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*Mathematics*
**2019**,
*7*(1),
44;
https://doi.org/10.3390/math7010044

Article

The A

_{α}-Spectral Radii of Graphs with Given Connectivity^{1}

School of Mathematics and Statistics and Hubei key Laboratory Mathematics Sciences, Central China Normal University, Wuhan 430079, China

^{2}

Department of Mathematics, Savannah State University, Savannah, GA 31419, USA

*

Correspondence: [email protected]; Tel.: +1-352-665-3381

^{†}

These authors contributed equally to this work.

Received: 22 November 2018 / Accepted: 24 December 2018 / Published: 4 January 2019

## Abstract

**:**

The ${A}_{\alpha}$-matrix is ${A}_{\alpha}(G)=\alpha D(G)+(1-\alpha )A(G)$ with $\alpha \in [0,1]$, given by Nikiforov in 2017, where $A(G)$ is adjacent matrix, and $D(G)$ is its diagonal matrix of the degrees of a graph G. The maximal eigenvalue of ${A}_{\alpha}(G)$ is said to be the ${A}_{\alpha}$-spectral radius of G. In this work, we determine the graphs with largest ${A}_{\alpha}(G)$-spectral radius with fixed vertex or edge connectivity. In addition, related extremal graphs are characterized and equations satisfying ${A}_{\alpha}(G)$-spectral radius are proposed.

Keywords:

adjacent matrix; signless Laplacian; spectral radius; connectivity## 1. Introduction

We consider simple finite connected graph G with the vertex set $V(G)$ and the edge set $E(G)$. The number of vertices $|V(G)|=n$ is the order of a graph, and the number of edges $|E(G)|$ is the size of a graph. Denote the neighborhood of $v\in V(G)$ by $N(v)=\{u\in V(G),vu\in E(G)\}$, and the degree of v by ${d}_{G}(v)=\left|N(v)\right|$ (or briefly ${d}_{v}$). For $L\subseteq V(G)$ and $R\subseteq E(G)$, let $w(G-L)$ or $w(G-R)$ be the number of components of $G-L$ or $G-R$. $L($ or $R)$ be a vertex(edge) cut set if $w(G-L$ (or $R)$) $\ge 2$ and $E(w,L)=\{wu\in E(G),u\in L\}$. For $U\subseteq V(G)$, $G[U]$ denote the induced subgraph of G, that is, $V(G[U])=U$ and $E(G[U])=\{uv|uv\in E(G),u,v\in U\}$.

If $A(G)$ is adjacency matrix of a graph G, and $D(G)$ is its diagonal matrix of the degrees of G, then the signless Laplacian matrix of G is $D(G)+A(G).$ With the successful studies of these matrices, Nikiforov [1] proposed the ${A}_{\alpha}$-matrix
with $\alpha \in [0,1]$. Obviously, ${A}_{0}(G)$ is the adjacent matrix and ${A}_{\frac{1}{2}}$ is the half of signless Laplacian matrix of G, respectively. For undefined terminologies and notations, we refer to [2].

$${A}_{\alpha}(G)=\alpha D(G)+(1-\alpha )A(G)$$

The research of (adjacency, signless Laplacian) spectral radius is an intriguing topic during past decades [3,4,5,6,7,8,9]. For instances, Lovász and J. Pelikán studied the spectral radius of trees [10]. The minimal Laplacian spectral radius of trees with given matching number is given by Feng et al. [7]. The properties of spectra of graphs and their line graphs are studied by Chen [11]. The signless Laplacian spectra of graphs is explored by Cvetković et al. [12]. Zhou [13] found bounds of signless Laplacian spectral radius and its hamiltonicity. Graphs having none or one signless Laplacian eigenvalue larger than three are obtained by Lin and Zhou [14]. At the same time, the maximal adjacency or signless Laplacian spectral radius have attracted many interests among the mathematical literature including algebra and graph theory. Ye et al. [6] gave the maximal adjacency or signless Laplacian spectral radius of graphs subject to fixed connectivity.

Inspired by these outcomes, we determine the graphs with largest ${A}_{\alpha}(G)$-spectral radius with given vertex or edge connectivity. In addition, the corresponding extremal graphs are provided and the equations satisfying the ${A}_{\alpha}(G)$-spectral radius are obtained.

## 2. Preliminary

In this section, we provide some important concepts and lemmas that will be used in the main proofs.

Denote by G a graph such that $V(G)=\{{v}_{1},{v}_{2},\cdots ,{v}_{n}\}$ is its vertex set and $E(G)$ is its edge set. The ${A}_{\alpha}$-matrix of G has the $(i,j)$-entry of ${A}_{\alpha}(G)$ is $1-\alpha $ if ${v}_{i}{v}_{j}\in E(G)$; $\alpha d({v}_{i})$ if $i=j$, and otherwise 0. For $\alpha \in [0,1]$, let ${\lambda}_{1}({A}_{\alpha}(G))\ge {\lambda}_{2}({A}_{\alpha}(G))\ge \cdots \ge {\lambda}_{n}({A}_{\alpha}(G))$ be the eigenvalues of ${A}_{\alpha}(G)$. The ${A}_{\alpha}$-spectral radius of G is considered as the maximal eigenvalue $\rho :={\lambda}_{1}({A}_{\alpha}(G))$. Let $X={({x}_{{v}_{1}},{x}_{{v}_{2}},\cdots ,{x}_{{v}_{n}})}^{T}$ be a real vector of $\rho $.

By ${A}_{\alpha}(G)=\alpha D(G)+(1-\alpha )A(G)$, we have the quadratic formula of ${X}^{T}{A}_{\alpha}(G)X$ can be expressed that

$$\begin{array}{c}\hfill {X}^{T}{A}_{\alpha}(G)X=\alpha \sum _{{v}_{i}\in V(G)}{x}_{{v}_{i}}^{2}{d}_{{v}_{i}}+2(1-\alpha )\sum _{{v}_{i}{v}_{j}\in E(G)}{x}_{{v}_{i}}{x}_{{v}_{j}}.\end{array}$$

Because ${A}_{\alpha}(G)$ is a real symmetric matrix, and by Rayleigh principle, we have the formula

$$\rho (G)=ma{x}_{X\ne 0}\frac{{X}^{T}{A}_{\alpha}(G)X}{{X}^{T}X}.$$

As we know that once X is an eigenvector of $\rho (G)$ for a connected graph G, X should be unique and positive. The corresponding eigenequations for ${A}_{\alpha}(G)$ is rewritten as

$$\begin{array}{c}\hfill \rho (G){x}_{{v}_{i}}=\alpha {d}_{{v}_{i}}{x}_{{v}_{i}}+(1-\alpha )\sum _{{v}_{i}{v}_{j}\in E(G)}{x}_{{v}_{j}}.\end{array}$$

As ${A}_{1}(G)=D(G)$, we study the ${A}_{\alpha}$-matrix for $\alpha \in [0,1)$ below. Based on the definition of ${A}_{\alpha}$-spectral radius, we have

**Lemma**

**1.**

[4,15] Let ${A}_{\alpha}(G)$ be the ${A}_{\alpha}$-matrix of a connected graph G $(\alpha \in [0,1))$, $v,w\in V(G)$, $u\in T\subset V(G)$ such that $T\subset N(v)\backslash (N(w)\cup \{w\})$. Let ${G}^{*}$ be a graph with vertex set $V(G)$ and edge set $E(G)\backslash \{uv,u\in T\}\cup \{uw,u\in T\}$, and X a unit eigenvector to $\rho ({A}_{\alpha}(G))$. If ${x}_{w}\ge {x}_{v}$ and $\left|T\right|\ne 0$, then $\rho ({G}^{*})>\rho (G).$

If G is a connected graph, then ${A}_{\alpha}(G)$ is a nonnegative irreducible symmetric matrix. By the results of [1,16,17] and adding extra edges to a connected graph, then ${A}_{\alpha}$-spectral radius will increase and the following lemma is straightforward.

**Lemma**

**2.**

$(i)$ If ${G}^{*}$ is any proper subgraph of connected graph G, and ρ is the ${A}_{\alpha}$-spectral radius, then $\rho ({G}^{*})<\rho (G).$

$(ii)$ If X is a positive vector and r is a positive number such that ${A}_{\alpha}(G)X<rX$, then $\rho (G)<r$.

Recall that the vertex connectivity (respectively, edge connectivity) of a graph G is the smallest number of vertices (respectively, edges) such that if we remove them, the graph will be disconnected or be a single vertex. For convenience, let ${\mathcal{F}}_{n}$ be the set of all graphs of order n, and ${\mathcal{F}}_{n}^{k}$ (respectively, ${\overline{\mathcal{F}}}_{n}^{k}$) $(k\ge 0)$ be the set of such graphs with order n and vertex (resp., edge) connectivity k. Note that ${\mathcal{F}}_{n}^{0}$ = ${\overline{\mathcal{F}}}_{n}^{0}$ having some disconnected graphs of order n, and ${\mathcal{F}}_{n}^{n-1}$ = ${\overline{\mathcal{F}}}_{n}^{n-1}$ consisting of the unique graph ${K}_{n}$. Obviously, ${\mathcal{F}}_{n}$ = ${\cup}_{k}{\mathcal{F}}_{n}^{k}$ = ${\cup}_{k}{\overline{\mathcal{F}}}_{n}^{k}$.

Recall the graph $K(p,q)(p\ge q\ge 0)$ obtained from ${K}_{p}$ by attaching a vertex together with edges connecting this vertex to q vertices of ${K}_{p}$. $K(p,q)$ is was found by Brualdi and Solehid in terms of stepwise adjacency matrix, but it is Peter Rowlinson who gives the purely combinatorial definition of such graph. For the property of $K(p,q)$, we refer to [18,19,20]. Clearly, $K(p,0)$ is ${K}_{p}$ with an additional isolated vertex. It’s not hard to see that $K(p,q)$ is of vertex (resp., edge) connectivity q. Let $\delta ,\Delta $ be the smallest and largest degrees of vertices in the graph G, respectively.

**Lemma**

**3.**

The graph ${K}_{n}$ is the graph in ${\mathcal{F}}_{n}$ having the largest ${A}_{\alpha}$-spectral radius, and ${K}_{n-1}\cup {K}_{1}=K(n-1,0)$ is the graph in ${\mathcal{F}}_{n}^{0}$ or ${\overline{\mathcal{F}}}_{n}^{0}$ having the smallest ${A}_{\alpha}$-spectral radius.

**Proof.**

By Lemma 2, the first statement is clear. For the second one, let G be a graph which attains the maximum ${A}_{\alpha}$-spectral radius in ${\mathcal{F}}_{n}^{0}$, then G only has two unique connected components: ${K}_{n-1}$, ${K}_{1}$; if not, any component of G will be a proper subgraph of ${K}_{n-1}$. Then $\rho (G)<\rho ({K}_{n-1})=\rho ({K}_{n-1}\cup {K}_{1})$, a contradiction. Then this lemma is proved. □

**Lemma**

**4.**

For $k\in [1,n-2]$, $K(n-1,k)$ is the graph having the largest ${A}_{\alpha}$-spectral radius in ${\mathcal{F}}_{n}^{k}$.

**Proof.**

Denote by G a graph having the largest ${A}_{\alpha}$-spectral radius in ${\mathcal{F}}_{n}^{k}$. x is a unit (positive) Perron vector of ${A}_{\alpha}$. Let U be the vertex cut of G having k vertices, and these components of $G-U$ be ${G}_{1},{G}_{2},\cdots ,{G}_{s}$, for $s\ge 2$. We declare that $s=2$; if not, adding all possible edges within the graph ${G}_{1}\cup {G}_{2}\cup \cdots \cup {G}_{s-1}$, we would get a graph belonging to ${\mathcal{F}}_{n}^{k}$ (because U is the smallest vertex cut set) and with a larger ${A}_{\alpha}$-spectral radius. Similarly, induced subgraph $G[U]$, the subgraphs ${G}_{1}$ and ${G}_{2}$ are complete subgranph, and every vertex of U connects these vertices of ${G}_{1}$ and ${G}_{2}$. Next we prove that one of ${G}_{1},{G}_{2}$ will be a singleton, which has a unique vertex. If not, suppose that ${G}_{1},{G}_{2}$ have orders greater than one. Without loss of generality, denote by u a vertex of ${G}_{1}$ having a smallest value for x among vertices in ${G}_{1}\cup {G}_{2}$. Deleting these edges of ${G}_{1}$ incident to u, and connecting all possible edges between ${G}_{1}-u$ and ${G}_{2}$, we get a graph $\tilde{G}=K(n-1,k)$ still in ${\mathcal{F}}_{n}^{k}$. By Lemma 1, $\rho (\tilde{G})>\rho (G)$, which yields a contradiction. So one of ${G}_{1},{G}_{2}$ is a singleton, and G is the desired graph $K(n-1,k)$. □

**Lemma**

**5.**

For $k\in [1,n-2]$, $K(n-1,k)$ is the graph having maximum ${A}_{\alpha}$-spectral radius in ${\overline{\mathcal{F}}}_{n}^{k}$.

**Proof.**

Denote by G a graph having the largest ${A}_{\alpha}$-spectral radius in ${\mathcal{F}}_{n}^{k}$. x is a unit (positive) Perron vector of ${A}_{\alpha}$. We know that each vertex of G has degree greater than or equal to k. Otherwise $G\notin {\overline{\mathcal{F}}}_{n}^{k}$. If there is a vertex u in G with degree k, then the edges adjacent to u are an edge cut such that $G-u$ is complete. The statement follows in this case. Then we will suppose that all vertices in G have degrees greater than k. Let ${E}_{c}$ be an edge cut set of G having k edges. So $G-{E}_{c}$ consists of only two components ${G}_{1},{G}_{2}$, respectively, of order ${n}_{1},{n}_{2}$. Obviously ${G}_{1},{G}_{2}$ are both complete. In addition, neither of ${G}_{1},{G}_{2}$ is a singleton. Otherwise G would contain a vertex of degree k, which contradicted to the above assumption. So ${G}_{1},{G}_{2}$ contain more than 1 vertex, i.e., ${n}_{1}\ge 2$ and ${n}_{2}\ge 2$.

Without loss of generality, suppose that ${G}_{1}$ contains a vertex ${w}_{1}$ having a minimal value given by x within all vertices of ${G}_{1}\cup {G}_{2}$, and consists of vertices ${w}_{1},{w}_{2},\cdots ,{w}_{{n}_{1}}$ such that $x({w}_{1})\le x({w}_{2})\le \cdots \le x({w}_{{n}_{1}})$. Assume that ${w}_{1}$ joins t vertices of ${G}_{2}$. Surely $t\le min\{k,{n}_{2}\}$.

If $t=k$, there exist no edges joining ${G}_{1}-{w}_{1}$ and ${G}_{2}$, and ${n}_{2}\ge k+2$ otherwise ${G}_{2}$ contains a vertex of degree k. Denote by ${G}^{\prime}$ a new graph with vertex set $V(G)$ and edge set $E(G)\backslash E({w}_{1},N)\cup E(N,{v}^{\prime})$, where $N=N({w}_{1})\cap V({G}_{1})$, and ${v}^{\prime}\in V({G}_{2})-N({w}_{1})\cap V({G}_{2})$, by Lemma 1, we have $\rho ({G}^{\prime})>\rho (G)$. Let ${G}^{\u2033}$ be another new graph with vertex set $V({G}^{\prime})$ and adding all possible edges between ${G}_{1}-{w}_{1}$ and ${G}_{2}$. Note that ${G}^{\u2033}=K(n-1,k)$, and ${G}^{\prime}$ is a proper subgraph of ${G}^{\u2033}$. By Lemma 2, we have $\rho ({G}^{\u2033})>\rho ({G}^{\prime})$. Thus, $\rho ({G}^{\u2033})>\rho (G)$, a contradiction.

If $t<k$. Partition the set $V({G}_{1})-{w}_{1}$ as: ${V}_{11}=\{{w}_{i}:i=2,3,\cdots ,{n}_{1}-(k-t)\}$, ${V}_{12}=\{{w}_{j}:j={n}_{1}-(k-t)+1,\cdots ,{n}_{1}\}$. Thus, $|{V}_{11}|={n}_{1}-(k-t)-1$; $|{V}_{12}|=k-t$.

Let $N=N({w}_{1})\cap {V}_{11}$, then $N\ne \mathsf{\varnothing}$ since $d({w}_{1})>k$. Note there is vertex ${v}^{\prime}\in V({G}_{2})-N({w}_{1})\cap V({G}_{2})$ since ${n}_{2}\ge k+2$. Let ${G}^{\prime}$ be a new graph having vertex set $V(G)$ and edge set $E(G)\backslash E({w}_{1},N)\cup E(N,{v}^{\prime})$, where $N=N({w}_{1})\cap {V}_{11}$, and ${v}^{\prime}\in V({G}_{2})-N({w}_{1})\cap V({G}_{2})$, by Lemma 1, we have $\rho ({G}^{\prime})>\rho (G)$. Let ${G}^{\u2033}$ be another new graph having vertex set $V({G}^{\prime})$ and adding all possible edges between ${G}_{1}-{w}_{1}$ and ${G}_{2}$, adding all edges between ${w}_{1}$ and ${V}_{12}$. Note that ${G}^{\prime \prime}=K(n-1,k)$, and ${G}^{\prime}$ is a proper subgraph of ${G}^{\u2033}$. Lemma 2 implies that $\rho ({G}^{\prime \prime})>\rho ({G}^{\prime})$. Thus, $\rho ({G}^{\prime \prime})>\rho (G)$, a contradiction. The result follows. □

## 3. Main Results

In this section, we will determine maximizing ${A}_{\alpha}$-spectral radius of of graphs with given connectivity. By Lemma 4 and Lemma 5, we obtain the following Theorem:

**Theorem**

**1.**

The graph ${K}_{n}$ is the graph in ${\mathcal{F}}_{n}$ with ${A}_{\alpha}$-spectral radius, and ${K}_{n-1}\cup {K}_{1}=K(n-1,0)$ is the unique one in ${\mathcal{F}}_{n}^{0}$ or ${\overline{\mathcal{F}}}_{n}^{0}$ with ${A}_{\alpha}$-spectral radius. For $k\in [1,n-2]$, $K(n-1,k)$ is the graph with maximum ${A}_{\alpha}$-spectral radius in ${\mathcal{F}}_{n}^{k}$ or ${\overline{\mathcal{F}}}_{n}^{k}$.

**Proof.**

By the Lemmas 3–5, we obtain the results. □

**Lemma**

**6.**

[20] Given a partition $\{1,2,\cdots ,n\}$ =${\Delta}_{1}\cup {\Delta}_{2}\cup \cdots \cup {\Delta}_{m}$ with $|{\Delta}_{i}|={n}_{i}>0$, A be any matrix partitioned into blocks ${A}_{ij}$, where ${A}_{ij}$ is an ${n}_{i}\times {n}_{j}$ block. Suppose that the block ${A}_{ij}$ has constant row sums ${b}_{ij}$, and let $B=({b}_{ij})$. Then the spectrum of B is contained in the spectrum of A (taking into account the multiplicities of the eigenvalues).

Since $K(n-1,k)$ contains ${K}_{n-1}$, we can partition $K(n-1,k)$ into three different subsets: $\{u\},T,S$, in which u is the vertex connecting a complete subgraph ${K}_{n-1}$ with k edges, a subset S is in ${K}_{n-1}$ connecting u, and $T=V({K}_{n-1}\backslash S)$. Let x be a Perron vector of $K(n-1,k)$. $S=\{{u}_{1},{u}_{2},\cdots ,{u}_{k}\}$ and $T=\{{v}_{1},{v}_{2},\cdots ,{v}_{t}\}$. Note that $k+t+1=n$.

**Theorem**

**2.**

Label the vertices of $K(n-1,k)$ as $u,{u}_{1},{u}_{2},\cdots ,{u}_{k},{v}_{1},{v}_{2}\cdots ,{v}_{t}$ with $k,t\ge 0$. The maximum eigenvalues of ${A}_{\alpha}(K(n-1,k))$ satisfy the equation: $f(\rho )=(\rho -k\alpha )(\rho -k\alpha -n+k+2)(\rho -n\alpha +1)-k(1-\alpha )(\rho -k\alpha -\alpha +1)(\rho -n\alpha +\alpha +1)+k{(1-\alpha )}^{3}(n-k-1)=0.$

**Proof.**

Since the matrix ${A}_{\alpha}=\alpha D+(1-\alpha )A$, where D has on the diagonal the vector $(k,n-1,n-2)$ and A consists of the following three row-vectors, in the order: $(0,k,0)$; $(1,k-1,n-k-1)$; $(0,k,n-k-2)$. Thus, by the Lemma 6, x is a constant value ${\beta}_{2}$ on the vertex set S, and constant value ${\beta}_{3}$ on the vertex set T. Defining $x(u)=:{\beta}_{1}$, $\rho (K(n-1,k))=:\rho $, also by (1), we get

$$\begin{array}{c}\hfill (\rho -\alpha k){\beta}_{1}=k(1-\alpha ){\beta}_{2}\end{array}$$

$$\begin{array}{c}\hfill (\rho -\alpha (n-1)){\beta}_{2}=(1-\alpha )({\beta}_{1}+(k-1){\beta}_{2}+t{\beta}_{3}),\phantom{\rule{4.pt}{0ex}}\mathrm{and}\phantom{\rule{4.pt}{0ex}}\end{array}$$

$$\begin{array}{c}\hfill (\rho -\alpha (n-2)){\beta}_{3}=(1-\alpha )(k{\beta}_{2}+(t-1){\beta}_{3}).\end{array}$$

Then we get

$$\begin{array}{c}\hfill (\rho -\alpha (n-1))=\frac{k{(1-\alpha )}^{2}}{\rho -k\alpha}+\frac{kt{(1-\alpha )}^{2}}{\rho -k\alpha -t+1}+(k-1)(1-\alpha ).\end{array}$$

Note that for $n=t+k+1$, that is, $n-1=k+t$. Then we have:

$$\begin{array}{c}\hfill (\rho -k\alpha )=\frac{k{(1-\alpha )}^{2}}{\rho -k\alpha}+\frac{kt{(1-\alpha )}^{2}}{\rho -k\alpha -t+1}+(k-1)(1-\alpha )+t\alpha .\end{array}$$

Then we obtain that

$$\begin{array}{c}\hfill (\rho -k\alpha )(\rho -k\alpha -n+k+2)(\rho -n\alpha +1)-k(1-\alpha )(\rho -k\alpha \\ \hfill {-\alpha +1)(\rho -n\alpha +\alpha +1)+k(1-\alpha )}^{3}(n-k-1)=0.\end{array}$$

Thus, our proof is finished. □

**Corollary**

**1.**

Let G be a graph of order n having vertex/edge connectivity k, where $1\le k\le n-2$, the maximum adjacency spectral radius is the largest root of the $f(\lambda )={\lambda}^{3}-(n-3){\lambda}^{2}-(n+k-2)\lambda +k(n-k-2)=0$.

**Proof.**

By Theorem 2, let $\alpha =0$, then $f(\lambda )={\lambda}^{3}-(n-3){\lambda}^{2}-(n+k-2)\lambda +k(n-k-2)=0$. It is obvious since ${A}_{0}=A(G)$. □

By letting the special values for $\alpha $, we have the following corollary.

**Corollary**

**2.**

Let G be a graph of order n having vertex/edge connectivity k, where $1\le k\le n-2$, the signless Laplacian spectral radius ${\lambda}_{1}=\frac{2n+k-4+\sqrt{{(2n-k-4)}^{2}+8k}}{2}$.

**Proof.**

By Theorem 2, let $\alpha =\frac{1}{2}$, then $f(\lambda )={\lambda}^{3}-\frac{1}{2}(3n+k-6){\lambda}^{2}+(\frac{1}{4}(n-4)(2n+3k)+k+2)\lambda -\frac{1}{4}k({n}^{2}-5n+6)=0$. It is obvious since $2{A}_{\frac{1}{2}}=D+Q$. Thus,

$$\begin{array}{cc}\hfill 8f(\lambda )& =8[{\lambda}^{3}-\frac{1}{2}(3n+k-6){\lambda}^{2}+(\frac{1}{4}(n-4)(2n+3k)+k+2)\lambda \hfill \\ & \hspace{1em}-\frac{1}{4}k({n}^{2}-5n+6)]\hfill \\ & ={(2\lambda )}^{3}-(3n+k-6){(2\lambda )}^{2}+((n-4)(2n+3k)+4k+8)(2\lambda )\hfill \\ & \hspace{1em}-2k({n}^{2}-5n+6)\hfill \\ & ={({\lambda}_{1})}^{3}-(3n+k-6){({\lambda}_{1})}^{2}+((n-4)(2n+3k)+4k+8)({\lambda}_{1})\hfill \\ & \hspace{1em}-2k({n}^{2}-5n+6).\hfill \end{array}$$

Let ${\lambda}_{1}=2\lambda $ and

$$\begin{array}{cc}\hfill F({\lambda}_{1})=& {({\lambda}_{1})}^{3}-(3n+k-6){({\lambda}_{1})}^{2}+((n-4)(2n+3k)+4k+8)({\lambda}_{1})\hfill \\ & -2k({n}^{2}-5n+6)=0.\hfill \end{array}$$

Then we get:
□

$$\begin{array}{ccc}\hfill {\lambda}_{1}& =& \frac{2n+k-4+\sqrt{{(2n-k-4)}^{2}+8k}}{2}.\hfill \end{array}$$

The above result is the same as [6].

## Author Contributions

All authors have contributed equally to this work. Investigation and Methodology: C.W.; Methodology and Correction: S.W.

## Funding

This work was partially supported by the National Natural Science Foundation of China under Grants 11771172 and 11571134.

## Conflicts of Interest

The authors declare no conflict of interest.

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