The A_α-Spectral Radii of Graphs with Given Connectivity

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.

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

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

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].

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

$$\begin{array}{c}\hfill \rho (G)x_{v_i}=\alpha d_{v_i}{x}_{v_i}+(1-\alpha )\sum _{v_i{v}_j\in E(G)}{x}_{v_j}.\end{array}$$

(1)

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 $G^{″}$ 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 $G^{″}=K(n-1,k)$, and $G^{\prime }$ is a proper subgraph of $G^{″}$. By Lemma 2, we have $\rho (G^{″})>\rho (G^{\prime })$. Thus, $\rho (G^{″})>\rho (G)$, 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 $G^{″}$ 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 $G^{″}$. 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

$$\begin{array}{c}\hfill (\rho -\alpha k){\beta }_1=k(1-\alpha ){\beta }_2\end{array}$$

$$\begin{array}{c}\hfill (\rho -\alpha (n-1)){\beta }_2=(1-\alpha )({\beta }_1+(k-1){\beta }_2+t{\beta }_3),\mathrm{and}\end{array}$$

$$\begin{array}{c}\hfill (\rho -\alpha (n-2)){\beta }_3=(1-\alpha )(k{\beta }_2+(t-1){\beta }_3).\end{array}$$

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 $2A_{\frac{1}{2}}=D+Q$. 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].