \cup T_{2}\). By Lemma 2.6, it suffices to show that \(S_{2}(T_{1} \cup T_{2}) < n(T)-1\). If \(\mu_{1}(T)= \mu_{1}(T_{1})\) and \(\mu_{2}(T)= \mu_{2}(T_{1})\) (or \(\mu_{1}(T)= \mu_{1}(T_{2})\) and \(\mu_{2}(T)= \mu_{2}(T_{2})\)), then the result follows since \(S_{2}(T_{1} \cup T_{2})=S_{2}(T_{i}) < m(T_{i})+3 \leq m(T)=n(T)-1\) for \(i=1, 2\). In what follows, we assume that \(S_{2}(T_{1} \cup T_{2})=\mu_{1}(T_{1})+\mu_{1}(T_{2})\). If \(\min\{d(T_{1}), d(T_{2})\} \geq3\), then Lemma 2.3 implies that \(S_{2}(T_{1} \cup T_{2}) < (n(T_{1})-0.5)+ (n(T_{2})-0.5) = n(T)-1\). If \(\max\{d(T_{1}), d(T_{2})\} \geq4\), say \(d(T_{2}) \geq4\), then Lemmas 2.1 and 2.4 imply that \(S_{2}(T_{1} \cup T_{2}) < n(T_{1})+ (n(T_{2})-1)=n(T)-1\). This completes the proof. □
Lemma 2.10
For any tree T of order n with \(d(T)=5\), \(S_{2}(T)< n(T)+1\).
Proof
Without loss of generality, we assume that \(v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}\) is a path of length 5 in T. We now consider the following three cases.
Case 1. \(\min\{d(v_{3}), d(v_{4})\} \geq3\).
Let \(T_{1}\), \(T_{2}\) be the two components of \(T-v_{3}v_{4}\). Then the result follows from Lemma 2.9.
Case 2. \(d (v_{3}) = d (v_{4}) = 2\).
Case 3. \(d(v_{3}) \geq3\) and \(d(v_{4}) = 2\) (or \(d(v_{4}) \geq3\) and \(d(v_{3}) = 2\)).
Without loss of generality, we assume that \(d(v_{3}) \geq3\) and \(d(v_{4}) = 2\). If \(d(v_{2}) \geq3\), let \(T-v_{2}v_{3}= T_{1} \cup T_{2}\); if there is \(P_{3}=uvv_{3}\) attached to \(v_{3}\), let \(T-v_{3}v_{4}= T_{1} \cup T_{2}\), where \(u, v \neq v_{i}\) (\(i=1,2,\ldots,6\)). Then the result follows from Lemma 2.9. We now assume that \(d(v_{2})=d(v_{4})=2\) and all the neighbors of \(v_{3}\) except for \(v_{2}\) and \(v_{4}\) are pendent vertices. That is, T is isomorphic to \(H_{a,b}\), where \(H_{a,b}\) (see Figure 2) is the tree of order n obtained from \(v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}\) by attaching a and \(b-1\) pendent vertices to \(v_{3}\) and \(v_{5}\), respectively, where \(a, b \geq1\) and \(a+b+5=n\). Note that the matrix \(1\cdot I_{n}-L(H_{a,b})\) has a and b different identical rows. Then the multiplicity of eigenvalue 1 is at least \(n-7\). Let \(\lambda_{1} \geq\lambda_{2} \geq\lambda_{3} \geq \lambda_{4}\geq\lambda_{5} \geq\lambda_{6} > \lambda_{7}=0\) be the other seven eigenvalues. Then \(\lambda_{1} + \lambda_{2} + \lambda_{3} + \lambda_{4}+\lambda_{5} + \lambda_{6}+ \lambda_{7}=n+5\) since \(\sum_{i=1}^{i=n}\mu_{i}=2m\). For \(a, b\geq 2\), \(H_{a,b}\) contains \(H_{2,2}\) as a subgraph. Then by Lemma 2.7 we have \(\lambda_{3} \geq\mu_{3}(H_{2,2})=2.44\) and \(\lambda_{4} \geq\mu_{4}(H_{2,2})=1.59\). Therefore, \(S_{2}(T)=\lambda_{1} + \lambda_{2} < n+5-(\lambda_{3}+\lambda_{4})< n+1\), as required. If \(a=1\), then by Lemmas 2.1 and 2.8 we have \(\mu_{1}(H_{1,b}) \leq(n-5)+\frac{n-4}{n-5}\) and \(\mu_{2}(H_{1,b}) \leq\mu_{1}(L_{v_{5}}(H_{1,b}))=4.26\). That is, \(S_{2}(T)=S_{2}(H_{1,b})=\mu_{1}(H_{1,b})+\mu_{2}(H_{1,b})< n+1\), as required. Similarly, if \(b=1\), then by Lemmas 2.1 and 2.8 we have \(\mu_{1}(H_{a,1}) \leq(n-4)+\frac{n-2}{n-4}\) and \(\mu_{2}(H_{a,1}) \leq\mu_{1}(L_{v_{3}}(H_{a,1}))=3.0\). It follows that \(S_{2}(T)=S_{2}(H_{a,1})=\mu_{1}(H_{a,1})+\mu _{2}(H_{a,1})< n+1\), as required.
From the discussion above, the proof is completed. □
Lemma 2.11
Let T be a tree of order n with \(d(T)\geq6\). Then \(S_{2}(T)< n(T)+1\).
Proof
We now consider the following two cases.
Case 1. \(d(T)\geq7\).
Let \(v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}v_{7}v_{8}\) be a path of length 7 in T and \(T-v_{4}v_{5}=T_{1} \cup T_{2}\). Then the result follows from Lemma 2.9.
Case 2. \(d(T)=6\).
If \(T= P_{7}\), then the result follows since \(S_{2}(P_{7}) < 8\). If \(T\neq P_{7}\), let \(v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}v_{7}\) be a path of length 6 in T. If \(d(v_{4})\geq3\), let \(T-v_{3}v_{4}=T_{1} \cup T_{2}\); if \(d(v_{3})\geq3\) (or \(d(v_{5})\geq3\)), let \(T-v_{3}v_{4}=T_{1} \cup T_{2}\) (or \(T-v_{4}v_{5}=T_{1} \cup T_{2}\)); if \(d(v_{2})\geq3\) (or \(d(v_{6})\geq3\)), let \(T-v_{2}v_{3}=T_{1} \cup T_{2}\) (or \(T-v_{5}v_{6}=T_{1} \cup T_{2}\)). In each of the above cases, by Lemma 2.9, we have \(S_{2}(T)< n(T)+1\). This completes the proof. □
Lemma 2.12
[6] The second largest Laplacian eigenvalue of \(F_{1, t, n-2t-3}\) satisfies \(\mu_{2}(F_{1, t, n-2t-3})=3\).
Note that for \(t \geq1\), the complement of \(F_{1, t, n-2t-3}\) is connected. Hence Lemma 2.1 implies that \(\mu_{1}(F_{1, t, n-2t-3})< n(F_{1, t, n-2t-3})\). This together with Lemma 2.12 implies the following lemma.
Lemma 2.13
For \(t \geq1\), \(S_{2}(F_{1, t, n-2t-3}) < n(F_{1, t, n-2t-3})+3\).
3 Main result
A unicyclic graph is a connected graph whose number of edges m is equal to the number of vertices n. It is easy to see that each unicyclic graph can be obtained by attaching rooted trees to the vertices of a cycle \(C_{k}\) for some k. Thus if \(R_{1},\ldots, R_{k}\) are k rooted trees (of orders \(n_{1},\ldots, n_{k}\), say), then we adopt the notation \(U(R_{1},\ldots, R_{k})\) to denote the unicyclic graph G (of order \(n=n_{1}+\cdots+n_{k}\)) obtained by attaching the rooted tree \(R_{i}\) to the vertex \(v_{i}\) of a cycle \(C_{k}=v_{1}v_{2} \cdots v_{k}v_{1}\) (i.e., by identifying the root of \(R_{i}\) with the vertex \(v_{i}\) for \(i=1,\ldots, k\)). Denote by \(e(v_{i})\) the maximum distance between \(v_{i}\) and any vertex of \(R_{i}\). In the special case when \(R_{i}\) is a rooted star \(K_{1,a_{i}}\) with the center of the star as its root (that is, \(e(v_{i})=1\)), we will simplify the notation by replacing \(R_{i}\) by the number \(a_{i}\).
The following lemma is immediate from Lemmas 2.6, 2.10 and 2.11.
Lemma 3.1
For any unicyclic graph G of order n with m edges, if there exists an edge \(e\in E(G)\) such that \(G-e\) is a tree with \(d(G-e) \geq5\), then \(S_{2}(G)< m(G)+3\).
Lemma 3.2
For \(T_{a,b,c}\), \(S_{2}(T_{a,b,c}) \leq m(T_{a,b,c})+3\) with equality if and only if \(a=n-3\) (that is, \(T_{a,b,c} \cong G_{n,n}\)).
Proof
Lemma 3.3
For \(Q_{a,b}\), \(S_{2}(Q_{a,b}) < m(Q_{a,b})+3\).
Proof
By a direct calculation, we have \(\phi(Q_{a, b},\lambda)=\lambda(\lambda-1)^{n-6}f_{3}(\lambda)\), where \(f_{3}(\lambda)=(\lambda-2)(\lambda^{4}-(n+4)\lambda ^{3}+(5n+ab+1)\lambda^{2}-(6n+2ab-2)\lambda+2n)\). Let \(\lambda_{1} \geq\lambda_{2} \geq\lambda_{3} \geq\lambda_{4}\geq \lambda_{5}>0 \) be the five roots of \(f_{3}(\lambda)=0\). Then we have \(\lambda_{1} + \lambda_{2} + \lambda_{3} + \lambda _{4}+\lambda_{5} =n+6 \) since \(\sum_{i=1}^{i=n}\mu_{i}=2m(G)\). For \(b \geq1\), \(Q_{a,b}\) contains \(Q_{1,1}\) as a subgraph. Then by Lemma 2.7 we have \(\lambda_{3} \geq\mu_{3}(Q_{1,1})=2\) and \(\lambda_{4} \geq\mu_{4}(Q_{1,1})=1.26\). Therefore, \(S_{2}(Q_{a,b}) = \lambda_{1} + \lambda_{2} = n+6- \lambda _{3} - \lambda_{4} - \lambda_{5} < n+3\). In what follows, we assume \(b=0\). Since \(f_{3}(1)=0\), we can rewrite \(\phi(Q_{a, 0},\lambda )=\lambda(\lambda-1)^{n-5}f_{4}(\lambda)\), where \(f_{3}(\lambda )=(\lambda-1)f_{4}(\lambda)\). Let \(\lambda_{1}' \geq\lambda_{2}' \geq\lambda_{3}' \geq\lambda _{4}'> 0\) be the four roots of \(f_{4}(\lambda)=0 \). Then we have \(\lambda_{1}' + \lambda_{2}' + \lambda_{3}' + \lambda _{4}' = n+5\) since \(\sum_{i=1}^{i=n}\mu_{i}=2m(G)\). For \(a \geq1\), \(Q_{a, 0}\) contains \(Q_{1,0}\) as a subgraph. Then by Lemma 2.7 we have \(\lambda_{3}' \geq\mu_{3}(Q_{1,0})=2\). Thus, \(S_{2}(Q_{a,b}) = \lambda_{1}' + \lambda_{2}'= n+5-\lambda_{3}' - \lambda_{4}' < n+3\). If \(a=0\), then \(S_{2}(Q_{a,b}) = S_{2}(C_{4}) < m(C_{4})+3\) by a straight calculation. This completes the proof. □
Now, we come to the main results of this paper.
Theorem 3.4
For any unicyclic graph G, \(S_{2}(G) \leq m(G)+3 \) with equality if and only if \(G \cong T(n-3,0,0)\).
Proof
For any unicyclic graph G, we assume that \(C_{k}=v_{1}v_{2} \cdots v_{k}v_{1}\) is the unique cycle in G (for some k) and G has the form \(U(R_{1},\ldots, R_{k})\). For \(k \geq5\), \(G-v_{1}v_{2}\) is a tree with \(d (G-v_{1}v_{2}) \geq 5\). Then by Lemma 3.1 we have \(S_{2}(G) < m(G)+3\). We now consider the following two cases.
Case 1. \(k=4 \).
Let \(C_{4}=v_{1}v_{2}v_{3}v_{4}v_{1}\) be the unique cycle in G. If there exist \(e(v_{i}) \geq2\), say \(e(v_{1})\geq2\), then \(G-v_{1}v_{2}\) is a tree with \(d(G-v_{1}v_{2}) \geq5\); if there are two adjacent vertices in \(C_{4}=v_{1}v_{2}v_{3}v_{4}v_{1}\), say \(v_{1}\) and \(v_{2}\), such that \(e(v_{1}) \geq1\) and \(e(v_{2}) \geq1\), then \(G-v_{1}v_{2}\) is a tree with \(d (G-v_{1}v_{2}) \geq5\). Then by Lemma 3.1 we have \(S_{2}(G) < m(G)+3\). We now assume \(G \cong Q_{a, b}\) (see Figure 4). Then the result follows from Lemma 3.3.
Case 2. \(k=3 \).
If \(\max \{e(v_{1}), e(v_{2}), e(v_{3}) \} \geq3 \), say \(e(v_{1})\geq 3 \), then \(G-v_{1}v_{2}\) is a tree with \(d(G-v_{1}v_{2}) \geq5\); if there are two vertices in \(C_{3}=v_{1}v_{2}v_{3}v_{1}\), say \(v_{1}\) and \(v_{2}\), such that \(e(v_{1})=2\) and \(e(v_{2}) \geq1\), then \(G-v_{1}v_{2}\) is a tree with \(d (G-v_{1}v_{2}) \geq5\). Therefore, Lemma 3.1 implies that \(S_{2}(G) < m(G)+3\). We now assume \(G \cong F_{1,t,n-2t-3}\) (\(t \geq1\)) or \(G \cong T_{a,b,c}\). Then the result follows from Lemma 2.13 or Lemma 3.2.
This completes the proof. □
Declarations
Acknowledgements
The authors would like to thank the anonymous referees for their constructive corrections and valuable comments on this paper, which have considerably improved the presentation of this paper. This project is supported by NSF of China (Nos. 11471077, 11301440), China Postdoctoral Science Foundation (No. 2014M551831), NSF of Fujian (No. 2014J01020), the Foundation to the Educational Committee of Fujian (JA13240, JA15381, JB13155).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
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