On graphs with $m(\partial^L_1)=n-3$

TL;DR

This paper characterizes connected graphs with a specific spectral property, namely those whose largest distance Laplacian eigenvalue has multiplicity exactly n-3, extending previous classifications for multiplicities n-1 and n-2.

Contribution

It provides a complete characterization of graphs where the largest distance Laplacian eigenvalue has multiplicity n-3, filling a gap in spectral graph theory.

Findings

Graphs with $m(oundary^L_1)=n-3$ are fully characterized.

Extends previous results for multiplicities n-1 and n-2.

Contributes to understanding the spectral properties of graphs.

Abstract

Let $\partial^L_1\ge\partial^L_2\ge\cdots\ge\partial^L_n$ be the distance Laplacian eigenvalues of a connected graph $G$ and $m(\partial^L_i)$ the multiplicity of $\partial^L_i$. It is well known that the graphs with $m(\partial^L_1)=n-1$ are complete graphs. Recently, the graphs with $m(\partial^L_1)=n-2$ have been characterized by Celso et al. In this paper, we completely determine the graphs with $m(\partial^L_1)=n-3$.

Equations

$$\{n+m,m+{\mu }_1,m+{\mu }_2,\dots ,m+{\mu }_n,n+{\mu }_1^{\prime },n+{\mu }_2^{\prime },\dots ,n+{\mu }_m^{\prime },0\}.$$

$$M_{1}=\left(\begin{array}[]{ccccc}t_{1}&-1&-2&-3&-2\\ -1&t_{2}&-1&-2&-1\\ -2&-1&t_{3}&-1&-2\\ -3&-2&-1&t_{4}&-2\\ -2&-1&-2&-2&t_{5}\end{array}\right)\begin{array}[]{l}v_{1}\\ v_{2}\\ v_{3}\\ v_{4}\\ u\end{array},M_{1}^{\prime}=\left(\begin{array}[]{ccccc}t_{1}&-1&-2&-3&-2\\ -1&t_{2}&-1&-2&-1\\ -2&-1&t_{3}&-1&-2\\ -3&-2&-1&t_{4}&-3\\ -2&-1&-2&-3&t_{5}\end{array}\right)\begin{array}[]{l}v_{1}\\ v_{2}\\ v_{3}\\ v_{4}\\ u\end{array}.$$

$$y_2+y_3+y_4+y_5=0$$

$$M_1^{\prime }y={\partial }_1^{L}y.$$

$$-y_2-2y_3-3y_4-2y_5=0.$$

$$M_{2}=\left(\begin{array}[]{ccccc}t_{1}&-1&-2&-3&-1\\ -1&t_{2}&-1&-2&-1\\ -2&-1&t_{3}&-1&-2\\ -3&-2&-1&t_{4}&-3\\ -1&-1&-2&-3&t_{5}\end{array}\right)\begin{array}[]{l}v_{1}\\ v_{2}\\ v_{3}\\ v_{4}\\ u\end{array},M_{2}^{\prime}=\left(\begin{array}[]{ccccc}t_{1}&-1&-2&-3&-1\\ -1&t_{2}&-1&-2&-1\\ -2&-1&t_{3}&-1&-2\\ -3&-2&-1&t_{4}&-2\\ -1&-1&-2&-2&t_{5}\end{array}\right)\begin{array}[]{l}v_{1}\\ v_{2}\\ v_{3}\\ v_{4}\\ u\end{array}.$$

$$M_{4}=\left(\begin{array}[]{ccccc}t_{1}&-1&-2&-3&-2\\ -1&t_{2}&-1&-2&-1\\ -2&-1&t_{3}&-1&-1\\ -3&-2&-1&t_{4}&-2\\ -2&-1&-1&-2&t_{5}\end{array}\right)\begin{array}[]{l}v_{1}\\ v_{2}\\ v_{3}\\ v_{4}\\ u\end{array}.$$

$$M_{5}=\left(\begin{array}[]{ccccc}t_{1}&-1&-2&-3&-1\\ -1&t_{2}&-1&-2&-1\\ -2&-1&t_{3}&-1&-1\\ -3&-2&-1&t_{4}&-2\\ -1&-1&-1&-2&t_{5}\end{array}\right)\begin{array}[]{l}v_{1}\\ v_{2}\\ v_{3}\\ v_{4}\\ u\end{array}.$$

$$M_{3}=\left(\begin{array}[]{ccccc}t_{1}&-1&-2&-3&-1\\ -1&t_{2}&-1&-2&-2\\ -2&-1&t_{3}&-1&-1\\ -3&-2&-1&t_{4}&-2\\ -1&-2&-1&-2&t_{5}\end{array}\right)\begin{array}[]{l}v_{1}\\ v_{2}\\ v_{3}\\ v_{4}\\ u\end{array}.$$

$$Tr(v_2)=\underset{v\in V(G)}{\max }Tr(v).$$

$$Tr(v_2)=2a+b+1<2a+b+3=Tr(v_4),$$

$$N_{1}=\left(\begin{array}[]{ccccc}t_{1}&-1&-2&-2&-1\\ -1&t_{2}&-1&-2&-2\\ -2&-1&t_{3}&-1&-2\\ -2&-2&-1&t_{4}&-1\\ -1&-2&-2&-1&t_{5}\end{array}\right)\begin{array}[]{l}v_{1}\\ v_{2}\\ v_{3}\\ v_{4}\\ u\end{array}.$$

$$\left(\begin{array}[]{ccccc}t_{1}&-1&-2&-2&-1\\ -1&t_{2}&-1&-2&-2\\ -2&-1&t_{3}&-1&-2\\ -2&-2&-1&t_{4}&-1\\ -1&-2&-2&-1&t_{5}\end{array}\right)\left(\begin{array}[]{c}0\\ a\\ 1\\ -1\\ -a\end{array}\right)=\partial^{L}_{1}\left(\begin{array}[]{c}0\\ a\\ 1\\ -1\\ -a\end{array}\right).$$

$${\partial }_1^L=t_4+a+1.$$

$${\partial }_1^L=t_5+\frac{1}{a}+2.$$

$$t_4=t_5+1.$$

$$\left(\begin{array}[]{ccccc}t_{1}&-1&-2&-2&-1\\ -1&t_{2}&-1&-2&-2\\ -2&-1&t_{3}&-1&-2\\ -2&-2&-1&t_{4}&-1\\ -1&-2&-2&-1&t_{5}\end{array}\right)\left(\begin{array}[]{c}b\\ 0\\ -b\\ 1\\ -1\end{array}\right)=\partial^{L}_{1}\left(\begin{array}[]{c}b\\ 0\\ -b\\ 1\\ -1\end{array}\right).$$

$${\partial }_1^L=t_4-b+1=t_5-b+1.$$

$$N_{2}=\left(\begin{array}[]{ccccc}t_{1}&-1&-2&-2&-1\\ -1&t_{2}&-1&-2&-1\\ -2&-1&t_{3}&-1&-2\\ -2&-2&-1&t_{4}&-1\\ -1&-1&-2&-1&t_{5}\end{array}\right)\begin{array}[]{l}v_{1}\\ v_{2}\\ v_{3}\\ v_{4}\\ u\end{array}.$$

$$N_{3}=\left(\begin{array}[]{ccccc}t_{1}&-1&-2&-2&-1\\ -1&t_{2}&-1&-2&-1\\ -2&-1&t_{3}&-1&-1\\ -2&-2&-1&t_{4}&-1\\ -1&-1&-1&-1&t_{5}\end{array}\right)\begin{array}[]{l}v_{1}\\ v_{2}\\ v_{3}\\ v_{4}\\ u\end{array}.$$

$${\mathcal{H}}_1(n)=\{G\in \mathcal{G}(n,n-3)\mid {\text{Spec}}_{\mathcal{L}}(G)=[({\partial }_1^L{)}^{n-3},{\partial }_{n-2}^L,{\partial }_{n-1}^L=n,{\partial }_n^L=0]\},$$

$${\mathcal{H}}_2(n)=\{G\in \mathcal{G}(n,n-3)\mid {\text{Spec}}_{\mathcal{L}}(G)=[({\partial }_1^L{)}^{n-3},{\partial }_{n-2}^L={\partial }_{n-1}^L=n,{\partial }_n^L=0]\}.$$

$$\mathcal{G}(n,n-3)={\mathcal{H}}_1(n)\cup {\mathcal{H}}_2(n).$$

$$\left\{\begin{array}[]{l}\emph{Spec}_{\mathcal{L}}(K_{2,n-2})=\{(2n-2)^{n-3},n+2,n,0\}\\ \emph{Spec}_{\mathcal{L}}(K_{n/2,n/2}+e)=\{(\frac{3n}{2})^{n-3},\frac{3n}{2}-2,n,0\}\\ \emph{Spec}_{\mathcal{L}}(K_{1,n-1}+e)=\{(2n-1)^{n-3},2n-3,n,0\}\end{array}\right.$$

$$\left\{\begin{array}[]{l}\emph{Spec}_{\mathcal{L}}(K_{2}\nabla(n-2)K_{1})=\{(2n-2)^{n-3},n^{2},0\}\\ \emph{Spec}_{\mathcal{L}}(K_{1}\nabla K_{\frac{n-1}{2},\frac{n-1}{2}})=\{((3n-1)/2)^{n-3},n^{2},0\}\\ \emph{Spec}_{\mathcal{L}}(K_{\frac{n}{3},\frac{n}{3},\frac{n}{3}})=\{(4n/3)^{n-3},n^{2},0\}\end{array}\right.$$

$$\mathcal{G}(n,n-3)=\{K_{2,n-2},K_{1,n-1}+e,K_{n/2,n/2}+e,K_2\nabla (n-2)K_1,K_1\nabla K_{\frac{n-1}{2},\frac{n-1}{2}},K_{\frac{n}{3},\frac{n}{3},\frac{n}{3}}\}.$$

$$\left\{\begin{array}[]{l}\mathcal{G}(4,1)=\{P_{4},K_{1,3}+e,K_{2}\nabla 2K_{1}\}\\ \mathcal{G}(5,2)=\{K_{2,3},K_{1,4}+e,K_{2}\nabla 3K_{1},K_{1}\nabla K_{2,2},C_{5}\}\end{array}\right..$$

Full text

\floatsetup

[table]capposition=top

On graphs with $m(\partial^{L}_{1})=n-3$ 111Supported

by the National Natural Science Foundation of China (Grant Nos. 11671344, 11531011, 11626205).

Lu Lu,   Qiongxiang Huang222 Corresponding author. Email: huangqx@xju.edu.cn,  Xueyi Huang

College of Mathematics and Systems Science, Xinjiang University, Urumqi, Xinjiang 830046, P.R. China

Abstract

Let $\partial^{L}_{1}\geq\partial^{L}_{2}\geq\cdots\geq\partial^{L}_{n}$ be the distance Laplacian eigenvalues of a connected graph $G$ and $m(\partial^{L}_{i})$ the multiplicity of $\partial^{L}_{i}$. It is well known that the graphs with $m(\partial^{L}_{1})=n-1$ are complete graphs. Recently, the graphs with $m(\partial^{L}_{1})=n-2$ have been characterized by Celso et al. In this paper, we completely determine the graphs with $m(\partial^{L}_{1})=n-3$.

Keywords: Distance Laplacian matrix; Laplacian matrix; Largest eigenvalue; Determined by distance Laplacian spectrum

AMS subject classifications: 05C50; 05C12; 15A18

1 Introduction

In this paper we only consider simple connected graphs. Let $G=(V,E)$ be a connected graph with vertex set $V=\{v_{1},v_{2},\ldots,v_{n}\}$ and edge set $E=\{e_{1},e_{2},\ldots,e_{m}\}$. The distance between $v_{i}$ and $v_{j}$, denoted by $d_{G}(v_{i},v_{j})$, is defined as the length of a shortest path between them. The diameter of $G$, denoted by $d(G)$, is the maximum distance between any two vertices of $G$. The distance matrix of $G$, denoted by $\mathcal{D}(G)$, is the $n\times n$ matrix whose $(i,j)$-entry is equal to $d_{G}(v_{i},v_{j})$, $i,j=1,2,\ldots,n$. The transmission $Tr(v_{i})$ of a vertex $v_{i}$ is defined as the sum of the distances between $v_{i}$ and all other vertices in $G$, that is, $Tr(v_{i})=\sum_{j=1}^{n}d_{G}(v_{i},v_{j})$. For more details about the distance matrix we refer the readers to [1]. Aouchiche and Hansen [2] introduced the Laplacian for the distance matrix of $G$ as $\mathcal{D}^{L}(G)=Tr(G)-\mathcal{D}(G)$, where $Tr(G)=diag(Tr(v_{1}),Tr(v_{2}),\ldots,Tr(v_{n}))$ is the diagonal matrix of the vertex transmissions in $G$. The eigenvalues of $\mathcal{D}^{L}(G)$, listed by $\partial^{L}_{1}\geq\partial^{L}_{2}\geq\cdots\geq\partial_{n}=0$, are called the distance Laplacian eigenvalues of $G$. The multiplicity of $\partial^{L}_{i}$ is denoted by $m(\partial^{L}_{i})$. The distance eigenvalues together with their multiplicities is called the distance Laplacian spectrum of $G$, denoted by $\emph{Spec}_{\mathcal{L}}(G)$.

The distance Laplacian matrix aroused many active studies, such as [11, 1, 10, 6]. Graphs with few distinct eigenvalues form an interesting class of graphs and possess nice combinatorial properties. With respect to distance Laplacian eigenvalues, we focus on the graphs with $m(\partial^{L}_{1})$ being large. Denote by $\mathcal{G}(n)$ the set of connected graphs of order $n$. Let $\mathcal{G}(n,k)=\{G\in\mathcal{G}(n)\mid m(\partial^{L}_{1})=k\}$ be the set of connected graphs with $m(\partial^{L}_{1})=k$. Aouchiche and Hansen [1] proved that $\mathcal{G}(n,n-1)=\{K_{n}\}$ and conjectured that $\mathcal{G}(n,n-2)=\{K_{1,n-1},K_{n/2,n/2}\}$, which has been confirmed by Celso et al. [6]. Motivated by their work, we try to characterize $\mathcal{G}(n,n-3)$. In this paper, we completely determine the graphs in $\mathcal{G}(n,n-3)$ (Theorem 3.3). By the way, we show that all these graphs are determined by their distance Laplacian spectra (Corollary 3.3).

2 Preliminaries

Let $G$ be a connected graph, we always denote by $N_{G}(v)$ the neighbuor set of $v$ in $G$, that is, $N_{G}(v)=\{u\in V(G)\mid u\sim v\}$. The $i$-th largest distance Laplacian eigenvalue of $G$ is denoted by $\partial^{L}_{i}(G)$, whose multiplicity is denoted by $m(\partial^{L}_{i}(G))$. When it is clear from the context which graph $G$ we mean, we delete $G$ from the notations like $d_{G}(v_{i},v_{j})$, $N_{G}(v)$, $\partial^{L}_{i}(G)$ and $m(\partial^{L}_{i}(G))$. For a subset $S\subseteq V(G)$, let $G[S]$ denote the subgraph of $G$ induced by $S$.

As usual, we always write, respectively, $K_{n}$, $P_{n}$ and $C_{n}$ for the complete graph, the path and the cycle on $n$ vertices. For integers $a_{1},a_{2},\ldots,a_{k}\geq 1$, let $K_{a_{1},a_{2},\ldots,a_{k}}$ denote the complete $k$-partite graph on $a_{1}+a_{2}+\cdots+a_{k}$ vertices. Let $G$ be a connected graph, denote by $\bar{G}$ the complement of $G$, which is a graph with vertex set $V(\bar{G})=V(G)$ and two vertices are adjacent whenever they are not adjacent in $G$. Let $G_{1}=(V_{1},E_{1})$ and $G_{2}=(V_{2},E_{2})$ be two connected graphs, the (disjoint-)union of $G_{1}$ and $G_{2}$ is the graph $G_{1}\cup G_{2}$, whose vertex set is $V_{1}\cup V_{2}$ and edge set is $E_{1}\cup E_{2}$. The join of $G_{1}$ and $G_{2}$ is the graph $G_{1}\nabla G_{2}$, which is obtained from $G_{1}\cup G_{2}$ by joining each vertex of $G_{1}$ with every vertex of $G_{2}$. Moreover, we write $mG=\underbrace{G\cup G\cup\cdots\cup G}_{m}$ for an integer $m\geq 2$.

At first, we introduce the famous Cauchy interlacing theorem.

Theorem 2.1** ([8]).**

Let $A$ be a real symmetric matrix of order $n$ with eigenvalues $\lambda_{1}(A)\geq\lambda_{2}(A)\geq\cdots\geq\lambda_{n}(A)$ and let $M$ be a principal submatrix of $A$ with order $m\leq n$ and eigenvalues $\lambda_{1}(M)\geq\lambda_{2}(M)\geq\cdots\geq\lambda_{m}(M)$. Then $\lambda_{i}(A)\geq\lambda_{i}(M)\geq\lambda_{n-m+i}(A)$, for all $1\leq i\leq m$.

Let $G$ be a graph on $n$ vertices, denote by $\mu_{1}\geq\mu_{2}\geq\cdots\geq\mu_{n-1}\geq\mu_{n}=0$ the Laplacian eigenvalues of $G$ and $m(\mu_{i})$ the multiplicity of $\mu_{i}$. There are many pretty properties for Laplacian eigenvalues.

Lemma 2.1** ([4]).**

*Let $G$ be a graph on $n$ vertices with Laplacian eigenvalues $\mu_{1}\geq\mu_{2}\geq\cdots\geq\mu_{n-1}\geq\mu_{n}=0$. Then we have the following results.

(i) Denote by $m(0)$ the multiplicity of [math] as a Laplacian eigenvalue and $w(G)$ the number of connected components of $G$. Then $w(G)=m(0)$.

(ii) $G$ has exactly two distinct Laplacian eigenvalues if and only if $G$ is a union of complete graphs of the same order and isolate vertices.

(iii) The Laplacian eigenvalues of $\bar{G}$ are given by $\mu_{i}(\bar{G})=n-\mu_{n-i}$ for $i=1,2,\ldots,n-1$ and $\mu_{n}(\bar{G})=0$.

(iv) Let $H$ be a graph on $m$ vertices with Laplacian eigenvalues $\mu_{1}^{\prime}\geq\mu_{2}^{\prime}\geq\cdots\geq\mu_{m}^{\prime}=0$, then the Laplacian spectrum of $G\nabla H$ is*

[TABLE]

With respect to distance Laplacian eigenvalues, there are some similar results. The following results are given by Aouchiche and Hansen.

Theorem 2.2** ([2]).**

Let $G$ be a connected graph on $n$ vertices with $d(G)\leq 2$. Let $\mu_{1}\geq\mu_{2}\geq\cdots\geq\mu_{n-1}\geq\mu_{n}=0$ be the Laplacian spectrum of $G$. Then the distance Laplacian spectrum of $G$ is $2n-\mu_{n-1}\geq 2n-\mu_{n-2}\geq\cdots\geq 2n-\mu_{1}\geq\partial^{L}_{n}=0$. Moreover, for every $i\in\{1,2,\ldots,n-1\}$ the eigenspaces corresponding to $\mu_{i}$ and to $2n-\mu_{i}$ are the same.

Theorem 2.3** ([2]).**

Let $G$ be a connected graph on $n$ vertices. Then $\partial^{L}_{n-1}\geq n$ and $\partial^{L}_{n-1}=n$ if and only if $\bar{G}$ is disconnected. Furthermore, the multiplicity of $n$ as a distance Laplacian eigenvalue is one less than the number of connected components of $\bar{G}$.

Theorem 2.4** ([2]).**

Let $G$ be a connected graph on $n$ vertices and $m\geq n$ edges. Consider the connected graph $G^{\prime}$ obtained from $G$ by the deletion of an edge. Let $\partial^{L}_{1},\partial^{L}_{2},\ldots,\partial^{L}_{n}$ and $\partial^{\prime L}_{1},\partial^{\prime L}_{2},\ldots,\partial^{\prime L}_{n}$ denote the distance Laplacian spectra of $G$ and $G^{\prime}$ respectively. Then $\partial^{\prime L}_{i}\geq\partial^{L}_{i}$ for all $i=1,\ldots,n$.

A graph $G$ is said to be a cograph if it contains no induced $P_{4}$. There’s a pretty result about cographs.

Lemma 2.2** ([5]).**

*Given a graph $G$, the following statements are equivalent:

1. $G$ is a cograph.

2. The complement of any connected subgraph of $G$ with at least two vertices is disconnected.

3. Every connected induced subgraph of $G$ has diameter less than or equal to $2$.*

3 Main results

Recall that $\mathcal{G}(n,k)=\{G\in\mathcal{G}(n)\mid m(\partial^{L}_{1})=k\}$. Aouchiche and Hansen [1] proved that $\mathcal{G}(n,n-1)=\{K_{n}\}$. Recently, Celso et al. [6] proved that $\mathcal{G}(n,n-2)=\{K_{1,n-1},K_{n/2,n/2}\}$. They also made efforts to characterize $\mathcal{G}(n,n-3)$. Though they did not give a complete characterization, their ideas are enlightening. Especially, they proved that the graphs in $\mathcal{G}(n,n-3)$ contain no induced $P_{5}$.

Lemma 3.1** ([6], Theorem 4.1).**

Let $G\in\mathcal{G}(n,n-3)$ with $n\geq 5$ then $G$ does not contain induced $P_{5}$.

Remark 1**.**

If $G$ does not contain induced $P_{5}$, then $d(G)\leq 3$. Note that $K_{n}\not\in\mathcal{G}(n,n-3)$. We obtain that $d(G)=2$ or $d(G)=3$ for any graph $G\in\mathcal{G}(n,n-3)$ with $n\geq 5$.

Lemma 3.2** ([6], Theorem 3.3).**

If $G$ is a connected graph then $\partial^{L}_{1}\geq\max_{v\in V(G)}Tr(v)+1$ with equality holds if and only if $G\cong K_{n}$.

Lemma 3.3**.**

Let $G\in\mathcal{G}(n,n-3)$ with $n\geq 6$, then $\partial^{L}_{1}$ is integral.

Proof.

Let $f(x)$ be the characteristic polynomial of $\mathcal{D}^{L}(G)$. As $\mathcal{D}^{L}(G)$ only contains integral entries, we obtain that $f(x)$ is a monic polynomial with integral coefficients. Let $p(x)$ be the minimal polynomial of $\partial^{L}_{1}$, then $p(x)\in Z[x]$ is irreducible in $Q[x]$ and $(p(x))^{n-3}|f(x)$. We assume that $p(x)$ is a polynomial of degree at least $2$. Therefore, $p(x)$ has another root $\partial\neq 0$, which is also a distance Laplacian eigenvalue of $G$ with multiplicity $n-3$. It leads to that $n\leq 2(n-3)\leq n-1$, a contradiction. Thus, we have $p(x)=x-\partial^{L}_{1}$ and the result follows. ∎

From Lemmas 3.2 and 3.3, we get the following result.

Corollary 3.1**.**

Let $G=(V,E)\in\mathcal{G}(n,n-3)$ with $n\geq 6$, then we have $\partial^{L}_{1}\geq\max_{v\in V}Tr(v)+2$. Furthermore, if there exists $v_{0}\in V$ such that $\partial^{L}_{1}=Tr(v_{0})+2$, then $Tr(v_{0})=\max_{v\in V}Tr(v)$.

Proof.

Obviously, $G\neq K_{n}$. By Lemma 3.2, we have that $\partial^{L}_{1}>\max_{v\in V}Tr(v)+1$. Besides, we get that $\partial^{L}_{1}$ is integral from Lemma 3.3. Therefore, we have that $\partial^{L}_{1}\geq\max_{v\in V}Tr(v)+2$. Furthermore, if $\partial^{L}_{1}=Tr(v_{0})+2$, then we have $Tr(v_{0})+2\geq\max_{v\in V}Tr(v)+2$. It follows that $Tr(v_{0})=\max_{v\in V}Tr(v)$. ∎

We say that a graph $G$ is $P_{5}$-free if it does not contain induced $P_{5}$. From Lemma 3.1, all graphs in $\mathcal{G}(n,n-3)$ are $P_{5}$-free. By Remark 1, a $P_{5}$-free graph may have diameter $2$ or $3$. Now we discuss $P_{5}$-free graphs with diameter $3$.

Lemma 3.4**.**

Let $G$ be a connected $P_{5}$-free graph on $n\geq 5$ vertices with $d(G)=3$. Then at least one of $I_{i}$ for $i=1,2,3,4,5$ (shown in Fig. 1) is an induced subgraph of $G$.

Proof.

Suppose that $d(v_{1},v_{4})=3$ and $P=v_{1}v_{2}v_{3}v_{4}$ is a shortest path from $v_{1}$ to $v_{4}$. Since $n\geq 5$ and $G$ is connected, there exists $u\in V(G)\setminus V(P)$ such that $N(u)\cap V(P)\neq\emptyset$, where $N(u)=\{v\in V(G)\mid v\sim u\}$ is the neighbour set of $u$ in $G$. Moreover, since $d(v_{1},v_{4})=3$, we have that $v_{1}$ and $v_{4}$ cannot be adjacent to $u$ simultaneously, that is, $\{v_{1},v_{4}\}\not\subseteq N(u)$. Therefore, we have $1\leq|N(u)\cap V(P)|\leq 3$.

Assume that $|N(u)\cap V(P)|=1$. We claim that $N(u)\cap V(P)=\{v_{2}\}$ or $\{v_{3}\}$ since $G$ is $P_{5}$-free. Both of them lead to the induced subgraph $I_{1}$.

Assume that $|N(u)\cap V(P)|=2$. We claim that $N(u)\cap V(P)=\{v_{1},v_{2}\}$, $\{v_{3},v_{4}\}$, $\{v_{1},v_{3}\}$, $\{v_{2},v_{4}\}$, or $\{v_{2},v_{3}\}$ because $\{v_{1},v_{4}\}\not\subseteq N(u)$. The former two cases lead to the induced subgraph $I_{2}$, the next two cases lead to the induced subgraph $I_{3}$ and the last case leads to the induced subgraph $I_{4}$.

Assume that $|N(u)\cap V(P)|=3$. We claim that $N(u)\cap V(P)=\{v_{1},v_{2},v_{3}\}$ or $\{v_{2},v_{3},v_{4}\}$ because $\{v_{1},v_{4}\}\not\subseteq N(u)$. Both cases lead to the induced subgraph $I_{5}$. ∎

Next we introduce a tool which will be used frequently.

Lemma 3.5**.**

Let $G\in\mathcal{G}(n,n-3)$ with $n\geq 5$ and $M$ a principal submatrix of $\mathcal{D}^{L}(G)$ of order $5$. Then $\partial^{L}_{1}$ is also an eigenvalue of $M$ with multiplicity at least two. Furthermore, for each $1\leq k\leq 5$, there exists an eigenvector $z=(z_{1},z_{2},\ldots,z_{5})^{T}$ of $M$ with respect to $\partial^{L}_{1}$ such that $z_{k}=0$ and $\sum_{i=1}^{5}z_{i}=0$.

Proof.

Let $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{5}$ be the eigenvalues of $M$. By Theorem 2.1, we have $\partial^{L}_{1}=\partial^{L}_{n-4}\leq\lambda_{1}\leq\partial^{L}_{1}$ and $\partial^{L}_{1}=\partial^{L}_{n-3}\leq\lambda_{2}\leq\partial^{L}_{2}=\partial^{L}_{1}$. Therefore, we have $\lambda_{1}=\lambda_{2}=\partial^{L}_{1}$. Suppose that $x=(x_{1},\ldots,x_{5})^{T}$ and $y=(y_{1},\ldots,y_{5})^{T}$ are two independent eigenvectors of $M$ with respect to $\partial^{L}_{1}$. For each fixed integer $1\leq k\leq 5$, by linear combination of $x$ and $y$, we get the eigenvector $z=(z_{1},\ldots,z_{5})^{T}$ satisfying $z_{k}=0$. Let $z*=(z_{1},\ldots,z_{5},0,\ldots,0)^{T}$. Note that $\partial^{L}_{1}\geq\frac{{z*}^{T}\mathcal{D}^{L}(G)z*}{{z*}^{T}z*}=\frac{z^{T}Mz}{z^{T}z}=\partial^{L}_{1}$. We get that $z*$ is an eigenvector of $\mathcal{D}^{L}(G)$ with respect to $\partial^{L}_{1}\neq 0$. Note that the all-ones vector $j$ is an eigenvector of $\mathcal{D}^{L}(G)$ with respect to [math]. We have ${z*}^{T}j=\sum_{i=1}^{5}z_{i}=0$. ∎

Denote by $J(a,b)$ the graph obtained from $K_{1,a}\cup K_{1,b}$ by joining each pendent vertex of $K_{1,a}$ with every pendent vertex of $K_{1,b}$ (shown in Fig. 2). The non-pendent vertices of $K_{1,a}$ and $K_{1,b}$ are called the roots of $J(a,b)$.

Lemma 3.6**.**

Let $G$ be a connected $P_{5}$-free graph on $n\geq 5$ vertices with diameter $d(G)=3$. If none of $I_{1}$, $I_{2}$, $I_{4}$ and $I_{5}$ is an induced subgraph of $G$, then $G=J(a,b)$ for some positive integers $a,b\geq 1$ and $a+b+2=n$.

Proof.

Let $d(v_{1},v_{4})=3$ and $P=v_{1}v_{2}v_{3}v_{4}$ a shortest path between $v_{1}$ and $v_{4}$. By Lemma 3.4, at least one of $I_{i}$ (shown in Fig. 1) is an induced subgraph of $G$ for $i=1,2,\ldots,5$. Since none of $I_{1}$, $I_{2}$, $I_{4}$ or $I_{5}$ is an induced subgraph of $G$, we obtain that $G$ contains induced $I_{3}$.

Note that $I_{3}=J(2,1)$ with roots $v_{1}$ and $v_{4}$ is an induced subgraph of $G$. We may assume that $G^{\prime}=J(a,b)$ with roots $v_{1}$ and $v_{4}$ is the maximal induced subgraph of $G$ including $J(2,1)$. Denote by $U_{1}=N_{G^{\prime}}(v_{1})=\{x_{1},x_{2},\ldots,x_{a}\}$ and $U_{2}=N_{G^{\prime}}(v_{4})=\{y_{1},y_{2},\ldots,y_{b}\}$. Obviously, $v_{2},u\in U_{1}$ and $v_{3}\in U_{2}$. In what follows we will show that $G=J(a,b)$ with roots $v_{1}$ and $v_{4}$.

By the way of contradiction, assume that $G\not=J(a,b)$. Then there exists $v\in V(G)\setminus V(G^{\prime})$ such that $N_{G}(v)\cap V(G^{\prime})\neq\emptyset$. Since $d(v_{1},v_{4})=3$, $v$ is adjacent to at most one of $v_{1}$ and $v_{4}$. We claim that $v$ is exactly adjacent to one of $v_{1}$ and $v_{4}$. Otherwise, we have $v\not\sim v_{1},v_{4}$. Then $N_{G}(v)\cap U_{1}\neq\emptyset$ or $N_{G}(v)\cap U_{2}\neq\emptyset$. If $v$ is adjacent to some vertex in $N_{G}(v)\cap U_{1}$ and some vertex in $N_{G}(v)\cap U_{2}$, say $v\sim x_{1}$ and $v\sim y_{1}$ (see Fig. 2 (1)), then we get the induced subgraph $G[v_{1},x_{1},y_{1},v_{4},v]=I_{4}$, a contradiction. If $v$ is only adjacent to some vertex in $N_{G}(v)\cap U_{1}$, say $v\sim x_{1}$ (see Fig. 2 (2)), then we get the induced subgraph $G[v_{1},x_{1},y_{1},v_{4},v]=I_{1}$, a contradiction. If $v$ is only adjacent to some vertex in $N_{G}(v)\cap U_{2}$, say $v\sim y_{1}$, then we also get the induced subgraph $G[v_{1},x_{1},y_{1},v_{4},v]=I_{1}$, a contradiction. Now we need to consider the following two situations.

Case 1.

$v\sim v_{1}$ and $v\not\sim v_{4}$;

First, we will show that $U_{2}\subseteq N_{G}(v)$. Otherwise, there exists some vertex in $U_{2}$ not adjacent to $v$, say $v\not\sim y_{1}$. Now, if $v\not\sim x_{1}$ (see Fig. 2 (3)), then we get the induced subgraph $G[v,v_{1},x_{1},y_{1},v_{4}]=P_{5}$, a contradiction; if $v\sim x_{1}$ (see Fig. 2 (4)), then we get the induced subgraph $G[v_{1},x_{1},y_{1},v_{4},v]=I_{2}$, a contradiction.

Next we will show that $N_{G}(v)\cap U_{1}=\emptyset$. Otherwise, there exists some vertex in $U_{1}$ adjacent to $v$, say $v\sim x_{1}$. Recall that $v\sim y_{1}$ (see Fig. 2 (5)) according to the above arguments, we get the induced subgraph $G[v_{1},x_{1},y_{1},v_{4},v]=I_{5}$, a contradiction.

Summariszing the above discussion, we know that $V(G^{\prime})\cup\{v\}$ induces a subgraph $J(a+1,b)$ of $G$. This is impossible since $G^{\prime}=J(a,b)$ is assumed to be the maximal induced subgraph including $J(2,1)$.

Case 2.

$v\sim v_{4}$ and $v\not\sim v_{1}$;

As similar as Case 1, by symmetry we can also deduce that $G[V(G^{\prime})\cup\{v\}]=J(a,b+1)$. This is also impossible.

We complete this proof. ∎

After the completion of the preparations, we get one of our main results.

Theorem 3.1**.**

Let $G\in\mathcal{G}(n,n-3)$ with $n\geq 6$, then $d(G)=2$.

Proof.

By Lemma 3.1 and Remark 1, we get that $G$ is $P_{5}$-free and $d(G)=2$ or $d(G)=3$. Assume by contradiction that $d(G)=3$. Let $d(v_{1},v_{4})=3$ and $P=v_{1}v_{2}v_{3}v_{4}$ a shortest path between $v_{1}$ and $v_{4}$. By Lemma 3.4, $G$ contains at least one of $I_{i}$ (labelled as Fig. 1) as an induced subgraph for $i=1,2,\ldots,5$.

Suppose that $I_{1}$ is an induced subgraph of $G$. Note that $d_{G}(v_{4},u)=d_{I_{1}}(v_{4},u)-1=2$ or $d_{G}(v_{4},u)=d_{I_{1}}(v_{4},u)=3$. We get that either $M_{1}$ or $M_{1}^{\prime}$ is a principal submatrix of $\mathcal{D}^{L}(G)$ with respect to $I_{1}$, where

[TABLE]

If $M_{1}$ is a principal submatrix of $\mathcal{D}^{L}(G)$, by Lemma 3.5, there exists an eigenvector $x=(x_{1},x_{2},x_{3},x_{4},0)$ satisfying $x_{1}+x_{2}+x_{3}+x_{4}=0$ such that $M_{1}x=\partial^{L}_{1}x$. Consider the fifth entry of both sides of $M_{1}x=\partial^{L}_{1}x$, we have $-2x_{1}-x_{2}-2x_{3}-2x_{4}=0$. It follows that $x_{2}=0$ and $x_{1}+x_{3}+x_{4}=0$. Next we consider the second entry of both sides of $M_{1}x=\partial^{L}_{1}x$, we have $-x_{1}-x_{3}-2x_{4}=0$. It follows that $x_{4}=0$ and $x_{1}+x_{3}=0$. We consider the fourth entry of both sides of $M_{1}x=\partial^{L}_{1}x$, we have $-3x_{1}-x_{3}=0$. It follows that $x_{1}=x_{3}=0$. Thus, we have $x=0$, a contradiction. If $M_{1}^{\prime}$ is a principal submatrix of $\mathcal{D}^{L}(G)$, by Lemma 3.5, there exists an eigenvector $y=(0,y_{2},y_{3},y_{4},y_{5})^{T}$ satisfying

[TABLE]

such that

[TABLE]

Consider the first entry of both sides of Eq. (2), we have

[TABLE]

Combining (1) and (3), we have $y_{2}=y_{4}$. If $y_{2}=y_{4}=0$, we consider the fourth entry of both sides of (2) and we get that $y_{5}=0$. It follows that $y=0$, a contradiction. If $y_{2}=y_{4}\neq 0$, we consider the second entry of both sides of (2) and we get that $\partial^{L}_{1}=t_{2}-\frac{y_{3}+2y_{4}+y_{5}}{y_{2}}=t_{2}-\frac{y_{2}+y_{3}+y_{4}+y_{5}}{y_{2}}$. From (1), we have $\partial^{L}_{1}=t_{2}$. It contradicts Corollary 3.1.

Suppose that $I_{2}$ is an induced subgraph of $G$. Note that $d_{G}(v_{4},u)=d_{I_{2}}(v_{4},u)=3$ or $d_{G}(v_{4},u)=d_{I_{2}}(v_{4},u)-1=2$. We get that the matrix $M_{2}$ or $M_{2}^{\prime}$ is a principal submatrix of $\mathcal{D}^{L}(G)$ with respect to $I_{2}$, where

[TABLE]

If $M_{2}$ is a principal submatrix of $\mathcal{D}^{L}(G)$, by Lemma 3.5, there exists an eigenvector $x=(x_{1},0,x_{3},x_{4},x_{5})^{T}$ satisfying $x_{1}+x_{3}+x_{4}+x_{5}=0$ such that $M_{2}x=\partial^{L}_{1}x$. We successively consider the second, the fourth and the third entries of both sides of $M_{2}x=\partial^{L}_{1}x$, we get that $x_{3}=x_{4}=0$ and $x_{1}+x_{5}=0$. If $x_{1}=0$, then $x_{5}=0$ and $x=0$, a contradiction. If $x_{1}\neq 0$, consider the first entry of both sides of $M_{2}x=\partial^{L}_{1}x$, we get that $\partial^{L}_{1}=t_{1}-\frac{x_{5}}{x_{1}}=t_{1}+1$. It contradicts Corollary 3.1. If $M_{2}^{\prime}$ is a principal submatrix of $\mathcal{D}^{L}(G)$, by Lemma 3.5, there exists an eigenvector $y=(y_{1},y_{2},y_{3},y_{4},0)^{T}$ satisfying $y_{1}+y_{2}+y_{3}+y_{4}=0$ such that $M_{2}^{\prime}y=\partial^{L}_{1}y$. Consider the fifth entry of both sides of $M_{2}^{\prime}y=\partial^{L}_{1}y$, we have $-y_{1}-y_{2}-2y_{3}-2y_{4}=0$. It leads to that $y_{1}+y_{2}=y_{3}+y_{4}=0$. If $y_{3}=y_{4}=0$, we consider the third entry of both sides of $M_{2}^{\prime}y=\partial^{L}_{1}y$ and we get that $y_{1}=y_{2}=0$. It leads to that $y=0$, a contradiction. If $y_{1}=y_{2}=0$, we consider the second entry of both sides of $M_{2}^{\prime}y=\partial^{L}_{1}y$ and we get that $y_{3}=y_{4}=0$. It leads to that $y=0$, a contradiction. If $y_{1},y_{2},y_{3},y_{4}\neq 0$, without loss of generality, we may suppose that $y=(a,-a,1,-1,0)$. Consider the third entry of both sides of $M_{2}^{\prime}y=\partial^{L}_{1}y$, we have $\partial^{L}_{1}=t_{3}+1-a$. By Corollary 3.1, we have $a<0$. Consider the fourth entry of both sides of $M_{2}^{\prime}y=\partial^{L}_{1}y$, we have $\partial^{L}_{1}=t_{4}+1+a$. By Corollary 3.1, we have $a>0$, a contradiction.

Suppose that $I_{4}$ is an induced subgraph of $G$. We get that the matrix $M_{4}$ is a principal submatrix of $\mathcal{D}^{L}(G)$ with respect to $I_{4}$, where

[TABLE]

By Lemma 3.5, there exists an eigenvector $x=(x_{1},0,x_{3},x_{4},x_{5})^{T}$ satisfying $x_{1}+x_{3}+x_{4}+x_{5}=0$ such that $M_{4}x=\partial^{L}_{1}x$. Consider the second and the fourth entries of both sides of $M_{4}x=\partial^{L}_{1}x$ successively, we get that $x_{4}=0$, $x_{1}=x_{3}$ and $x_{5}=-2x_{1}$. If $x_{1}=x_{3}=0$, then $x=0$, a contradiction. If $x_{1}=x_{3}\neq 0$, consider the third entry of both sides of $M_{4}x=\partial^{L}_{1}x$ and we get that $\partial^{L}_{1}=t_{3}$. It contradicts Corollary 3.1.

Suppose that $I_{5}$ is an induced subgraph of $G$. We get that the matrix $M_{5}$ is a principal submatrix of $\mathcal{D}^{L}(G)$ with respect to $I_{5}$, where

[TABLE]

By Lemma 3.5, there exists an eigenvector $x=(x_{1},0,x_{3},x_{4},x_{5})^{T}$ satisfying $x_{1}+x_{3}+x_{4}+x_{5}=0$ such that $M_{5}x=\partial^{L}_{1}x$. We successively consider the second and the fourth entries of both sides of $M_{5}x=\partial^{L}_{1}x$, we have that $x_{4}=0$, $x_{1}=x_{3}$ and $x_{5}=-2x_{3}$. If $x_{3}=0$, we have $x=0$, a contradiction. If $x_{3}\neq 0$, consider the third entry of both sides of $M_{5}x=\partial^{L}_{1}x$, we have $\partial^{L}_{1}=t_{3}$. It contradicts Corollary 3.1.

Suppose that $I_{3}$ is an induced subgraph of $G$. On the one hand, we get that the matrix $M_{3}$ is a principal submatrix of $\mathcal{D}^{L}(G)$ with respect to $I_{3}$, where

[TABLE]

By Lemma 3.5, there exists an eigenvector $x=(x_{1},x_{2},0,x_{4},x_{5})^{T}$ satisfying $x_{1}+x_{2}+x_{4}+x_{5}=0$ such that $M_{3}x=\partial^{L}_{1}x$. We successively consider the third, the first and the fourth entries of both sides of $M_{3}x=\partial^{L}_{1}x$, then we get that $x_{1}=x_{4}=0$ and $x_{2}+x_{5}=0$. If $x_{2}=x_{5}=0$, then $x=0$, a contradiction. If $x_{2}\neq 0$, without loss of generality, we may suppose that $x=(0,1,0,0,-1)^{T}$. Consider the second entry of $M_{3}x=\partial^{L}_{1}x$, we get that $\partial^{L}_{1}=t_{2}+2$. By Corollary 3.1, we get that

[TABLE]

On the other hand, recall that $G$ is $P_{5}$-free. Moreover, by the arguments above, we have that $G$ contains no induced $I_{1}$, $I_{2}$, $I_{4}$ or $I_{5}$. Therefore, by Lemma 3.6, we have that $G=J(a,b)$ with roots $v_{1}$ and $v_{4}$. By simple calculation, we have $Tr(v_{1})=a+2b+3$, $Tr(v_{4})=2a+b+3$, $Tr(x)=2a+b+1$ for every $x\in N(v_{1})$ and $Tr(y)=2b+a+1$ for every $y\in N(v_{4})$. Note that $v_{2}\in N(v_{1})$. We get that

[TABLE]

which contradicts (4).

We complete the proof. ∎

The result above showed that the graphs in $\mathcal{G}(n,n-3)$ have diameter $2$. In fact, we can further obtain that $G$ is the join of two graphs. To prove this, we need the following result.

Lemma 3.7**.**

Let $G\in\mathcal{G}(n,n-3)$ with $n\geq 6$, then none of $J_{1}(=C_{5})$, $J_{2}$ or $J_{3}$ (shown in Fig. 3) can be an induced subgraph of $G$.

Proof.

Assume by contradiction that $J_{1}=C_{5}$ is an induced subgraph of $G$. We get that the matrix $N_{1}$ is a principal submatrix of $\mathcal{D}^{L}(G)$ with respect to $J_{1}$, where

[TABLE]

By Lemma 3.5, there exists an eigenvector $x=(0,x_{2},x_{3},x_{4},x_{5})^{T}$ satisfying $x_{2}+x_{3}+x_{4}+x_{5}=0$ such that $N_{1}x=\partial^{L}_{1}x$. From the first entry of $N_{1}x=\partial^{L}_{1}x$, we have $-x_{2}-2x_{3}-2x_{4}-x_{5}=0$. Therefore, we have $x_{3}+x_{4}=0$ and $x_{2}+x_{5}=0$. If $x_{3}=x_{4}=0$, consider the third entry of both sides of $N_{1}x=\partial^{L}_{1}x$ and we get that $x_{2}=x_{5}=0$. It leads to that $x=0$, a contradiction. If $x_{2}=x_{5}=0$, consider the second entry of both sides of $N_{1}x=\partial^{L}_{1}x$ and we get that $x_{3}=x_{4}=0$. It leads to that $x=0$, a contradiction. If $x_{2},x_{3},x_{4},x_{5}\neq 0$, without loss of generality, we may suppose that $x=(0,a,1,-1,-a)^{T}$. Thus, we have

[TABLE]

Consider the fourth entry of both sides of (5), we have

[TABLE]

By Corollary 3.1, we have $a\geq 1$. Consider the fifth entry of both sides of Eq. (5), we have

[TABLE]

By Lemma 3.3, we get that $\partial^{L}_{1}$ is integral. Therefore, $a$ and $\frac{1}{a}$ are both integral. Thus, we have $a=1$ and $\partial^{L}_{1}=t_{4}+2=t_{5}+3$. It follows that

[TABLE]

On the other hand, by Lemma3.5, there also exists an eigenvector $y=(y_{1},0,y_{3},y_{4},y_{5})^{T}$ satisfying $y_{1}+y_{3}+y_{4}+y_{5}=0$ such that $N_{1}y=\partial^{L}_{1}y$. From the second entry of $N_{1}y=\partial^{L}_{1}y$, we have $-y_{1}-y_{3}-2y_{4}-2y_{5}=0$. Therefore, we have $y_{4}+y_{5}=0$ and $y_{1}+y_{3}=0$. If $y_{1}=y_{3}=0$ or $y_{4}=y_{5}=0$, we also get $y=0$, a contradiction. If $y_{1},y_{3},y_{4},y_{5}\neq 0$, without loss of generality, we may suppose that $y=(b,0,-b,1,-1)^{T}$. Thus, we have

[TABLE]

Consider the fourth and the fifth entries of both sides of Eq. (7), we have

[TABLE]

It follows that $t_{4}=t_{5}$, which contradicts (6).

Assume by contradiction that $J_{2}$ is an induced subgraph of $G$. We get that the matrix $N_{2}$ is a principal submatrix of $\mathcal{D}^{L}$ with respect to $J_{2}$, where

[TABLE]

By Lemma 3.5, there exists an eigenvector $x=(x_{1},x_{2},x_{3},x_{4},0)^{T}$ satisfying $x_{1}+x_{2}+x_{3}+x_{4}=0$ such that $N_{2}x=\partial^{L}_{1}x$. We successively consider the fifth, the third, the first and the fourth entries of both sides of $N_{2}x=\partial^{L}_{1}x$, then we get that $x=0$, a contradiction.

Assume by contradiction that $J_{3}$ is an induced subgraph of $G$. We get that the matrix $N_{3}$ is a principal submatrix of $\mathcal{D}^{L}$ with respect to $J_{3}$, where

[TABLE]

By Lemma 3.5, there exists an eigenvector $x=(x_{1},x_{2},0,x_{4},x_{5})^{T}$ satisfying $x_{1}+x_{2}+x_{4}+x_{5}=0$ such that $N_{3}x=\partial^{L}_{1}x$. Consider the third, the first, the fourth and the second entries of both sides of $N_{3}x=\partial^{L}_{1}x$ succesively, we get that $x=0$, a contradiction. ∎

Using the above tools, we get the following result.

Theorem 3.2**.**

Let $G\in\mathcal{G}(n,n-3)$ with $n\geq 6$, then $\bar{G}$ is disconnected. It means that $G$ is the join of some connected graphs.

Proof.

By Lemma 2.2, it suffices to show that $G$ contains no induced $P_{4}$. Assume by contradiction that $G$ contains an induced $P_{4}=v_{1}v_{2}v_{3}v_{4}$. By Theorem 3.1, we have $d(G)=2$. Therefore, there exists a vertex $u\in V(G)$ such that $u\sim v_{1},v_{4}$. It follows that at least one of $J_{1}$, $J_{2}$ and $J_{3}$ will be an induced subgraph of $G$, contradicts Lemma 3.7. ∎

For any graph $G\in\mathcal{G}(n,n-3)$, we see that $G$ has at most four distinct eigenvalues, and we also have $\partial^{L}_{n-1}(G)=n$ by Theorems 2.3 and 3.2. Denote by

[TABLE]

and

[TABLE]

Therefore, $\mathcal{H}_{1}(n)$ and $\mathcal{H}_{2}(n)$ are the sets of graphs with four and three distinct eigenvalues in $\mathcal{G}(n,n-3)$, respectively. Thus we have the disjoint decomposition

[TABLE]

Mohammadian [9] gave the following result.

Lemma 3.8** ([9], Theorem 8).**

Let $G$ be a graph on $n\geq 5$ vertices whose distinct Laplacian eigenvalues are $0<\alpha<\beta<\gamma$. Then the multiplicity of $\alpha$ is $n-3$ if and only if $G$ is one of the graphs $K_{2,n-2}$, $K_{n/2,n/2}+e$ or $K_{1,n-1}+e$, where $K_{n/2,n/2}+e$ and $K_{1,n-1}+e$ are the graphs obtained from $K_{n/2,n/2}$ and $K_{1,n-1}$, respectively, by adding an edge $e$ joining any two non-adjacent vertices.

Note that, when $d(G)=2$, there exists a correspondence between the distance Laplacian spectrum and the Laplacian spectrum of $G$. We have the following result.

Corollary 3.2**.**

For an integer $n\geq 6$, we have $\mathcal{H}_{1}(n)=\{K_{2,n-2},K_{n/2,n/2}+e,K_{1,n-1}+e\}$, and their distance Laplacian spectra are given by

[TABLE]

Proof.

Let $G\in\mathcal{H}_{1}(n)$ and $\emph{Spec}_{\mathcal{L}}(G)=\{(\partial^{L}_{1})^{n-3},\partial^{L}_{n-2},n,0\}$ where $\partial^{L}_{1}>\partial^{L}_{n-2}>n$. By Theorem 3.1, we have $d(G)=2$. Therefore, by Theorem 2.2, the Laplacian spectrum of $G$ is $\{n,2n-\partial^{L}_{n-2},(2n-\partial^{L}_{1})^{n-3},0\}$. Thus, we get that $G\in\{K_{2,n-2},K_{n/2,n/2}+e,K_{1,n-1}+e\}$ from Lemma 3.8. Conversely, note that all of $K_{2,n-2}$, $K_{n/2,n/2}+e$ and $K_{1,n-1}+e$ are the join of two graphs, by Lemma 2.1 (iv) and Theorem 2.2, we obtain their distance Laplacian spectra, which are shown in (8). Therefore, $K_{2,n-2},K_{n/2,n/2}+e,K_{1,n-1}+e\in\mathcal{H}_{1}(n)$, and the result follows. ∎

In what follows we characterise $\mathcal{H}_{2}(n)$.

Lemma 3.9**.**

For an integer $n\geq 6$, we have $\mathcal{H}_{2}(n)=\{K_{2}\nabla(n-2)K_{1},K_{1}\nabla K_{\frac{n-1}{2},\frac{n-1}{2}},K_{\frac{n}{3},\frac{n}{3},\frac{n}{3}}\}$, and their distance Laplacian spectra are given by

[TABLE]

Proof.

Let $G\in\mathcal{H}_{2}(n)$ and $\emph{Spec}_{\mathcal{L}}(G)=\{(\partial^{L}_{1})^{n-3},n^{2},0\}$ where $\partial^{L}_{1}>n$. By Theorem 3.1, we get that $d(G)=2$. Therefore, by Theorem 2.2, the Laplacian spectrum of $G$ is $\{n^{2},(2n-\partial^{L}_{1})^{n-3},0\}$. By Lemma 2.1 (iii), the Laplacian spectrum of $\bar{G}$ is $\{(\partial^{L}_{1}-n)^{n-3},0^{3}\}$. By Lemma 2.1 (i), $\bar{G}$ has exactly three components, denoted by $G_{1}$, $G_{2}$ and $G_{3}$. Moreover, by Lemma 2.1 (ii), $G_{1}$, $G_{2}$ and $G_{3}$ are either complete graphs of the same order or isolate vertices. If none of them is an isolate vertex, then $G_{1}\cong G_{2}\cong G_{3}\cong K_{n/3}$. It follows that $G=\overline{3K_{n/3}}=K_{\frac{n}{3},\frac{n}{3},\frac{n}{3}}$. If there’s exactly one of them is an isolate vertex, say $G_{3}$, then $G_{1}\cong G_{2}\cong K_{(n-1)/2}$. It follows that $G=\overline{2K_{(n-1)/2}\cup K_{1}}=K_{1}\nabla K_{\frac{n-1}{2},\frac{n-1}{2}}$. If there are exactly two of them are isolate vertices, say $G_{2}$ and $G_{3}$, then $G_{1}\cong K_{n-2}$. It follows that $G=\overline{K_{n-2}\cup 2K_{1}}=K_{2}\nabla(n-2)K_{1}$. Conversely, note that all of $K_{2}\nabla(n-2)K_{1}$, $K_{1}\nabla K_{\frac{n-1}{2},\frac{n-1}{2}}$ and $K_{\frac{n}{3},\frac{n}{3},\frac{n}{3}}$ are the join of two graphs, by Lemma 2.1 (iv) and Theorem 2.2, we obtain their distance Laplacian spectra, which are shown in (9). Therefore, $K_{2}\nabla(n-2)K_{1},K_{1}\nabla K_{\frac{n-1}{2},\frac{n-1}{2}},K_{\frac{n}{3},\frac{n}{3},\frac{n}{3}}\in\mathcal{H}_{2}(n)$, and the result follows. ∎

Recall that $\mathcal{G}(n,n-3)=\mathcal{H}_{1}(n)\cup\mathcal{H}_{2}(n)$. Combining Corollary 3.2 and Lemma 3.9, we completely determine $\mathcal{G}(n,n-3)$ in the following result.

Theorem 3.3**.**

For an integer $n\geq 6$, we have

[TABLE]

Remark 2**.**

By using the software SageMath, we get the graphs with $m(\partial^{L}_{1})=n-3$ for $n=4$ and $n=5$. That is,

[TABLE]

We end up this paper by the following result.

Corollary 3.3**.**

Let $G\in\mathcal{G}(n,n-3)$ with $n\geq 5$ then $G$ is determined by its distance Laplacian spectrum.

Proof.

Let $H\in\mathcal{G}(n)$ with $\emph{Spec}_{\mathcal{L}}(H)=\emph{Spec}_{\mathcal{L}}(G)$. We get that $H\in\mathcal{G}(n,n-3)$. Then, the result follows by pairwise comparing the distance Laplacian spectra of graphs in $\mathcal{G}(n,n-3)$, which are presented in (8) and (9). ∎