Properties and Applications of the Eigenvector Corresponding to the Laplacian Spectral Radius of a Graph

Abstract

We study the properties of the eigenvector corresponding to the Laplacian spectral radius of a graph and show some applications. We obtain some results on the Laplacian spectral radius of a graph by grafting and adding edges. We also determine the structure of the maximal Laplacian spectrum tree among trees with 𝑛 vertices and 𝑘 pendant vertices (𝑛, 𝑘 fixed), and the upper bound of the Laplacian spectral radius of some trees.

1. Introduction

The theory of graph spectra has been established in the 1950s and 1960s. Chung has taken the investigation of the theory of graph spectra to a new level by a 45-minute report [1] presented at the World Congress of Mathematicians in 1994, and in monographs [2]. Applications of the theory of graph spectra have also been found in the fields of electrical networks and vibration theory [3, 4]. The wide range of the applications of the theory of graph spectra has led it to become a very active field of research of graph theory over the last thirty to forty years, and large numbers of results are continuously emerging.

There are many results on the (adjacency) spectral radius for different classes of graphs. Guo et al. [5] have studied the largest and the second largest spectral radius of trees with 𝑛 vertices and diameter 𝑑. Guo and Shao [6] have studied the first ⌊𝑑/2⌋+1 (where ⌊𝑥⌋ represents the maximal integer not more than 𝑥) spectral radii of graphs with 𝑛 vertices and diameter 𝑑. Wu et al. [7] have studied the spectral radius of trees on 𝑘 pendant vertices.

There are also many results on the Laplacian spectral radius for different classes of graphs. Hong and Zhang [8] have studied the upper and lower bounds for the Laplacian spectral radius of trees. Guo [9] has studied the second largest Laplacian eigenvalues of trees. In the paper, we further study the properties of the eigenvector corresponding to the Laplacian spectral radius of a graph and obtain some applications on the Laplacian spectral radius of trees with 𝑘 pendant vertices.

Let 𝐺=(𝑉(𝐺),𝐸(𝐺)) be a simple connected graph with vertex set 𝑉(𝐺)={𝑣₁,𝑣₂,…,𝑣_𝑛} and edge set 𝐸(𝐺). Let 𝑑_𝐺(𝑣_𝑖) denote the degree of vertex 𝑣_𝑖 (1≤𝑖≤𝑛) of graph 𝐺 sometimes for convenience we denote by 𝑑(𝑣_𝑖). If 𝐷(𝐺)=diag(𝑑(𝑣₁),𝑑(𝑣₂),…,𝑑(𝑣_𝑛)) is the diagonal matrix of the vertex degree and 𝐴(𝐺) is the (0,1) adjacency matrix of 𝐺, then the matrix 𝐿(𝐺)=𝐷(𝐺)−𝐴(𝐺) is called the Laplacian matrix of graph 𝐺. When 𝐺 is connected, 𝐴(𝐺) is irreducible, and, by the Perron-Frobenius theorem [10–12], the spectral radius 𝜌(𝐺) is simple and there is a unique positive unit eigenvector. We refer to such an eigenvector as the Perron vector of 𝐺. It is easy to see that 𝐿(𝐺) is a positive semidefinite 𝑀 matrix and its rows sum to 0,  𝐿(𝐺) is singular. Hence, the eigenvalues of 𝐿(𝐺) can be ordered as 𝜇₁(𝐺)≥𝜇₂(𝐺)≥⋯≥𝜇_𝑛(𝐺)=0. The largest eigenvalue 𝜇₁(𝐺) of 𝐿(𝐺) is called the Laplacian spectral radius of the graph 𝐺, denoted by 𝜇(𝐺). This has been used extensively in fields such as theoretical chemistry, combinatorial optimization, and communication networks [13, 14].

A pendant vertex of 𝐺 is a vertex of degree 1. Let 𝑁_𝐺(𝑣) denote the adjacent vertex set of 𝑣 in 𝐺 and 𝑑_𝑣 the degree of 𝑣. 𝒯_𝑛,𝑘 denotes the set of trees for 𝑛 vertices and 𝑘 pendant vertices (𝑛, 𝑘 fixed). The tree 𝑇_𝑛,𝑘 is obtained from a star 𝐾_1,𝑘 and 𝑘 paths of almost equal length by joining each pendant vertex of 𝐾_1,𝑘 to an end of one path. Obviously, 2≤𝑘≤𝑛−1. Assume 𝑞=[(𝑛−1)/𝑘], 𝑛−1=𝑘𝑞+𝑟  (0≤𝑟≤𝑘−1); 𝑇_𝑛,𝑘 is shown in Figure 1.

Figure 1 

In general, 𝑇_{𝑛,𝑛−1}=𝐾_1,𝑛−1, 𝑇_𝑛,2=𝑃_𝑛, and 𝑇_{𝑛,𝑛−2}=𝑆(1,𝑛−3), a tree is obtained from two stars 𝐾_1,𝑛−3 and 𝐾_1,1 by joining their central vertices. For example, 𝑇_6,5=𝐾_1,5, 𝑇_6,4=𝑆(1,3), and 𝑇_6,2 is a path of length 6 (Figure 2).

Figure 2 

In Section 2, we describe some properties of the eigenvector corresponding to the Laplacian spectral radius of a graph. Section 3 presents some applications of the eigenvector corresponding to the Laplacian spectral radius of a graph, including some results on the Laplacian spectral radius of a graph by grafting and adding edges. Then we obtain the structure of the maximal Laplacian spectrum tree among trees with 𝑛 vertices and 𝑘 pendant vertices (𝑛, 𝑘 fixed), and the upper bound of the Laplacian spectral radius for some trees.

2. Properties of the Eigenvector Corresponding to the Laplacian Spectral Radius of a Graph

For convenience, we denote 𝐿(𝐺) by 𝐿 sometimes. Since the Laplacian matrix 𝐿(𝐺) is a real symmetric matrix, we have the following theorem.

Theorem 1. Let 𝐺 be a simple connected graph of 𝑛 vertices. 𝐿(𝐺) is the Laplacian matrix of 𝐺, with vector 𝑥∈ℛ^𝑛×1. Then (1)𝜇(𝐺)=max_{1≤𝑖≤𝑛}𝜇_𝑖, where 𝜇_𝑖 is the eigenvalue of 𝐿(𝐺); (2)𝜇(𝐺)=max_‖𝑥‖=1(𝑥^′𝐿𝑥); (3)if ‖𝑥‖=1 and 𝑥^′𝐿(𝐺)𝑥=𝜇(𝐺), then 𝐿(𝐺)𝑥=𝜇(𝐺)𝑥.

Proof. (1) It is proved by the definition.
(2) Since 𝐿(𝐺) is a real symmetric matrix, there exists an orthogonal matrix 𝑃 for which 𝑃⁻¹𝐿𝑃=diag(𝜇₁,𝜇₂,…,𝜇_𝑛)≜𝐷, where 𝑃=(𝑊₁,𝑊₂,…,𝑊_𝑛). Let 𝑥=𝑃𝑦, so 𝑥^′𝐿𝑥=𝑦^′𝑃^′𝐿𝑃𝑦=𝑦^′𝑃⁻¹𝐿𝑃𝑦=𝑦^′𝐷𝑦=∑^𝑛_𝑖=1𝜇_𝑖𝑦²_𝑖.
𝑃 is an orthogonal matrix and 𝑥=𝑃𝑦. Then, when ‖𝑥‖=1, we have ‖𝑦‖=1. Assume, without loss of generality, that 𝜇₁≤𝜇₂≤⋯≤𝜇_𝑛. Thus, max_‖𝑥‖=1𝑥^′𝐿𝑥=max_𝑦^𝑛_=1𝑖=1𝜇_𝑖𝑦²_𝑖≤max_𝑦^𝑛_=1𝑖=1𝜇_𝑛𝑦²_𝑖=𝜇_𝑛max_𝑦^𝑛_=1𝑖=1𝑦²_𝑖=𝜇_𝑛=𝜇(𝐺). (1)
Let 𝑦=𝑒_𝑛=(0,0,…,0,1)^′_𝑛×1, so we have 𝑥=𝑃𝑦=𝑊_𝑛, and then the above equality holds. Therefore, 𝜇(𝐺)=max_‖𝑥‖=1(𝑥^′𝐿𝑥).
(3) Since 𝐿(𝐺) is a real symmetric matrix, there exists an orthogonal matrix 𝑃 for which 𝑃⁻¹𝐿𝑃=diag(𝜇₁,𝜇₂,…,𝜇_𝑛)≜𝐷, where 𝑃=(𝑊₁,𝑊₂,…,𝑊_𝑛), and 𝑊_𝑖 is the eigenvector corresponding to 𝜇_𝑖 of 𝐿(𝐺). Assume, without loss of generality, that 𝜇₁≤𝜇₂≤⋯≤𝜇_𝑛.
Let 𝑥=𝑃𝑦, and then 𝑥^′𝐿(𝐺)𝑥=𝑦^′𝑃^′𝐿𝑃𝑦=𝑦^′𝐷𝑦. Since 𝑥^′𝐿(𝐺)𝑥=𝜇(𝐺), we have 𝑥^′𝐿(𝐺)𝑥=𝑦^′𝐷𝑦=𝜇_𝑛=𝜇(𝐺). Thus, 𝑦=𝑒_𝑛=(0,0,…,0,1)^′_𝑛×1 and then 𝑥=𝑃𝑦=𝑃𝑒_𝑛=𝑊_𝑛. However, 𝑊_𝑛 is the eigenvector corresponding to 𝜇_𝑛 of 𝐿(𝐺), that is, 𝐿(𝐺)𝑊_𝑛=𝜇_𝑛𝑊_𝑛. Thus, 𝐿(𝐺)𝑥=𝜇(𝐺)𝑥.

Definition 2. Let 𝐺 be a simple connected graph, 𝐺=(𝑉,𝐸). If vector 𝑥∈ℛ^𝑛×1 satisfies ‖𝑥‖=1 and 𝑥^′𝐿(𝐺)𝑥=𝜇(𝐺), then we call 𝑥 the standard Laplacian spectral vector of 𝐺.

Theorem 3. Let 𝑇 be a tree, and 𝑉(𝑇)={𝑣₁,𝑣₂,…,𝑣_𝑛}. Let 𝐿 be the Laplacian matrix of 𝑇 and 𝑥 be the standard Laplacian spectral vector of 𝑇. Then(1)𝑥 is a real vector; (2)|𝑥|>0.

Proof. (1) Since 𝐿 is a real symmetric matrix, we have 𝜇(𝑇)∈𝑅. Thus, 𝑥 is a real vector.
(2) Suppose that |𝑥|>0 is not established. Then there exists vertex set 𝐻={𝑣_𝑗1,𝑣_𝑗2,…,𝑣_𝑗𝑡}, which means 𝑥_𝑗𝑖=0(𝑖=1,2,…,𝑡), where 𝑥_𝑗 corresponds to vertex 𝑣_𝑗. Since 𝑥 is the standard Laplacian spectral vector of 𝑇, we have ‖𝑥‖=1. Thus, 𝑥≠0 and there exists vertex 𝑣∈𝑉(𝑇) that makes 𝑥_𝑣=0 and vertex 𝑢∈𝑁_𝑇(𝑣) that makes 𝑥_𝑢≠0.
Let 𝑇 form a root tree with 𝑣 as its root vertex. 𝑁_𝑇(𝑣)={𝑤₁,𝑤₂,…,𝑤_𝑠}≜𝑊. Let 𝑇_𝑖 be the subtree of 𝑇 obtained from vertex 𝑤_𝑖 and all its descendants. Let 𝑦_𝑣=𝑥_𝑣=0, 𝑦_𝑤𝑖=|𝑥_𝑤𝑖|(𝑖=1,2,…,𝑠),𝑦_𝑗=𝑥_𝑗 when 𝑥_𝑤𝑖≥0, and let 𝑣_𝑗 be the vertex of subtree 𝑇_𝑖, 𝑦_𝑗=−𝑥_𝑗 when 𝑥_𝑤𝑖<0 and 𝑣_𝑗 be the vertex of subtree 𝑇_𝑖(𝑖=1,2,…,𝑠).
Thus, we have 𝑦_𝑤𝑖≥0(𝑖=1,2,…,𝑠), ‖𝑦‖=1 and 𝑦^′𝐿𝑦=𝑥^′𝐿𝑥=𝜇(𝐿(𝑇))≜𝜇(𝑇). By Definition 2, we know that 𝑦 is the standard Laplacian spectral vector of 𝑇. Then (𝐷−𝐴)𝑦=𝜇(𝑇)𝑦. We have ((𝐷−𝐴)𝑦)_𝑣=(𝜇(𝑇)𝑦)_𝑣. Thus, ∑^𝑠_𝑖=1𝑦_𝑤𝑖=0.
However, 𝑦_𝑤𝑖≥0 and there exists 𝑖₀∈{1,2,…,𝑠} that makes 𝑤_𝑖0=𝑢, 𝑦_𝑢>0. Therefore, ∑^𝑠_𝑖=1𝑦_𝑤𝑖>0. This contradiction completes the proof.

Theorem 4. Let 𝑣 be a vertex of tree 𝑇 and 𝑣₁,𝑣₂,…,𝑣_𝑠 be the pendant vertices of 𝑇 that are adjacent to 𝑣. If 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑛)^′ is the standard Laplacian spectral vector of 𝑇, where 𝑥_𝑖 corresponds to the vertex 𝑣_𝑖(1≤𝑖≤𝑛). Then 𝑥_𝑣𝑗=𝑥_𝑣𝑖, 1≤𝑖<𝑗≤𝑠.

Proof. 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑛)^′ is the standard Laplacian spectral vector of 𝑇. By Definition 2 and Theorem 1, we have 𝐿(𝑇)𝑥=𝜇(𝑇)𝑥. 𝐿(𝑇)=𝐷(𝑇)−𝐴(𝑇), so (𝐷(𝑇)−𝐴(𝑇))𝑥=𝜇(𝑇)𝑥, and then we have (𝑑_𝑇(𝑣_𝑖)−𝜇(𝑇))𝑥_𝑣𝑖=𝑥_𝑣,1≤𝑖≤𝑠. Therefore, we have (1−𝜇(𝑇))𝑥_𝑣1=𝑥_𝑣, (1−𝜇(𝑇))𝑥_𝑣2=𝑥_𝑣,…,(1−𝜇(𝑇))𝑥_𝑣𝑠=𝑥_𝑣. Thus, 𝑥_𝑣𝑗=𝑥_𝑣𝑖, 1≤𝑖<𝑗≤𝑠.

Theorem 5. Let 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑛)^′ be the standard Laplacian spectral vector of tree 𝑇, where 𝑥_𝑖 corresponds to vertex 𝑣_𝑖(1≤𝑖≤𝑛), 𝑇=(𝑉,𝐸). Then (1)for ∀𝑢𝑣∈𝐸(𝑇), we have 𝑥_𝑢𝑥_𝑣<0; (2)∑^𝑛_𝑖=1𝑥_𝑖=0.

Proof. (1) Since 𝑥 is the standard Laplacian spectral vector of 𝑇, by Definition 2 we have 𝜇(𝑇)=𝑥^′𝐿(𝑇)𝑥=∑_{(𝑖,𝑗)∈𝐸}(𝑥_𝑖−𝑥_𝑗)². By Theorem 3, we have |𝑥_𝑢|>0 and |𝑥_𝑣|>0. Thus, 𝑥_𝑢𝑥_𝑣≠0.
Suppose that there exists edge 𝑢𝑣∈𝐸(𝑇) such that 𝑥_𝑢𝑥_𝑣>0. Let tree 𝑇 form a root tree with 𝑟 as its root vertex. Assume, without loss of generality, that vertex 𝑢 is the parent node of the vertex 𝑣 (Figure 3). Let 𝑇₁ be the subtree of 𝑇 obtained from vertex 𝑣 and all its descendants, and let 𝑇₁ be a 𝑘-layer tree, 𝑇₁=(𝑉₁,𝐸₁) (Figure 4). Without loss of generality, we assume that 𝑥_𝑢>0, so 𝑥_𝑣>0.
Let 𝑦_𝑖=𝑥_𝑖 (𝑣_𝑖∈𝑉⧵𝑉(𝑇₁)), 𝑦_𝑣=−𝑥_𝑣, and 𝑦_𝑤1=|𝑥_𝑤1| (∀𝑤₁ are the vertices in the second layer of 𝑇₁), 𝑦_𝑤2=(−1)³|𝑥_𝑤2|(∀𝑤₂aretheverticesinthethirdlayerof𝑇₁),…,𝑦_𝑤𝑘−1=(−1)^𝑘|𝑥_𝑤𝑘−1| (∀𝑤_𝑘−1 are the vertices in the 𝑘th layer of 𝑇₁). Then ‖𝑦‖=‖𝑥‖=1 and we have 𝑦^′𝐿(𝑇)𝑦=_{𝑖,𝑗∈𝐸}𝑦_𝑖−𝑦_𝑗²=_{𝑖,𝑗∈𝐸⧵𝐸1}𝑥_𝑖−𝑥_𝑗²+_{𝑖,𝑗∈𝐸1⧵𝑢𝑣}𝑦_𝑖−𝑦_𝑗²+𝑥_𝑢+𝑥_𝑣²>_{𝑖,𝑗∈𝐸⧵𝐸1}𝑥_𝑖−𝑥_𝑗²+_{𝑖,𝑗∈𝐸1⧵𝑢𝑣}𝑥_𝑖−𝑥_𝑗²+𝑥_𝑢−𝑥_𝑣²=_{𝑖,𝑗∈𝐸}𝑥_𝑖−𝑥_𝑗²=𝜇(𝑇). (2) Thus, there exists a unit vector 𝑦=(𝑦₁,𝑦₂,…,𝑦_𝑛)^′ such that 𝜇(𝑇)<𝑦^′𝐿(𝑇)𝑦. This implies a contradiction with Theorem 1. Therefore, 𝑥_𝑢𝑥_𝑣<0.
(2) Since 𝑥 is the standard Laplacian spectral vector of 𝑇, we have (𝐷(𝑇)−𝐴(𝑇))𝑥=𝜇(𝑇)𝑥. (3) Then 𝑑_𝑇𝑣_𝑖−𝜇(𝑇)𝑥_𝑖=_{𝑣𝑗∈𝑁𝑇𝑣𝑖}𝑥_𝑗. (4) Thus, we have ^𝑛_𝑖=1𝑑_𝑇𝑣_𝑖−𝜇(𝑇)𝑥_𝑖=^𝑛_{𝑖=1𝑣𝑗∈𝑁𝑇𝑣𝑖}𝑥_𝑗. (5) Therefore, ∑^𝑛_𝑖=1𝑑_𝑇(𝑣_𝑖)𝑥_𝑖−𝜇(𝑇)∑^𝑛_𝑖=1𝑥_𝑖=∑^𝑛_𝑖=1𝑑_𝑇(𝑣_𝑖)𝑥_𝑖. So ∑^𝑛_𝑖=1𝑥_𝑖=0.

Figure 3 

Figure 4 

3. Applications of the Eigenvector Corresponding to the Laplacian Spectral Radius

In this section, we present some applications of the eigenvector of the Laplacian spectral radius of a graph.

Theorem 6. Let 𝑢 and 𝑣 be the vertices of tree T. Denote the distance of vertices 𝑢 and 𝑣 by 𝑑(𝑢,𝑣) and 𝑑(𝑢,𝑣)=𝑘, where 𝑘 is an odd number and 𝑘≥3. Let 𝐺^′ be the graph obtained from 𝑇 by adding edge (𝑢,𝑣). Then 𝜇(𝐺^′)>𝜇(𝑇).

Proof. Let 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑛)^′ be the standard Laplacian spectral vector of tree 𝑇, where 𝑥_𝑖 corresponds to vertex 𝑣_𝑖(1≤𝑖≤𝑛). Since 𝑘 is an odd number and 𝑘≥3, by Theorem 5 we have 𝑥_𝑢𝑥_𝑣<0. Thus, 𝜇𝐺^′=max_𝑦=1𝑦^′𝐿𝐺^′𝑦=max_{𝑦=1𝐺𝑖,𝑗∈𝐸′}𝑦_𝑖−𝑦_𝑗²≥_{𝐺𝑖,𝑗∈𝐸′}𝑥_𝑖−𝑥_𝑗²>_{(𝑇)𝑖,𝑗∈𝐸}𝑥_𝑖−𝑥_𝑗²=𝜇(𝑇). (6) Therefore, 𝜇(𝐺^′)>𝜇(𝑇).

According to the well-known Courant-Weyl inequalities, we have the following lemma.

Lemma 7 (see [15]). Let 𝐺 be a graph with 𝑛 vertices and let 𝐺^′=𝐺+𝑒 be the graph obtained by inserting a new edge 𝑒 into 𝐺. Let 𝜇_𝑖(𝐺) and 𝜇_𝑖(𝐺^′) be the eigenvalues of the Laplacian matrix of 𝐺 and 𝐺^′(𝑖=1,2,…,𝑛), respectively. Then 𝜇₁(𝐺^′)≥𝜇₁(𝐺)≥𝜇₂(𝐺^′)≥𝜇₂(𝐺)≥⋯≥𝜇_𝑛(𝐺^′)=𝜇_𝑛(𝐺)=0.

Lemma 8 (see [16]). Let 𝐺 be a connected graph for 𝑛 vertices with at least one edge. Then 𝜇(𝐺)≥Δ(𝐺)+1, where Δ(𝐺) is the largest degree of the graph 𝐺, and equality holds if and only if Δ(𝐺)=𝑛−1.

Theorem 9. Let 𝐺 be a connected graph for 𝑛 vertices, and let 𝑢_𝑗 be the vertex of 𝐺 and 𝑣^𝑗₁,𝑣^𝑗₂,…,𝑣^𝑗_𝑚𝑗 be the pendant vertices of 𝐺 adjacent to vertex 𝑢_𝑗(1≤𝑗≤𝑘). Let 𝐺₁ be the graph obtained from 𝐺 by adding ∑^𝑘_𝑗=1𝑞_𝑗(1≤𝑞_𝑗≤𝑚_𝑗(𝑚_𝑗−1)/2) edges, where 𝑡_𝑗 edges are among 𝑣^𝑗₁,𝑣^𝑗₂,…,𝑣^𝑗_𝑚𝑗(𝑗=1,2,…,𝑘). Then 𝜇(𝐺)=𝜇(𝐺₁).

Proof. If Δ(𝐺)=𝑛−1, by Lemma 8 we have 𝜇(𝐺)=Δ(𝐺)+1=𝑛. However, it is clear that 𝜇(𝐺₁)=Δ(𝐺₁)+1=𝑛−1+1=𝑛, so 𝜇(𝐺)=𝜇(𝐺₁). We suppose that Δ(𝐺)≤𝑛−2.
Let 𝐺^′ be the graph obtained from 𝐺 by adding ∑^𝑘_𝑗=1𝑡_𝑗(𝑡_𝑗=𝑚_𝑗(𝑚_𝑗−1)/2) edges, where 𝑡_𝑗 edges are among 𝑣^𝑗₁,𝑣^𝑗₂,…,𝑣^𝑗_𝑚𝑗(𝑗=1,2,…,𝑘). By Lemma 7, we have 𝜇(𝐺)≤𝜇(𝐺₁)≤𝜇(𝐺^′). To obtain the result, we only need to prove that 𝜇(𝐺)≥𝜇(𝐺^′).
Let 𝑋=(𝑋₁,𝑋₂,…,𝑋_𝑛)^′ be the standard Laplacian spectral vector of 𝐺^′, then 𝐿(𝐺^′)𝑋=𝜇(𝐺^′)𝑋. For 𝐿(𝐺^′)=𝐷(𝐺^′)−𝐴(𝐺^′), we have 𝑑_𝐺′𝑣^𝑙_𝑖−𝜇𝐺^′𝑋_𝑣^𝑙_𝑖=𝑋_𝑢𝑙+^𝑚𝑙_{𝑗=1,𝑗≠𝑖}𝑋_𝑣^𝑙_𝑗,1≤𝑖≤𝑚_𝑙,1≤𝑙≤𝑘. (7) Then, for 1≤𝑖<𝑗≤𝑚_𝑙 and 1≤𝑙≤𝑘, we have 𝑑_𝐺′𝑣^𝑙_𝑖−𝜇𝐺^′𝑋_𝑣^𝑙_𝑖−𝑑_𝐺′𝑣^𝑙_𝑗−𝜇𝐺^′𝑋_𝑣^𝑙_𝑗=𝑋_𝑣^𝑙_𝑗−𝑋_𝑣^𝑙_𝑖 (8) since 𝑑_𝐺′(𝑣^𝑙₁)=𝑑_𝐺′(𝑣^𝑙₂)=⋯=𝑑_𝐺′(𝑣^𝑙_𝑞𝑙)=𝑚_𝑙≤𝑛−2. Thus, 𝑚_𝑙−𝜇𝐺^′𝑋_𝑣^𝑙_𝑖−𝑋_𝑣^𝑙_𝑗=𝑋_𝑣^𝑙_𝑗−𝑋_𝑣^𝑙_𝑖,1≤𝑖<𝑗≤𝑚_𝑙,1≤𝑙≤𝑘. (9)
For Δ(𝐺^′)≤𝑛−2, by Lemma 8 we have 𝜇(𝐺^′)>Δ(𝐺^′)+1≥𝑚_𝑙+1. So 𝑋_𝑣^𝑙_𝑖=𝑋_𝑣^𝑙_𝑗, 1≤𝑖<𝑗≤𝑚_𝑙, 1≤𝑙≤𝑘. Thus, we have 𝜇(𝐺)≥𝑋^′𝐿(𝐺)𝑋=_{(𝑖,𝑗)∈𝐸(𝐺)}𝑋_𝑖−𝑋_𝑗²=_{𝐺𝑖,𝑗∈𝐸′}𝑋_𝑖−𝑋_𝑗²=𝑋^′𝐿𝐺^′𝑋=𝜇𝐺^′. (10) That is, 𝜇(𝐺)≥𝜇(𝐺^′). This completes the proof.

Theorem 10. Let 𝑢, 𝑣 be two vertices of tree 𝑇. Suppose that 𝑣₁,𝑣₂,…,𝑣_𝑠(1≤𝑠≤𝑑_𝑣) are some vertices of 𝑁_𝑇(𝑣)⧵(𝑁_𝑇(𝑢)∪{𝑢}), and 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑛)^′ is the standard Laplacian spectral vector of 𝑇, where 𝑥_𝑖 corresponds to vertex 𝑣_𝑖(1≤𝑖≤𝑛). Let 𝑇^∗ be the tree obtained from 𝑇 by deleting the edges (𝑣,𝑣_𝑖) and adding the edges (𝑢,𝑣_𝑖)(1≤𝑖≤𝑠) (Figure 5). If |𝑥_𝑢|≥|𝑥_𝑣|, then 𝜇(𝑇)<𝜇(𝑇^∗).

Figure 5 

Proof. For tree 𝑇, let 𝑇 form a root tree with vertex 𝑣 as the parent node of vertex 𝑣_𝑖(1≤𝑖≤𝑠), 𝑇=(𝑉,𝐸). Let 𝑇₁ be the subtree of 𝑇 obtained from the parent node 𝑣 and 𝑣₁,𝑣₂,…,𝑣_𝑠 and the 𝑣₁ descendants, 𝑣₂ descendants,…, 𝑣_𝑠 descendants, 𝑇₁=(𝑉₁,𝐸₁). For tree 𝑇^∗,   let 𝑇^∗ form a root tree with vertex 𝑢 as the parent node of vertex 𝑣_𝑖(1≤𝑖≤𝑠), 𝑇^∗=(𝑉^∗,𝐸^∗). Let 𝑇^∗₁ be the subtree of 𝑇^∗ obtained from the parent node 𝑢 and 𝑣₁,𝑣₂,…,𝑣_𝑠 and the 𝑣₁ descendants, 𝑣₂ descendants,…, and 𝑣_𝑠 descendants. Let subtree 𝑇^∗₁ be a 𝑘-layer tree with vertex 𝑢 as the root node. Consider 𝐸^′₁=𝐸₁⧵{𝑣𝑣₁,𝑣𝑣₂,…,𝑣𝑣_𝑠}.
Let 𝑦_𝑖=𝑥_𝑖(𝑣_𝑖∈𝑉⧵𝑉(𝑇₁)), 𝑦_𝑣=𝑥_𝑣, 𝑦_𝑤1=−(sign𝑥_𝑢)|𝑥_𝑤1| (forall𝑤₁ are the vertices in the second layer of 𝑇^∗₁), 𝑦_𝑤2=(−1)²(sign𝑥_𝑢)|𝑥_𝑤2| (forall𝑤₂ are the vertices in the third layer of 𝑇^∗₁),…, and 𝑦_𝑤𝑘−1=(−1)^𝑘−1(sign𝑥_𝑢)|𝑥_𝑤𝑘−1| (forall𝑤_𝑘−1 are the vertices in the 𝑘th layer of 𝑇^∗₁). Then we have ‖𝑦‖=1 and 𝑦^′𝐿𝑇^∗𝑦−𝑥^′𝐿(𝑇)𝑥=_{𝑖,𝑗∈𝐸∗}𝑦_𝑖−𝑦_𝑗²−_{𝑖,𝑗∈𝐸}𝑥_𝑖−𝑥_𝑗²=_{𝑖,𝑗∈𝐸′1}𝑦_𝑖−𝑦_𝑗²+_{𝑖,𝑗∈𝐸∗⧵𝐸′1}𝑦_𝑖−𝑦_𝑗²−_{𝑖,𝑗∈𝐸′1}𝑥_𝑖−𝑥_𝑗²+_{𝑖,𝑗∈𝐸⧵𝐸′1}𝑥_𝑖−𝑥_𝑗²=_{𝑖,𝑗∈𝐸∗⧵𝐸′1}𝑦_𝑖−𝑦_𝑗²−_{𝑖,𝑗∈𝐸⧵𝐸′1}𝑥_𝑖−𝑥_𝑗²=^𝑠_𝑖=1𝑥_𝑢+𝑥_𝑣𝑖²−^𝑠_𝑖=1𝑥_𝑣+𝑥_𝑣𝑖²=^𝑠_𝑖=1𝑥_𝑢²−𝑥_𝑣²+2^𝑠_𝑖=1𝑥_𝑣𝑖𝑥_𝑢+𝑥_𝑣≥0. (11) That is, 𝑦^′𝐿(𝑇^∗)𝑦≥𝑥^′𝐿(𝑇)𝑥. Thus, 𝜇(𝑇)=𝑥^′𝐿(𝑇)𝑥≤𝑦^′𝐿𝑇^∗𝑦≤max_𝑦=1𝑦^′𝐿𝑇^∗𝑦=𝜇𝑇^∗. (12) Therefore, 𝜇(𝑇)≤𝜇(𝑇^∗).
If 𝜇(𝑇)=𝜇(𝑇^∗), then the equalities in (12) hold. Thus, 𝜇(𝑇)=𝑦^′𝐿(𝑇^∗)𝑦=𝜇(𝑇^∗). By Theorem 1, we have 𝐿(𝑇^∗)𝑦=𝜇(𝑇^∗)𝑦. Thus, 𝜇𝑇^∗𝑦_𝑣=𝐿𝑇^∗𝑦_𝑣=𝑑_𝑣∗𝑦_𝑣−_{𝑣𝑖∈𝑁𝑇∗(𝑣)}𝑦_𝑖, (13) where 𝑑_𝑣∗ is the degree of vertex 𝑣 of tree 𝑇^∗.
In addition, from 𝐿(𝑇)𝑥=𝜇(𝑇)𝑥 we have 𝜇(𝑇)𝑥_𝑣=(𝐿(𝑇)𝑥)_𝑣=𝑑_𝑣𝑥_𝑣−_{𝑣𝑖∈𝑁𝑇(𝑣)}𝑥_𝑖=𝑑_𝑣𝑦_𝑣−_{𝑣𝑖∈𝑁𝑇∗(𝑣)}𝑥_𝑖−^𝑠_𝑖=1𝑥_𝑖, (14) where 𝑑_𝑣 is the degree of vertex 𝑣 of tree 𝑇.
By (13) and (14), we have 𝜇𝑇^∗−𝜇(𝑇)𝑥_𝑣=^𝑠_𝑖=1𝑥_𝑖−𝑠𝑥_𝑣+_{𝑣𝑖∈𝑁𝑇∗(𝑣)}𝑥_𝑖−𝑦_𝑖. (15) Assume, without loss of generality, that 𝑥_𝑣>0.
If 𝑥_𝑢>0, by Theorem 5, we have 𝜇𝑇^∗−𝜇(𝑇)𝑥_𝑣=^𝑠_𝑖=1𝑥_𝑖−𝑠𝑥_𝑣<0, (16) which implies the contradiction 𝜇(𝑇^∗)<𝜇(𝑇).
If 𝑥_𝑢<0, by Theorem 5, we have 𝜇𝑇^∗−𝜇(𝑇)𝑥_𝑣=^𝑠_𝑖=1𝑥_𝑖−𝑠𝑥_𝑣+_{𝑣𝑖∈𝑁𝑇∗(𝑣)}𝑥_𝑖−𝑥_𝑖<0, (17) which implies the contradiction 𝜇(𝑇^∗)<𝜇(𝑇).
Thus, if |𝑥_𝑢|≥|𝑥_𝑣|, then 𝜇(𝑇)<𝜇(𝑇^∗).

Corollary 11. Suppose that 𝑇^∗ is as defined as in Theorem 10  𝑦=(𝑦₁,𝑦₂,…,𝑦_𝑛)^𝑇 is the standard Laplacian spectral vector of 𝑇^∗, and then |𝑦_𝑢|>|𝑦_𝑣|.

Proof. Otherwise, if |𝑦_𝑢|≤|𝑦_𝑣|, by Theorem 10, we have 𝜇(𝑇^∗)<𝜇(𝑇). This contradiction completes the proof.

Corollary 12. Let 𝐺 be a simple connected graph with 𝑛 vertices. Let 𝑟, 𝑠, and 𝑡 be three distinct vertices, where 𝑟𝑠∈𝐸(𝐺) and 𝑟𝑡∉𝐸(𝐺). Let 𝐺^∗ be the simple connected graph obtained from 𝐺 by deleting the edge 𝑟𝑠 and adding the edge 𝑟𝑡. Suppose that 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑛)^′ is the standard Laplacian spectral vector of 𝐺, where 𝑥_𝑖 corresponds to the vertex 𝑣_𝑖(1≤𝑖≤𝑛). 𝑥^∗=(𝑥^∗₁,𝑥^∗₂,…,𝑥^∗_𝑛)^′ is the standard Laplacian spectral vector of 𝐺^∗, where 𝑥^∗_𝑖 corresponds to the vertex 𝑣^∗_𝑖(1≤𝑖≤𝑛). If |𝑥_𝑡|≥|𝑥_𝑠|, then |𝑥^∗_𝑡|>|𝑥^∗_𝑠|.

Proof. Obviously, 𝐺 is obtained from 𝐺^∗ by deleting the edge 𝑟𝑡 and adding the edge 𝑟𝑠. Assume that |𝑥^∗_𝑡|≤|𝑥^∗_𝑠|. Since 𝐺^∗ is a simple connected graph, by Theorem 10 we have 𝜇(𝐺^∗)<𝜇(𝐺). If |𝑥_𝑡|≥|𝑥_𝑠|, by Theorem 10 we have 𝜇(𝐺)<𝜇(𝐺^∗). This contradiction completes the proof.

Lemma 13 (see [17]). Let 𝑢 be a vertex of tree 𝐺, and for positive integers 𝑘 and 𝑙 let 𝐺_𝑘,𝑙 denote the tree obtained from 𝐺 by adding pendant paths of length 𝑘 and 𝑙 at 𝑢 (Figure 6). If 𝑘≥𝑙≥1, then 𝜇(𝐺_𝑘,𝑙)>𝜇(𝐺_{𝑘+1,𝑙−1}).

Figure 6 

Theorem 14 (see [7, 8]). Among all the trees for 𝑛 vertices and 𝑘 pendant vertices, the maximal Laplacian spectral radius is obtained uniquely at 𝑇_𝑛,𝑘.

Proof. We have to prove that if 𝑇∈𝒯_𝑛,𝑘, then 𝜇(𝑇)≤𝜇(𝑇_𝑛,𝑘) with equality only when 𝑇=𝑇_𝑛,𝑘. Let 𝑡 be the cardinality of the vertices of degree 3 or greater.
Case  1. 𝑡=0. In this case, 𝑇 is a path of length 𝑛, and hence 𝑇=𝑇_𝑛,2. We have 𝜇(𝑇)=𝜇(𝑇_𝑛,2).
Case  2. 𝑡=1. The result 𝜇(𝑇)≤𝜇(𝑇_𝑛,𝑘) follows from repeated use of Lemma 13, and equality holds if and only if 𝑇=𝑇_𝑛,𝑘.
Case  3. 𝑡>1. Let 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑛)^𝑇 be the standard Laplacian spectral vector of 𝑇, where 𝑥_𝑖 corresponds to the vertex 𝑣_𝑖 (1≤𝑖≤𝑛). Suppose that 𝑢,𝑣 are two vertices of 𝑇 of degree 3 or greater, and |𝑥_𝑢|≥|𝑥_𝑣|. Since 𝑇 is a tree, there is a unique path between 𝑢 and 𝑣, and only one of the neighbors of 𝑣, say 𝑤, is on the path. Assume {𝑣₁,𝑣₂,…,𝑣_𝑑𝑣−2}⊂𝑁_𝑇(𝑣)⧵{𝑤}. Delete the edges (𝑣,𝑣_𝑖) and add the edges (𝑢,𝑣_𝑖) (1≤𝑖≤𝑑_𝑣−2). Then we obtain a new tree 𝑇^∗₁. Obviously, 𝑇^∗₁ still has 𝑘 pendant vertices. By Theorem 10, we have 𝜇(𝑇)<𝜇(𝑇^∗₁) and the cardinality of the vertices of degree 3 or greater decreases to 𝑡−1.
If 𝑡−1>1, repeat the above step to 𝑇^∗₁ until the cardinality is only 1. Thus, we obtain trees 𝑇^∗₂,𝑇^∗₃,…,𝑇^∗_𝑡−1 and 𝜇(𝑇^∗₁)<𝜇(𝑇^∗₂)<⋯<𝜇(𝑇^∗_𝑡−1). Moreover, each 𝑇^∗_𝑖 has 𝑘 pendant vertices. Referring to Case 2, we have 𝜇(𝑇^∗_𝑡−1)≤𝜇(𝑇_𝑛,𝑘). Therefore, 𝜇(𝑇)<𝜇(𝑇_𝑛,𝑘).
The cases complete the proof.

Theorem 15 (see [8]). 𝜇(𝑇_{𝑛,𝑘−1})<𝜇(𝑇_𝑛,𝑘),3≤𝑘≤𝑛−1.

Proof. The proof is obvious from Lemma 13.

For the sake of clarity, we identify a graph by its characteristic polynomial. Let Φ(𝐿(𝐺)) be the characteristic polynomial of graph 𝐺. For convenience, we denote Φ(𝐿(𝐺)) by Φ(𝐺) sometimes. First, we require some lemmas.

Lemma 16 (see [9]). Let 𝑒=𝑢𝑣 be a cut edge of the simple connected graph 𝐺. 𝐺₁ and 𝐺₂ are two connected branches of 𝐺−𝑢𝑣, where 𝑢∈𝑉(𝐺₁), 𝑣∈𝑉(𝐺₂). Then Φ(𝐺)=Φ(𝐺₁)Φ(𝐺₂)−Φ(𝐺₁)Φ(𝐿_𝑣(𝐺₂))−Φ(𝐺₂)Φ(𝐿_𝑢(𝐺₁)). Here, Φ(𝐿_𝑣(𝐺₂)) denotes the determinant obtained by deleting the row and column for vertex 𝑣 of 𝐺₂ from the determinant for 𝐿(𝐺₂), and Φ(𝐿_𝑢(𝐺₁)) denotes the determinant obtained by deleting the row and column for vertex 𝑢 of 𝐺₁ from the determinant for 𝐿(𝐺₁).

Lemma 17 (see [18]). Let 𝑃_𝑛 be a path with 𝑛 vertices, let 𝑃⁰_𝑛 be the path obtained by adding a loop on a pendant vertex in 𝑃_𝑛, and let 𝑃_𝑛⁰⁰ be the path obtained by adding loops on the two pendant vertices in 𝑃_𝑛. Suppose that the degree of contribution corresponding to each loop is 1 and Φ(𝑃₀)=0, Φ(𝑃⁰₀)=1, Φ(𝑃₀⁰⁰)=1. Then (1)Φ(𝑃_𝑛+1)=(𝑥−2)Φ(𝑃_𝑛)+Φ(𝑃_𝑛−1),(𝑛≥1), (2)𝑥Φ(𝑃⁰_𝑛)=Φ(𝑃_𝑛+1)+Φ(𝑃_𝑛), (3)Φ(𝑃_𝑛)=𝑥Φ(𝑃⁰⁰_𝑛−1), (𝑛≥1),(4)Φ(𝑃_𝑛)=(𝑥/√𝑥²−4𝑥)(𝑎^𝑛−𝑏^𝑛), where 𝑎=(𝑥−2+√𝑥²−4𝑥)/2 and 𝑏=(𝑥−2−√𝑥²−4𝑥)/2.

From Lemma 16, by induction we can obtain the following lemma.

Lemma 18. Let 𝐺₁,𝐺₂,…,𝐺_𝑠,𝐺 be 𝑠+1 disjoint simple connected graphs, where 𝑣_𝑖∈𝐺_𝑖(𝑖=1,2,…,𝑠) and 𝑣∈𝐺. Let 𝐻_𝑠 be the graph obtained from 𝐺₁,𝐺₂,…,𝐺_𝑠 and 𝐺 by adding new edges 𝑣𝑣₁,𝑣𝑣₂,…,𝑣𝑣_𝑠. Then Φ𝐻_𝑠=Φ(𝐺)^𝑠_𝑖=1Φ𝐺_𝑖−Φ𝐿_𝑣𝑖𝐺_𝑖−Φ𝐿_𝑣(𝐺)×^𝑠_𝑖=1Φ𝐺_𝑖^𝑠_{𝑗=1,𝑗≠𝑖}Φ𝐺_𝑗−Φ𝐿_𝑣𝑗𝐺_𝑗. (18)

From Lemmas 16, 17, and 18, we have the following theorem.

Theorem 19. The characteristic polynomial of 𝑇_𝑛,𝑘(2≤𝑘≤𝑛−1) is Φ𝑇_𝑛,𝑘=Φ𝑃_𝑞+1−Φ𝑃⁰_𝑞^𝑟−1Φ𝑃_𝑞−Φ𝑃⁰_𝑞−1^{𝑘−𝑟−1}×𝑥Φ𝑃_𝑞+1−Φ𝑃⁰_𝑞Φ𝑃_𝑞−Φ𝑃⁰_𝑞−1−𝑟Φ𝑃_𝑞−Φ𝑃⁰_𝑞−1Φ𝑃_𝑞+1−(𝑘−𝑟)Φ𝑃_𝑞Φ𝑃_𝑞+1−Φ𝑃⁰_𝑞, (19) where 𝑞=[(𝑛−1)/𝑘], 𝑛−1=𝑘𝑞+𝑟(0≤𝑟≤𝑘−1), and 𝑃_𝑞 is a path of length 𝑞.

As for the Laplacian spectral radius of 𝑇_𝑛,𝑘, we only need to consider the greatest root of 𝑥Φ𝑃_𝑞+1−Φ𝑃⁰_𝑞Φ𝑃_𝑞−Φ𝑃⁰_𝑞−1−𝑟Φ𝑃_𝑞−Φ𝑃⁰_𝑞−1Φ𝑃_𝑞+1−(𝑘−𝑟)Φ𝑃_𝑞Φ𝑃_𝑞+1−Φ𝑃⁰_𝑞=0. (20)

In general, the Laplacian spectral radius is difficult to calculate although the characteristic polynomial can be identified by Chebvshev polynomials. Thus, we only give some results in some cases.

Lemma 20. For a univariate cubic equation 𝑥³+𝑘𝑥²+𝑚𝑥+𝑛=0, we can obtain the roots as follows: 𝑥₁=³−𝑞2+𝑞2²+𝑝3³+³−𝑞2−𝑞2²+𝑝3³−𝑘3,𝑥₂=𝜔³−𝑞2+𝑞2²+𝑝3³+𝜔²³−𝑞2−𝑞2²+𝑝3³−𝑘3,𝑥₃=𝜔²³−𝑞2+𝑞2²+𝑝3³+𝜔³−𝑞2−𝑞2²+𝑝3³−𝑘3, (21) where 𝑝=−(1/3)𝑘²+𝑚, 𝑞=(2/27)𝑘³−(1/3)𝑘𝑚+𝑛 and 𝜔=𝑒^2𝜋𝑖/3.

Proof. The lemma is easy to prove.

Theorem 21. If [𝑛/2]≤𝑘≤𝑛−1, then 𝜇𝑇_𝑛,𝑘=𝑘+43+2𝑘²−𝑘+43𝜃cos3,𝜃=arccos2𝑘³−3𝑘²−48𝑘+27𝑛−162𝑘²−𝑘+4^3/2,(0<𝜃<𝜋). (22) Moreover, if 𝑘=𝑛−1, then 𝜇(𝑇_{𝑛,𝑛−1})=𝑛.

Proof. In this case, for [𝑛/2]≤𝑘≤𝑛−1,𝑞=1. By (20), we have 𝑥Φ𝑃_𝑞+1−Φ𝑃⁰_𝑞Φ𝑃_𝑞−Φ𝑃⁰_𝑞−1−𝑟Φ𝑃_𝑞−Φ𝑃⁰_𝑞−1Φ𝑃_𝑞+1−(𝑘−𝑟)Φ𝑃_𝑞Φ𝑃_𝑞+1−Φ𝑃⁰_𝑞=0. (23) Hence, 𝑥³−(𝑘+4)𝑥²+(3𝑘+4)𝑥−(𝑘+𝑟+1)=0.
Let 𝑓(𝑥)=𝑥³−(𝑘+4)𝑥²+(3𝑘+4)𝑥−(𝑘+𝑟+1), and then we have1^𝑜 for 𝑓(0)=−(𝑘+𝑟+1)<0 and 𝑓(1)=𝑘−𝑟>0, and thus 𝑓(𝑥) has a root in (0,1),2^𝑜 for 𝑓(1)>0 and 𝑓(𝑘)=−𝑘(𝑘−3)−𝑟−1<0, and thus 𝑓(𝑥) has a root in (1,𝑘),3^𝑜 for 𝑓(𝑘)<0 and 𝑓(+∞)>0, and thus 𝑓(𝑥) has a root in (𝑘,+∞). Therefore, the equation has three different real roots.
Let 𝑥=𝑦+(𝑘+4)/3, and then the equation can translate to 𝑦³+𝑝𝑦+𝑞=0, where 𝑝=−(1/3)(𝑘²−𝑘+4) and 𝑞=−(1/27)(2𝑘³−3𝑘²−21𝑘+27𝑟+11).
Let 𝑠=−𝑞/2+√−Δ𝑖=‖𝑠‖𝑒^𝑖𝜃, where Δ=(𝑞/2)²+(𝑝/3)³<0. Then‖𝑠‖=−𝑞2+√=−Δ𝑖−𝑞2²+(−Δ)=𝑞²4−𝑞²4−𝑝³=127𝑘27²−𝑘+4^3/2,𝜃=arccos−𝑞/2‖𝑠‖=arccos2𝑘³−3𝑘²−21𝑘+27𝑟+112𝑘²−𝑘+4^3/2,(0<𝜃<𝜋). (24) Thus, based on the Lemma 20, we can obtain three roots: 𝑥₁=³−𝑞2+√−Δ𝑖+³−𝑞2−√−Δ𝑖+𝑘+43=³‖𝑠‖𝑒^𝑖𝜃+³‖𝑠‖𝑒^−𝑖𝜃+𝑘+43=𝑘²−𝑘+43𝑒^𝑖𝜃/3+𝑘²−𝑘+43𝑒^{−𝑖𝜃/3}+𝑘+43=𝑘+43+2𝑘²−𝑘+43𝜃cos3,𝑥₂=𝜔³−𝑞2+√−Δ𝑖+𝜔²³−𝑞2−√−Δ𝑖+𝑘+43=𝜔³‖𝑠‖𝑒^𝑖𝜃+𝜔²³‖𝑠‖𝑒^−𝑖𝜃+𝑘+43=𝑘²−𝑘+43𝑒^𝑖𝜃/3𝑒^2𝑖𝜋/3+𝑘²−𝑘+43𝑒^{−𝑖𝜃/3}𝑒^4𝑖𝜋/3+𝑘+43=𝑘+43+2𝑘²−𝑘+43𝜃cos3+2𝜋3,𝑥₃=𝜔²³−𝑞2+√−Δ𝑖+𝜔³−𝑞2−√−Δ𝑖+𝑘+43=𝜔²³‖𝑠‖𝑒^𝑖𝜃+𝜔³‖𝑠‖𝑒^−𝑖𝜃+𝑘+43=𝑘²−𝑘+43𝑒^𝑖𝜃/3𝑒^4𝑖𝜋/3+𝑘²−𝑘+43𝑒^{−𝑖𝜃/3}𝑒^2𝑖𝜋/3+𝑘+43=𝑘+43+2𝑘²−𝑘+43cos2𝜋3−2𝜃3. (25)(26)(27) For 0<𝜃<𝜋, then cos(𝜃/3)>cos(𝜃/3+2𝜋/3) and cos(𝜃/3)>cos(2𝜋/3−𝜃/3). Thus, 𝑥₁>𝑥₂, 𝑥₁>𝑥₃. So the largest root of the univariate cubic equation is 𝜇𝑇_𝑛,𝑘=𝑘+43+2𝑘²−𝑘+43𝜃cos3,𝜃=arccos2𝑘³−3𝑘²−21𝑘+27𝑟+112𝑘²−𝑘+4^3/2,(0<𝜃<𝜋). (28) For 𝑟=𝑛−1−𝑘, we have 𝜇𝑇_𝑛,𝑘=𝑘+43+2𝑘²−𝑘+43𝜃cos3,𝜃=arccos2𝑘³−3𝑘²−48𝑘+27𝑛−162𝑘²−𝑘+4^3/2,(0<𝜃<𝜋). (29) If 𝑘=𝑛−1, then it is easy to prove that 𝑓(𝑛)=0. Thus, 𝜇(𝑇_{𝑛,𝑛−1})=𝑛.

From Theorems 14, 15, and 21, we have the following corollaries.

Corollary 22. Let 𝑇 be a tree for 𝑛 vertices. Then 𝜇(𝑇)≤𝑛 and equality holds if and only if 𝑇=𝑇_{𝑛,𝑛−1}(𝐾_1,𝑛−1), the star with 𝑛 vertices.

Corollary 23. Let 𝑇 be a tree for 𝑛 vertices with 𝑛≥4 and 𝑇≇𝑇_{𝑛,𝑛−1}. Then 𝜇(𝑇)≤𝑛+23+2√𝑛²−5𝑛+103𝜃cos3,𝜃=arccos2𝑛³−15𝑛²+15𝑛+522𝑛²−5𝑛+10^3/2,(0<𝜃<𝜋) (30) and equality holds if and only if 𝑇=𝑇_{𝑛,𝑛−2}(𝑆(1,𝑛−3)).

Corollary 24. Let 𝑇 be a tree for 𝑛 vertices. Let 𝑀(𝑇) denote one maximum matching of 𝑇 and suppose that |𝑀(𝑇)|=𝑖. Then 𝜇(𝑇)≤𝑛+4−𝑖3+2𝑛²−(2𝑖+1)𝑛+𝑖²+𝑖+43𝜃cos3,𝜃=arccos2𝑛³−(6𝑖+3)𝑛²+6𝑖²+6𝑖−21𝑛−2𝑖³−3𝑖²+48𝑖−162𝑛²−(2𝑖+1)𝑛+𝑖²+𝑖+4^3/2(0<𝜃<𝜋), (31)(32) and equality holds if and only if 𝑇=𝑇_{𝑛,𝑛−𝑖}.

Proof. Let 𝑘 denote the cardinality of the pendant vertices of 𝑇. For |𝑀(𝑇)|=𝑖, we have 𝑘≤𝑖+𝑛−2𝑖=𝑛−𝑖. By Theorems 14 and 15, we have 𝜇(𝑇)≤𝜇(𝑇_𝑛,𝑘)≤𝜇(𝑇_{𝑛,𝑛−𝑖}), and equality holds if and only if 𝑇=𝑇_{𝑛,𝑛−𝑖}.
𝑇_{𝑛,𝑛−𝑖} is shown in Figure 7. Since 𝑖≤𝑛/2,𝑛−𝑖≥𝑛/2≥[𝑛/2], by Theorem 21 we have 𝜇(𝑇)≤𝑛+4−𝑖3+2𝑛²−(2𝑖+1)𝑛+𝑖²+𝑖+43𝜃cos3,𝜃=arccos2𝑛³−(6𝑖+3)𝑛²+6𝑖²+6𝑖−21𝑛−2𝑖³−3𝑖²+48𝑖−162𝑛²−(2𝑖+1)𝑛+𝑖²+𝑖+4^3/2,(0<𝜃<𝜋). (33)(34)

Figure 7 

Corollary 25. Let 𝑇 be any tree for 𝑛=2𝑖 vertices and |𝑀(𝑇)|=𝑖. Then 𝜇(𝑇)≤𝑖+43+2𝑖²−𝑖+43𝜃cos3,𝜃=arccos2𝑖³−3𝑖²+6𝑖−162𝑖²−𝑖+4^3/2,(0<𝜃<𝜋) (35) and equality holds if and only if 𝑇=𝑇_{𝑛,𝑛−𝑖}.

Acknowledgments

The authors would like to thank the anonymous referees for valuable suggestions and corrections, which have improved the paper. This research was supported by the Scientific Research Foundation of Huaqiao University (10HZR26) and the Natural Science Foundation of Fujian Province (Z0511028).