The multiplicity of the Laplacian eigenvalue $2$ in some bicyclic graphs

TL;DR

This paper examines the conditions under which the Laplacian eigenvalue 2 appears in graphs formed by connecting two unicyclic graphs with an edge, focusing on cases where only one or both graphs have 2 as an eigenvalue.

Contribution

It extends previous work by analyzing the presence of the eigenvalue 2 in combined graphs when only one of the component graphs has this eigenvalue.

Findings

Identifies structural conditions for eigenvalue 2 in combined graphs

Provides Laplacian characteristic polynomial-based criteria

Analyzes cases with one or both graphs having eigenvalue 2

Abstract

The Laplacian matrix of a graph $G$ is denoted by $L(G)=D(G)-A(G)$, where $D(G)=diag(d(v_{1}),\ldots , d(v_{n}))$ is a diagonal matrix and $A(G)$ is the adjacency matrix of $G$. Let $G_1$ and $G_2$ be two graphs. A one-edge connection of two graphs $G_1$ and $G_2$ is a graph $G=G_1\odot_{uv} G_2$ with $V(G)=V(G_1)\cup V(G_2)$ and $E(G)= E(G_1)\cup E(G_2)\cup \{e=uv\}$, where $u\in V(G_1)$ and $v\in V(G_2)$. We investigate the multiplicity of the Laplacian eigenvalue $2$ of $G_1\odot_{uv} G_2$, while the unicyclic graphs $G_1$ and $G_2$ have $2$ among their Laplacian eigenvalues, by using their Laplacian characteristic polynomials. Some structural conditions ensuring the presence of the existence $2$ in the $G=G_1\odot_{uv} G_2$ where both $G_1$ and $G_2$ have $2$ as Laplacian eigenvalue, have been investigated, while, here we study the existence Laplacian eigenvalue $2$ in $G=G_1\odot_{uv} G_2$ where at most one of $G_1$ or $G_2$ has $2$ as Laplacian eigenvalue.

Equations

$$(d(v_i)-\lambda )x_i=\sum _{v_j\in N(v_i)}{x}_j\mathrm{for}\mathrm{all}i=1,\dots ,n.$$

$$det\left(\begin{array}[]{cc}M&N\\ P&Q\\ \end{array}\right)=\begin{vmatrix}M&N\\ P&Q\end{vmatrix}=\mid M\mid.\mid Q-PM^{-1}N\mid$$

$$L_G(\lambda )=M.Q-M.Q_v-Q.M_u$$

$$s_i=|\{T_k^{(i)}:|V(T_k^{(i)})|\text{mbox}is\text{mbox}odd;1\le k\le g_i\}|;i=1,2,$$

$$L_{G_1}(\lambda )=|M|=(\lambda -2{)}^{2}f(\lambda ),L_{G_2}(\lambda )=|Q|=(\lambda -2{)}^{2}g(\lambda ).$$

$$M_u=\left|\begin{array}{ccc}\lambda -2& 1& 0\\ 1& \lambda -2& 1\\ 0& 1& \lambda -2\end{array}\right|=(\lambda -2)({\lambda }^2-4\lambda +2).$$

$$M_u=\left|\begin{array}{cc}{M}^{\prime }& N\\ N^T& Q^{\prime }\end{array}\right|$$

$$M^{\prime}=\left(\begin{array}[]{cccc}\lambda-2&1&0&0\\ 1&\lambda-2&1&0\\ 0&1&\lambda-2&1\\ 0&0&1&\lambda-2\\ \end{array}\right),$$

$$Q^{\prime }=(q_{ij}^{\prime });q_{ij}^{\prime }=\{\begin{array}{ll}\lambda -2,& i=j;\\ a_{ij},& i\ne j;\end{array};1\le i,j\le m-5$$

$$N_{4\times (m-5)}=(n_{ij});n_{ij}=\{\begin{array}{ll}1,& i=4,j=1;\\ 0,& o.w;\end{array};1\le i\le 4and1\le j\le m-5.$$

$$[N^{T}M^{\prime-1}N]_{m-5\times m-5}=\left(\begin{array}[]{ccccc}\frac{M^{\prime}_{4,4}}{|M^{\prime}|}&0&\cdots&0&0\\ 0&\ddots&\ddots&&0\\ \vdots&\ddots&0&\ddots&\vdots\\ 0&&\ddots&\ddots&0\\ 0&0&\cdots&0&0\\ \end{array}\right).$$

$$Q^{\prime }-N^T{M}^{\prime -1}N=(b_{ij}^{\prime });b_{ij}^{\prime }=\{\begin{array}{ll}\lambda -2-\frac{M_{4,4}^{\prime }}{|M^{\prime }|},& i=j=1;\\ \lambda -2,& i=j\ne 1;\\ a_{ij},& i\ne j;\end{array};1\le i,j\le m-5.$$

$$H=(h_{ij});h_{ij}=\{\begin{array}{ll}-\frac{M_{4,4}^{\prime }}{|M^{\prime }|},& i=j=1;\\ 0,& i=1\ne j;\\ \lambda -2,& i=j\ne 1;\\ a_{ij},& i\ne 1,i\ne j;\end{array};1\le i,j\le m-5.$$

$$|Q^{\prime }-N^T{M}^{\prime -1}N|=|Q^{\prime }|-\frac{M_{4,4}^{\prime }}{|M^{\prime }|}{Q}_{1,1}^{\prime }$$

$$\Rightarrow M_u=|M^{\prime }||Q^{\prime }|-M_{4,4}^{\prime }{Q}_{1,1}^{\prime }.$$

$$M_{4,4}^{\prime }=\left|\begin{array}{ccc}\lambda -2& 1& 0\\ 1& \lambda -2& 1\\ 0& 1& \lambda -2\end{array}\right|=(\lambda -2)({\lambda }^2-4\lambda +2).$$

$$u$$

$$(d_G(u)-2)x(u)=\sum _{v_j\in N_G(u)}x(v_j)$$

$$\displaystyle\Rightarrow(d_{G_{1}}(u)-2)x(u)+x(u)=\sum_{v_{j}\in N_{G_{1}}(u)}x(v_{j})+x(v).\$$

$$(d_G(v)-2)x(v)=\sum _{v_j\in N_G(v)}x(v_j)$$

$$\Rightarrow (d_{G_2}(v)-2)x(v)+x(v)=\sum _{v_j\in N_{G_2}(v)}x(v_j)+x(u)$$

$$\Rightarrow (d_{G_2}(v)-2)x(v)=\sum _{v_j\in N_{G_2}(v)}x(v_j).$$

$$(d_{G_2}(w)-2)x_{G_2}(w)=\sum _{v_i\in N_{G_2}(w)}x(v_i)\forall w\in V(G_2).$$

$$X=(-1,0,0,1,0,0,0,0,0,0,0{)}^t$$

$$1$$

Full text

AMS 2010 Subject Classification: 05C50; 11C08; 15A18.

Keywords: Laplacian eigenvalue, characteristic polynomial, multiplicity, unicyclic graph, bicyclic graph.

1 Introduction

All graphs in this paper are finite and undirected with no loops or multiple edges. Let $G$ be a graph with $n$ vertices. The vertex set and the edge set of $G$ are denoted by $V(G)$ and $E(G)$, respectively. The Laplacian matrix of $G$ is $L(G)=D(G)-A(G)$, where $D(G)=diag(d(v_{1}),\ldots,d(v_{n}))$ is a diagonal matrix and $d(v)$ denotes the degree of the vertex $v$ in $G$ and $A(G)$ is the adjacency matrix of $G$. We shall use the notation $\lambda_{k}(G)$ to denote the $k^{\mathrm{th}}$ Laplacian eigenvalue of the graph $G$ and we assume that $\lambda_{1}(G)\geq\cdots\geq\lambda_{n}(G)=0$. Also, the multiplicity of the eigenvalue $\lambda$ of $L(G)$ is denoted by $m_{G}(\lambda)$. A vertex of degree one is called a leaf vertex and a vertex is said quasi leaf (support vertex) if it is incident to a leaf vertex. Connected graphs in which the number of edges equals the number of vertices are called unicyclic graphs. Therefore, a unicyclic graph is either a cycle or a cycle with some attached trees. Let $\mathfrak{U}_{n,g}$ be the set of all unicyclic graphs of order $n$ with girth $g$. Throughout this paper, we suppose that the vertices of the cycle $C_{g}$ are labeled by $v_{1},...,v_{g}$, ordered in a natural way around $C_{g}$, say in the clockwise direction. A rooted tree is a tree in which one vertex has been designated the root. Furthermore, assume that $T_{i}$ is a rooted tree of order $n_{i}\geq 1$ attached to $v_{i}\in V(T_{i})\cap V(C_{g})$, where $\sum_{i=1}^{g}n_{i}=n$. This unicyclic graph is denoted by $C(T_{1},...,T_{g})$. The sun graph of order $2n$ is a cycle $C_{n}$ with an edge terminating in a leaf vertex attached to each vertex that is the corona of $C_{n}\circ K_{1}$. A broken sun graph is a unicyclic subgraph of a sun graph, so one can assume a sun graph is a broken sun graph too. A one-edge connection of two graphs $G_{1}$ and $G_{2}$ is a graph $G=G_{1}\odot_{uv}G_{2}$ with $V(G)=V(G_{1})\cup V(G_{2})$ and $E(G)=E(G_{1})\cup E(G_{2})\cup\{e=uv\}$, where $u\in V(G_{1})$ and $v\in V(G_{2})$. We use the notation $u\sim v$ when $e=uv\in E(G)$.

By [7, Theorem 13], due to Kelmans and Chelnokov, the Laplacian coefficient, $\xi_{n-k}$, can be expressed in terms of subtree structures of $G$, for $0\leq k\leq n$. Suppose that $F$ is a spanning forest of $G$ with components $T_{i}$ of order $n_{i}$, and $\gamma(F)=\Pi_{i=1}^{k}n_{i}$. The Laplacian characteristic polynomial of $G$ is denoted by $L_{G}(\lambda)=det(\lambda I-L(G))=\Sigma_{i=0}^{n}(-1)^{i}\xi_{i}\lambda^{n-i}$. If $M$ is a square matrix, then the determinant of $M$ is denoted by $|M|$ and the minor of the entry in the $i^{\mathrm{th}}$ row and $j^{\mathrm{th}}$ column is the determinant of the submatrix formed by deleting the $i^{\mathrm{th}}$ row and $j^{\mathrm{th}}$ column. This number is often denoted by $M_{i,j}$. A square matrix is non-singular if its determinant is non-zero or it has an inverse.

Let $G$ be a graph with $n$ vertices. It is convenient to adopt the following terminology from [4]: for a vector $X=(x_{1},\ldots,x_{n})^{t}\in\mathbb{R}^{n}$, we say $X$ gives a valuation of the vertex of $V$, and with each vertex $v_{i}$ of $V$, we associate the number $x_{i}$, which is the value of the vertex $v_{i}$, that is $x(v_{i})=x_{i}$. Then $\lambda$ is an eigenvalue of $L(G)$ with the corresponding eigenvector $X=(x_{1},\ldots,x_{n})$ if and only if $X\neq 0$ and

[TABLE]

In this article, we would like to study the eigenvalue $2$ in bicyclic graphs with just $2$ cycles. The main Theorem is about the multiplicity of the Laplacian eigenvalue $2$ of $G_{1}\odot_{uv}G_{2}$, while the unicyclic graphs $G_{1}$ and $G_{2}$ have $2$ among their Laplacian eigenvalues, by using their Laplacian characteristic polynomials. By addition, we investigate the existence Laplacian eigenvalue $2$ in $G=G_{1}\odot_{uv}G_{2}$ where at most one of $G_{1}$ or $G_{2}$ has $2$ as Laplacian eigenvalue, by using Equation (1).

2 Main results

We start this section by studying the multiplicity of the Laplacian eigenvalue $2$ of $G_{1}\odot_{uv}G_{2}$, where the unicyclic graphs $G_{1}$ and $G_{2}$ have $2$ among their Laplacian eigenvalues. For this, we need the following results.

Lemma 1

.[2, Lemma 2.2]* If $M$ is a non-singular square matrix, then*

[TABLE]

We use the notation $M_{u}$ when in $M_{i,i},i$ is the row and column corresponding to the vertex $u$ in a graph.

Lemma 2

.[6, Lemma 8]* Let $G_{1}$ and $G_{2}$ be two graphs of order $n$ and $m$, respectively. Then the Laplacian characteristic polynomial of $G=G_{1}\odot_{uv}G_{2}$ is*

[TABLE]

where $M=\lambda I-L(G_{1})$ and $Q=\lambda I-L(G_{2})$.

Corollary 3

.* Let $G$ be a unicyclic graph on $n$ vertices. If $\lambda_{1},\ldots,\lambda_{n}$ are the Laplacian eigenvalues of $G$, then $\lambda_{1},\ldots,\lambda_{n}$ are the Laplacian eigenvalues of $G^{\prime}=G\odot_{uv}G$.*

**Proof. **

Suppose $u_{1},\ldots,u_{n}$ and $v_{1},\ldots,v_{n}$ are the vertices of $G$ and the copy of $G$, respectively. Let $G^{\prime}=G\odot_{u_{i}v_{j}}G$ and $M=L_{G}(\lambda)$. Therefore $L_{G^{\prime}}(\lambda)=M^{2}-M.M_{v_{j}}-M.M_{u_{i}}$, by Equation (3). So $L_{G^{\prime}}(\lambda)=M(M-M_{u_{i}}-M_{v_{j}})$ and the result follows.* $\Box$*

In [3], we have shown the upper bound for $m_{G}(\lambda)$, where $\lambda>1$.

Lemma 4

.[3, Lemma 2]* Let $G$ be a bicyclic graph and $\lambda>1$ be an integral eigenvalue of $L(G)$. It holds that $m_{G}(\lambda)\leq 3$.*

Now, we identified some bicyclic graphs that having $2$ among their Laplacian eigenvalues with multiplicity $3$.

Theorem 5

.[Main Theorem]* Let $G_{1}$ and $G_{2}$ be unicyclic graphs containing a perfect matching with $m_{G_{1}}(2)=m_{G_{2}}(2)=2$. It holds that $m_{G_{1}\odot_{uv}G_{2}}(2)=3$ where $u\in C_{g_{1}}$ and $v\in C_{g_{2}}$.*

**Proof. **

If $m_{G_{1}}(2)=m_{G_{2}}(2)=2$, then there are two situations.

If $G_{1}=C_{m}$ and $G_{2}=C_{n}$ then $m\equiv n\equiv 0(\mathrm{mod}\,4)$, by [1, Theorem 12]. 2. 2.

If $G_{1}=C(T^{(1)}_{1},\ldots,T^{(1)}_{g_{1}})$ and $G_{2}=C(T^{(2)}_{1},\ldots,T^{(2)}_{g_{2}})$ such that $\sum^{k=g_{1}}_{k=1}|V(T^{(1)}_{k})|=m$, $\sum^{k=g_{2}}_{k=1}|V(T^{(2)}_{k})|=n$ and there exists at least one $i$ (or $j$) so that $|V(T^{(1)}_{i})|\geq 3$ (or $|V(T^{(2)}_{j})|\geq 3)$. Let

[TABLE]

so $s_{i}=g_{i}$ and $g_{i}\equiv 0\ (\mathrm{mod}\,4)$, for $i=1,2$, by [1, Theorem 13].

In addition, $G=G_{1}\odot{uv}G_{2}$ is a bicyclic graph so $m_{G}(2)\leq 3$, by Lemma 4. On the other hand,

[TABLE]

According to Equation (3), it is enough to show that $M_{u}$ and $Q_{v}$ have $\lambda-2$ as a factor.

Case $1$. Let $G_{1}=C_{m}$, $G_{2}=C_{n}$ and $m\equiv n\equiv 0(\mathrm{mod}\,4)$.

We use induction on $m$. If $G_{1}=C_{4}$ then the result follows and so the induction basis holds.

[TABLE]

Suppose in the cyclic graph $C_{m-4}$, $M_{u}(C_{m-4})$ with $m-4\equiv 0(\mathrm{mod}\,4)$ has $\lambda-2$ as a factor.

Let $G_{1}=C_{m}$ and without loss of generality, assume that $u=u_{m}$. So

[TABLE]

such that

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

Therefore, $M_{u}=|M^{\prime}||Q^{\prime}-N^{T}M^{\prime-1}N|$, by Lemma 2.

[TABLE]

Also

[TABLE]

Then

[TABLE]

Furthermore, $|Q^{\prime}-N^{T}M^{\prime-1}N|=|Q^{\prime}|+|H|$, such that

[TABLE]

Consequently,

[TABLE]

[TABLE]

On the other hand

[TABLE]

Also, $|Q^{\prime}_{(m-5)\times(m-5)}|$ according to the induction hypothesis has $\lambda-2$ as a factor therefore, $M_{u}$ has $\lambda-2$ as a factor. With a similar method, one can check that $Q_{v}$ has $\lambda-2$ as a factor. Thus, $m_{G}(2)=3$ and we are done.

Case $2$. Let $G_{1}=C(T^{(1)}_{1},\ldots,T^{(1)}_{g_{1}})$ and $G_{2}=C(T^{(2)}_{1},\ldots,T^{(2)}_{g_{2}})$ so $s_{i}=g_{i}$ and $g_{i}\equiv 0\ (\mathrm{mod}\,4)$, for $i=1,2$. In addition, there exists some $i$ ($j$), such that $|V(T^{(1)}_{i})|\geq 3$ ($|V(T^{(2)}_{j})|\geq 3$). Let $u^{\prime}\in V(T^{(1)}_{i})$ and $d(u^{\prime},u_{i})=max_{x\in V(T^{(1)}_{i})}d(x,u_{i})$, where $u_{i}$ is the root of $T^{(1)}_{i}$. Since $G_{1}$ has a perfect matching, $u^{\prime}$ is a pendent vertex and its neighbor, say $u^{\prime\prime}$, has degree $2$. Thus, $G=(G\setminus\{u^{\prime},u^{\prime\prime}\})\odot S_{2}$. Using [5, Theorem 2.5] we obtain $m_{G}(2)=m_{G\setminus\{u^{\prime},u^{\prime\prime}\}}(2)$. So by repeating this method in the graph $G\setminus\{u^{\prime},u^{\prime\prime}\}$ and omitting all leaf vertices and quasi leaf vertices of degree $2$ and using [5, Theorem 2.5], the obtained graph is two cycles like graph in the case $1$. So $m_{G}(2)=3$ and the proof is complete. * $\Box$*

Example 1

. $G_{1}$ and $G_{2}$ are two unicyclic graphs with $m_{G_{1}}(2)=m_{G_{2}}(2)=2$. Also, there exists at least one $k$ such that $|T^{(i)}_{k}|\geq 3$ for $i=1$ or i=$2$. The bicyclic graph $G_{1}\odot_{uv}G_{2}$ has $2$ among its Laplacian eigenvalues with multiplicity $3$.

The following example shows that the converse of Theorem 5 does not necessarily true in general.

Example 2

. Let $G$ be a unicyclic graph same as Figure $2$. $G$ has $2$ among its Laplacian eigenvalues with multiplicity $1$. But $G^{\prime}=G\odot_{uv}G$ has $2$ among its Laplacian eigenvalues with multiplicity $3$ (Figure $3$).

In [3] authors have considered a necessary and sufficient condition in the bicyclic graph $G=G_{1}\odot_{uv}G_{2}$ for having the Laplacian eigenvalue $2$, where $G_{1}$ and $G_{2}$ are unicyclic graphs and have $2$ among their Laplacian eigenvalues. We study the Laplacian eigenvalue $2$ of $G=G_{1}\odot_{uv}G_{2}$, where $G_{1}$ and $G_{2}$ are unicyclic graphs and $G_{1}$ or $G_{2}$ dose not have $2$ among its Laplacian eigenvalues by using Equation (1) without having the Laplacian characteristic polynomial.

Theorem 6

.* Let $G_{1}=C(T_{1},\ldots,T_{g_{1}})$ and $G_{2}=C(T_{1},\ldots,T_{g_{2}})$ be unicyclic graphs such that $G_{1}$ has a perfect matching and $2$ among its Laplacian eigenvalues. Also, $G_{2}$ dose not have $2$ among its Laplacian eigenvalues. Then $G=G_{1}\odot_{uv}G_{2}$ has $2$ among its Laplacian eigenvalues if and only if $x(u)=0$.*

**Proof. **

Let $G_{1}$ be a unicyclic graph containing a perfect matching such that has $2$ among its Laplacian eigenvalues. So $s\equiv 0\ (mod\,4)$, where $s$ is the number of trees of odd orders in $G_{1}$, by [1, Theorem 9].

Case $1$. If $g_{1}\not\equiv 0\ (mod\,4)$ or $g_{1}\equiv 0\ (mod\,4)$ and $s\neq g$ then $m_{G_{1}}(2)=1$ by [1, Theorem 13] and there exists the eigenvector corresponding to the eigenvalue $2$ like $Y=(x_{1},\ldots,x_{n_{1}})^{t}$, such that $x_{i}\in\{-1,1\}$ for $i=1\cdots n_{1}$, by [3, Theorem 4], so $x(u)\neq 0$. By contrary, if $G$ has $2$ among its Laplacian eigenvalues, then we can assume that $X=(x_{1},\ldots,x_{n_{1}+n_{2}})^{t}$ is an eigenvector of $L(G)$ corresponding to the eigenvalue $2$. All vertices of $G$ satisfy in Equation (1). Therefore

[TABLE]

On the other hand, $G_{1}$ has $2$ among its Laplacian eigenvalues. So, we have $x(v)=x(u)\neq 0$. Also

[TABLE]

Thus, by noting the fact that $d_{G_{2}}(w)=d_{G}(w)$ for the other vertices of $G_{2}$, we have

[TABLE]

Hence, there exist an eigenvector opposed to zero of $L(G_{2})$ corresponding to the eigenvalue $2$ and this is a contraction. Therefore, the proof is complete.

Case $2$. If $s=g_{1}$ and $g_{1}\equiv 0\ (mod\,4)$ then $m_{G_{1}}(2)=2$ by [1, Theorem 13] and there exists an eigenvector of $L(G_{1})$ like $X=(x_{1},\ldots,x_{n_{1}})^{t}$ corresponding to the eigenvalue $2$, such that $x_{i}\in\{-1,0,1\}$. For proving the “if part”, let $x(u)=0$. Then by assigning [math] to all vertices of $G_{2}$, one can easily check that $Y=(X,0)^{t}$ is an eigenvector of $L(G)$ corresponding to the eigenvalue $2$ (all vertices of $G$ satisfy in Equation (1)). Conversely, suppose $Y$ is an eigenvector of $L(G)$ corresponding to the eigenvalue $2$ such that $y(u)\neq 0$. By a similar method in case $1$, an eigenvector like $X\neq 0$ exists such that satisfy in Equation (1) for $\lambda=2$ in $L(G_{2})$ and this is a contradiction. Therefore, the result follows.* $\Box$*

Theorem 7

.* Let $G_{1}$ be a broken sun graph of order $n_{1}$ which has no perfect matching and has $2$ among its Laplacian eigenvalues. Also, $G_{2}$ is a unicyclic graph of order $n_{2}$ that does not have $2$ among its Laplacian eigenvalues. So, $G=G_{1}\odot_{uv}G_{2}$ has $2$ among its Laplacian eigenvalues if and only if $x(u)=0$.*

**Proof. **

First, assume that $g_{1}\equiv 0\ (mod\,4)$ and there exist odd number of vertices of degree 2 between any pair of consecutive vertices of degree $3$, by [1, Theorem 10]. We may assign $\{-1,0,1\}$ to the vertices of $C_{g_{1}}$, by the pattern $0,1,0,-1$ consecutively starting with a vertex of degree $3$, and assign to each pendent vertex the negative of value of its neighbor, to obtain an eigenvector of $L(G)$ corresponding to the eigenvalue $2$ like $X=(x_{1},\ldots,x_{n_{1}})^{t}$. Now, if $x(u)=0$ then by assigning [math] to all vertices of $G_{2}$, one can easily check that $Y=(X,0)^{t}$ is an eigenvector of $L(G)$ corresponding to the eigenvalue $2$ (all vertices of $G$ satisfy in Equation (1)). Conversely, suppose $X$ is an eigenvector of $L(G)$ corresponding to the eigenvalue $2$ such that $x(u)\neq 0$. By a similar method in Theorem 6 case $1$, there exists an eigenvector like $Z\neq 0$ such that $Z$ satisfy in Equation (1) for $\lambda=2$ in $L(G_{2})$ and this is a contradiction. Therefore, the proof is complete.* $\Box$*

Example 3

. In the following Figure $(4)$, $G_{1}$ and $G_{2}$ are broken graphs such that just one of them has $2$ among its Laplacian eigenvalues. By Theorem 7 and by Equation (1),

[TABLE]

is an eigenvector of $L(G)$ corresponding to the eigenvalue $2$.

In the follow, we show that two broken sun graphs $G_{1}$ and $G_{2}$ do not have $2$ among their Laplacian eigenvalues, but $G=G_{1}\odot_{uv}G_{2}$ may have $2$ among Laplacian eigenvalues.

Theorem 8

.* Let $G_{1}\in\mathfrak{U}_{n_{1},g_{1}}$ and $G_{2}\in\mathfrak{U}_{n_{2},g_{2}}$, and $m_{1}$, $m_{2}$ be the numbers of leaf vertices in $G_{1}$ and $G_{2}$ for which $m_{1}=g_{1}-1$ and $m_{2}=g_{2}-1$. Let $d_{G_{1}}(u)=d_{G_{2}}(v)=2$. Then $L(G=G_{1}\odot_{uv}G_{2})$ has $2$ among its Laplacian eigenvalues.*

**Proof. **

Let $G_{1}\in\mathfrak{U}_{n_{1},g_{1}}$ and $G_{2}\in\mathfrak{U}_{n_{2},g_{2}}$. We may assign $1$ and $-1$ to the vertices of $C_{g_{1}}$ and $C_{g_{2}}$, and assign $-1$ and $1$ to each leaf vertices of $G_{1}$ and $G_{2}$, respectively. One may check that every vertices of $G$ satisfies in Equation (1). Therefore $X=(x_{1},\ldots,x_{n_{1}+n_{2}})$ is an eigenvector corresponding to the Laplacian eigenvalue $2$, such that $x_{i}\in\{-1,1\}$, for $i=1,\ldots,n_{1}+n_{2}$ and we are done. * $\Box$*

Example 4

. Let $G_{1}$ and $G_{2}$ be broken sun graphs (see Figure $5$), then $G=G_{1}\odot_{uv}G_{2}$ has $2$ among its Laplacian eigenvalues by Theorem 8.

Now, we finish this article with extend this result for graphs with $n$ cycles.

Lemma 9

.* Let $G$ be a graph with $n$ cycles and $\lambda>1$ be an integral Laplacian eigenvalue of $G$. Therefore, $m_{G}(\lambda)\leq n+1$.*

**Proof. **

If $G$ be a unicyclic graph then $m_{G}(\lambda)\leq 2$, by [1, Lemma 4]. By induction on $n$, let $m_{G}(\lambda)\leq n$ for all graphs with cycles less than $n$. The graph $G^{\prime}=G-e$, where $e$ belongs to one of the cycles of $G$, is a graph with $n-1$ cycles and by using the induction assumption $m_{G^{\prime}}(\lambda)\leq n$. If $m_{G}(\lambda)>n+1$ then $m_{G^{\prime}}(\lambda)>n$, by [8, Theorem 3.2] and this contradicts. So, the result follows.* $\Box$*