On graphs whose third largest distance eigenvalue dose not exceed $-1$

Highlights

• In thispaper,the distance eigenvalues of chain graphs are originally discussed. We present the chain graphs whose third largest distance eigenvalues are atmost-1.

• Furthermore, using clique extension, we characterize all connected graphs whose third largest distance eigenvalue is atmost-1.

• As an application, it is proved that a graph is determined by its distance spectrum if its third largest distance eigenvalue is lessthan-1. This is our important goal and highlight.

Abstract

In this paper, the distance eigenvalues of chain graphs are discussed. Using clique extension, we characterize all connected graphs whose third largest distance eigenvalue is at most $-1$. As an application, it is proved that a graph is determined by its distance spectrum if its third largest distance eigenvalue is less than $-1$.

Introduction

Unless stated otherwise, we follow [1] for terminology and notations. Let $G$ be a connected graph with at least three vertices. The vertex set and edge set of $G$ are denoted by $V\left(G\right)$ and $E\left(G\right),$ respectively. The set of neighbors of a vertex $v$ in $G$ is denoted by $N_G\left(v\right)$. For any two vertices $u$ and $v$ in $G,$ the length of a short path from $u$ to $v$ is called the distance between these two vertices and denoted $d_G(u,v)$. The distance matrix of $G,$ denoted by $\mathcal{D}\left(G\right),$ is a symmetrical matrix whose entries indicate the distance between vertices of $G$. The third largest eigenvalue of $\mathcal{D}\left(G\right),$ denoted by ${\lambda }_3\left(G\right),$ is called the third largest distance eigenvalue of $G$. The distance characteristic polynomial of $G$ is $P\left(\lambda \right)=\det (\lambda I-\mathcal{D}(G\left)\right),$ where $I$ means the identity matrix. Clearly, the distance eigenvalues of $G$ are the roots of $P\left(\lambda \right)=0$. Let $G_1$ and $G_2$ be two disjoint graphs. The join graph $G_1\vee G_2$ is the graph with vertex set $V\left(G_1\right)\cup V\left(G_2\right)$ and edge set $E\left(G_1\right)\cup E\left(G_2\right)\cup \{uv:u\in V\left(G_1\right),v\in V\left(G_2\right)\}$. As usual, we use $P_n,$ $C_n,$ $K_n$ and $K_{s,n-s}$ to denote the path, the cycle, the complete graph and the complete bipartite graph with $n$ vertices, respectively.

For a complete graph $K_n$ with $n\ge 3$ vertices, it is easy to see that ${\lambda }_3\left(K_n\right)=-1$. If $G$ is not a complete graph, then $G$ contains a path $P_3$ as its induced subgraph, and $\mathcal{D}\left(P_3\right)$ obviously is a principal submatrix of $\mathcal{D}\left(G\right)$. The third largest eigenvalue of $\mathcal{D}\left(P_3\right)$ is $-2$. Thus, by Cauchy Interlacing Theorem (see, for example, [6]), a lower bound for ${\lambda }_3\left(G\right)$ is

$${\lambda }_3\left(G\right)\ge -2.$$

The distribution of ${\lambda }_3\left(G\right)$ is considered in the paper. Our aim is to determine all connected graphs whose third largest distance eigenvalue belongs to $[-2,-1]$.

Let $h_1\ge h_2\ge \cdots \ge h_s$ be a sequence of positive integers with $h_1\ge s\ge 1$. The graph $C(h_1,h_2,\dots ,h_s)$ is a connected bipartite graph satisfying the following conditions: the vertices are partitioned into $\{v_1,v_2,\dots ,v_s\}$ and $\{u_1,u_2,\dots ,u_{h_1}\},$ and the neighbors of $v_i$ are $\{u_1,\dots ,u_{h_i}\}$ for $1\le i\le s$. The graph $C(h_1,h_2,\dots ,h_s)$ is known as a chain graph [3] or a double nested graph [4]. Clearly, any complete bipartite graph is a chain graph. Recall that a connected graph is a chain graph if it is $\{2K_2,C_3,C_5\}$-free (see, for example, [5]).

We now define a family of chain graphs:

$$\mathcal{G}=\{K_{s_1,t_1},C(t_2,t_2-1),C(t_3,1),C(3,1,1),C(3,2,1),C(3,2,2)\},$$

where $s_1+t_1\ge 3,$ $t_2\ge 2$ and $t_3\ge 3$. The first main result of this paper presents the chain graphs whose third largest distance eigenvalues are at most $-1$.

Theorem 1.1

Let $G$ be a chain graph. Then ${\lambda }_3\left(G\right)\le -1$ if and only if $G\in \mathcal{G}$.

Let $G$ be a connected graph with vertex set $V\left(G\right)=\{v_1,v_2,\dots ,v_n\}$. Construct a new graph as follows. We replace the vertex $v_i$ by a clique $V_i$ for $1\le i\le n,$ and add edges joining each vertex of $V_i$ to each vertex of $V_j$ if $v_i{v}_j\in E\left(G\right)$. The resulting graph is called a clique extension of the graph $G$. The number $\max \left\{\right|V_i|:1\le i\le n\}$ is called the extension index of the graph. Moreover, if $|V_1|=|V_2|=\cdots =|V_n|=k,$ then we call it a $k$-clique extension of $G$. Note that any graph is the 1-clique extension of itself.

Using Theorem 1.1 and the tool of clique extension, all connected graphs whose third largest distance eigenvalues are at most $-1$ are determined.

Theorem 1.2

Let $G$ be a connected graph with at least three vertices. Then ${\lambda }_3\left(G\right)\le -1$ if and only if $G$ is either a complete graph or a clique extension of a graph in $\mathcal{G}$.

It is well-known that there is only one positive distance eigenvalue for any tree (see [2], [10]). In [11], Xing and Zhou showed that the path and star are the only two graphs whose the second largest distance eigenvalue is less than $-1/2$. A generalized result was presented in [10] by considering the block graphs. In this paper, the third largest distance eigenvalue of trees is also discussed.

Given a connected graph $G$ with at least two vertices, one can see that the clique extension of $G$ cannot be a tree if its extension index is greater than 1. Using this fact and Theorem 1.2, we immediately determine the trees whose third largest distance eigenvalue is at most $-1$.

Corollary 1.3

Let $T$ be a tree with at least three vertices. Then ${\lambda }_3\left(T\right)\le -1$ if and only if $T$ is isomorphic to $K_{1,s},$ $C(t,1)$ or $C(3,1,1),$ where $s\ge 2$ and $t\ge 2$.

A graph $G$ is said to be determined by its distance spectrum if there is no other non-isomorphic graph which has the same distance spectrum as $G$. According to Theorem 1.2, we show that the graph is determined by its distance spectrum if its third largest distance eigenvalue is less than $-1$.

Theorem 1.4

Let $G$ be a connected graph with ${\lambda }_3\left(G\right)<-1$. Then $G$ is determined by its distance spectrum.

In addition, $C(3,1,1)$ is the only tree whose third largest distance eigenvalue equals $-1$. Note also that $C(3,1,1)$ is a double star, and it was proved that double stars are determined by their distance spectra [9]. Therefore, the following result is an immediate consequence of Theorem 1.4.

Corollary 1.5

Let $T$ be a tree with ${\lambda }_3\left(T\right)\le -1$. Then $T$ is determined by its distance spectrum.

The proofs of the main results are presented in the next section.