# 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.

## 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 . Arisen from a data communication problem, the spectrum of the distance matrix was studied by Graham and Pollack  in 1971, early related work may be found also in [10, 11], and now it has been studied extensively, see the recent survey  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 , 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 . 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  and $$2\mu _{1/2}(G)$$ is just the distance signless Laplacian spectral radius of G .

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.

## 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 , 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 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

()

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

()

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.

## 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  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 .

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 , 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  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., .

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.  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 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 , 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 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 , 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. □

## 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 .

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 .

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

## 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  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

()

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  showed that $$P_{n}$$ is the unique connected graph of order n with maximum distance 0-spectral radius, and it was proved in  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 .

### 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 , and the unique odd-cycle unicyclic graph of order $$n\ge 3$$ with maximum distance $$\frac{1}{2}$$-spectral radius .

### 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 $$\alpha =\frac{1}{2}$$.

## 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 , 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 . Thus, it is possible to study the distance energy and the distance signless Laplacian energy in a unified way.

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### Acknowledgements

We would like to thank the referees and the editor for their suggestions and comments.

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## Funding

This work was supported by the National Natural Science Foundation of China (No. 11671156).

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Correspondence to Bo Zhou.

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