On the Generalized Adjacency Spread of a Graph

Abstract

For a simple finite graph G, the generalized adjacency matrix is defined as ${\mathcal{A}}_{\alpha }\left(G\right)=\alpha \mathcal{D}\left(G\right)+(1-\alpha )A\left(G\right),\alpha \in [0,1]$, where $A\left(G\right)$ and $\mathcal{D}\left(G\right)$ are respectively the adjacency matrix and diagonal matrix of the vertex degrees. The $A_{\alpha }$-spread of a graph G is defined as the difference between the largest eigenvalue and the smallest eigenvalue of the ${\mathcal{A}}_{\alpha }\left(G\right)$. In this paper, we answer the question posed in (Lin, Z.; Miao, L.; Guo, S. Bounds on the $A_{\alpha }$-spread of a graph. Electron. J. Linear Algebra 2020, 36, 214–227). Furthermore, we show that the path graph, $P_n$, has the smallest $\mathcal{S}\left({\mathcal{A}}_{\alpha }\right)$ among all trees of order n. We establish a relationship between $\mathcal{S}\left({\mathcal{A}}_{\alpha }\right)$ and $\mathcal{S}\left(A\right).$ We obtain several bounds for $\mathcal{S}\left({\mathcal{A}}_{\alpha }\right)$.

1. Introduction

Throughout this work, we consider finite, simple, and connected graphs with order $\left|V\right|=n$ and size $\left|E\right|=m$. In graph G, the neighborhood $N_G\left(v\right)$ or, briefly, the $N\left(v\right)$ of a vertex v is the set of all vertices adjacent to v. The degree $d\left(v_i\right)$ (or simply $d_i$) of a vertex $v_i$ is the number of edges of the incident on $v_i$. The smallest and the largest degrees of graph G are denoted by $\delta \left(G\right)$ and $\Delta \left(G\right)$, respectively. For two vertices $u,v\in V\left(G\right)$, the term $d_{uv}$ (or $d(u,v)$) shows the distance between them. The diameter of G is ${max}_{u,v\in V\left(G\right)}{d}_{uv}$. Moreover, $m_i=\frac{1}{d_i}{\sum }_{j\sim i}{d}_j$.

In the current work, we denote the adjacency matrix of G by $A\left(G\right)$ and $\mathcal{D}\left(G\right)$ indicates the diagonal matrix of degrees. The positive semi-definite matrices $\mathcal{L}\left(G\right)=\mathcal{D}\left(G\right)-A\left(G\right)$ and $\mathcal{Q}\left(G\right)=\mathcal{D}\left(G\right)+A\left(G\right)$ are, respectively, the Laplacian matrix and the signless Laplacian matrix of G. The eigenvalues of $A\left(G\right)$, $\mathcal{L}\left(G\right)$, and $\mathcal{Q}\left(G\right)$ are, respectively, the eigenvalues, Laplacian eigenvalues, and the signless Laplacian eigenvalues of G.

In [1], Nikiforov proposed studying the generalized adjacency matrix ${\mathcal{A}}_{\alpha }\left(G\right)$ (or, briefly, ${\mathcal{A}}_{\alpha }$) of G, defined as

$${\mathcal{A}}_{\alpha }={\mathcal{A}}_{\alpha }\left(G\right)=\alpha \mathcal{D}\left(G\right)+(1-\alpha )A\left(G\right),0\le \alpha \le 1.$$

Some properties of this matrix were studied in [2,3,4,5,6,7,8]. Clearly, ${\mathcal{A}}_0\left(G\right)$ is the adjacency matrix and $2{\mathcal{A}}_{\frac{1}{2}}\left(G\right)$ is the signless Laplacian matrix. Moreover, ${\mathcal{A}}_{\alpha }\left(G\right)-{\mathcal{A}}_{\beta }\left(G\right)=(\alpha -\beta )\mathcal{L}\left(G\right).$ For $\alpha \ne 1$, we note that

$${\mathcal{A}}_{\alpha }\left(G\right)=(\alpha -1)\left(\frac{\alpha }{\alpha -1}\mathcal{D}\left(G\right)-A\left(G\right)\right).$$

Let ${\lambda }_i$ be the i-th largest eigenvalue of a symmetric matrix $\mathcal{M}$, $\overline{\lambda }=\left({\sum }_{i=1}^n{\lambda }_i\left(\mathcal{M}\right)\right)/n$ and c denote the center of the smallest disc D, which contains all eigenvalues of $\mathcal{M}$. The spread of $\mathcal{M}$, Reference [9], is $\mathcal{S}\left(\mathcal{M}\right)={\lambda }_{max}\left(\mathcal{M}\right)-{\lambda }_{min}\left(\mathcal{M}\right)$. Several properties of this quantity for the adjacency matrix and some other graph matrices were studied in [10,11,12,13,14,15,16,17,18,19,20].

For the ${\mathcal{A}}_{\alpha }$-spread, the natural question arises, as follows. Which graphs have the minimum (or maximum) ${\mathcal{A}}_{\alpha }$-spreads among all graphs with n vertices? In the literature, there are some results related to this question. For example, Lin et al. in [4] proved that among all trees, if $\frac{5+\sqrt{5}}{10}\le \alpha \le 1$, then $\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(P_n\right)\right)\le \mathcal{S}\left({\mathcal{A}}_{\alpha }\left(T\right)\right)$ and equality hold only for the path graph $P_n$. In [21], the authors proved that for a graph with a leaf, if $\alpha \in (\frac{23}{32},1)$ then the ${\mathcal{A}}_{\alpha }$-spread of graph $P_n$ is less than or equal to the ${\mathcal{A}}_{\alpha }$-spread of an arbitrary graph G and equality holds only for the path graph $P_n$.

Lin, Miao, and Guo [4] proposed the following open problem.

Problem 1.

For a real number $\alpha \in R$, characterize all graphs that satisfy the following equality.

$$\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)={\left(4{(1-\alpha )}^{2}m+2{\alpha }^2\sum _{i=1}^n{d}_i^2-\frac{8}{n}{\alpha }^2{m}^2\right)}^{\frac{1}{2}}.$$

One of our aims in this paper is to solve Problem 1, which will be given as Theorem 1.

The rest of the paper is organized as follows. In Section 2, we provide the solution to Problem 1. Further, we show that the path graph, $P_n$, has the smallest $\mathcal{S}\left({\mathcal{A}}_{\alpha }\right)$ among all trees of order n. In Section 3, we establish a relationship between $\mathcal{S}\left({\mathcal{A}}_{\alpha }\right)$ and $\mathcal{S}\left(A\right)$. We obtain several bounds for $\mathcal{S}\left({\mathcal{A}}_{\alpha }\right)$.

2. Solution of Problem 1

The Frobenius norm of a matrix $\mathcal{M}={\left(m_{ij}\right)}_{m\times n}$ (see [22]) is given by

$${\parallel \mathcal{M}\parallel }_F={\left(\sum _{i=1}^m\sum _{j=1}^n{\left|m_{ij}\right|}^2\right)}^{\frac{1}{2}}.$$

Lemma 1

([23]). If ${\mathcal{M}}_{n\times n}$ is a normal matrix of order n with eigenvalues $m_1\ge m_2\ge \cdots \ge m_n$, then

$$\mathcal{S}{\left(\mathcal{M}\right)}^2\le 2\left({\parallel \mathcal{M}\parallel }_F^2-\frac{{\left(tr\mathcal{M}\right)}^2}{n}\right),$$

with equality if and only if $m_2=m_3=\cdots =m_{n-1}=\frac{m_1+m_n}{2}$.

Lemma 2

([22,24]). Let $\mathcal{X}$ and $\mathcal{Y}$ be two Hermitian matrices with, respectively, eigenvalues ${\mu }_1\ge {\mu }_2\ge \cdots \ge {\mu }_n$ and ${\nu }_1\ge {\nu }_2\ge \cdots \ge {\nu }_n$ such that $\mathcal{Z}=\mathcal{X}+\mathcal{Y}$, and ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_n$ be all eigenvalues of $\mathcal{Z}$. Then

$${\lambda }_k\le {\mu }_j+{\nu }_{k-j+1},k\ge j,{\lambda }_k\ge {\mu }_j+{\nu }_{k-j+n},j\ge k.$$

Moreover, the equality holds if and only if there exists a unit vector, which is an eigenvector to each of the three eigenvalues involved.

Now, we give the complete solution to Problem 1.

Theorem 1.

If $n\ge 6$ and $\alpha \in [0,1)$, then

$$\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\le {\left(4{(1-\alpha )}^{2}m+2{\alpha }^2\sum _{i=1}^n{d}_i^2-\frac{8}{n}{\alpha }^2{m}^2\right)}^{\frac{1}{2}},$$

(1)

and the equality holds if and only if n is even and $G\cong K_{\frac{n}{2},\frac{n}{2}}$.

Proof. 

Let G be a graph and ${\mu }_1\ge {\mu }_2\ge \cdots \ge {\mu }_n$ and ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_n$, be, respectively, the eigenvalues and generalized eigenvalues of G. Since $\parallel {\mathcal{A}}_{\alpha }{\left(G\right)\parallel }_F^2=2{(1-\alpha )}^{2}m+{\alpha }^2{\displaystyle \sum _{i=1}^n}{d}_i^2$ and $tr{\mathcal{A}}_{\alpha }\left(G\right)=2\alpha m$, putting $\mathcal{M}={\mathcal{A}}_{\alpha }\left(G\right)$, the result follows from Lemma 1 and, thus, we have

$${\lambda }_2={\lambda }_3=\cdots ={\lambda }_{n-1}=\frac{{\lambda }_1+{\lambda }_n}{2}.$$

Thus, $tr{\mathcal{A}}_{\alpha }\left(G\right)={\sum }_{i=1}^n{\lambda }_i=n{\lambda }_2=2\alpha m$. This implies that ${\lambda }_2={\lambda }_3=\cdots ={\lambda }_{n-1}=\frac{2\alpha m}{n}$. From the definition of ${\mathcal{A}}_{\alpha }\left(G\right)$ we have

$$A\left(G\right)=\frac{1}{1-\alpha }({\mathcal{A}}_{\alpha }\left(G\right)-\alpha \mathcal{D}\left(G\right)),0\le \alpha <1.$$

By taking $k=j=2$ in the left inequality in Lemma 2 we have

$${\mu }_2\le \frac{1}{1-\alpha }({\lambda }_2-\alpha \Delta )=\frac{\alpha }{1-\alpha }\left(\frac{2m}{n}-\Delta \right)\le 0.$$

(2)

Moreover, taking $k=j=n-1$ in the right inequality in Lemma 2 yields

$${\mu }_{n-1}\ge \frac{1}{1-\alpha }({\lambda }_{n-1}-\alpha \delta )=\frac{\alpha }{1-\alpha }\left(\frac{2m}{n}-\delta \right)\ge 0.$$

(3)

From inequalities (2) and (3), we have ${\mu }_2={\mu }_3=\cdots ={\mu }_{n-1}=0$, $-{\mu }_n={\mu }_1$ and $\delta =\Delta =\frac{2m}{n}$. So G is a regular bipartite graph, i.e., $G\cong K_{\frac{n}{2},\frac{n}{2}}$.

Conversely, if $G=K_{\frac{n}{2},\frac{n}{2}}$, then the adjacency spectrum of G consists of eigenvalue $\frac{n}{2}$ with multiplicity 1, eigenvalue 0 with multiplicity $n-2$, and eigenvalues $\frac{-n}{2}$ with multiplicity 1. On the other hand, for a k-regular graph G we have ${\lambda }_i=\alpha k+(1-\alpha ){\mu }_i$, for all $i=1,2,\cdots ,n$. So, equality occurs in (1). □

We require the following observation.

Lemma 3

([25]). For the Hermitian matrix $\mathcal{M}=\left(m_{ij}\right)$, it holds

$$\mathcal{S}\left(\mathcal{M}\right)\ge \underset{i,j}{max}{\left({(m_{ii}-m_{jj})}^2+2\sum _{s\ne j}|m_{js}{|}^2+2\sum _{s\ne i}{|m_{is}|}^2\right)}^{\frac{1}{2}}.$$

Theorem 2.

Consider an arbitrary graph G.

$\left(i\right)$

If $\Delta -\delta \ge 2,$ then for $\alpha \ge \frac{1}{2},$

$$\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge {\left({\alpha }^2{(\Delta -\delta )}^2+2{(1-\alpha )}^2(\Delta +\delta )\right)}^{\frac{1}{2}}.$$

$\left(ii\right)$

If $\Delta -\delta \le 1,$ then for $\alpha \le 2-\sqrt{2},$

$$\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge 2(1-\alpha )\sqrt{\Delta }.$$

In particular, for $\alpha =\frac{1}{2},$ the equality holds for $G\cong K_2.$

Proof. 

Let ${\mathcal{A}}_{\alpha }\left(G\right)=\left(a_{ij}\right),i,j\in \{1,\cdots ,n\}$. Lemma 3 states that

$$\begin{array}{ccc}\hfill \mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)& \ge & \underset{i,j}{max}{\left({(a_{ii}-a_{jj})}^2+2\sum _{s\ne j}|a_{js}{|}^2+2\sum _{s\ne i}{|a_{is}|}^2\right)}^{\frac{1}{2}}\hfill \\ & =& \underset{i,j}{max}{\left({\alpha }^2{(d_j-d_i)}^2+2{(1-\alpha )}^2(d_j+d_i)\right)}^{\frac{1}{2}}.\hfill \end{array}$$

Suppose $d_j\ge d_i$ and $d_j-d_i=k\in \{0,1,\dots ,\Delta -\delta \},$ we have

$$\begin{array}{ccc}\hfill \mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)& \ge & \underset{k}{max}\underset{d_j-d_i=k}{max}{\left({\alpha }^2{(d_j-d_i)}^2+2{(1-\alpha )}^2(d_j+d_i)\right)}^{\frac{1}{2}}\hfill \\ & =& \underset{k}{max}\underset{d_j-d_i=k}{max}{\left({\alpha }^2{k}^2+2{(1-\alpha )}^2(2d_i+k)\right)}^{\frac{1}{2}}\hfill \\ & =& \underset{k}{max}{\left({\alpha }^2{k}^2+2{(1-\alpha )}^2(2(\Delta -k)+k)\right)}^{\frac{1}{2}}\hfill \\ & =& \underset{k}{max}{\left({\alpha }^2{k}^2+2{(1-\alpha )}^2(2\Delta -k)\right)}^{\frac{1}{2}},\hfill \end{array}$$

as ${\alpha }^2{k}^2+2{(1-\alpha )}^2(2d_i+k)$ is increasing in $d_i$. Thus,

$$\begin{array}{ccc}\hfill \mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)& \ge & max\left\{2(1-\alpha )\sqrt{\Delta },{\left({\alpha }^2{(\Delta -\delta )}^2+2{(1-\alpha )}^2(\Delta +\delta )\right)}^{\frac{1}{2}}\right\}.\hfill \end{array}$$

Now, if $\Delta -\delta \ge 2$, then for $\alpha \ge \frac{1}{2}$, we have ${\alpha }^2{(\Delta -\delta )}^2+2{(1-\alpha )}^2(\Delta +\delta )\ge 4{(1-\alpha )}^2\Delta$, i.e., ${\alpha }^2(\Delta -\delta )\ge 2{(1-\alpha )}^2$, i.e., ${\alpha }^2\ge {(1-\alpha )}^2.$ If $\Delta -\delta \le 1$, then for $\alpha \le 2-\sqrt{2}$, we have $4{(1-\alpha )}^2\Delta \ge {\alpha }^2{(\Delta -\delta )}^2+2{(1-\alpha )}^2(\Delta +\delta )$, that is, $2{(1-\alpha )}^2\ge {\alpha }^2(\Delta -\delta )$. This implies that $2{(1-\alpha )}^2\ge {\alpha }^2$ and the result follows.

For $G=K_2$ and $\alpha =\frac{1}{2}$, it is clear that $\mathcal{S}\left(A_{\frac{1}{2}}\left(K_2\right)\right)={\lambda }_1\left(A_{\frac{1}{2}}\left(K_2\right)\right)=1$. Moreover,

$${\left({\alpha }^2{(\Delta -\delta )}^2+2{(1-\alpha )}^2(\Delta +\delta )\right)}^{\frac{1}{2}}=2(1-\alpha )\sqrt{\Delta }=1,$$

and this shows that equality occurs. This completes the proof. □

Now, we show that the path graph, $P_n$, has the smallest value of $\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)$ among all trees of order n.

Theorem 3.

Let T be a tree of order n having the maximum degree Δ. If $\alpha \ge 0.5$ and $\Delta \ge 5$, then $\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(T\right)\right)>\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(P_n\right)\right).$

Proof. 

Suppose that the tree T satisfies in part (i) of Theorem 2. Then for $\alpha \ge 0.5$ and $\Delta \ge 5$, we have

$$\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(T\right)\right)\ge {\left({\alpha }^2{(\Delta -1)}^2+2{(1-\alpha )}^2(\Delta +1)\right)}^{\frac{1}{2}}\ge 4-3\alpha ,$$

(4)

or equivalently

$$({\Delta }^2-6){\alpha }^2-4(\Delta -5)\alpha +2\Delta -14\ge 0.$$

(5)

Consider the polynomial $f\left(\alpha \right)=({\Delta }^2-6){\alpha }^2-4(\Delta -5)\alpha +2\Delta -14$, for $\alpha \in [0.5,1]$. It is easy to see that discriminant ▽ of polynomial $f\left(\alpha \right)$ is negative for all $\Delta \ge 8$. This in turn gives that inequality (5) holds for all $\Delta \ge 8$. For $\Delta =5,6,7$, the inequality (5) holds for all $\alpha \ge 0.5$. Since $\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(P_n\right)\right)\le \alpha +4(1-\alpha )cos\left(\frac{\pi }{n+1}\right)\le 4-3\alpha$, it follows that $\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(T\right)\right)>\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(P_n\right)\right)$ holds for all $\Delta \ge 5$ and $\alpha \ge 0.5$. Therefore, the result follows. □

For $\alpha \ge 0.5$ and $\Delta -\delta \ge 2$, we have

$${\left({\alpha }^2{(\Delta -\delta )}^2+2{(1-\alpha )}^2(\Delta +\delta )\right)}^{\frac{1}{2}}\ge 4-3\alpha$$

which gives

$${\alpha }^2[{(\Delta -\delta )}^2-9]+2{(1-\alpha )}^2(\Delta +\delta )+24\alpha -16\ge 0.$$

(6)

If $\Delta -\delta \ge 3$, then it is easy to see that inequality (6) holds for all $\alpha \ge \frac{2}{3}$. If $\frac{1}{2}\le \alpha <\frac{2}{3}$ and $\Delta -\delta \ge 3$, it can be verified that inequality (6) holds for all $\Delta +\delta \ge \frac{8-2\alpha }{{(1-\alpha )}^2}$. Thus, we have the following observation.

Theorem 4.

Let G be a graph of order n having a maximum degree Δ and minimum degree δ.

$\left(i\right)$

If $\alpha \ge \frac{2}{3}$ and $\Delta -\delta \ge 3$, then $\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge \mathcal{S}\left({\mathcal{A}}_{\alpha }\left(P_n\right)\right)$.

$\left(ii\right)$

If $\frac{1}{2}\le \alpha <\frac{2}{3}$, $\Delta -\delta \ge 3$ and $\Delta +\delta \ge \frac{8-2\alpha }{{(1-\alpha )}^2}$, then $\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge \mathcal{S}\left({\mathcal{A}}_{\alpha }\left(P_n\right)\right)$.

3. Bounds for $\mathcal{S}\left({\mathcal{A}}_{\alpha }\right)$ of a Graph G

As can be seen in [1], for $\frac{1}{2}<\alpha \le 1$, we can obtain that $\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)$ is less than or equal with the largest eigenvalue of ${\mathcal{A}}_{\alpha }\left(G\right)$, with equality if and only if ${\lambda }_n\left({\mathcal{A}}_{\alpha }\right)=0$. The following theorem gives a relationship between $\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)$ and $\mathcal{S}\left(A\right(G\left)\right)$, showing that $\mathcal{S}\left(A\right(G\left)\right)$ lies in the closed interval as

$$\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\in [(1-\alpha )\mathcal{S}\left(A\left(G\right)\right)-\alpha (\Delta -\delta ),\alpha (\Delta -\delta )+(1-\alpha )\mathcal{S}\left(A\left(G\right)\right)].$$

Theorem 5.

For a graph G, we have

$$(1-\alpha )\mathcal{S}\left(A\left(G\right)\right)-\alpha (\Delta -\delta )\le \mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\le \alpha (\Delta -\delta )+(1-\alpha )\mathcal{S}\left(A\left(G\right)\right).$$

(7)

The equality holds on both sides only for regular graphs.

Proof. 

By taking $k=j=1$ in the left inequality and $k=j=n$ in the right inequality in Lemma 2 we have that ${\lambda }_1\le \alpha \Delta +(1-\alpha ){\mu }_1$ and ${\lambda }_n\ge \alpha \delta +(1-\alpha ){\mu }_n$. So

$$\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\le \alpha (\Delta -\delta )+(1-\alpha )\mathcal{S}\left(A\left(G\right)\right).$$

Moreover, taking $k=n$ and $j=1$ in the left inequality and $k=1$ and $j=n$ in the right inequality yields ${\lambda }_n\le \alpha \Delta +(1-\alpha ){\mu }_n$ and ${\lambda }_1\ge \alpha \delta +(1-\alpha ){\mu }_1$. So

$$\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge (1-\alpha )\mathcal{S}\left(A\left(G\right)\right)-\alpha (\Delta -\delta ).$$

Suppose that either of the equalities holds. Then Lemma 2 implies that there exists a unit vector, which is an eigenvector to each of the three eigenvalues ${\lambda }_1$, $\Delta$, and ${\mu }_1$; thus, G is a ${\lambda }_1$-regular graph. □

Taking $G=P_n$, $\Delta =2$, $\delta =1$ in Theorem 5, for all $\alpha \in [0,1]$ we obtain

$$\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(P_n\right)\right)\le \alpha +4(1-\alpha )cos\frac{\pi }{n+1}.$$

(8)

Theorem 6.

For $\alpha \in [0,1)$, we have

$$\begin{array}{c}\hfill (n-1)\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge n{\lambda }_1\left({\mathcal{A}}_{\alpha }\left(G\right)\right)-2\alpha m,\end{array}$$

(9)

with equality if and only if $G\cong K_n$.

Proof. 

Since ${\sum }_{i=1}^n{\lambda }_i\left({\mathcal{A}}_{\alpha }\left(G\right)\right)=2\alpha m$, we conclude $2\alpha m\ge {\lambda }_1\left({\mathcal{A}}_{\alpha }\right)+(n-1){\lambda }_n\left({\mathcal{A}}_{\alpha }\right)$ and thus ${\lambda }_n\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\le \frac{(2\alpha m-{\lambda }_1\left({\mathcal{A}}_{\alpha }\left(G\right)\right))}{n-1}.$ In other words, the spread of ${\mathcal{A}}_{\alpha }\left(G\right)$ is bounded below by

$$\begin{array}{c}\hfill {\lambda }_1-\frac{2\alpha m-{\lambda }_1}{n-1}=\frac{n}{n-1}{\lambda }_1-\frac{2\alpha m}{n-1}.\end{array}$$

The equality in (9) holds if and only if $2\alpha m={\lambda }_1\left({\mathcal{A}}_{\alpha }\right)+(n-1){\lambda }_n\left({\mathcal{A}}_{\alpha }\right)$, and this holds if and only if ${\lambda }_2\left({\mathcal{A}}_{\alpha }\left(G\right)\right)=\cdots ={\lambda }_n\left({\mathcal{A}}_{\alpha }\left(G\right)\right)$, i.e., if and only if $G\cong K_n$. □

Lemma 4

([26]). Let $A=\left(a_{ij}\right)$ be an $n\times n$ Hermitian matrix, with $n\ge 2$ with eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_n$. Then

$${\lambda }_1-{\lambda }_n\ge \frac{|{\sum }_{i\ne j}{a}_{ij}|}{n-1}.$$

Theorem 7.

Let G be a graph of order $n\ge 2,$ then

$$\begin{array}{c}\hfill \mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge \frac{2m(1-\alpha )}{n-1}.\end{array}$$

Proof. 

Using Lemma 4, since ${\mathcal{A}}_{\alpha }\left(G\right)$ is a Hermitian matrix of order $n\ge 2$ with eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_n$, then

$$\begin{array}{c}\hfill \mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)={\lambda }_1\left({\mathcal{A}}_{\alpha }\left(G\right)\right)-{\lambda }_n\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge \frac{|{\sum }_{i\ne j}{\left({\mathcal{A}}_{\alpha }\left(G\right)\right)}_{ij}|}{n-1}=\frac{2m(1-\alpha )}{n-1}.\end{array}$$

The proof is completed. □

Lemma 5

([27]). Let $M=\left(m_{ij}\right)$ be an $n\times n$ non-negative matrix. Let S be the sum of entries and d (resp., f) be any number that is larger than or equal to the largest diagonal element (resp., the largest off-diagonal element) of M. Then

$${\lambda }_1\left(M\right)\le \frac{d-f+\sqrt{{(d-f)}^2+4Sf}}{2}.$$

Theorem 8.

Let G be a graph of order n with m edges and the maximum degree Δ.

$\left(i\right)$

If $\alpha \ge \frac{1}{\Delta +1}$ then

$$\begin{array}{c}\hfill \mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\le \frac{(n+1)\alpha \Delta +\alpha -1+\sqrt{{(\alpha (\Delta +1)-1)}^2-8m(1-\alpha )}}{2}.\end{array}$$

$\left(ii\right)$

If $\alpha <\frac{1}{\Delta +1}$ then

$$\begin{array}{c}\hfill \mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\le \frac{(n-1)(1-\alpha )+\alpha \Delta +\sqrt{{(\alpha (\Delta +1)-1)}^2-8m(1-\alpha )}}{2}.\end{array}$$

Proof. 

For the graph G with n vertices, m edges and the maximum degree $\Delta$, the largest diagonal element of ${\mathcal{A}}_{\alpha }\left(G\right)$ is $\alpha \Delta$ and its largest off-diagonal element is $1-\alpha$. So from Lemma 5, we have

$${\lambda }_1\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\le \frac{\alpha (\Delta +1)-1+\sqrt{{(\alpha (\Delta +1)-1)}^2-8m(1-\alpha )}}{2}.$$

Moreover, from ([28], Theorem 1) we have the following cases.

$\left(i\right)$

If $\alpha \ge \frac{1}{\Delta +1}$, we have that the largest element of ${\mathcal{A}}_{\alpha }\left(G\right)$ is $\alpha \Delta$; therefore, we have that

$${\lambda }_n\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge \frac{-n\alpha \Delta }{2}.$$

$\left(ii\right)$

If $\alpha <\frac{1}{\Delta +1}$, the largest element of ${\mathcal{A}}_{\alpha }\left(G\right)$ is $(1-\alpha )$. So

$${\lambda }_n\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge \frac{-n(1-\alpha )}{2}.$$

Combining the above arguments, we have the proof. □

Theorem 9.

Let G be a graph of order n with m edges, then

$$\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge \sqrt{\frac{4mn{(1-\alpha )}^2+2n{\alpha }^2{\sum }_1^n{d}_i^2-8{\alpha }^2{m}^2}{n(n-1)}}.$$

If G is a complete bipartite graph $K_{\frac{n}{2},\frac{n}{2}}$, then the equality holds.

Proof. 

Let ${\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n$ be the eigenvalues of ${\mathcal{A}}_{\alpha }\left(G\right)$. Taking $z_i={\lambda }_i$ and $z_j={\lambda }_j$ in ([29], Lemma 2.2), it follows that

$$\frac{1}{n}\sum _{1\le i<j\le n}^{}{|{\lambda }_i-{\lambda }_j|}^2=\sum _{i=1}^n{|{\lambda }_i|}^2-\frac{|tr{\mathcal{A}}_{\alpha }{\left(G\right)|}^2}{n}.$$

On the other hand $|{\lambda }_i-{\lambda }_j{|}^2\le {\mathcal{S}}^2\left({\mathcal{A}}_{\alpha }\left(G\right)\right)$, for $1\le i<j\le n$. So we have

$$\sum _{1\le i<j\le n}^{}{|{\lambda }_i-{\lambda }_j|}^2\le \frac{n(n-1)}{2}{\mathcal{S}}^2\left({\mathcal{A}}_{\alpha }\left(G\right)\right).$$

Therefore

$${\mathcal{S}}^2\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge \frac{2}{n-1}\left(\sum _{i=1}^n{|{\lambda }_i|}^2-\frac{|tr{\mathcal{A}}_{\alpha }{\left(G\right)|}^2}{n}\right).$$

Using ${\sum }_{i=1}^n{|{\lambda }_i|}^2=2{(1-\alpha )}^{2}m+{\alpha }^2{\displaystyle \sum _{i=1}^n}{d}_i^2$ and $tr{\mathcal{A}}_{\alpha }\left(G\right)=2\alpha m$, The desired result is obtained immediately. □

Theorem 10.

Let G be a graph with n vertices, and m edges, then

$$\frac{R}{\sqrt{n}}\le \mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\le \frac{R}{\sqrt{2}},$$

where $R=\sqrt{16m^2{\alpha }^2\frac{(n-1)}{n}-8{\sum }_{i<j}{\lambda }_i{\lambda }_j}$.

Proof. 

In [30], R. A. Smith and L. Mirsky show that for arbitrary matrix A, ${\mathcal{S}}^2\left(A\right)\le 2{\sum }_{i=1}^n{\lambda }_i^2$. Since ${\mathcal{A}}_{\alpha }$ is invariant under translation, then

$$\begin{array}{c}\hfill {\mathcal{S}}^2\left({\mathcal{A}}_{\alpha }\right)\le 2\sum _{i=1}^n{({\lambda }_i-\overline{\lambda })}^2,\end{array}$$

(10)

where $\overline{\lambda }=\left({\sum }_{i=1}^n{\lambda }_i\right)/n$. On the other hand, ${({\lambda }_i-c)}^2\le {(\mathcal{S}\left({\mathcal{A}}_{\alpha }\right)/2)}^2$, so

$$\begin{array}{c}\hfill \frac{4}{n}\sum _{i=1}^n{({\lambda }_i-c)}^2\le {\left(\mathcal{S}\left({\mathcal{A}}_{\alpha }\right)\right)}^2.\end{array}$$

(11)

Moreover,

$$\begin{array}{ccc}\hfill \sum _{i=1}^n{({\lambda }_i-c)}^2& =& \sum _{i=1}^n({\lambda }_i^2-2c{\lambda }_i+c^2)\hfill \\ & =& \sum _{i=1}^n{\lambda }_i^2-\frac{{\left(tr{\mathcal{A}}_{\alpha }\left(G\right)\right)}^2}{n}+n{\left(c-\frac{tr{\mathcal{A}}_{\alpha }\left(G\right)}{n}\right)}^2.\hfill \end{array}$$

Since $\overline{\lambda }=\left({\sum }_{i=1}^n{\lambda }_i\right)/n=tr{\mathcal{A}}_{\alpha }\left(G\right)/n$, substituting c with $\overline{\lambda }$ results that ${\sum }_{i=1}^n{({\lambda }_i-\overline{\lambda })}^2={\sum }_{i=1}^n{\lambda }_i^2-{\left(tr{\mathcal{A}}_{\alpha }\left(G\right)\right)}^2/n,$ which results that ${\sum }_{i=1}^n{({\lambda }_i-\overline{\lambda })}^2\le {\sum }_{i=1}^n{({\lambda }_i-c)}^2$. Therefore, from the inequality (11), we have

$$\begin{array}{c}\hfill \frac{4}{n}\sum _{i=1}^n{({\lambda }_i-\overline{\lambda })}^2\le {\mathcal{S}}^2\left({\mathcal{A}}_{\alpha }\right).\end{array}$$

(12)

Finally, we have

$$\begin{array}{ccc}\hfill \sum _{i=1}^n{({\lambda }_i-\overline{\lambda })}^2& =& \sum _{i=1}^n({\lambda }_i^2-2\overline{\lambda }{\lambda }_i+{\overline{\lambda }}^2)\hfill \\ & =& \sum _{i=1}^n{\lambda }_i^2-\frac{{(tr{\mathcal{A}}_{\alpha }\left(G\right))}^2}{n}\hfill \\ & =& {(\sum _{i=1}^n{\lambda }_i)}^2-2\sum _{i<j}{\lambda }_i{\lambda }_j-\frac{{\left(tr{\mathcal{A}}_{\alpha }\left(G\right)\right)}^2}{n}\hfill \\ & =& (\frac{n-1}{n}){\left(tr{\mathcal{A}}_{\alpha }\left(G\right)\right)}^2-2\sum _{i<j}{\lambda }_i{\lambda }_j\hfill \\ & =& 4m^2{\alpha }^2(\frac{n-1}{n})-2\sum _{i<j}{\lambda }_i{\lambda }_j.\hfill \end{array}$$

(13)

By using (10) and (13) we have that

$$\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\le \sqrt{8m^2{\alpha }^2\frac{(n-1)}{n}-4{\sum }_{i<j}{\lambda }_i{\lambda }_j},$$

and (12) and (13) yield that

$$\mathcal{S}\left({\mathcal{A}}_{\alpha }\left(G\right)\right)\ge \sqrt{\frac{16m^2{\alpha }^2\frac{(n-1)}{n}-8{\sum }_{i<j}{\lambda }_i{\lambda }_j}{n}}.$$

The proof is complete. □

4. Summary and Conclusions

A question asked by Lin et al. was solved. Furthermore, it is proved that $P_n$ has the smallest spread among all trees; some new bounds for $\mathcal{S}\left({\mathcal{A}}_{\alpha }\right)$ were investigated. These results can be extended for the generalized distance matrix of a graph, which will be looked at in future work. As mentioned in the introduction, ${\mathcal{A}}_0\left(G\right)$ is the adjacency matrix and $2{\mathcal{A}}_{\frac{1}{2}}\left(G\right)$ is the signless Laplacian matrix. So we have $\mathcal{S}\left(A\right)=\mathcal{S}\left({\mathcal{A}}_0\right)$ and $\mathcal{S}\left(\mathcal{Q}\right)=2\mathcal{S}\left({\mathcal{A}}_{\frac{1}{2}}\right)$. Therefore, from the obtained bounds for $\mathcal{S}\left({\mathcal{A}}_{\alpha }\right)$, the bounds for $\mathcal{S}\left(A\right)$ and $\mathcal{S}\left(\mathcal{Q}\right)$ can be reached. This means that for two special cases $\alpha =0$ and $\alpha =\frac{1}{2}$, we obtain two well-known matrices, namely the adjacency matrix and signless Laplacian matrix. Hence, for example, Theorem 6 gives the following new bounds

$$\mathcal{S}\left(A\right)\ge \frac{n{\mu }_1}{n-1},\mathrm{and}\mathcal{S}\left(\mathcal{Q}\right)\ge \frac{n{\lambda }_1\left(\mathcal{Q}\right)-m}{n-1}.$$

Moreover, from Theorem 7, we have

$$\mathcal{S}\left(A\right)\ge \frac{2m}{n-1},\mathrm{and}\mathcal{S}\left(\mathcal{Q}\right)\ge \frac{2m}{n-1},$$

and Theorem 9 yields that

$$\mathcal{S}\left(A\right)\ge 2\sqrt{\frac{m}{n-1}},\mathrm{and}\mathcal{S}\left(\mathcal{Q}\right)\ge 2\sqrt{\frac{4mn+2n{\sum }_1^n{d}_i^2-8m^2}{n(n-1)}}.$$