On the distance α-spectral radius of a connected graph

Abstract

For a connected graph G and $\alpha \in [0,1)$, the distance α-spectral radius of G is the spectral radius of the matrix $D_{\alpha }(G)$ defined as $D_{\alpha }(G)=\alpha T(G)+(1-\alpha )D(G)$, where $T(G)$ is a diagonal matrix of vertex transmissions of G and $D(G)$ is the distance matrix of G. We give bounds for the distance α-spectral radius, especially for graphs that are not transmission regular, propose local graft transformations that decrease or increase the distance α-spectral radius, and determine the graphs that minimize and maximize the distance α-spectral radius among several families of graphs.

1 Introduction

We consider simple and undirected graphs. Let G be a connected graph of order n with vertex set $V(G)$ and edge set $E(G)$. For $u,v\in V(G)$, the distance between u and v in G, denoted by $d_{G}(u,v)$ or simply $d_{uv}$ if the graph G is clear from the context, is the length of a shortest path from u to v in G. The distance matrix of G is the $n\times n$ matrix $D(G)=(d_{G}(u,v))_{u,v\in V(G)}$. For $u\in V(G)$, the transmission of u in G, denoted by $T_{G}(u)$, is defined as the sum of distances from u to all other vertices of G, i.e., $T_{G}(u)=\sum_{v\in V(G)}d_{G}(u,v)$. The transmission matrix $T(G)$ of G is the diagonal matrix of transmissions of G. Then $Q(G)=T(G)+D(G)$ is the distance signless Laplacian matrix of G, proposed recently in [1]. Arisen from a data communication problem, the spectrum of the distance matrix was studied by Graham and Pollack [12] in 1971, early related work may be found also in [10, 11], and now it has been studied extensively, see the recent survey [2] and the very recent papers [4, 5, 17, 18, 26]. The distance signless Laplacian spectrum has also received much attention, see, e.g., [1, 3, 4, 7, 15, 16, 29].

Throughout this paper we assume that $\alpha \in [0,1)$. Motivated by the work of Nikiforov [22], we consider the convex combinations $D_{\alpha }(G)$ of $T(G)$ and $D(G)$, defined as

$$ D_{\alpha }(G)=\alpha T(G)+(1-\alpha )D(G), $$

see [6]. Evidently, $D_{0}(G)=D(G)$ and $2D_{1/2}(G)=Q(G)$. We call the eigenvalues of $D_{\alpha }(G)$ the distance α-eigenvalues of G. As $D_{\alpha }(G)$ is a symmetric matrix, the distance α-eigenvalues of G are all real, which are denoted by $\mu ^{(1)}_{\alpha } (G), \ldots , \mu ^{(n)}_{\alpha }(G)$, arranged in nonincreasing order, where $n=|V(G)|$. The largest distance α-eigenvalue $\mu ^{(1)}_{\alpha } (G)$ of G is called the distance α-spectral radius of G, written as $\mu _{\alpha } (G)$. Obviously, $\mu ^{(1)}_{0} (G), \ldots , \mu ^{(n)}_{0}(G)$ are the distance eigenvalues of G, and $2\mu ^{(1)}_{1/2} (G), \ldots , 2\mu ^{(n)}_{1/2}(G)$ are the distance signless Laplacian eigenvalues of G. Particularly, $\mu _{0}(G)$ is just the distance spectral radius [2] and $2\mu _{1/2}(G)$ is just the distance signless Laplacian spectral radius of G [1].

In this paper, we give sharp bounds for the distance α-spectral radius, and particularly an upper bound for the distance α-spectral radius of connected graphs that are not transmission regular, and propose some types of graft transformations that decrease or increase the distance α-spectral radius. We also determine the unique graphs with minimum distance α-spectral radius among trees and unicyclic graphs, respectively, as well as the unique graphs (trees) with maximum and second maximum distance α-spectral radii, and the unique graph with maximum distance α-spectral radius among connected graphs with given clique number, and among odd-cycle unicyclic graphs, respectively.

2 Preliminaries

Let G be a connected graph with $V(G)=\{v_{1},\ldots ,v_{n}\}$. A column vector $x=(x_{v_{1}},\ldots , x_{v_{n}})^{\top }\in \mathbb{R}^{n}$ can be considered as a function defined on $V(G)$ which maps vertex $v_{i}$ to $x_{v_{i}}$, i.e., $x(v_{i})=x_{v_{i}}$ for $i=1,\ldots ,n$. Then

$$ x^{\top }D_{\alpha }(G)x=\alpha \sum _{u\in V(G)}T_{G}(u)x_{u}^{2}+2 \sum_{\{u,v\}\subseteq V(G)}(1-\alpha )d_{G}(u,v)x_{u}x_{v}, $$

or equivalently,

$$ x^{\top }D_{\alpha }(G)x=\sum_{\{u,v\}\subseteq V(G)}d_{G}(u,v) \bigl( \alpha \bigl(x_{u}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{u}x_{v} \bigr). $$

Since $D_{\alpha }(G)$ is a nonnegative irreducible matrix, by the Perron–Frobenius theorem, $\mu _{\alpha } (G)$ is simple and there is a unique positive unit eigenvector corresponding to $\mu _{\alpha } (G)$, which is called the distance α-Perron vector of G. If x is the distance α-Perron vector of G, then for each $u\in V(G)$,

$$ \mu _{\alpha }(G)x_{u}=\sum_{v\in V(G)}d_{G}(u,v) \bigl(\alpha x_{u}+(1- \alpha )x_{v}\bigr), $$

which is called the α-equation of G at u. For a unit column vector $x\in \mathbb{R}^{n}$ with at least one nonnegative entry, by Rayleigh’s principle, we have $\mu _{\alpha } (G)\ge x^{\top }D_{\alpha }(G)x$ with equality if and only if x is the distance α-Perron vector of G.

As in [27], we have the following result.

Lemma 2.1

Suppose thatGis a connected graph, ηis an automorphism ofG, andxis the distanceα-Perron vector ofG. Then for$u,v\in V(G)$, $\eta (u)=v$implies that$x_{u}=x_{v}$.

Proof

Let $P=(p_{uv})_{u,v\in V(G)}$ be the permutation matrix such that $p_{vu}=1$ if and only if $\eta (u)=v$ for $u,v\in V(G)$. We have $D_{\alpha }(G) = P^{\top }D_{\alpha }(G) P$ and Px is a positive unit vector. Thus $\mu _{\alpha }(G)=x^{\top }D_{\alpha }(G)x=(Px)^{\top }D_{\alpha }(G)(Px)$, implying Px is also the distance α-Perron vector of G. Thus $P x = x$, and the result follows. □

Let G be a graph. For $v\in V(G)$, let $N_{G}(v)$ be the set of neighbors of v in G, and $\operatorname{deg}_{G}(v)$ be the degree of v in G. Let $G-v$ be the subgraph of G obtained by deleting v and all edges containing v. For $S\subseteq V(G)$, let $G[S]$ be the subgraph of G induced by S. For a subset $E'$ of $E(G)$, $G-E'$ denotes the graph obtained from G by deleting all the edges in $E'$, and in particular, we write $G-xy$ instead of $G-\{xy\}$ if $E'=\{xy\}$. Let G̅ be the complement of G. For a subset $E'$ of $E(\overline{G})$, denote $G+E'$ the graph obtained from G by adding all edges in $E'$, and in particular, we write $G+xy$ instead of $G+\{xy\}$ if $E'=\{xy\}$.

For a nonnegative square matrix A, the Perron–Frobenius theorem implies that A has an eigenvalue that is equal the maximum modulus of all its eigenvalues; this eigenvalue is called the spectral radius of A, denoted by $\rho (A)$. Note that $\mu _{\alpha }(G)=\rho (D_{\alpha }(G))$ for a connected graph G.

Restating Corollary 2.2 in [20, p. 38], we have

Lemma 2.2

([20])

Suppose thatAandBare square nonnegative matrices, Ais irreducible, and$A-B$is nonnegative but nonzero. Then$\rho (A)> \rho (B)$.

By Lemma 2.2, we have

Lemma 2.3

Suppose thatGis a connected graph with$u,v\in V(G)$, anduandvare not adjacent. Then$\mu _{\alpha }(G+uv)< \mu _{\alpha }(G)$.

The transmission of a connected graph G, denoted by $\sigma (G)$, is the sum of distances between all unordered pairs of vertices in G. Clearly, $\sigma (G)=\frac{1}{2}\sum_{v\in V(G)} T_{G}(v)$. A graph is said to be transmission regular if $T_{G}(v)$ is a constant for each $v\in V(G)$. By Rayleigh’s principle, we have

Lemma 2.4

Suppose thatGis a connected graph of ordern. Then$\mu _{\alpha }(G)\ge \frac{2\sigma (G)}{n}$with equality if and only ifGis transmission regular.

For an $n\times n$ nonnegative matrix $A=(a_{ij})$, let $r_{i}$ be the ith row sum of A, i.e., $r_{i}=\sum_{j=1}^{n} a_{ij}$ for $i=1, \ldots , n$, and let $r_{\min }$ and $r_{\max }$ be the minimum and maximum row sums of A, respectively.

Lemma 2.5

([3])

Let$A=(a_{ij})$be an$n\times n$nonnegative matrix with row sums$r_{1},\ldots ,r_{n}$. Let$S=\{1,\ldots ,n\}$, $r_{\min }=r_{p}$, $r_{\max }=r_{q}$for somepandqwith$1 \le p$, $q\le n$, $\ell =\max \{r_{i}-a_{ip}: i\in S\setminus \{p\}\}$, $m=\min \{r_{i}-a_{iq}: i\in S\setminus \{q\}\}$, $s=\max \{a_{ip}: i\in S\setminus \{p\}\}$and$t=\min \{a_{iq}: i\in S\setminus \{q\}\}$. Then

$$\begin{aligned}& \frac{a_{qq}+m+\sqrt{(m-a_{qq})^{2}+4t(r_{\max }-a_{qq})}}{2} \\& \quad \le \rho (A) \\& \quad \le \frac{a_{pp}+\ell +\sqrt{(\ell -a_{pp})^{2}+4s(r_{\min }-a_{pp})}}{2}. \end{aligned}$$

Moreover, the first equality holds if$r_{i}-a_{iq}=m$and$a_{iq}=t$for all$i\in S\setminus \{q\}$, and the second equality holds if$r_{i}-a_{ip}=\ell$and$a_{ip}=s$for all$i\in S\setminus \{p\}$.

Let $J_{s\times t}$ be the $s\times t$ matrix of all 1’s, $0_{s\times t}$ the $s\times t$ matrix of all 0’s, and $I_{s}$ the identity matrix of order s.

Let $K_{n}$, $P_{n}$, and $S_{n}$ be the complete graph, the path, and the star of order n, respectively. Let $C_{n}$ denote the cycle of order $n\ge 3$.

For a connected graph G, let $T_{\min }(G)$ and $T_{\max }(G)$ be the minimum and maximum transmissions of G, respectively.

3 Bounds for the distance α-spectral radius

Let G be a connected graph of order n. Note that $D_{\alpha }(K_{n})=\alpha (n-1) I_{n}+(1-\alpha )(J_{n\times n}-I_{n})$, and thus $\mu _{\alpha }(K_{n})=n-1$. By Lemma 2.3, we have $\mu _{\alpha }(G)\ge n-1$ with equality if and only if $G\cong K_{n}$.

If $(d_{1},\ldots , d_{n})$ is the nonincreasing degree sequence of a graph G of order at least 2, then $d_{1}$ (resp. $d_{2}$) is the maximum (resp. second maximum) degree, $d_{n}$ (resp. $d_{n-1}$) is the minimum (resp. second minimum) degree of G. The diameter of G is the maximum distance between all vertex pairs of G. Using techniques from [33] by considering the first two minima or maxima of the entries of the distance α-Perron vector, we may prove the following lower and upper bounds: If G is a connected graph of order $n\ge 2$ with maximum degree Δ and second maximum degree $\Delta '$, then

$$\begin{aligned} \mu _{\alpha }(G) \ge & \frac{1}{2} \bigl( \alpha \bigl(4n-4- \Delta -\Delta '\bigr) \\ &{} +\sqrt{\alpha ^{2}\bigl(4n-4-\Delta -\Delta ' \bigr)^{2}-4(2\alpha -1) (2n-2- \Delta ) \bigl(2n-2-\Delta '\bigr)} \bigr) \end{aligned}$$

with equality if and only if G is regular with diameter at most 2. If G is a connected graph of order $n\ge 2$ with minimum degree δ and second minimum degree $\delta '$, then

$$\begin{aligned} \mu _{\alpha }(G) \le & \frac{1}{2}\large \bigl( \alpha \bigl(2dn-2-(d-1) \bigl(d+\delta +\delta '\bigr)\bigr) \\ &{}+\sqrt{\alpha ^{2}\bigl(2dn-2-(d-1) \bigl(d+\delta +\delta '\bigr)\bigr)^{2}-4(2 \alpha -1) S S'} \bigr) \end{aligned}$$

with equality if and only if G is regular with $d\le 2$, where d is the diameter of G, $S=dn-\frac{d(d-1)}{2}-1-\delta (d-1)$ and $S'=dn-\frac{d(d-1)}{2}-1-\delta ' (d-1)$. The proof of the above bounds may be found in the early version of this paper at arXiv:1901.10180.

Similarly, bounds for the distance α-spectral radius for connected bipartite graphs may be obtained as in [33].

A connected graph G of order n is distinguished vertex deleted regular (DVDR) if there is a vertex v of degree $n-1$ such that $G-v$ is regular. By the techniques in [3], we have the following bounds. For completeness, we include a proof here.

Theorem 3.1

LetGbe a connected graph anduandvbe vertices such that$T_{G}(u)=T_{\min }(G)$and$T_{G}(v)=T_{\max }(G)$. Let$m_{1}=\max \{T_{G}(w)-(1-\alpha )d(u,w): w\in V(G)\setminus \{u\}\}$, $m_{2}=\min \{T_{G}(w)-(1-\alpha )d(v,w): w\in V(G)\setminus \{v\}\}$, and$e(w)=\max \{d(w,z): z\in V(G)\}$for$w\in V(G)$. Then

$$\begin{aligned}& \frac{m_{2}+\alpha T_{\max }(G)+\sqrt{(m_{2}-\alpha T_{\max }(G))^{2}+4(1-\alpha )^{2}T_{\max }(G)}}{2} \\& \quad \le \mu _{\alpha }(G) \\& \quad \le \frac{m_{1}+\alpha T_{\min }(G)+\sqrt{(m_{1}-\alpha T_{\min }(G))^{2}+4(1-\alpha )^{2}e(u)T_{\min }(G)}}{2}. \end{aligned}$$

The first equality holds if and only ifGis a complete graph and the second equality holds if and only ifGis a DVDR graph.

Proof

Let M be the submatrix of $D_{\alpha }(G)$ obtained by deleting the row and column corresponding to vertex v. Let $M'$ be the matrix obtained from M by reducing some nondiagonal entries of each row with row sum greater than $m_{2}$ in M such that $M'$ is nonnegative and each row sum in $M'$ is $m_{2}$.

Let $D^{(1)}$ be the matrix obtained from $D_{\alpha }(G)$ by replacing all $(w,v)$-entries by $1-\alpha$ for $w\in V(G)\setminus \{v\}$, and replacing the submatrix M by $M'$. Obviously, $D_{\alpha }(G)$ and $D^{(1)}$ are nonnegative and irreducible, and $D_{\alpha }(G)\ge D^{(1)}$. By Lemma 2.2, we have $\mu _{\alpha }(G)\ge \rho (D^{(1)})$ with equality if and only if $D_{\alpha }(G)=D^{(1)}$. By applying Lemma 2.5 to $D^{(1)}$, we obtain the lower bound for $\mu _{\alpha }(G)$. Suppose that this lower bound is attained. Then $D_{\alpha }(G)=D^{(1)}$. As all $(w,v)$-entries are equal to $1-\alpha$ for $w\in V(G)\setminus \{v\}$, implying $\operatorname{deg}_{G}(v)=n-1$. As $T_{G}(v)=T_{\max }(G)$, G is a complete graph. Conversely, if G is a complete graph, then it is obvious that the lower bound for $\mu _{\alpha }(G)$ is attained.

Let C be the submatrix of $D_{\alpha }(G)$ obtained by deleting the row and column corresponding to vertex u. Let $C'$ be the matrix obtained from C by adding positive numbers to nondiagonal entries of each row with row sum less than $m_{1}$ in C such that each row sum in $C'$ is $m_{1}$. Let $D^{(2)}$ be the matrix obtained from $D_{\alpha }(G)$ by replacing all $(w,u)$-entries by $(1-\alpha )e(u)$ for $w\in V(G)\setminus \{u\}$, and replacing the submatrix C by $C'$. Note that $D_{\alpha }(G)$ and $D^{(2)}$ are nonnegative and irreducible, and $D^{(2)}\ge D_{\alpha }(G)$. By Lemma 2.2, $\mu _{\alpha }(G)\le \rho (D^{(2)})$ with equality if and only if $D_{\alpha }(G)=D^{(2)}$. By applying Lemma 2.5 to $D^{(2)}$, we obtain the upper bound for $\mu _{\alpha }(G)$.

Suppose that this upper bound is attained. By Lemma 2.2, $D_{\alpha }(G)=D^{(2)}$. As all $(w,u)$-entries are equal to $(1-\alpha )e(u)$ for $w\in V(G)\setminus \{u\}$, implying $e(u)=1$, i.e., $\operatorname{deg}_{G}(u)=n-1$. Note that $T_{G}(w)=m_{1}+1-\alpha$ for all $w\in V(G)\setminus \{u\}$ and $T_{\min }(G)=T_{G}(u)=n-1$. If $m_{1}+1-\alpha =n-1$, then G is a complete graph, which is a DVDR graph. Otherwise, $m_{1}+1-\alpha >n-1$.

Recall from [3] that an incomplete connected graph of order n is a DVDR graph if and only if except one vertex of degree $n-1$ each other vertex has the same transmission. Thus, the upper bound for $\mu _{\alpha }(G)$ is attained if and only if G is a DVDR graph. □

We mention that more bounds for $\mu _{\alpha }(G)$ may be derived even from some known bounds for nonnegative matrices, see, e.g., [9].

Let G be a connected graph of order n. Let $\varLambda =T_{\max }(G)$. As $\mu _{\alpha }(G)\le \varLambda$ with equality if and only if G is transmission regular. For a connected non-transmission-regular graph G of order n, Liu et al. [19] showed that

$$ \mu _{0}(G)< \varLambda - \frac{n\varLambda -2\sigma (G)}{(n\varLambda -2\sigma (G)+1)n} $$

and

$$ \mu _{1/2}(G)< \varLambda - \frac{n\varLambda -2\sigma (G)}{(2(n\varLambda -2\sigma (G))+1)n}. $$

Note that $4\sigma (G)< n^{2}\varLambda$. We show new bounds as follows:

$$ \mu _{0}(G)< \varLambda - \frac{n\varLambda -2\sigma (G)}{(n\varLambda -2\sigma (G))\frac{4\sigma (G)}{n\varLambda }+n} $$

and

$$ \mu _{1/2}(G)< \varLambda - \frac{n\varLambda -2\sigma (G)}{(n\varLambda -2\sigma (G))\frac{8\sigma (G)}{n\varLambda }+n}. $$

Instead of proving the two inequalities, we prove the following somewhat general result.

Theorem 3.2

LetGbe a connected non-transmission-regular graph of ordern. Then

$$ \mu _{\alpha }(G)< \varLambda - \frac{(1-\alpha )n\varLambda (n\varLambda -2\sigma (G))}{4\sigma (G)(n\varLambda -2\sigma (G))+(1-\alpha )n^{2}\varLambda }, $$

where$\varLambda =T_{\max }(G)$.

Proof

Let x be the α-Perron vector of G. Denote by $x_{u}=\max \{x_{w}: w\in V(G)\}$ and $x_{v}=\min \{x_{w}: w\in V(G)\}$. Since G is not transmission regular, we have $x_{u}> x_{v}$, and thus

$$\begin{aligned} \mu _{\alpha }(G) =&x^{\top }D_{\alpha }(G)x \\ =&\alpha \sum_{w\in V(G)}T_{G}(w)x_{w}^{2}+2(1- \alpha )\sum_{\{w,z \}\subseteq V(G)}d_{wz}x_{w}x_{z} \\ < & 2\alpha \sigma (G)x_{u}^{2}+2(1-\alpha )\sigma (G)x_{u}^{2}, \end{aligned}$$

implying that $x_{u}^{2}>\frac{\mu _{\alpha }(G)}{2\sigma (G)}$. Note that

$$\begin{aligned}& \varLambda -\mu _{\alpha }(G) \\& \quad = \varLambda -\alpha \sum_{w\in V(G)}T_{G}(w)x_{w}^{2}-2(1- \alpha ) \sum_{\{w,z\}\subseteq V(G)}d_{wz}x_{w}x_{z} \\& \quad = \sum_{w\in V(G)}\bigl(\varLambda -T_{G}(w) \bigr)x_{w}^{2}+(1-\alpha )\sum _{ \{w,z\}\subseteq V(G)}d_{wz}(x_{w}-x_{z})^{2} \\& \quad \ge \sum_{w\in V(G)}\bigl(\varLambda -T_{G}(w)\bigr)x_{v}^{2}+(1-\alpha )\sum _{ \{w,z\}\subseteq V(G)}d_{wz}(x_{w}-x_{z})^{2} \\& \quad = \bigl(n\varLambda -2\sigma (G)\bigr)x_{v}^{2}+(1- \alpha )\sum_{\{w,z\} \subseteq V(G)}d_{wz}(x_{w}-x_{z})^{2}. \end{aligned}$$

We need to estimate $\sum_{\{w,z\}\subseteq V(G)}d_{wz}(x_{w}-x_{z})^{2}$. Let $P=w_{0}w_{1}\ldots w_{\ell}$ be a shortest path connecting u and v, where $w_{0}=u$, $w_{\ell}=v$, and $\ell\ge 1$. Obviously,

$$ \sum_{\{w,z\}\subseteq V(G)}d_{wz}(x_{w}-x_{z})^{2} \ge N_{1}+N_{2}, $$

where $N_{1}=\sum_{w\in V(G)\setminus V(P)}\sum_{z\in V(P)}d_{wz}(x_{w}-x_{z})^{2}$ and $N_{2}=\sum_{\{w,z\}\subseteq V(P)}d_{wz}(x_{w}-x_{z})^{2}$. For $w\in V(G)\setminus V(P)$, by the Cauchy–Schwarz inequality, we have

$$ d_{wu}(x_{w}-x_{u})^{2}+d_{wv}(x_{w}-x_{v})^{2} \ge (x_{w}-x_{u})^{2}+(x_{w}-x_{v})^{2} \ge \frac{1}{2}(x_{u}-x_{v})^{2}, $$

and thus

$$\begin{aligned} N_{1} \ge & \sum_{w\in V(G)\setminus V(P)} \bigl(d_{wu}(x_{w}-x_{u})^{2}+d_{wv}(x_{w}-x_{v})^{2} \bigr) \\ \ge & \sum_{w\in V(G)\setminus V(P)} \frac{1}{2}(x_{u}-x_{v})^{2} \\ = &\frac{n-\ell -1}{2}(x_{u}-x_{v})^{2}. \end{aligned}$$

For $1\le i \le \ell -1$ and $\ell \ge 2$, by the Cauchy–Schwarz inequality, we have

$$\begin{aligned}& d_{w_{0}w_{i}}(x_{w_{0}}-x_{w_{i}})^{2}+d_{w_{i} w_{\ell }}(x_{w_{i}}-x_{w_{\ell }})^{2} \\& \quad \ge \min \{i,\ell -i\} \bigl((x_{w_{0}}-x_{w_{i}})^{2}+(x_{w_{i}}-x_{w_{\ell }})^{2} \bigr) \\& \quad \ge \min \{i,\ell -i\}\cdot \frac{1}{2}(x_{w_{0}}-x_{w_{\ell }})^{2} \\& \quad = \frac{1}{2}\min \{i,\ell -i\}(x_{u}-x_{v})^{2}, \end{aligned}$$

and thus

$$\begin{aligned} N_{2} \ge & d_{uv}(x_{u}-x_{v})^{2}+ \sum_{i=1}^{\ell -1} \bigl(d_{w_{0}w_{i}}(x_{w_{i}}-x_{w_{0}})^{2}+d_{w_{i} w_{\ell }}(x_{w_{i}}-x_{w_{\ell }})^{2} \bigr) \\ \ge & \ell (x_{u}-x_{v})^{2}+\sum _{i=1}^{\ell -1}\frac{1}{2}\min \{i, \ell -i\}(x_{u}-x_{v})^{2} \\ =& \Biggl(\ell +\frac{1}{2}\sum_{i=1}^{\ell -1} \min \{i,\ell -i\} \Biggr) (x_{u}-x_{v})^{2} \\ =& \textstyle\begin{cases} \frac{\ell ^{2}+8\ell }{8}(x_{u}-x_{v})^{2} & \text{if $\ell $ is even}, \\ \frac{\ell ^{2}+8\ell -1}{8}(x_{u}-x_{v})^{2} & \text{if $\ell $ is odd}. \end{cases}\displaystyle \end{aligned}$$

Case 1.u and v are adjacent, i.e., $\ell =1$.

In this case, we have

$$\begin{aligned} \sum_{\{w,z\}\subseteq V(G)}d_{wz}(x_{w}-x_{z})^{2} \geq & N_{1}+N_{2} \\ \ge & \frac{n-1-1}{2}(x_{u}-x_{v})^{2}+(x_{u}-x_{v})^{2} \\ =& \frac{n}{2}(x_{u}-x_{v})^{2}. \end{aligned}$$

Thus

$$\begin{aligned} \varLambda -\mu _{\alpha }(G) \ge & \bigl(n\varLambda -2\sigma (G) \bigr)x_{v}^{2}+(1- \alpha )\sum _{\{w,z\}\subseteq V(G)}d_{wz}(x_{w}-x_{z})^{2} \\ \ge & \bigl(n\varLambda -2\sigma (G)\bigr)x_{v}^{2}+(1- \alpha )\frac{n}{2}(x_{u}-x_{v})^{2}. \end{aligned}$$

Viewed as a function of $x_{v}$, $(n\varLambda -2\sigma (G))x_{v}^{2}+(1-\alpha )\frac{n}{2}(x_{u}-x_{v})^{2}$ achieves its minimum value $\frac{(1-\alpha )n(n\varLambda -2\sigma (G))}{2(n\varLambda -2\sigma (G))+(1-\alpha )n}x_{u}^{2}$. Recall that $x_{u}^{2}>\frac{\mu _{\alpha }(G)}{2\sigma (G)}$. Then we have

$$\begin{aligned} \varLambda -\mu _{\alpha }(G) >&\frac{(1-\alpha )n(n\varLambda -2\sigma (G))}{2(n\varLambda -2\sigma (G))+(1-\alpha )n} \cdot \frac{\mu _{\alpha }(G)}{2\sigma (G)} \\ =&\frac{(1-\alpha )n(n\varLambda -2\sigma (G))\varLambda }{2\sigma (G)(2(n\varLambda -2\sigma (G))+(1-\alpha )n)} \\ &{}- \frac{(1-\alpha )n(n\varLambda -2\sigma (G))(\varLambda -\mu _{\alpha }(G))}{2\sigma (G)(2(n\varLambda -2\sigma (G))+(1-\alpha )n)}, \end{aligned}$$

which implies that

$$ \varLambda -\mu _{\alpha }(G)> \frac{(1-\alpha )n\varLambda (n\varLambda -2\sigma (G))}{4\sigma (G)(n\varLambda -2\sigma (G))+(1-\alpha )n^{2}\varLambda }. $$

Case 2.u and v are not adjacent, i.e., $\ell \ge 2$.

Suppose first that ℓ is even. Then

$$\begin{aligned} \sum_{\{w,z\}\subseteq V(G)}d_{wz}(x_{w}-x_{z})^{2} \geq &N_{1}+N_{2} \\ \ge & \frac{n-\ell -1}{2}(x_{u}-x_{v})^{2}+ \frac{\ell ^{2}+8\ell }{8}(x_{u}-x_{v})^{2} \\ =& \frac{\ell ^{2}+4\ell +4n-4}{8}(x_{u}-x_{v})^{2}. \end{aligned}$$

Thus

$$\begin{aligned} \varLambda -\mu _{\alpha }(G) \ge & \bigl(n\varLambda -2\sigma (G) \bigr)x_{v}^{2}+(1- \alpha )\sum _{\{w,z\}\subseteq V(G)}d_{wz}(x_{w}-x_{z})^{2} \\ \ge & \bigl(n\varLambda -2\sigma (G)\bigr)x_{v}^{2}+(1- \alpha ) \frac{\ell ^{2}+4\ell +4n-4}{8}(x_{u}-x_{v})^{2}. \end{aligned}$$

Viewed as a function of $x_{v}$, $(n\varLambda -2\sigma (G))x_{v}^{2}+(1-\alpha ) \frac{\ell ^{2}+4\ell +4n-4}{8}(x_{u}-x_{v})^{2}$ achieves its minimum value $\frac{(1-\alpha )(n\varLambda -2\sigma (G))(\ell ^{2}+4\ell +4n-4)}{8(n\varLambda -2\sigma (G))+(1-\alpha )(\ell ^{2}+4\ell +4n-4)}x_{u}^{2}$. As $x_{u}^{2}>\frac{\mu _{\alpha }(G)}{2\sigma (G)}$, we have

$$ \varLambda -\mu _{\alpha }(G) > \frac{(1-\alpha )(n\varLambda -2\sigma (G))(\ell ^{2}+4\ell +4n-4)}{(1-\alpha )(\ell ^{2}+4\ell +4n-4)+8(n\varLambda -2\sigma (G))} \cdot \frac{\mu _{\alpha }(G)}{2\sigma (G)}, $$

i.e.,

$$ \varLambda -\mu _{\alpha }(G)> \frac{(1-\alpha )(n\varLambda -2\sigma (G))(\ell ^{2}+4\ell +4n-4)\varLambda }{16\sigma (G)(n\varLambda -2\sigma (G))+(1-\alpha )(\ell ^{2}+4\ell +4n-4)n\varLambda }. $$

As a function of ℓ, the expression on the right-hand side in the above inequality is strictly increasing for $\ell \ge 2$. Thus we have

$$\begin{aligned} \varLambda -\mu _{\alpha }(G) >&\frac{(1-\alpha )(n\varLambda -2\sigma (G))(n+2)\varLambda }{4\sigma (G)(n\varLambda -2\sigma (G))+(1-\alpha )(n+2)n\varLambda } \\ >&\frac{(1-\alpha )n\varLambda (n\varLambda -2\sigma (G))}{4\sigma (G)(n\varLambda -2\sigma (G))+(1-\alpha )n^{2}\varLambda }. \end{aligned}$$

Now suppose that ℓ is odd. Then

$$\begin{aligned}& \sum_{\{w,z\}\subseteq V(G)}d_{wz}(x_{w}-x_{z})^{2} \\& \quad \geq N_{1}+N_{2} \\& \quad \ge \frac{n-\ell -1}{2}(x_{u}-x_{v})^{2}+ \frac{\ell ^{2}+8\ell -1}{8}(x_{u}-x_{v})^{2} \\& \quad = \frac{\ell ^{2}+4\ell +4n-5}{8}(x_{u}-x_{v})^{2}. \end{aligned}$$

Thus, as early, we have

$$\begin{aligned}& \varLambda -\mu _{\alpha }(G) \\& \quad \ge \bigl(n\varLambda -2\sigma (G)\bigr)x_{v}^{2}+(1- \alpha ) \frac{\ell ^{2}+4\ell +4n-5}{8}(x_{u}-x_{v})^{2} \\& \quad \ge \frac{(1-\alpha )(\ell ^{2}+4\ell +4n-5)(n\varLambda -2\sigma (G))}{8(n\varLambda -2\sigma (G))+(1-\alpha )(\ell ^{2}+4\ell +4n-5)}x_{u}^{2} \\& \quad > \frac{(1-\alpha )(\ell ^{2}+4\ell +4n-5)(n\varLambda -2\sigma (G))}{8(n\varLambda -2\sigma (G))+(1-\alpha )(\ell ^{2}+4\ell +4n-5)} \cdot \frac{\mu _{\alpha }(G)}{2\sigma (G)}, \end{aligned}$$

implying

$$ \varLambda -\mu _{\alpha }(G)> \frac{(1-\alpha )(n\varLambda -2\sigma (G))(\ell ^{2}+4\ell +4n-5)\varLambda }{16\sigma (G)(n\varLambda -2\sigma (G))+(1-\alpha )(\ell ^{2}+4\ell +4n-5)n\varLambda }. $$

As a function of ℓ, the expression on the right-hand side in the above inequality is strictly increasing for $\ell \ge 3$. Thus we have

$$\begin{aligned} \varLambda -\mu _{\alpha }(G) >&\frac{(1-\alpha )(n\varLambda -2\sigma (G))(4+n)\varLambda }{4\sigma (G)(n\varLambda -2\sigma (G))+(1-\alpha )(4+n)n\varLambda } \\ >&\frac{(1-\alpha )n\varLambda (n\varLambda -2\sigma (G))}{4\sigma (G)(n\varLambda -2\sigma (G))+(1-\alpha )n^{2}\varLambda }. \end{aligned}$$

The result follows by combining Cases 1 and 2. □

4 Effect of graft transformations on distance α-spectral radius

In this section, we study the effect of some local graft transformations on distance α-spectral radius.

A path $u_{0}\cdots u_{r}$ (with $r\geq 1$) in a graph G is called a pendant path (of length r) at $u_{0}$ if $\operatorname{deg}_{G}(u_{0})\geq 3$, the degrees of $u_{1},\ldots ,u_{r-1}$ (if any exists) are all equal to 2 in G, and $\operatorname{deg}_{G}(u_{r})=1$. A pendant path of length 1 at $u_{0}$ is called a pendant edge at $u_{0}$.

A vertex of a graph is a pendant vertex if its degree is 1. A cut edge of a connected graph is an edge whose removal yields a disconnected graph.

If P is a pendant path of G at u with length $r\geq 1$, then we say G is obtained from H by attaching a pendant path P of length r at u with $H=G[V(G)\setminus (V(P)\setminus \{u\})]$. If the pendant path of length 1 is attached to a vertex u of H, then we also say that a pendant vertex is attached to u.

Theorem 4.1

Suppose thatGis a connected graph, uvis a cut edge with$\operatorname{deg}_{G}(u)\ge 2$, andvis adjacent to a pendant vertex$v'$. Let

$$ G_{uv}=G-\bigl\{ uw: w\in N_{G}(u)\setminus \{v\}\bigr\} +\bigl\{ vw: w\in N_{G}(u) \setminus \{v\}\bigr\} . $$

Then$\mu _{\alpha }(G)>\mu _{\alpha }(G_{uv})$.

Proof

Let $G_{1}$ and $G_{2}$ be the components of $G-uv$ containing u and v, respectively. Let x be the distance α-Perron vector of $G_{uv}$. By Lemma 2.1, $x_{u}=x_{v'}$. As we pass from G to $G_{uv}$, the distance between a vertex in $V(G_{1})\setminus \{u\}$ and a vertex in $V(G_{2})$ is decreased by 1, the distance between a vertex $V(G_{1})\setminus \{u\}$ and u is increased by 1, and the distances between all other vertex pairs remain unchanged. Thus

$$\begin{aligned}& \mu _{\alpha } (G)-\mu _{\alpha } (G_{uv}) \\& \quad \geq x^{\top }\bigl(D_{\alpha }(G)-D_{\alpha }(G_{uv}) \bigr)x \\& \quad = \sum_{w\in V(G_{1})\setminus \{u\}}\sum_{z\in V(G_{2}) } \bigl( \alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr) \\& \qquad {} -\sum_{ w\in V(G_{1})\setminus \{u\}} \bigl(\alpha \bigl(x_{w}^{2}+x_{u}^{2} \bigr)+2(1-\alpha )x_{w}x_{u} \bigr) \\& \quad \geq \sum_{w\in V(G_{1})\setminus \{u\}} \bigl(\alpha \bigl(x_{w}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{w}x_{v} \bigr) \\& \qquad {} +\sum_{w\in V(G_{1})\setminus \{u\}} \bigl(\alpha \bigl(x_{w}^{2}+x_{v'}^{2} \bigr)+2(1-\alpha )x_{w}x_{v'} \bigr) \\& \qquad {} -\sum_{ w\in V(G_{1})\setminus \{u\}} \bigl(\alpha \bigl(x_{w}^{2}+x_{u}^{2} \bigr)+2(1-\alpha )x_{w}x_{u} \bigr) \\& \quad = \sum_{w\in V(G_{1})\setminus \{u\}} \bigl(\alpha \bigl(x_{w}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{w}x_{v} \bigr) \\& \quad > 0, \end{aligned}$$

implying $\mu _{\alpha } (G)-\mu _{\alpha } (G_{uv})>0$, i.e., $\mu _{\alpha }(G)>\mu _{\alpha }(G_{uv})$. □

The previous theorem has been established for $\alpha =0,\frac{1}{2}$ in [16, 25].

Theorem 4.2

Suppose thatGis a connected graph withkedge-disjoint nontrivial induced subgraphs$G_{1}, \ldots , G_{k}$such that$V(G_{i})\cap V(G_{j})=\{u\}$for$1\le i< j\le k$and$\bigcup_{i=1}^{k}V(G_{i})=V(G)$, where$k\ge 3$. Let$\emptyset \ne K\subseteq \{3, \ldots , k\}$and let$N_{K}=\bigcup_{i\in K}N_{G_{i}}(u)$. For$v'\in V(G_{1})\setminus \{u\}$and$v''\in V(G_{2})\setminus \{u\}$, let

$$ G' =G-\{uw: w\in N_{K}\}+\bigl\{ v'w: w\in N_{K}\bigr\} $$

and

$$ G'' =G-\{uw: w\in N_{K}\}+\bigl\{ v''w: w\in N_{K}\bigr\} . $$

Then$\mu _{\alpha }(G)< \max \{\mu _{\alpha }(G'), \mu _{\alpha }(G'')\}$.

Proof

Let x be the distance α-Perron vector of G. Let $V_{K}= (\bigcup_{i\in K} V(G_{i}) )\setminus \{u\}$. Let

$$\begin{aligned} \varGamma =&\sum_{w\in V(G_{2})\setminus \{u\}}\sum _{z\in V_{K}} \bigl(\alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr) \\ &{}-\sum_{w\in V(G_{1})\setminus \{u\}}\sum_{z\in V_{K}} \bigl( \alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr). \end{aligned}$$

As we pass from G to $G'$, the distance between a vertex in $V(G_{2})$ and a vertex in $V_{K}$ is increased by $d_{G}(u,v')$, the distance between a vertex w in $V(G_{1})\setminus \{u\}$ and a vertex in $V_{K}$ is decreased by $d_{G}(w,u)-d_{G}(w,v')$, which is at most $d_{G}(u,v')$, and the distances between all other vertex pairs are increased or remain unchanged. Thus

$$\begin{aligned}& \mu _{\alpha } \bigl(G'\bigr)-\mu _{\alpha } (G) \\& \quad \geq x^{\top }\bigl(D_{\alpha }\bigl(G' \bigr)-D_{\alpha }(G)\bigr)x \\& \quad \geq \sum_{w\in V(G_{2})}\sum _{z\in V_{K}} \bigl(d_{G}\bigl(u,v' \bigr) \bigl(\alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr) \bigr) \\& \qquad {} -\sum_{w\in V(G_{1})\setminus \{u\}}\sum_{z\in V_{K}} \bigl(d_{G}\bigl(u,v'\bigr) \bigl(\alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr) \bigr) \\& \quad = d_{G}\bigl(u,v'\bigr) \biggl(\varGamma +\sum _{z\in V_{K}} \bigl(\alpha \bigl(x_{u}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{u}x_{z} \bigr) \biggr) \\& \quad > d_{G}\bigl(u,v'\bigr)\varGamma . \end{aligned}$$

If $\varGamma \ge 0$, then $\mu _{\alpha } (G')-\mu _{\alpha } (G)> d_{G}(u,v')\varGamma \ge 0$, implying $\mu _{\alpha } (G)<\mu _{\alpha } (G')$. Suppose that $\varGamma <0$. As we pass from G to $G''$, the distance between a vertex in $V(G_{1})$ and a vertex in $V_{K}$ is increased by $d_{G}(u,v'')$, the distance between a vertex w in $V(G_{2})\setminus \{u\}$ and a vertex in $V_{K}$ is decreased by $d_{G}(w,u)-d_{G}(w,v'')$, which is at most $d_{G}(u,v'')$, and the distances between all other vertex pairs are increased or remain unchanged. Thus

$$\begin{aligned}& \mu _{\alpha } \bigl(G''\bigr)-\mu _{\alpha } (G) \\& \quad \geq x^{\top }\bigl(D_{\alpha }\bigl(G'' \bigr)-D_{\alpha }(G)\bigr)x \\& \quad \geq \sum_{w\in V(G_{1})}\sum _{z\in V_{K}} \bigl(d_{G}\bigl(u,v'' \bigr) \bigl(\alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr) \bigr) \\& \qquad {} -\sum_{w\in V(G_{2})\setminus \{u\}}\sum_{z\in V_{K}} \bigl(d_{G}\bigl(u,v''\bigr) \bigl( \alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr) \bigr) \\& \quad = d_{G}\bigl(u,v''\bigr) \biggl(- \varGamma + \sum_{z\in V_{K}} \bigl(\alpha \bigl(x_{u}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{u}x_{z} \bigr) \biggr) \\& \quad > d_{G}\bigl(u,v''\bigr) (-\varGamma ) \\& \quad > 0, \end{aligned}$$

implying $\mu _{\alpha } (G'')-\mu _{\alpha } (G)> 0$, i.e., $\mu _{\alpha }(G)<\mu _{\alpha }(G'')$. □

Weak versions of previous theorem for $\alpha =0$ have been given in [28, 30] and a weak version for $\alpha =\frac{1}{2}$ may be found in [16].

For positive integer p and a graph G with $u\in V(G)$, let $G(u;p)$ be the graph obtained from G by attaching a pendant path of length p at u. Let $G(u;0)=G$, and in this case a pendant path of length 0 is understood the trivial path consisting of a single vertex u.

For nonnegative integers p, q and a graph G, let $G_{u}(p,q)$ be the graph $H(u;q)$ with $H=G(u;p)$. The following corollary has been known for $\alpha =0$ in [24, 28] and $\alpha =\frac{1}{2}$ in [15, 16].

Corollary 4.1

LetHbe a nontrivial connected graph with$u\in V(H)$. If$p\ge q\ge 1$, then$\mu _{\alpha } (H_{u}(p,q))<\mu _{\alpha } (H_{u}(p+1,q-1))$.

Proof

Let $G=H_{u}(p,q)$. Let $P=uu_{1}\cdots u_{p}$ and $Q=uv_{1}\cdots v_{q}$ be two pendant paths of lengths p and q, respectively, in G. Using the notations in Theorem 4.2 with $k=3$, $G_{1}=P$, $G_{2}=Q$, $G_{3}=H$, $v'=u_{p-q+1}$ and $v''=v_{1}$, we have $G'\cong G'' \cong H_{u}(p+1,q-1)$, and thus by Theorem 4.2, we have $\mu _{\alpha } (H_{u}(p,q))<\mu _{\alpha } (H_{u}(p+1,q-1))$. □

Theorem 4.3

Suppose thatGis a connected graph with three edge-disjoint induced subgraphs$G_{1}$, $G_{2}$and$G_{3}$such that$V(G_{1})\cap V(G_{3})=\{u\}$, $V(G_{2})\cap V(G_{3})=\{v\}$, $\bigcup_{i=1}^{3}V(G_{i})=V(G)$, and$G_{1}-u$, $G_{2}-v$, and$G_{3}-u-v$are all nontrivial. Suppose that$uv\in E(G_{3})$. For$u'\in N_{G_{1}}(u)$and$v'\in N_{G_{2}}(v)$, let

$$ G'=H+\bigl\{ u'w: w\in N_{G_{3}-uv}(u)\bigr\} +\bigl\{ uw: w\in N_{G_{3}-uv}(v)\bigr\} $$

and

$$ G''=H+\bigl\{ vw: w\in N_{G_{3}-uv}(u)\bigr\} +\bigl\{ v'w: w\in N_{G_{3}-uv}(v)\bigr\} , $$

where$H=G-\{uw: w\in N_{G_{3}-uv}(u)\}-\{vw: w\in N_{G_{3}-uv}(v)\}$. Then$\mu _{\alpha } (G)<\mu _{\alpha } (G')$or$\mu _{\alpha } (G)<\mu _{\alpha } (G'')$.

Proof

Let x be the distance α-Perron vector of G. Let

$$\begin{aligned} \varGamma =&\sum_{w\in V(G_{2})}\sum _{z\in V(G_{3})\setminus \{u,v\} } \bigl(\alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr) \\ &{}- \sum_{w\in V(G_{1})}\sum_{z\in V(G_{3})\setminus \{u,v\}} \bigl( \alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr). \end{aligned}$$

As we pass from G to $G'$, the distance between a vertex in $V(G_{2})$ and a vertex in $V(G_{3})\setminus \{u,v\}$ is increased by 1, the distance between a vertex in $V(G_{1})$ and a vertex in $V(G_{3})\setminus \{u,v\}$ may be increased, unchanged, or decreased by 1, and the distances between any other vertex pairs remain unchanged. Thus

$$\begin{aligned} \mu _{\alpha } \bigl(G'\bigr)-\mu _{\alpha } (G) \ge & x^{\top }\bigl(D_{\alpha }\bigl(G'\bigr)- D_{\alpha }(G)\bigr)x \\ \ge &\sum_{w\in V(G_{2})}\sum_{z\in V(G_{3})\setminus \{u,v\} } \bigl(\alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr) \\ &{}-\sum_{w\in V(G_{1})}\sum_{z\in V(G_{3})\setminus \{u,v\} } \bigl( \alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr) \\ =&\varGamma . \end{aligned}$$

If $\varGamma \ge 0$, then $\mu _{\alpha } (G')-\mu _{\alpha } (G)\ge 0$, i.e., $\mu _{\alpha } (G)\le \mu _{\alpha } (G')$. If $\mu _{\alpha } (G)=\mu _{\alpha } (G')$, then $\mu _{\alpha } (G')=x^{\top }D_{\alpha }(G')x$, implying x is the distance α-Perron vector of $G'$. By the α-equations of G and $G'$ at v, we have

$$\begin{aligned} 0 =& \mu _{\alpha }\bigl(G'\bigr)x_{v}-\mu _{\alpha }(G)x_{v} \\ =&\sum_{w\in V(G_{3})\setminus \{u,v\}} \bigl(d_{G'}(v,w)-d_{G}(v,w) \bigr) \bigl(\alpha x_{v}+(1-\alpha )x_{w}\bigr) \\ =&\sum_{w\in V(G_{3})\setminus \{u,v\}} \bigl(\alpha x_{v}+(1- \alpha )x_{w}\bigr) \\ >& 0, \end{aligned}$$

a contradiction. Thus, if $\varGamma \ge 0$, then $\mu _{\alpha } (G)<\mu _{\alpha } (G')$.

Suppose that $\varGamma <0$. As earlier, we have

$$\begin{aligned} \mu _{\alpha }\bigl(G''\bigr)-\mu _{\alpha }(G) \ge & x^{\top }\bigl(D_{\alpha } \bigl(G''\bigr)-D_{ \alpha }(G)\bigr)x \\ \ge &\sum_{w\in V(G_{1})}\sum_{z\in V(G_{3})\setminus \{u,v\} } \bigl(\alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr) \\ &{}-\sum_{w\in V(G_{2})}\sum_{z\in V(G_{3})\setminus \{u,v\} } \bigl( \alpha \bigl(x_{w}^{2}+x_{z}^{2} \bigr)+2(1-\alpha )x_{w}x_{z} \bigr) \\ =& -\varGamma \\ >&0, \end{aligned}$$

and thus $\mu _{\alpha }(G)<\mu _{\alpha }(G'')$. □

A weak version of previous theorem for $\alpha =\frac{1}{2}$ has been established in [16].

For nonnegative integers p, q and a graph G with $u,v\in V(G)$, let $G_{u,v}(p,q)$ be the graph $H(v;q)$ with $H=G(u;p)$. The following corollary has been known for $\alpha =0,\frac{1}{2}$ in [15, 32].

Corollary 4.2

LetHbe a connected graph of order at least 3 with$uv\in E(H)$. Suppose that$\eta (u)=v$for some automorphismηofG. For$p\ge q\ge 1$, we have$\mu _{\alpha }(H_{u,v}(p,q))<\mu _{\alpha }(H_{u,v}(p+1,q-1))$.

Proof

Let $G=H_{u,v}(p,q)$. Let $P=uu_{1}\cdots u_{p}$ and $Q=vv_{1}\cdots v_{q}$ be two pendant paths of lengths p and q in G at u and v, respectively. Using the notations of Theorem 4.3 with $G_{1}=P$, $G_{2}=Q$, $G_{3}=H$, $u'=u_{1}$ and $v'=v_{1}$, we have $G'\cong H_{u,v}(p-1,q+1)$ and $G''\cong H_{u,v}(p+1,q-1)$, and thus by Theorem 4.3, we have $\mu _{\alpha } (H_{u,v}(p,q))<\max \{\mu _{\alpha }(H_{u,v}(p-1,q+1)), \mu _{\alpha }(H_{u,v}(p+1,q-1))\}$. If $p=q$ ($p=q+1$, respectively), then $H_{u,v}(p-1,q+1)\cong H_{u,v}(p+1,q-1)$ ($H_{u,v}(p,q)\cong H_{u,v}(p-1,q+1)$, respectively) as $\eta (u)=v$ for some automorphism η of G, and thus from the above inequality, we have $\mu _{\alpha } (G)<\mu _{\alpha }(H_{u,v}(p+1,q-1))$. Suppose that $p\ge q+2$ and $\mu _{\alpha } (G)<\mu _{\alpha }(H_{u,v}(p-1,q+1))$. If $p\not \equiv q \pmod{2}$, then we have

$$\begin{aligned} \mu _{\alpha } (G) \le &\mu _{\alpha } \biggl(H_{u,v} \biggl( \frac{p+q+3}{2},\frac{p+q-3}{2} \biggr) \biggr) \\ < &\mu _{\alpha } \biggl(H_{u,v} \biggl(\frac{p+q+1}{2}, \frac{p+q-1}{2} \biggr) \biggr) \\ < & \mu _{\alpha } \biggl(H_{u,v} \biggl(\frac{p+q+3}{2}, \frac{p+q-3}{2} \biggr) \biggr), \end{aligned}$$

which is impossible. If $p\equiv q \pmod{2}$, then we have

$$\begin{aligned} \mu _{\alpha } (G) \le & \mu _{\alpha } \biggl(H_{u,v} \biggl( \frac{p+q}{2}+1,\frac{p+q}{2}-1 \biggr) \biggr) \\ < &\mu _{\alpha } \biggl(H_{u,v} \biggl(\frac{p+q}{2}, \frac{p+q}{2} \biggr) \biggr) \\ < & \mu _{\alpha } \biggl(H_{u,v} \biggl(\frac{p+q}{2}-1, \frac{p+q}{2}+1 \biggr) \biggr), \end{aligned}$$

which is also impossible. Therefore $\mu _{\alpha } (H_{u,v}(p,q))<\mu _{\alpha } (H_{u,v}(p+1,q-1))$. □

5 Graphs with small or large distance α-spectral radius

First we determine the graphs with minimum distance α-spectral radius among trees and unicyclic graphs.

Theorem 5.1

LetGbe a tree of ordern. Then$\mu _{\alpha }(G)\geq \mu _{\alpha }(S_{n})$with equality if and only if$G\cong S_{n}$.

Proof

The result is trivial if $n=1,2,3$. Suppose that $n\ge 4$. Let G be a tree of order n such that $\mu _{\alpha }(G)$ is as small as possible. Let d be the diameter of G. Evidently, $d\geq 2$. Suppose that $d\ge 3$. Let $v_{0}v_{1}\cdots v_{d}$ be a diametral path of G. By Theorem 4.1, $\mu _{\alpha }(G_{v_{1}v_{2}})<\mu _{\alpha }(G)$, a contradiction. Thus $d= 2$, i.e., $G\cong S_{n}$. □

In Theorem 5.1, the case $\alpha =0$ has been known in [24] and the case $\alpha =\frac{1}{2}$ has been known in [16, 29].

For $n-1\ge 3$ and $1\le a\le \lfloor \frac{n-2}{2} \rfloor$, let $D_{n,a}$ be the tree obtained from vertex-disjoint $S_{a+1}$ with center u and $S_{n-a-1}$ with center v by adding an edge uv. Let T be a tree of order n with minimum distance α-spectral radius, where $T\ncong S_{n}$. Let d be the diameter of T. Then $d\geq 3$. Suppose that $d\geq 4$. Let $v_{0}v_{1}\cdots v_{d}$ be a diametral path of T. Note that $T_{v_{1}v_{2}} \ncong S_{n}$. By Theorem 4.1, $\mu _{\alpha }(T_{v_{1}v_{2}})<\mu _{\alpha }(T)$, a contradiction. Thus $d=3$, implying $T\cong D_{n,a}$ for some a with $1\leq a\leq \lfloor \frac{n-2}{2}\rfloor$.

Let $S_{n}^{+}$ is the graph obtained from $S_{n}$ by adding an edge between two vertices of degree one.

Lemma 5.1

([29])

LetGbe a unicyclic graph of order$n\ge 6$. If$G\ncong S_{n}^{+}$, then

$$ \sigma (G)\ge n^{2}-n-4>\sigma \bigl(S_{n}^{+} \bigr)=n^{2}-2n. $$

Note that for $n=5$, we have $\sigma (C_{n})=\sigma (S_{n}^{+})$. So, in the above lemma, the condition $n\ge 6$ is necessary.

Theorem 5.2

LetGbe a unicyclic graph of order$n\ge 8$. Then$\mu _{\alpha }(G)\ge \mu _{\alpha }(S_{n}^{+})$with equality if and only if$G\cong S_{n}^{+}$.

Proof

Suppose that $G\ncong S_{n}^{+}$. We only need to show that $\mu _{\alpha }(G)>\mu _{\alpha }(S_{n}^{+})$.

By Lemmas 2.4 and 5.1, we have

$$ \mu _{\alpha }(G)\ge \frac{2\sigma (G)}{n}\ge \frac{2(n^{2}-n-4)}{n}. $$

By [20, p. 24, Theorem 1.1] or by Theorem 3.2, we have

$$ \mu _{\alpha }\bigl(S_{n}^{+}\bigr)< T_{\max }\bigl(S_{n}^{+}\bigr)=2n-3. $$

Since $n\ge 8$, we have

$$ \mu _{\alpha }(G)\ge \frac{2(n^{2}-n-4)}{n}\ge 2n-3>\mu _{\alpha } \bigl(S_{n}^{+}\bigr), $$

as desired. □

The result in Theorem 5.2 for $\alpha =0, \frac{1}{2}$ has been known in [29, 31].

In the following, we determine the graphs with maximum distance α-spectral radius among some classes of graphs.

For $2\le \Delta \le n-1$, let $B_{n,\Delta }$ be a tree obtained by attaching $\Delta -1$ pendant vertices to a terminal vertex of the path $P_{n-\Delta +1}$. In particular, $B_{n,2}=P_{n}$ and $B_{n, n-1}=S_{n}$. The following theorem for $\alpha =0,\frac{1}{2}$ was given in [16, 24] for trees.

Theorem 5.3

LetGbe a connected graph of ordernwith maximum degree Δ, where$2\le \Delta \le n-1$. Then$\mu _{\alpha } (G)\leq \mu _{\alpha } (B_{n,\Delta })$with equality if and only if$G\cong B_{n,\Delta }$.

Proof

Let G be a graph among connected graphs of order n with maximum degree Δ such that $\mu _{\alpha } (G)$ is as large as possible. Then G has a spanning tree T with maximum degree Δ. By Lemma 2.3, $\mu _{\alpha }(G)\le \mu _{\alpha }(T)$ with equality if and only if $G\cong T$. Thus G is a tree.

The result is trivial if $n=3,4$ and if $\Delta =2, n-1$. Suppose that $3\leq \Delta \leq n-2$. We only need to show that $G\cong B_{n,\Delta }$.

Let $u\in V(G)$ with $\operatorname{deg}_{G}(u)=\Delta$. Suppose that there exists a vertex different from u with degree at least 3. Then we may choose such a vertex w of degree at least 3 such that $d_{G}(u,w)$ is as large as possible. Obviously, there are two pendant paths, say P and Q, at w of lengths at least 1. Let p and q be the lengths of P and Q, respectively. Assume that $p\ge q$. Let $H=G[V(G)\setminus ((V(P)\cup V(Q))\setminus \{w\})]$. Then $G\cong H_{w}(p,q)$. Note that $G'=H_{w}(p+1,q-1)$ is a tree of order n with maximum degree Δ. By Corollary 4.1, $\mu _{\alpha }(G)<\mu _{\alpha }(G')$, a contradiction. Then u is the unique vertex of G with degree at least 3, and thus G consists of Δ pendant paths, say $Q_{1}, \ldots , Q_{\Delta }$ at u. If two of them, say $Q_{i}$ and $Q_{j}$ with $i\ne j$ are of lengths at least 2, then $G\cong H'_{u}(r,s)$, where $H'=G[V(G)\setminus ((V(Q_{i})\cup V(Q_{j}))\setminus \{u\})]$, and r and s are the lengths of $Q_{i}$ and $Q_{j}$, respectively. Assume that $r\ge s$. Obviously, $G''=H'_{u}(r+1,s-1)$ is a tree of order n with maximum degree Δ. By Corollary 4.1, $\mu _{\alpha } (G)< \mu _{\alpha } (G'')$, also a contradiction. Thus there is exactly one pendant path at u of length at least 2, implying $G\cong B_{n,\Delta }$. □

If G is a connected graph of order 1 or 2, then $G\cong P_{n}$. If G is a connected graph of order 3, then $G\cong P_{3}$, $K_{3}$, and by Lemma 2.3, $\mu _{\alpha }(K_{3})<\mu _{\alpha }(P_{3})$.

Ruzieh and Powers [23] showed that $P_{n}$ is the unique connected graph of order n with maximum distance 0-spectral radius, and it was proved in [25] that $B_{n,3}$ is the unique tree of order n different from $P_{n}$ with maximum distance 0-spectral radius. For $\alpha =\frac{1}{2}$, the following theorem was given in [16].

Theorem 5.4

LetGbe a connected graph of order$n\geq 4$, where$G\ncong P_{n}$. Then$\mu _{\alpha }(G)\leq \mu _{\alpha }(B_{n,3})<\mu _{\alpha }(P_{n})$with equality if and only if$G\cong B_{n,3}$.

Proof

First suppose that G is a tree. If $n=4$, then the result follows from Theorem 4.1. Suppose that $n\geq 5$. Let Δ be the maximum degree of G. Since $G\ncong P_{n}$, we have $\Delta \geq 3$. By Theorem 5.3, $\mu _{\alpha }(G)\le \mu _{\alpha }(B_{n,\Delta })$ with equality if and only if $G\cong B_{n, \Delta }$. By Corollary 4.1, $\mu _{\alpha }(G)\le \mu _{\alpha }(B_{n,\Delta })\le \mu _{\alpha }(B_{n,3})< \mu _{\alpha }(P_{n})$ with equalities if and only if $\Delta =3$ and $G\cong B_{n, \Delta }$, i.e., $G\cong B_{n,3}$.

Now suppose that G is not a tree. Then G contains at least one cycle. If there is a spanning tree T with $T\ncong P_{n}$, then by Lemma 2.3 and the above argument, we have $\mu _{\alpha }(G)< \mu _{\alpha }(T)\leq \mu _{\alpha }(B_{n,3})$. If any spanning tree of G is a path, then G is a cycle $C_{n}$. Now we only need to show that $\mu _{\alpha }(C_{n})<\mu _{\alpha }(B_{n,3})$.

Let $C_{n}=u_{1}u_{2}\cdots u_{n}u_{1}$ and $T'=C_{n}-\{u_{1}u_{2}, u_{2}u_{3}\}+u_{2}u_{n}$. Then $T'\cong B_{n,3}$. Let x be the distance α-Perron vector of $C_{n}$. By Lemma 2.3, we have $x_{u_{1}}=\cdots =x_{u_{n}}$. As we pass from $C_{n}$ to $T'$, the distance between $u_{2}$ and $u_{1}$ is increased by 1, the distance between $u_{2}$ and $u_{i}$ with $3\leq i\leq \lceil \frac{n+1}{2} \rceil$ is increased by $n-2i+3$, the distance between $u_{2}$ and $u_{i}$ with $\lfloor \frac{n+1}{2} \rfloor +2\leq i\leq n$ is decreased by 1, and the distances between all other vertex pairs are increased or remain unchanged. Thus

$$\begin{aligned}& \mu _{\alpha }\bigl(T'\bigr)-\mu _{\alpha }(C_{n}) \\& \quad = x^{\top }\bigl(D_{\alpha }\bigl(T' \bigr)-D_{\alpha }(G)\bigr)x \\& \quad \ge \alpha \bigl(x_{u_{2}}^{2}+x_{u_{1}}^{2} \bigr)+2(1-\alpha )x_{u_{2}}x_{u_{1}} \\& \qquad {} -\sum_{i= \lfloor \frac{n+1}{2} \rfloor +2}^{n} \bigl( \alpha \bigl(x_{u_{2}}^{2}+x_{u_{i}}^{2} \bigr)+2(1-\alpha )x_{u_{2}}x_{u_{i}} \bigr) \\& \qquad {} +\sum_{i=3}^{ \lceil \frac{n+1}{2} \rceil }(n-2i+3) \bigl( \alpha \bigl(x_{u_{2}}^{2}+x_{u_{i}}^{2} \bigr)+2(1-\alpha )x_{u_{2}}x_{u_{i}} \bigr) \\& \quad = 2x_{u_{1}}^{2} \Biggl(1- \biggl(n- \biggl\lfloor \frac{n+1}{2} \biggr\rfloor -1 \biggr)+\sum_{i=3}^{ \lceil \frac{n+1}{2} \rceil }(n-2i+3) \Biggr) \\& \quad = 2x_{u_{1}}^{2} \biggl(1+ \biggl(n-1- \biggl\lceil \frac{n+1}{2} \biggr\rceil \biggr) \biggl( \biggl\lceil \frac{n+1}{2} \biggr\rceil -2 \biggr) \biggr) \\& \quad \ge 2x_{u_{1}}^{2} \\& \quad > 0, \end{aligned}$$

and therefore $\mu _{\alpha }(C_{n})<\mu _{\alpha }(B_{n,3})$, as desired. □

A clique of G is a subset of vertices whose induced subgraph is a complete graph, and the clique number of G is the maximum number of vertices in a clique of G. For $2\le \omega \le n$. Let $Ki_{n,\omega }$ be the graph obtained from a complete graph $K_{\omega }$ and a path $P_{n-\omega }$ by adding an edge between a vertex of $K_{\omega }$ and a terminal vertex of $P_{n-\omega }$ if $\omega < n$ and let $Ki_{n,\omega }=K_{n}$ if $\omega =n$. In particular, $Ki_{n,2}\cong P_{n}$ for $n\ge 2$. The following result for $\alpha =0,\frac{1}{2}$ was given in [15, 21].

Theorem 5.5

LetGbe a connected graph of order$n\geq 2$with clique number$\omega \geq 2$. Then$\mu _{\alpha } (G)\leq \mu _{\alpha } (Ki_{n,\omega })$with equality if and only if$G\cong Ki_{n,\omega }$.

Proof

It is trivial if $\omega =n$ and it follows from Theorem 5.4 if $\omega =2$.

Suppose that $3\le \omega \le n-1$. Let G be a graph among connected graphs of order n with clique number ω such that $\mu _{\alpha } (G)$ is as large as possible. We only need to show that $G\cong Ki_{n,\omega }$.

Let $S=\{v_{1},\ldots ,v_{\omega }\}$ be a clique of G. By Lemma 2.3, $G-E(G[S])$ is a forest. Let $T_{i}$ be the component of $G-E(G[S])$ containing $v_{i}$, where $1\leq i\leq \omega$. For $1\leq i\leq \omega$, by Corollary 4.1, if $T_{i}$ is nontrivial, then $T_{i}$ is a pendant path at $v_{i}$. Note that any two distinct vertices in $G[S]$ are adjacent. By Corollary 4.2, there is only one nontrivial $T_{i}$, and thus $G\cong Ki_{n,\omega }$. □

Recall that $Ki_{n,3}$ is the unique unicyclic graph of order $n\ge 3$ with maximum distance 0-spectral radius [31], and the unique odd-cycle unicyclic graph of order $n\ge 3$ with maximum distance $\frac{1}{2}$-spectral radius [15].

Theorem 5.6

LetGbe a unicyclic odd-cycle graph of order$n\ge 3$. Then$\mu _{\alpha } (G)\leq \mu (Ki_{n,3})$with equality if and only if$G\cong Ki_{n,3}$.

Proof

If $n=3,4$, the result is trivial. Suppose that $n\geq 5$. Let G be a graph with maximum distance α-spectral radius among unicyclic odd-cycle graphs of order n. We only need to show that $G\cong Ki_{n,3}$.

Let $C=v_{1} \cdots v_{2k+1}v_{1}$ be the unique cycle of G, where $k\ge 1$. Let $T_{i}$ be the component of $G-E(C)$ containing $v_{i}$ for $1\le i\le 2k+1$. Let $U_{1}= V(T_{2k})\cup V(T_{2k+1})$, $U_{2}=\bigcup_{k+1\le i\le 2k-1} V(T_{i})$ and $U_{3}=\bigcup_{1\le i\le k-1} V(T_{i})$. Let x be the distance α-Perron vector of G. Let

$$\begin{aligned} \varGamma =&\sum_{u\in U_{1} }\sum _{v\in U_{3}} \bigl(\alpha \bigl(x_{u}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{u} x_{v} \bigr) \\ &{}-\sum_{u\in U_{1} }\sum_{v\in U_{2}} \bigl(\alpha \bigl(x_{u}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{u} x_{v} \bigr). \end{aligned}$$

Suppose that $k\ge 2$. Let $G'=G-v_{1}v_{2k+1}+v_{2k+1}v_{2k-1}$. Note that the length of C is odd. As we pass from G to $G'$, the distance between a vertex in $S_{1}$ and a vertex in $S_{3}$ is increased by at least 1, the distance between $S_{2}$ and $V(T_{2k+1})$ is decreased by 1, and the distance between all other vertex pairs are increased or remain unchanged. Thus

$$\begin{aligned} \mu _{\alpha }\bigl(G'\bigr)-\mu _{\alpha }(G) \ge & x^{\top }\bigl(D_{\alpha }\bigl(G' \bigr)-D_{ \alpha }(G)\bigr)x \\ \ge &\sum_{u\in U_{1}}\sum _{v\in U_{3}} \bigl(\alpha \bigl(x_{u}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{u} x_{v} \bigr) \\ &{}-\sum_{u\in V(T_{2k+1})}\sum_{v\in U_{2}} \bigl(\alpha \bigl(x_{u}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{u} x_{v} \bigr) \\ >&\sum_{u\in U_{1}}\sum_{v\in U_{3}} \bigl(\alpha \bigl(x_{u}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{u} x_{v} \bigr) \\ &{}-\sum_{u\in U_{1}}\sum_{v\in U_{2}} \bigl(\alpha \bigl(x_{u}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{u} x_{v} \bigr). \end{aligned}$$

If $\varGamma \ge 0$, then $\mu _{\alpha }(G')>\mu _{\alpha }(G)$, a contradiction. Thus $\varGamma <0$. Let $G''=G-v_{2k}v_{2k-1}+v_{2k}v_{1}$. As we pass from G to $G''$, the distance between a vertex in $S_{1}$ and a vertex in $U_{2}$ is increased by at least 1, the distance between $U_{3}$ and $V(T_{2k})$ is decreased by 1, and the distance between all other vertex pairs are increased or remain unchanged. As above, we have

$$\begin{aligned} \mu _{\alpha }\bigl(G''\bigr)-\mu _{\alpha }(G) \ge & x^{\top }\bigl(D_{\alpha } \bigl(G''\bigr)-D_{ \alpha }(G)\bigr)x \\ \ge &\sum_{u\in U_{1}}\sum _{v\in U_{2}} \bigl(\alpha \bigl(x_{u}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{u} x_{v} \bigr) \\ &{}-\sum_{u\in V(T_{2k})}\sum_{v\in U_{3}} \bigl(\alpha \bigl(x_{u}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{u} x_{v} \bigr) \\ >&\sum_{u\in U_{1}}\sum_{v\in U_{2}} \bigl(\alpha \bigl(x_{u}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{u} x_{v} \bigr) \\ &{}-\sum_{u\in U_{1}}\sum_{v\in U_{3}} \bigl(\alpha \bigl(x_{u}^{2}+x_{v}^{2} \bigr)+2(1-\alpha )x_{u} x_{v} \bigr) \\ >& 0. \end{aligned}$$

Thus $\mu _{\alpha }(G'')>\mu _{\alpha }(G)$, also a contradiction. It follows that $k=1$, i.e., the unique cycle of G is of length 3.

Obviously, $T_{i}$ is a tree for $1\le i\le 3$. For $1\le i\le 3$, by Corollary 4.1, if $T_{i}$ is nontrivial, then it is a path with a terminal vertex $v_{i}$. Then by Corollary 4.2, only one $T_{i}$ is nontrivial. Thus $G\cong Ki_{n,3}$. □

Let G be a unicyclic graph of order $n\ge 4$ with maximum distance α-spectral radius. By Corollary 4.1, the maximum degree of G is 3 and all vertices of degree 3 lie on the unique cycle. Let u be a vertex of degree 3 and P be the pendant path at u. Let v and w be the two neighbors of u on the cycle, and z the neighbor of u on P. Let $G_{1}=G-uw+vw$ and $G_{2}=G-uw+wz$. Then $\mu _{\alpha }(G)<\max \{\mu _{\alpha }(G_{1}), \mu _{\alpha }(G_{2}) \}$ if the length of the cycle of G is odd, see [4, Lemma 6.11]. Note that the argument does not work when the length of the cycle of G is even. So we need other ways to determine the unicyclic graph(s) with maximum distance α-spectral radius even for $\alpha =\frac{1}{2}$.

6 Remarks

In this paper, we study the distance α-spectral radius of a connected graph. We consider bounds for the distance α-spectral radius, local transformations to change the distance α-spectral radius, and the characterizations for graphs with minimum and/or maximum distance α-spectral radius in some classes of connected graphs.

Besides the distance α-spectral radius, we may concern other eigenvalues of $D_{\alpha }(G)$ for a connected graph G. We give examples.

For an $n\times n$ Hermitian matrix C, let $\lambda _{1}(C), \ldots , \lambda _{n}(C)$ be the eigenvalues of C, arranged in a nonincreasing order. Let A, B be $n\times n$ Hermitian matrices. Weyl’s inequalities [13, p. 181] state that

$$ \lambda _{j}(A+B)\le \lambda _{i}(A)+\lambda _{j-i+1}(B)\quad \mbox{for }1\le i\le j\le n, $$

and

$$ \lambda _{j}(A+B)\ge \lambda _{i}(A)+\lambda _{j-i+n}(B) \quad \mbox{for }1\le j\le i\le n. $$

Using these inequalities, and as in the recent work of Atik and Panigrahi [3], we have

Theorem 6.1

LetGbe a connected graph andλbe any eigenvalue of$D_{\alpha }(G)$other than the distanceα-spectral radius. Then

$$ 2\alpha T_{\min }(G)-T_{\max }(G)+(1-\alpha ) (n-2)\le \lambda \le T_{ \max }(G)-(1-\alpha )n. $$

Proof

Let $D_{\alpha }(G)=A+B$, where $A=(\alpha T_{\min }(G)-(1-\alpha ))I_{n}+(1-\alpha )J_{n\times n}$. Then B is a nonnegative symmetric matrix with maximum row sum $T_{\max }(G)-\alpha T_{\min }(G)-(1-\alpha )(n-1)$. Thus $|\lambda _{n}(B)|\le \lambda _{1}(B)\le T_{\max }(G)-\alpha T_{\min }(G)-(1- \alpha )(n-1)$.

For matrix A, we have $\lambda _{1}(A)=\alpha T_{\min }(G)+(1-\alpha )(n-1)$ and $\lambda _{j}(A)=\alpha T_{\min }(G)-1+\alpha$ for $j=2,\ldots ,n$. Thus, for $j=2,\ldots , n$, we have by the above Weyl’s inequalities that

$$\begin{aligned} \lambda _{j}\bigl(D_{\alpha }(G)\bigr) \le & \lambda _{1}(B)+\lambda _{j}(A) \\ \le & T_{\max }(G)-\alpha T_{\min }(G)-(1-\alpha ) (n-1)+ \alpha T_{ \min }(G)-1+\alpha \\ =&T_{\max }(G)-(1-\alpha )n \end{aligned}$$

and

$$\begin{aligned} \lambda _{j}\bigl(D_{\alpha }(G)\bigr) \ge & \lambda _{n}(B)+\lambda _{j}(A) \\ \ge & -T_{\max }(G)+\alpha T_{\min }(G)+(1-\alpha ) (n-1)+ \alpha T_{ \min }(G)-1+\alpha \\ =& 2\alpha T_{\min }(G)-T_{\max }(G)+(1-\alpha ) (n-2). \end{aligned}$$

This completes the proof. □

Let G be a connected graph and λ be any eigenvalue of $D_{\alpha }(G)$ other than the distance α-spectral radius. By previous theorem, we have

$$ \vert \lambda \vert \le T_{\max }(G)-(1-\alpha ) (n-2). $$

The distance α-energy of a connected graph G of order n is defined as

$$ \mathcal{E}_{\alpha }(G)=\sum_{i=1}^{n} \biggl\vert \mu _{\alpha }^{(i)}(G)- \frac{2\alpha \sigma (G)}{n} \biggr\vert . $$

Then $\mathcal{E}_{0}(G)$ is the distance energy of G [14, 33], while

$$ \mathcal{E}_{1/2}(G)=\frac{1}{2}\sum _{i=1}^{n} \biggl\vert 2\mu _{1/2}^{(i)}(G)- \frac{2\sigma (G)}{n} \biggr\vert $$

is half of the distance signless Laplacian energy of G [8]. Thus, it is possible to study the distance energy and the distance signless Laplacian energy in a unified way.