On the $A_\alpha$-spectral Radius of Graphs Without Large Matchings

Abstract

Let G be an n-vertex graph. A matching in G is a set of independent edges, i.e., no two edges in the set are adjacent in G. The matching number is the maximal size of a matching in G. Nikiforov (Appl Anal Discrete Math 11(1):81–107, 2017) proposed the $A_{\alpha }$-matrix of a graph G, as follows:

$$\begin{aligned} A_{\alpha }(G)=\alpha D(G)+(1- \alpha )A(G),~\alpha \in [0,1], \end{aligned}$$

where D(G) and A(G) are the diagonal degree matrix and adjacency matrix of G, respectively. In this contribution, we establish some $A_\alpha$-spectral conditions to guarantee that there do not exist large matchings in a graph G,  recovering the previous results and obtaining similar results for a wide variety of spectral conditions from the $A_{\alpha }$-matrix. Our main tools are the Berge–Tutte formula and the double leading eigenvectors.

1 Introduction

Throughout the paper, we only consider simple, finite and undirected graphs. Let G be a graph with vertex set $V(G)=\{v_1,v_2,\ldots ,v_n\}$ and edge set E(G). The order of G is the number $n=|V(G)|$ of its vertices. For a given subset $V_1\subseteq V(G),$ let $G-V_1$ denote the graph obtained from G by deleting the vertices in $V_1$ and their incident edges and let $G[V_1]$ denote the subgraph of G induced by $V_1$. For two graphs $G_1$ and $G_2$, we define $G_1\cup G_2$ to be their disjoint union. The join $G_1\vee G_2$ is obtained from $G_1\cup G_2$ by joining every vertex of $G_1$ with every vertex of $G_2$ by an edge.

For a vertex $v\in V(G),$ let $d_v$ denote the degree of v in G, which is the number of vertices adjacent to v. A graph is r-regular if each vertex has the same degree r.

For a graph G with $V{(G)}=\{v_1,\ldots ,v_n\}$, its adjacency matrix A(G) is an $n\times n$  0-1 matrix whose (i, j)-entry is 1 if and only if $v_iv_j\in E{(G)}$. Let D(G) denote the diagonal matrix of vertex degrees of G. Then, the signless Laplacian matrix of G is defined as $Q(G)=D(G)+A(G)$.

In 2017, Nikiforov [27] introduced the $A_\alpha$-matrix of G as $A_\alpha (G)=\alpha D(G)+(1-\alpha )A(G)$, where $0\leqslant \alpha \leqslant 1$. Ever since then, the study on this novel matrix attracted more and more researchers’ attention. Nikiforov [27] gave some fundamentals on $A_\alpha$-spectra. Lin, Xue and Shu [22] characterized the eigenvalues of $A_\alpha (G)$ for $\alpha >\frac{1}{2}$. Lin, Liu and Xue [21] characterized some types of graphs determined by their $A_\alpha$-spectra. Some relationships between the $A_\alpha$-eigenvalues and some other graph parameters are established (see [4, 12]). For more advances on $A_\alpha$-spectra, we refer the readers to [14, 15, 17,18,19,20] and the references cited therein.

Note that $A_0(G)=A(G)$ and $A_{\frac{1}{2}}(G)=\frac{1}{2}Q(G)$. Since $A_\alpha (G)$ is a real symmetric nonnegative matrix, its eigenvalues are real, which can be arranged in non-increasing order as $\lambda _1(A_\alpha (G))\geqslant \cdots \geqslant \lambda _n(A_\alpha (G)).$ For connected graph G,  by the Perron–Frobenius theorem, $\lambda _1(A_\alpha (G))$ is always positive (unless G is trivial) and $\lambda _1(A_\alpha (G))>|\lambda _i(A_\alpha (G))|$ for $i=2,3,\ldots ,n$, and we call $\lambda _1(A_\alpha (G))$ the $A_\alpha$-spectral radius of G, written as $\rho _\alpha (G)$. In particular, $\rho _0(G)$ is just the adjacency spectral radius of graph G, whereas $2\rho _\frac{1}{2}(G)$ equals the signless Laplacian spectral radius, say q(G), of graph G.

A subset $E_1\subseteq E(G)$ is called a matching if any two members of $E_1$ are not adjacent in G. A matching with the maximum size in G is called a maximum matching. A perfect matching in a graph G is a matching covering all vertices. The matching number $\mu (G)$ is the size of a maximum matching in G. As usual, let $K_n$ be the complete graph on n vertices.

Matching theory is an important topic in combinatorics and mathematical chemistry [26]. In particular, the large matchings in graphs attract much attention, see [1, 9, 11, 24] and the references cited in. In this paper, we focus on graphs without large matchings via the spectral graph theory.

In recent years, the relationships between the spectral conditions and the existence of matchings in graphs with given order have been studied extensively. In 2005, Brouwer and Haemers [2] established a lower bound on the third-largest eigenvalue to guarantee that a regular graph contains a perfect matching. Cioabǎ, Gregory and Haemers [5] improved the lower bound. Recently, O [28] established an adjacency spectral condition to guarantee that a graph has a perfect matching. Zhang and Lin [33] presented a distance spectral condition to guarantee the existence of a perfect matching in a graph. Zhao, Huang and Wang [34] gave a sufficient condition for the existence of a perfect matching in a graph in terms of the $A_\alpha$-spectral radius. Kim, O, Sim and Shin [10] determined the best lower bound for the spectral radius of an n-vertex connected graph to guarantee that the matching number $\mu (G)>\frac{n-k}{2}$, where k is a positive integer. For more advances on the relationship between eigenvalues and the existence of certain subgraphs in graphs, one may be referred to [6, 7, 13, 16, 23, 25, 29] and the references cited therein.

In this paper, we establish sharp upper bounds on the $A_\alpha$-spectral radius of an n-vertex connected graph G to ensure that $\mu (G)\leqslant \lceil \frac{n-k}{2}\rceil ,$ where k is a positive integer. Note that if $k=1$, then the graph having the largest $A_\alpha$-spectral radius among all n-vertex graphs with matching number at most $\lceil \frac{n-1}{2}\rceil$ is the complete graph $K_n$, whereas if $k=2$ and n is odd, then the graph having the largest $A_\alpha$-spectral radius among all n-vertex graphs with matching number at most $\lceil \frac{n-2}{2}\rceil$ is also the complete graph $K_n$. Therefore, in what follows we consider even n with $k\geqslant 2$ and odd n with $k\geqslant 3$.

Theorem 1.1

Let $\alpha$ be in [0, 1) and let G be an n-vertex connected graph with matching number at most $\lceil \frac{n-k}{2}\rceil$ satisfying $n\geqslant f(\alpha ,k)$, where

$$\begin{aligned} f(\alpha ,k)={\left\{ \begin{array}{ll} 3k+2, &{}\text {if } \alpha =0;\\ 5k,&{}\text {if } 0<\alpha \leqslant \frac{1}{2};\\ 7k,&{}\text {if } \frac{1}{2}<\alpha \leqslant \frac{2}{3};\\ \frac{3k-1}{1-\alpha },&{}\text {if } \frac{2}{3}<\alpha <1. \end{array}\right. } \end{aligned}$$

(1.1)

Then, $\rho _\alpha (G)\leqslant \theta _0$ with equality if and only if $G\cong K_1\vee (K_{2\lceil \frac{n-k}{2}\rceil -1}\cup (n-2\lceil \frac{n-k}{2}\rceil )K_1)$, where $\theta _0$ is the largest root of $x^3-(\alpha n+\alpha -2+2\lceil \frac{n-k}{2}\rceil )x^2+(\alpha ^2n-n-2\alpha +1+2\alpha n\lceil \frac{n-k}{2}\rceil )x-(\alpha -2)(2\alpha -1)n-\alpha -4(\alpha -1)^2 \lceil \frac{n-k}{2}\rceil ^2+2(n-2\alpha n+3\alpha ^2-4\alpha +2)\lceil \frac{n-k}{2}\rceil =0$.

Theorem 1.2

Let $\alpha$ be in [0, 1) and let G be an n-vertex connected graph with matching number at most $\frac{n-k}{2}$. If n and k have the same parity with $n\leqslant 3k,$ then $\rho _\alpha (G)\leqslant \theta _1$ with equality if and only if $G\cong K_{\frac{n-k}{2}}\vee (\frac{n+k}{2}K_1)$, where $\theta _1 =\frac{1}{4}[(2\alpha +1)n-k-2+g(n,k,\alpha )]$ with

$$\begin{aligned} g(n,k,\alpha )=\sqrt{(4\alpha ^2-8\alpha +5)n^2+(4\alpha k-2k-4)n+(k-2)(4\alpha k-3k-2)}. \end{aligned}$$

The proof techniques in the paper for our main results follow the idea of O [28]. Our main tools are the Berge–Tutte formula and the double leading eigenvectors.

Our paper is organized as follows: In the next section, we introduce some preliminary results, which will be used in the subsequent sections. In Sect. 3, we give the proof of Theorem 1.1. In Sect. 4, we give the proof of Theorem 1.2. In the last section, we recover the previous results and deduce some new ones from Theorems 1.1 and 1.2. Some further research problems are given as well.

2 Some Preliminaries

In this section, we give some necessary preliminary results, which will be used to prove our main results.

Lemma 2.1

[30] For a graph G, the matching number

$$\begin{aligned} \mu (G)=\min _{S\subseteq V(G)}\frac{n-o(G-S)+|S|}{2}, \end{aligned}$$

where $o(G-S)$ is the number of odd components in $G-S$.

Given an $n \times n$ matrix M, a principal submatrix of M is obtained by deleting the same set of rows and columns from M.

Lemma 2.2

(Cauchy Interlacing Theorem) Let A be a symmetric $n\times n$ matrix, and let B be an $m\times m$ principal submatrix of A for some $m<n$. If the eigenvalues of A are $\lambda _1\geqslant \lambda _2\geqslant \cdots \geqslant \lambda _n$, and the eigenvalues of B are $\mu _1\geqslant \mu _2\geqslant \cdots \geqslant \mu _m$, then $\lambda _i\geqslant \mu _i\geqslant \lambda _{i+n-m}$ for all $1\leqslant i\leqslant m$.

Cauchy Interlacing Theorem is a direct consequence of the Courant–Fischer–Weyl min–max principle. Fisk [8] gave a direct proof.

Let M be a real symmetric matrix whose rows and columns are indexed by $V=\{1, \ldots , n\}.$ Assume that M can be written as

$$\begin{aligned} M=\left( \begin{array}{ccc} M_{11} &{} \cdots &{} M_{1s} \\ \vdots &{} \ddots &{} \vdots \\ M_{s1} &{} \cdots &{} M_{ss} \\ \end{array} \right) \end{aligned}$$

according to partition $\pi :\ V=V_1\cup \cdots \cup V_s,$ wherein $M_{ij}$ denotes the submatrix (block) of M formed by rows in $V_i$ and the columns in $V_j.$ Let $m_{ij}$ denote the average row sum of $M_{ij}.$ Then, matrix $M_{\pi }=(m_{ij})$ is called the quotient matrix of M. If the row sum of each block $M_{ij}$ is a constant, then the partition is equitable.

Lemma 2.3

[3, 32] Let M be a real matrix with an equitable partition $\pi ,$ and let $M_{\pi }$ be the corresponding quotient matrix. Then, every eigenvalue of $M_{\pi }$ is an eigenvalue of M. Furthermore, if M is nonnegative, then the largest eigenvalues of M and $M_{\pi }$ are equal.

Lemma 2.4

[27] If G is connected, and H is a proper subgraph of G, then $\rho _\alpha (H)<\rho _\alpha (G).$

Lemma 2.5

[27] The eigenvalues of $A_\alpha (K_n)$ are $\lambda _1(A_\alpha (K_n))=n-1$ and $\lambda _k(A_\alpha (K_n))=\alpha n-1$ for $2\leqslant k\leqslant n$.

3 Proof of Theorem 1.1

In this section, we give the proof of Theorem 1.1, which establishes an upper bound on the $A_\alpha$-spectral radius for a connected graph G ensuring the matching number $\mu (G)\leqslant \lceil \frac{n-k}{2}\rceil$.

Proof of Theorem 1.1

Assume that G is connected and $\mu (G)\leqslant \left\lceil \frac{n-k}{2}\right\rceil$. By Lemma 2.1, we have

$$\begin{aligned} \mu (G)=\min _{S\subseteq V(G)}\frac{n-o(G-S)+|S|}{2}\leqslant \left\lceil \frac{n-k}{2}\right\rceil , \end{aligned}$$

which deduces that there exists some nonempty subset S of V(G) such that $\frac{n-o(G-S)+|S|}{2}\leqslant \lceil \frac{n-k}{2}\rceil$, i.e., $o(G-S)-|S|\geqslant n-2\lceil \frac{n-k}{2}\rceil$ and all components of $G-S$ are odd. Otherwise, we can remove one vertex from each even component to the set S. Consequently, the number of odd components and the number of vertices in S increase simultaneously.

For convenience, let $|S|=s$ and $o(G-S)=\ell$. Then, denote all the odd components of $G-S$ by $H_1, H_2,\ldots ,H_\ell$. Without loss of generality, assume that $|V(H_1)|=n_1\geqslant |V(H_2)|=n_2\geqslant \cdots \geqslant |V(H_\ell )|=n_\ell$. Note that $n=s+n_1+n_2+\cdots +n_\ell$. Hence, $\ell$ has the same parity with that of $n+s.$ Consequently, $\ell$ and $s+n-2\lceil \frac{n-k}{2}\rceil$ have the same parity.

Let $G_1$ be the graph obtained from G by adding edges to connect every vertex of S with every vertex of $G-S$, and by adding edges in S and in all components of $G-S$ so that all components in $G-S$ and G[S] are complete graphs. It is clear that $G_1\cong K_s\vee (K_{n_1}\cup K_{n_2}\cup \cdots \cup K_{n_\ell })$. Clearly, G is a subgraph of $G_1.$ Hence, $\rho _{\alpha }(G)\leqslant \rho _{\alpha }(G_1)$ with equality if and only if $G=G_1$ (based on Lemma 2.4).

In order to characterize the extremal graph having the largest $A_\alpha$-spectral radius, we need the following three claims.

Claim 1

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

$$\begin{aligned} \rho _{\alpha }(K_s\vee (K_{n_1}\cup K_{n_2}\cup \cdots \cup K_{n_\ell }))\leqslant \rho _{\alpha }(K_s\vee (K_{n-s-\ell +1}\cup (\ell -1)K_1)). \end{aligned}$$

Equality holds if and only if $(n_1,n_2,\ldots ,n_\ell )=(n-s-\ell +1,1,\ldots ,1)$.

Proof of Claim 1

As $G_1=K_s\vee (K_{n_1}\cup K_{n_2}\cup \cdots \cup K_{n_\ell })$, we may let $V(K_s)=\{v_1,\ldots ,v_s\}, V(K_{n_1})=\{v_{s+1},v_{s+2},\ldots ,v_{s+n_1}\}, \ldots , V(K_{n_\ell })=\{v_t,\ldots ,v_{n-1},v_n\}$, where $t=s+n_1+n_2+\cdots +n_{\ell -1}+1$.

Note that $n_i$ is odd for all $i\in \{1,\ldots ,\ell \}$. If $n_1=n-s-\ell +1$, $n_2=\cdots =n_\ell =1$, then the result holds obviously. If not, then there exists some $j\in \{2,\ldots ,\ell \}$ such that $n_j\geqslant 3$. Without loss of generality, we assume that $j=\ell$. In this case, it suffices to prove that $\rho _{\alpha }(K_s\vee (K_{n_1}\cup K_{n_2}\cup \cdots \cup K_{n_\ell }))<\rho _{\alpha }(K_s\vee (K_{n_1+2}\cup K_{n_2}\cup \cdots \cup K_{n_\ell -2})).$

Recall that $t=s+n_1+n_2+\cdots +n_{\ell -1}+1$. Then, construct a new graph $G_2$ as

$$\begin{aligned} G_2=G_1-\sum _{i=t}^{n-2}v_{n-1}v_i-\sum _{i=t}^{n-2}v_nv_i+ \sum _{i=s+1}^{s+n_1}v_{n-1}v_i+\sum _{i=s+1}^{s+n_1}v_nv_i, \end{aligned}$$

i.e., $G_2\cong K_s\vee (K_{n_1+2}\cup K_{n_2}\cup \cdots \cup K_{n_\ell -2})$.

Assume that $\varvec{x}=(x_1,\ldots ,x_i,\ldots , x_n)^T$ is the Perron vector of $A_\alpha (G_1)$, in which $x_i$ denotes the entry of $\varvec{x}$ corresponding to the vertex $v_i\in V(G_1)$. By symmetry (see also [27, Proposition 16]), it is easy to see that all vertices of $V(K_s)$ (resp. $V(K_{n_1}),V(K_{n_2}),\ldots$, $V(K_{n_\ell })$) have the same entries in $\varvec{x}$. Thus, we may assume $x_{v_i}=x_0$ for any $v_i\in V(K_s)$, $x_{v_i}=x_1$ for any $v_i\in V(K_{n_1}), \ldots , x_{v_i}=x_\ell$ for any $v_i\in V(K_{n_\ell })$. Then, by $A_\alpha (G_1)\varvec{x}=\rho _\alpha (G_1)\varvec{x}$, we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \rho _\alpha (G_1)x_1=(1-\alpha )sx_0+(s+n_1-1)\alpha x_1+(n_1-1)(1-\alpha )x_1,\\ \rho _\alpha (G_1)x_\ell =(1-\alpha )sx_0+(s+n_\ell -1)\alpha x_\ell +(n_\ell -1) (1-\alpha )x_\ell . \end{array}\right. } \end{aligned}$$

This gives us that

$$\begin{aligned}&{[}\rho _\alpha (G_1)-(s+n_1-1)\alpha -(n_1-1)(1-\alpha )]x_1\\&=[\rho _\alpha (G_1)-(s+n_\ell -1)\alpha -(n_\ell -1)(1-\alpha )]x_\ell . \end{aligned}$$

Notice that $n_1\geqslant n_\ell$. Hence, $\rho _\alpha (G_1)-(s+n_1-1)\alpha -(n_1-1)(1-\alpha )\leqslant \rho _\alpha (G_1)-(s+n_\ell -1)\alpha -(n_\ell -1)(1-\alpha )$, which implies $x_1\geqslant x_\ell$.

Recall that $t=s+n_1+n_2+\cdots +n_{\ell -1}+1$. Then, by Rayleigh quotient, one has

$$\begin{aligned} \rho _\alpha (G_2)-\rho _\alpha (G_1)\geqslant&\varvec{x}^T(A_\alpha (G_2)-A_\alpha (G_1))\varvec{x}\\ =&\sum _{i=s+1}^{s+n_1}2(1-\alpha )(x_{v_{n-1}}+x_{v_n})x_{v_i}\\&+\sum _{i=s+1}^{s+n_1}2\alpha x_{v_i}^2+(n_1-n_\ell +2)\alpha (x_{v_{n-1}}^2+x_{v_n}^2)\\&-\sum _{i=t}^{n-2}2(1-\alpha )(x_{v_{n-1}}+x_{v_n})x_{v_i} -\sum _{i=t}^{n-2}2\alpha x_{v_i}^2\\ =&\sum _{i=s+1}^{s+n_1}4(1-\alpha )x_\ell x_1+\sum _{i=s+1}^{s+n_1}2\alpha x_1^2+2(n_1-n_\ell +2)\alpha x_\ell ^2\\&-\sum _{i=t}^{n-2}4(1-\alpha )x_\ell ^2-\sum _{i=t}^{n-2}2\alpha x_\ell ^2 \\ =&\ 4x_\ell (1-\alpha )(n_1x_1-(n_\ell -2)x_\ell )\\&+2\alpha (n_1x_1^2-(n_\ell -2)x_\ell ^2)+2(n_1-n_\ell +2)\alpha x_\ell ^2\\ >&\ 0. \end{aligned}$$

The last inequality follows by $n_1\geqslant n_\ell , x_1\geqslant x_\ell$ and $\alpha \in [0,1).$ Hence, Claim 1 holds.

Claim 2

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

$$\begin{aligned}&\rho _{\alpha }(K_s\vee (K_{n-s-\ell +1}\cup (\ell -1)K_1))\\&\qquad \quad \leqslant \rho _{\alpha }(K_s\vee (K_{2\lceil (n-k)/2\rceil -2s+1}\cup (s+n-2\left\lceil (n-k)/2\right\rceil -1)K_1)). \end{aligned}$$

Equality holds if and only if $\ell =s+n-2\left\lceil (n-k)/2\right\rceil$.

Proof of Claim 2

Recall that $\ell$ and $s+n-2\lceil (n-k)/2\rceil$ have the same parity. If $\ell =s+n-2\lceil (n-k)/2\rceil$, then $K_s\vee (K_{n-s-\ell +1}\cup (\ell -1)K_1\cong K_s\vee (K_{2\lceil (n-k)/2\rceil -2s+1}\cup (s+n-2\lceil (n-k)/2\rceil -1)K_1)$ and the result holds obviously. If not, then $\ell \geqslant s+n-2\lceil (n-k)/2\rceil +2$. To prove Claim 2, it suffices to show that $\rho _{\alpha }(K_s\vee (K_{n-s-\ell +1}\cup (\ell -1)K_1))<\rho _{\alpha }(K_s\vee (K_{n-s-\ell +3}\cup (\ell -3)K_1))$.

For convenience, let $G_3=K_s\vee (K_{n-s-\ell +1}\cup (\ell -1)K_1)$ and partition its vertex set as $V(G_3)=V(K_s)\cup V(K_{n-s-\ell +1})\cup V((\ell -1)K_1)$, where $V(K_s)=\{u_1,u_2,\ldots ,u_s\}$, $V(K_{n-s-\ell +1})=\{v_1,v_2,\ldots ,v_{n-s-\ell +1}\}$ and $V((\ell -1)K_1)=\{w_1,w_2,\ldots ,w_{\ell -1}\}$.

Construct a new graph $G_4$ as follows:

$$\begin{aligned} G_4=G_3+w_1w_2+\sum _{i=1}^{n-s-\ell +1}w_1v_i+\sum _{i=1}^{n-s-\ell +1}w_2v_i. \end{aligned}$$

Clearly, $G_4\cong K_s\vee (K_{n-s-\ell +3}\cup (\ell -3)K_1)$. It is straightforward to check that $G_3$ is a proper spanning subgraph of $G_4$. By Lemma 2.4, we have $\rho _\alpha (G_3)<\rho _\alpha (G_4)$. So Claim 2 holds.

Claim 3

Assume $n\geqslant f(\alpha ,k)$ defined in (1.1), where $\alpha$ is in [0, 1). Then,

$$\begin{aligned}&\rho _{\alpha }(K_s\vee (K_{2\lceil (n-k)/2\rceil -2s+1}\cup (s+n-2\lceil (n-k)/2\rceil -1)K_1))\\&\qquad \leqslant \rho _{\alpha }(K_1\vee (K_{2\lceil (n-k)/2\rceil -1}\cup (n-2\lceil (n-k)/2\rceil )K_1)) \end{aligned}$$

with equality if and only if $s=1$.

Proof of Claim 3

If $s=1$, then the result holds obviously. If not, then $2\leqslant s\leqslant \lceil (n-k)/2\rceil$ for $n\geqslant 2s+n-2\lceil (n-k)/2\rceil$ and it suffices to show that

$$\begin{aligned}&\rho _{\alpha }(K_s\vee (K_{2\lceil (n-k)/2\rceil -2s+1}\cup (s+n-2\lceil (n-k)/2\rceil -1)K_1))\\&\qquad \quad <\rho _{\alpha }(K_1\vee (K_{2\lceil (n-k)/2\rceil -1}\cup (n-2\lceil (n-k)/2\rceil )K_1)). \end{aligned}$$

In fact, we prove it by considering the following two possible cases.

Case 1. n and k have the same parity. In this case, $\lceil (n-k)/2\rceil =(n-k)/2.$ Then,

$$\begin{aligned}&K_s\vee (K_{2\lceil (n-k)/2\rceil -2s+1}\cup (s+n-2\lceil (n-k)/2\rceil -1)K_1)\\&\qquad \quad =K_s\vee (K_{n-2s-k+1}\cup (s+k-1)K_1) \end{aligned}$$

and

$$\begin{aligned} K_1\vee (K_{2\lceil (n-k)/2\rceil -1}\cup (n-2\lceil (n-k)/2\rceil )K_1)=K_1\vee (K_{n-k-1}\cup k K_1). \end{aligned}$$

Let $G_5^s:=K_s\vee (K_{n-2s-k+1}\cup (s+k-1)K_1)$. Then, partition the vertex set $V(G_5^s)$ as $V(K_s)\cup V(K_{n-2s-k+1})\cup V((s+k-1)K_1)$. The quotient matrix of $A_\alpha (G_5^s)$ can be written as

$$\begin{aligned} B_5^s= \left( \begin{array}{ccc} n\alpha -s\alpha +s-1 &{} (n-2s-k+1)(1-\alpha ) &{} (s+k-1)(1-\alpha )\\ s(1-\alpha ) &{} n+s\alpha -2s-k &{} 0\\ s(1-\alpha ) &{} 0 &{} s\alpha \end{array} \right) . \end{aligned}$$

(3.1)

By a direct calculation, the characteristic polynomial of $B_5^s$ is

$$\begin{aligned} \Phi _{B_5^s}(x)=&x^3+[(1-\alpha )s-(1+\alpha )n+k+1]x^2\\&-[s^2-(\alpha ^2n-k-2\alpha +2)s-(n-k)(\alpha n-1)]x\\&+(3\alpha -2)(1-\alpha )s^3+[(2\alpha ^2-2\alpha +1)n-(4k-3)\alpha ^2\\&+(7k-6)\alpha -3k+2]s^2\\&+(n-k)(-\alpha ^2n+\alpha ^2k-2\alpha k+k-\alpha ^2+3\alpha -1)s. \end{aligned}$$

Clearly, the vertex partition of $G_5^s$ is equitable. By Lemma 2.3, the largest root of $\Phi _{B_5^s}(x)=0$ equals the $A_\alpha$-spectral radius of $G_5^s$.

Let $\sigma _1=\rho _\alpha (G_5^s)\geqslant \sigma _2\geqslant \sigma _3$ be the three roots of $\Phi _{B_5^s}(x)=0$. We claim that $\sigma _2\leqslant n+s\alpha -2s-k$. In fact, let $\mathcal {D}=diag(s,n-2s-k+1,s+k-1)$. It is easy to check that $\mathcal {D}^\frac{1}{2}B_5^s\mathcal {D}^{-\frac{1}{2}}$ is symmetric, and also contains

$$\begin{aligned} \left( \begin{array}{cc} n+s\alpha -2s-k &{} 0\\ 0 &{} s\alpha \end{array} \right) \end{aligned}$$

as its principal submatrix. Obviously, the $A_\alpha$-spectrum of $\mathcal {D}^\frac{1}{2}B_5^s\mathcal {D}^{-\frac{1}{2}}$ is the same as that of $B_5^s$. By the Cauchy Interlacing Theorem, we obtain $\sigma _2\leqslant n+s\alpha -2s-k\leqslant \sigma _1.$ Note that $n+s\alpha -2s-k<n-s-k\leqslant n-k-1$. Then, $\sigma _2<n-k-1$. On the other hand, $K_{n-k}$ is a proper subgraph of $G_5^1=K_1\vee (K_{n-k-1}\cup kK_1)$. Together with Lemmas 2.4 and 2.5, one has $n-k-1=\rho _\alpha (K_{n-k})<\rho _\alpha (G_5^1).$ Therefore,

$$\begin{aligned} \sigma _{2}<n-k-1<\rho _{\alpha }(G_5^1). \end{aligned}$$

(3.2)

By a direct calculation, we obtain the spectral radius of $G_5^s$ for $s=\frac{n-k}{2}$ as follows:

$$\begin{aligned} \rho _\alpha (G_5^{\frac{n-k}{2}}) =\frac{1}{4}[(2\alpha +1)n-k-2+g(n,k,\alpha )], \end{aligned}$$

(3.3)

where $g(n,k,\alpha )=\sqrt{(4\alpha ^2-8\alpha +5)n^2+(4\alpha k-2k-4)n+(k-2)(4\alpha k-3k-2)}.$

In order to complete the proof of Claim 3 in this case, we proceed by considering the following four subcases.

Subcase 1.1. $\alpha =0$ and $n\geqslant 3k+2$. In this subcase, we are to show that $\rho _0(G_5^s)<\rho _0(G_5^1)$ for $2\leqslant s\leqslant (n-k)/2$.

Since $\sigma _{2}<\rho _0(G_5^1)$, it suffices to prove that $\Phi _{B_5^s}(\rho _0(G_5^1))>0$ for $2\leqslant s\leqslant (n-k)/2$. By a simple calculation, one has

$$\begin{aligned} \Phi _{B_5^s}(x)-\Phi _{B_5^1}(x)=(s-1)[x^2-(s+k-1)x+(s+k)n-k^2-(3s+2)k-2s^2]. \end{aligned}$$

Let $\phi _1(x)=x^2-(s+k-1)x+(s+k)n-k^2-(3s+2)k-2s^2$ be a real function in x. Then, $\phi '_1(x)=2x-(s+k-1)$. It is easy to see that

$$\begin{aligned} \phi '_1(n-k-1)= & {} 2n-3k-1-s\geqslant 2n-3k-1-\frac{n-k}{2}\\= & {} \frac{3}{2}n-\frac{5}{2}k-1\geqslant \frac{3}{2}(3k+2)-\frac{5}{2}k-1=2k+2>0. \end{aligned}$$

Clearly, $\phi '_1(x)\geqslant \phi '_1(n-k-1)>0$ for $x\geqslant n-k-1$. This implies that $\phi _1(x)$ is a monotonically increasing function for $x\geqslant n-k-1$. Hence, $\phi _1(\rho _0(G_5^1))>\phi _1(n-k-1)$. Note that

$$\begin{aligned} \phi _1(n-k-1)&=-2s^2-(2k-1)s+n^2-(2k+1)n+k^2\nonumber \\&\geqslant -2\left( (n-k)/2\right) ^2-(2k-1)(n-k)/2+n^2-(2k+1)n+k^2 \end{aligned}$$

(3.4)

$$\begin{aligned}&=\frac{1}{2}[n^2-(4k+1)n+3k^2-k]\nonumber \\&\geqslant \frac{1}{2}[(3k+2)^2-(4k+1)(3k+2)+3k^2-k]\\&=1,\nonumber \end{aligned}$$

(3.5)

where inequality in (3.4) follows by $\frac{2k-1}{4}>0$ and $2\leqslant s\leqslant (n-k)/2$, inequality in (3.5) follows by $\frac{4k+1}{2}<3k+2\leqslant n$. Thus, $\phi _1(\rho _0(G_5^1))>\phi _1(n-k-1)>0$. Since $\Phi _{B_5^s}(x)-\Phi _{B_5^1}(x)=(s-1)\phi _1(x)$ and $\Phi _{B_5^1}(\rho _0(G_5^1))=0$, we have $\Phi _{B_5^s}(\rho _0(G_5^1))=\Phi _{B_5^s}(\rho _0(G_5^1))-\Phi _{B_5^1}(\rho _0(G_5^1))=(s-1)\phi _1(\rho _0(G_5^1))>0$ for $2\leqslant s\leqslant (n-k)/2$. Hence, the result holds for $\alpha =0$ and $n\geqslant 3k+2$.

Subcase 1.2. $0<\alpha \leqslant \frac{1}{2}$ and $n\geqslant 5k$.

Firstly, we consider $2\leqslant s\leqslant \frac{n-k}{2}-1$. Note that $\sigma _{2}<n-k-1<\rho _{\alpha }(G_{5}^{1})$. Hence, for $2\leqslant s\leqslant \frac{n-k}{2}-1$, in order to show that $\rho _{\alpha }(G_5^s)<\rho _{\alpha }(G_{5}^{1}),$ it suffices to prove $\Phi _{B_{5}^{s}}(\rho _{\alpha }(G_{5}^{1}))>0$. A simple calculation yields that $\Phi _{B_5^s}(x)-\Phi _{B_5^1}(x)=(s-1)\phi _2(x),$ where

$$\begin{aligned} \phi _2(x)=&(1-\alpha )x^2+(n\alpha ^2-2\alpha -s-k+1)x-\alpha ^2n^2+((2s+2k+1)\alpha ^2\nonumber \\&-(2s+2k-1)\alpha +s+k)n-(3s^2+4ks+k^2+3k)\alpha ^2\nonumber \\&+(5s^2+7ks-s+2k^2+4k-1)\alpha -2s^2-3ks-k^2-2k \end{aligned}$$

(3.6)

is a real function in x. Then, $\phi '_2(x)=2(1-\alpha )x+n\alpha ^2-2\alpha -s-k+1$. So we obtain

$$\begin{aligned} \phi '_2(n-k-1)&=(\alpha ^2-2\alpha +2)n+(2\alpha -3)k-s-1\\&\geqslant (\alpha ^2-2\alpha +2)n+(2\alpha -3)k-\left( (n-k)/2-1\right) -1\\&=\left( \alpha ^2-2\alpha +\frac{3}{2}\right) n+\left( 2\alpha -\frac{5}{2}\right) k\\&\geqslant \left( \alpha ^2-2\alpha +\frac{3}{2}\right) \cdot 5k+\left( 2\alpha -\frac{5}{2}\right) k\\&=(5\alpha ^2-8\alpha +5)k\\&>0. \end{aligned}$$

Clearly, $\phi '_2(x)\geqslant \phi '_2(n-k-1)>0$ for $x\geqslant n-k-1$. Hence, we obtain that $\phi _2(x)$ is a monotonically increasing function for $x\geqslant n-k-1,$ and so $\phi _2(\rho _\alpha (G_5^1))>\phi _2(n-k-1)$. Let $\phi _2(n-k-1)=\phi _3(s)$, where

$$\begin{aligned} \phi _3(s)=&-(3\alpha ^2-5\alpha +2)s^2+((2\alpha ^2-2\alpha )n-4k\alpha ^2+(7k-1)\alpha -2k+1)s\nonumber \\&+(1-\alpha )n^2+(k\alpha ^2+\alpha -2k-1)n-(k^2+3k)\alpha ^2+(k^2+4k)\alpha +k^2. \end{aligned}$$

(3.7)

Now we consider the real function $\phi _3(x)$ with $2\leqslant x\leqslant \frac{n-k}{2}-1$. The derivative of $\phi _3(x)$ is given by

$$\begin{aligned} \phi '_3(x)=&-2(3\alpha ^2-5\alpha +2)x+(2\alpha ^2-2\alpha )n-4k\alpha ^2&+(7k-1)\alpha -2k+1.\qquad \quad \end{aligned}$$

(3.8)

Since $-2(3\alpha ^2-5\alpha +2)<0$ for $0<\alpha \leqslant \frac{1}{2}$, one has $\phi '_3(x)\leqslant \phi '_3(2)=2(n-2k-6)\alpha ^2-(2n-7k-19)\alpha -2k-7$. Let

$$\begin{aligned} \phi _4(\alpha ):=2(n-2k-6)\alpha ^2-(2n-7k-19)\alpha -2k-7 \end{aligned}$$

(3.9)

be a real function in $\alpha ,$ where $0\leqslant \alpha \leqslant \frac{1}{2}$. Note that $2(n-2k-6)\geqslant 2(5k-2k-6)\geqslant 0$ for $n\geqslant 5k$. Consequently, $\phi _4(\alpha )\leqslant \max \{\phi _4(0),\phi _4(\frac{1}{2})\}$. By a simple calculation, one has $\phi _4(0)=-2k-7<0$ and $\phi _4(\frac{1}{2})=-\frac{1}{2}(n-k+1)\leqslant -\frac{1}{2}(5k-k+1)<0$ for $n\geqslant 5k$. Then, $\phi '_3(2)=\phi _4(\alpha )<0$ for $0<\alpha \leqslant \frac{1}{2}$. Therefore, $\phi '_3(x)\leqslant \phi '_3(2)<0$. Consequently, $\phi _3(x)$ is a monotonically decreasing function for $2\leqslant x\leqslant \frac{n-k}{2}-1$. Hence, $\phi _3(\frac{n-k}{2}-1)\leqslant \phi _3(s)\leqslant \phi _3(2)$. Let $\phi _3(\frac{n-k}{2}-1)=\phi _5(\alpha ),$ where

$$\begin{aligned} \phi _5(\alpha )=&\frac{1}{4}[(n^2-(2k-4)n+k^2 -8k-12)\alpha ^2-(3n^2-(8k-10)n\nonumber \\&+5k^2-10k-24)\alpha +2(n^2-(4k-3)n+3k^2-k-6)]. \end{aligned}$$

(3.10)

Next we consider the real function $\phi _5(\alpha )$ in $\alpha$ with $0\leqslant \alpha \leqslant \frac{1}{2}$. The derivative of $\phi _5(\alpha )$ is $\phi '_5(\alpha )=\frac{1}{4}[2(n^2-(2k-4)n+k^2-8k-12)\alpha -(3n^2-(8k-10)n+5k^2-10k-24)]$. Since

$$\begin{aligned}&n^2-(2k-4)n+k^2-8k-12\geqslant (5k)^2\\&-(2k-4)5k+k^2-8k-12=4(4k^2+3k-3)>0, \end{aligned}$$

where the first inequality follows from $\frac{2k-4}{2}<5k\leqslant n$. So we obtain $\phi '_5(\alpha )\leqslant \phi '_5(\frac{1}{2})$. By a simple calculation, one has $\phi '_5(\frac{1}{2})=-\frac{1}{2}[n^2-(3k-3)n+2k^2-k-6]\leqslant -\frac{1}{2}[(5k)^2-(3k-3)5k+2k^2-k-6]=-(6k^2+7k-3)<0,$ where the first inequality holds since $\frac{3k-3}{2}<5k\leqslant n$. So we obtain $\phi '_5(\alpha )<0$. Consequently, $\phi _5(\alpha )$ is a monotonically decreasing function for $0\leqslant \alpha \leqslant \frac{1}{2}$. Thus, $\phi _5(\frac{1}{2})\leqslant \phi _5(\alpha )<\phi _5(0)$ for $0<\alpha \leqslant \frac{1}{2}$, where

$$\begin{aligned} \phi _5\left( \frac{1}{2}\right)&=\frac{1}{16}[3n^2-(18k-8)n+15k^2+4k-12]\nonumber \\&\geqslant \frac{1}{16}[3\cdot (5k)^2-(18k-8)\cdot 5k+15k^2+4k-12]\\&=\frac{1}{4}(11k-3)\nonumber \\&>0.\nonumber \end{aligned}$$

(3.11)

The inequality in (3.11) follows by $\frac{18k-8}{6}<5k\leqslant n$. It is easy to see that $\phi _3(\frac{n-k}{2}-1)=\phi _5(\alpha )\geqslant \phi _5(\frac{1}{2})>0$ for $0<\alpha \leqslant \frac{1}{2}$. On the other hand, $\phi _2(\rho _\alpha (G_5^1))>\phi _2(n-k-1)=\phi _3(s)\geqslant \phi _3(\frac{n-k}{2}-1)$. Therefore, $\phi _2(\rho _\alpha (G_5^1))>0$. Bearing in mind that $\Phi _{B_5^s}(x)-\Phi _{B_5^1}(x)=(s-1)\phi _2(x)$ and $\Phi _{B_5^1}(\rho _{\alpha }(G_5^1))=0$, one has $\Phi _{B_5^s}(\rho _{\alpha }(G_5^1))=\Phi _{B_5^s}(\rho _{\alpha }(G_5^1))- \Phi _{B_5^1}(\rho _{\alpha }(G_5^1))=(s-1)\phi _2(\rho _\alpha (G_5^1))>0$ for $2\leqslant s\leqslant \frac{n-k}{2}-1$. Hence, $\Phi _{B_5^s}(\rho _{\alpha }(G_5^1))>0$ holds for $2\leqslant s\leqslant \frac{n-k}{2}-1$.

Now we consider $s=\frac{n-k}{2}$. In what follows we show that $\rho _\alpha (G_5^{\frac{n-k}{2}})<n-k-1$ if $n\geqslant 5k+2$ and $\Phi _{B_5^1}(\rho _\alpha (G_5^{\frac{n-k}{2}}))<0$ if $n=5k$, where $\rho _\alpha (G_5^{\frac{n-k}{2}})$ is given in (3.3).

In order to show $\rho _\alpha (G_5^{\frac{n-k}{2}})<n-k-1$ for $n\geqslant 5k+2$, it suffices to show $\zeta (n,k,\alpha )>0,$ where

$$\begin{aligned} \zeta (n,k,\alpha )=(-n^2+(2k+2)n-k^2+2k)\alpha +n^2-(4k+2)n+3k^2+2k.\nonumber \\ \end{aligned}$$

(3.12)

In fact,

$$\begin{aligned} \zeta (n,k,\alpha )&\geqslant \frac{1}{2}(-n^2+(2k+2)n-k^2+2k)+n^2-(4k+2)n+3k^2+2k \end{aligned}$$

(3.13)

$$\begin{aligned}&=\frac{1}{2}(n^2-(6k+2)n+5k^2+6k)\nonumber \\&\geqslant \frac{1}{2}((5k+2)^2-(6k+2)(5k+2)+5k^2+6k)\\&=2k>0,\nonumber \end{aligned}$$

(3.14)

where inequality in (3.13) follows by $-n^2+(2k+2)n-k^2+2k\leqslant -(5k)^2+(2k+2)\cdot 5k-k^2+2k=-4k(4k-3)<0$ and the one in (3.14) follows by $\frac{6k+2}{2}<5k+2\leqslant n$. Together with (3.2), we obtain $\rho _\alpha (G_5^{\frac{n-k}{2}})< \rho _\alpha (G_5^1)$, as desired.

In view of (3.1) and by some calculations, we see that

$$\begin{aligned} \Phi _{B_5^1}(x)=&x^3-[(1+\alpha )n+\alpha -k-2]x^2+[\alpha n^2 +(\alpha ^2-\alpha k-1)n-2\alpha +1]x\nonumber \\&-\alpha ^2n^2+[(2k+1)\alpha ^2-(2k-1)\alpha +k]n-(3+k)k\alpha ^2\nonumber \\&+(2k^2+4k-1)\alpha -k(k+2). \end{aligned}$$

(3.15)

If $n=5k,$ then together with (3.3) and (3.15) we obtain $\Phi _{B_5^1}(\rho _\alpha (G_5^{2k}))=-\frac{1}{2}(2k-1)P(\alpha ,k),$ where

$$\begin{aligned} P(\alpha ,k)=&(3\alpha k-\alpha -k)\sqrt{25k^2 \alpha ^2-(44k^2+2k)\alpha +28k^2-4k+1}\\&+(15k^2-5k)\alpha ^2-(31k^2-13k+1)\alpha +10k^2-3k. \end{aligned}$$

With the help of Mathematica [31] we find that $\min \{P(\alpha ,k)|0<\alpha \leqslant \frac{1}{2},\ k\geqslant 2\}=\sqrt{10}-3>0$. Then, one has $\Phi _{B_5^1}(\rho _\alpha (G_5^{2k}))=-\frac{1}{2}(2k-1)P(\alpha ,k)<0$ for $0<\alpha \leqslant \frac{1}{2}$ and $k\geqslant 2$, which implies that $\rho _\alpha (G_5^{2k})<\rho _\alpha (G_5^1)$, as desired.

Hence, our result holds in this subcase.

Subcase 1.3. $\frac{1}{2}<\alpha \leqslant \frac{2}{3}$ and $n\geqslant 7k$. Along the line of Subcase 1.2, we may show our result in this subcase similarly.

Firstly, we consider $2\leqslant s \leqslant \frac{n-k}{2}-1$. Similar to Subcase 1.2, one has $\Phi _{B_5^s}(x)-\Phi _{B_5^1}(x)=(s-1)\phi _2(x),$ where $\phi _2(x)$ is given in (3.6), in which $\frac{1}{2}<\alpha \leqslant \frac{2}{3}$ and $n\geqslant 7k$. Therefore,

$$\begin{aligned} \phi '_2(n-k-1)&=(\alpha ^2-2\alpha +2)n+(2\alpha -3)k-s-1\\&\geqslant (\alpha ^2-2\alpha +2)n+(2\alpha -3)k-\left( \frac{n-k}{2}-1\right) -1\\&\qquad \left( \text {As } s \leqslant \frac{n-k}{2}-1\right) \\&\geqslant \left( \alpha ^2-2\alpha +\frac{3}{2}\right) \cdot 7k+\left( 2\alpha -\frac{5}{2}\right) k\\&\qquad \left( \text {As } \alpha ^2-2\alpha +\frac{3}{2}>0, n\geqslant 7k\right) \\&=(7\alpha ^2-12\alpha +8)k>0. \end{aligned}$$

Obviously, $\phi '_2(x)\geqslant \phi '_2(n-k-1)>0$ for $x\geqslant n-k-1$. Hence, under the condition $\frac{1}{2}<\alpha \leqslant \frac{2}{3}$ and $n\geqslant 7k$, one obtains $\phi _2(x)$ is a monotonically increasing function for $x\geqslant n-k-1$. So we obtain $\phi _2(\rho _\alpha (G_5^1))>\phi _2(n-k-1)=\phi _3(s)$, where $\phi _3(s)$ is given in (3.7). Let $\phi _3(x)$ be a real function in x with $2\leqslant x\leqslant \frac{n-k}{2}-1$. The derivative of $\phi _3(x)$, written as $\phi '_3(x)$, is given in (3.8). Since $-2(3\alpha ^2-5\alpha +2)\leqslant 0$ for $\frac{1}{2}<\alpha \leqslant \frac{2}{3}$, one has $\phi '_3(x)\leqslant \phi '_3(2)=\phi _4(\alpha )$, which is given in (3.9). Bear in mind all the $\phi _3(x),\, \phi '_3(x)$ and $\phi _4(\alpha )$ satisfy $\frac{1}{2}<\alpha \leqslant \frac{2}{3}$ and $n\geqslant 7k$.

In what follows we consider the real function $\phi _4(\alpha )$ with $\frac{1}{2}\leqslant \alpha \leqslant \frac{2}{3}$. In view of (3.9), one has $2(n-2k-6)\geqslant 2(7k-2k-6)>0$ for $n\geqslant 7k$. This gives $\phi _4(\alpha )\leqslant \max \{\phi _4(\frac{1}{2}),\phi _4(\frac{2}{3})\}$. By a simple calculation, one has

$$\begin{aligned} \phi _4\left( \frac{1}{2}\right) =-\frac{1}{2}(n-k+1)\leqslant -\frac{1}{2}(7k-k+1)<0 \end{aligned}$$

and

$$\begin{aligned} \phi _4\left( \frac{2}{3}\right) =-\frac{1}{9}(4n-8k-3)\leqslant -\frac{1}{9}(28k-8k-3)<0 \end{aligned}$$

for $n\geqslant 7k$. Then, $\phi '_3(2)=\phi _4(\alpha )<0$ for $\frac{1}{2}<\alpha \leqslant \frac{2}{3}$. Thus, $\phi '_3(x)\leqslant \phi '_3(2)<0$, which implies that $\phi _3(x)$ is a monotonically decreasing function for $2\leqslant x\leqslant \frac{n-k}{2}-1$. Hence, $\phi _5(\alpha )=\phi _3(\frac{n-k}{2}-1)\leqslant \phi _3(s)\leqslant \phi _3(2)$, where $\phi _5(\alpha )$ is given in (3.10) satisfying $\frac{1}{2}<\alpha \leqslant \frac{2}{3}$. It is straightforward to check that $\phi _5(\alpha )$ is a monotonically decreasing function for $\frac{1}{2}<\alpha \leqslant \frac{2}{3}$. Thus, $\phi _5(\frac{2}{3})\leqslant \phi _5(\alpha )<\phi _5(\frac{1}{2})$. Note that

It is easy to see that $\phi _3(\frac{n-k}{2}-1)=\phi _5(\alpha )\geqslant \phi _5(\frac{2}{3})>0$ and $\phi _2(\rho _\alpha (G_5^1))>\phi _2(n-k-1)=\phi _3(s)\geqslant \phi _3(\frac{n-k}{2}-1)$. Therefore, $\phi _2(\rho _\alpha (G_5^1))>0$. Since $\Phi _{B_5^s}(x)-\Phi _{B_5^1}(x)=(s-1)\phi _2(x)$ and $\Phi _{B_5^1}(\rho _{\alpha }(G_5^1))=0$, one has $\Phi _{B_5^s}(\rho _{\alpha }(G_5^1))=\Phi _{B_5^s}(\rho _{\alpha }(G_5^1)) -\Phi _{B_5^1}(\rho _{\alpha }(G_5^1))=(s-1)\phi _2(\rho _\alpha (G_5^1))>0$ for $2\leqslant s\leqslant \frac{n-k}{2}-1$. Hence, our result holds for $2\leqslant s\leqslant \frac{n-k}{2}-1$.

Now we consider $s=\frac{n-k}{2}$. In what follows, we shall prove that $\rho _\alpha (G_5^{\frac{n-k}{2}})<n-k-1$ if $n\geqslant 7k+2$ and $\Phi _{B_5^1}(\rho _\alpha (G_5^{\frac{n-k}{2}}))<0$ if $n=7k$, where $\rho _\alpha (G_5^{\frac{n-k}{2}})$ is given in (3.3).

In order to show $\rho _\alpha (G_5^{\frac{n-k}{2}})<n-k-1$ for $n\geqslant 7k+2,$ it suffices to show $\zeta (n,k,\alpha )>0,$ where $\zeta (n,k,\alpha )$ is given in (3.12). In fact,

$$\begin{aligned} \zeta (n,k,\alpha )\geqslant&\frac{2}{3}(-n^2+(2k+2)n-k^2+2k)+n^2-(4k+2)n+3k^2+2k \end{aligned}$$

(3.16)

$$\begin{aligned} =&\frac{1}{3}(n^2-(8k+2)n+7k^2+10k)\nonumber \\ \geqslant&\frac{1}{3}[(7k+2)^2-(8k+2)(7k+2)+7k^2+10k] \\ =&\frac{8k}{3}>0,\nonumber \end{aligned}$$

(3.17)

where inequality in (3.16) follows by $\alpha \leqslant \frac{2}{3}$ and $-n^2+(2k+2)n-k^2+2k\leqslant -(7k)^2+(2k+2)\cdot 7k-k^2+2k=-4k(9k-4)<0$ and inequality in (3.17) follows by $\frac{8k+2}{2}<7k+2\leqslant n$. Together with (3.2), we obtain $\rho _\alpha (G_5^{\frac{n-k}{2}})<\rho _\alpha (G_5^1)$, as desired.

If $n=7k,$ then together with (3.3) and (3.15) we obtain $\Phi _{B_5^1}(\rho _\alpha (G_5^{3k}))=-\frac{1}{2}(3k-1)H(\alpha ,k),$ where

$$\begin{aligned} H(\alpha ,k)&=(4\alpha k-\alpha -k)\sqrt{49k^2\alpha ^2-(90k^2+2k) \alpha +57k^2-6k+1}\\&\quad +(28k^2-5k)\alpha ^2-(61k^2-15k+1)\alpha +21k^2-3k. \end{aligned}$$

Similarly, with the help of Mathematica [31] we find that $\min \{H(\alpha ,k)|\frac{1}{2}<\alpha \leqslant \frac{2}{3},\ k\geqslant 2\}=\frac{4}{9}(2\sqrt{553}-45)\approx 0.9031>0$. Then, one has $\Phi _{B_5^1}(\rho _\alpha (G_5^{3k}))=-\frac{1}{2}(3k-1)H(\alpha ,k)<0$ for $\frac{1}{2}<\alpha \leqslant \frac{2}{3}$ and $k\geqslant 2$, which implies that $\rho _\alpha (G_5^{3k})<\rho _\alpha (G_5^1)$, as desired.

Hence, our result holds in this subcase.

Subcase 1.4. $\frac{2}{3}<\alpha <1$ and $n\geqslant \frac{3k-1}{1-\alpha }$. In this subcase, in order to show, for $2\leqslant s\leqslant (n-k)/2$, that $\rho _{\alpha }(G_5^s)<\rho _{\alpha }(G_5^1)$, it suffices to show $\Phi _{B_5^s}(n-k-1)>0$ (based on (3.2)). In fact, let $\Phi _{B_5^s}(n-k-1)=\psi (s),$ where

$$\begin{aligned} \psi (s)&=(3\alpha -2)(1-\alpha )s^{3}+[(2\alpha ^2-2\alpha )n-(4k-3)\alpha ^2+(7k-6)\alpha -2k+3]s^2\nonumber \\&\quad +[(1-\alpha )n^2+((k-2)\alpha ^2+3\alpha -2k-1)n\nonumber \\&\quad -(k^2-k)\alpha ^2+(k^2-3k+1)\alpha +k^2+2k-1]s-(1-\alpha )n^2\nonumber \\&\quad -[(k+1)\alpha -2k-1]n-k^2-k. \end{aligned}$$

(3.18)

Let $\psi (x)$ be a real function in x with $2\leqslant x\leqslant \frac{n-k}{2}$. Then, the first derivative of $\psi (x)$ is

$$\begin{aligned} \psi '(x)&=3(3\alpha -2)(1-\alpha )x^{2}+2[(2\alpha ^2-2\alpha )n-(4k-3)\alpha ^2\nonumber \\&\quad +(7k-6)\alpha -2k+3]x+[(1-\alpha )n^2+((k-2)\alpha ^2+3\alpha -2k-1)n\nonumber \\&\quad -(k^2-k)\alpha ^2+(k^2-3k+1)\alpha +k^2+2k-1]. \end{aligned}$$

(3.19)

In what follows, we show

$$\begin{aligned} \psi '(2)>0,\quad \ \psi '\left( \frac{n-k}{2}\right) <0. \end{aligned}$$

(3.20)

In view of (3.19), one has $\psi '(2)=(1-\alpha )n^{2}+[(k+6)\alpha ^2-5\alpha -2k-1]n-(k^2+15k+24)\alpha ^2+(k^2+25k+37)\alpha +k^2-6k-13.$ Then, let

$$\begin{aligned} \psi _1(x)= & {} (1-\alpha )x^{2}+[(k+6)\alpha ^2-5\alpha -2k-1]x-(k^2+15k+24) \alpha ^2\\&+(k^2+25k+37)\alpha +k^2-6k-13 \end{aligned}$$

be a real function in x, where $x\geqslant \frac{3k-1}{1-\alpha }$ with $\frac{2}{3}<\alpha <1$. Consequently,

$$\begin{aligned} \psi '_1(x)&=2(1-\alpha )x+(k+6)\alpha ^2-5\alpha -2k-1\\&\geqslant 2(1-\alpha )\cdot \frac{3k-1}{1-\alpha }+(k+6)\alpha ^2-5\alpha -2k-1\qquad \left( \text {As } \alpha<1,\,x\geqslant \frac{3k-1}{1-\alpha }\right) \\&=(k+6)\alpha ^2-5\alpha +4k-3\\&>(k+6)\cdot \left( \frac{2}{3}\right) ^2-5\cdot \frac{2}{3}+4k-3\quad \left( \text {As } k\geqslant 2,\alpha>\frac{2}{3} \hbox { and } \frac{5}{2(k+6)}<\frac{2}{3}<\alpha \right) \\&=\frac{1}{9}(40k-33)>0. \end{aligned}$$

Thus, $\psi _1(x)$ is a monotonically increasing function for $x\geqslant \frac{3k-1}{1-\alpha }$. Hence,

$$\begin{aligned} \psi _1(n)&\geqslant (1-\alpha )\cdot \left( \frac{3k-1}{1-\alpha }\right) ^{2}+[(k+6)\alpha ^2-5\alpha -2k-1]\cdot \frac{3k-1}{1-\alpha }\nonumber \\&\quad -(k^2+15k+24)\alpha ^2+(k^2+25k+37)\alpha +k^2-6k-13\nonumber \\&=\frac{1}{1-\alpha }[(k^2+15k+24)\alpha ^3+(k^2-23k-67)\alpha ^2\nonumber \\&\quad +(16k+55)\alpha +4k^2-13k-11]. \end{aligned}$$

(3.21)

Let $\psi _2(\alpha )=(k^2+15k+24)\alpha ^3+(k^2-23k-67)\alpha ^2+(16k+55)\alpha +4k^2-13k-11$ be a real function in $\alpha$ with $\frac{2}{3}\leqslant \alpha \leqslant 1$. Consider the second derivative of $\psi _2(\alpha )$, one has

$$\begin{aligned} \psi ''_2(\alpha )&=6(k^2+15k+24)\alpha +2(k^2-23k-67)\\&\geqslant 6(k^2+15k+24)\cdot \frac{2}{3}+2(k^2-23k-67)\qquad \qquad (\text {As } k\geqslant 2) \\&=6k^2+14k-38\\&>0.\qquad \qquad \qquad \qquad \qquad \qquad \qquad (\text {As } k\geqslant 2). \end{aligned}$$

Thus, the first derivative of $\psi _2(\alpha )$ is a monotonically increasing function for $\frac{2}{3}\leqslant \alpha \leqslant 1$. Consequently,

$$\begin{aligned} \psi '_2(\alpha )&=3(k^2+15k+24)\alpha ^2+2(k^2-23k-67)\alpha +(16k+55)\\&\geqslant 3(k^2+15k+24)\cdot \left( \frac{2}{3}\right) ^2+2(k^2-23k-67)\cdot \frac{2}{3}+(16k+55)\\&=\frac{1}{3}(8k^2+16k-7)\\&>0. \end{aligned}$$

Therefore, $\psi _2(\alpha )$ is a monotonically increasing function for $\frac{2}{3}\leqslant \alpha \leqslant 1$. This gives us that

$$\begin{aligned} \psi _2(\alpha )&>(k^2+15k+24)\cdot \left( \frac{2}{3}\right) ^3+(k^2-23k-67)\cdot \left( \frac{2}{3}\right) ^2\\&\quad +(16k+55)\cdot \frac{2}{3}+4k^2-13k-11\\&=\frac{128}{27}k^2-\frac{73}{9}k+3>0 \end{aligned}$$

for $\frac{2}{3}<\alpha <1$. In view of (3.21), we obtain $\psi _1(n)>0$. That is to say, $\psi '(2)>0.$

Next we show $\psi '(\frac{n-k}{2})<0$. In view of (3.19), we have $\psi '(\frac{n-k}{2})=\frac{1}{4}(1-\alpha )[(\alpha -2)n^2+((2k-4)\alpha -4k+8)n-(3k^2-8k)\alpha +6k^2-4k-4]$. Then, let

$$\begin{aligned} \psi _3(x)= & {} \frac{1}{4}(1-\alpha )[(\alpha -2)x^2+((2k-4)\alpha -4k+8)x\nonumber \\&-(3k^2-8k)\alpha +6k^2-4k-4] \end{aligned}$$

(3.22)

be a real function in x, where $x\geqslant \frac{3k-1}{1-\alpha }$. Considering the first derivative of $\psi _3(x)$ gives us

$$\begin{aligned} \psi '_3(x)&=\frac{1}{4}(1-\alpha )[2(\alpha -2)x+(2k-4)\alpha -4k+8]\nonumber \\&\leqslant \frac{1}{4}(1-\alpha )[2(\alpha -2)\cdot \frac{3k-1}{1-\alpha }+(2k-4)\alpha -4k+8]\quad \left( \text {As } \alpha <1,\, x\geqslant \frac{3k-1}{1-\alpha }\right) \nonumber \\&=-\frac{1}{2}[(k-2)\alpha ^2-(6k-7)\alpha +8k-6]. \end{aligned}$$

(3.23)

If $k=2$, then in view of (3.23), we obtain $\psi '_3(x)<0.$ Similarly, if $k\geqslant 3$ then

$$\begin{aligned} \psi '_3(x)<-\frac{1}{2}[(k-2)\cdot 1^2-(6k-7)\cdot 1+8k-6] =\frac{1}{2}(1-3k) <0, \end{aligned}$$

where the first inequality follows by $\frac{(6k-7)}{2(k-2)}>1$. So we obtain that $\psi _3(x)$ is a monotonically decreasing function for $x\geqslant \frac{3k-1}{1-\alpha }$. Consequently, together with (3.22), one has $\psi _3(n)\leqslant \psi _3(\frac{3k-1}{1-\alpha })$ with $\psi _3(\frac{3k-1}{1-\alpha })=\frac{1}{4(1-\alpha )}\psi _4(\alpha ),$ where

$$\begin{aligned} \psi _4(\alpha )&=(8-3k)k\alpha ^3+2(3k^2-3k-4)\alpha ^2+(2k-3)(6k-7)\alpha \\&\quad -24k^2+36k-14 \end{aligned}$$

is a real function in $\alpha$ for $\frac{2}{3}<\alpha <1$. In what follows, we consider the monotonicity of the function $\psi _4(\alpha )$ with $\alpha \in [\frac{2}{3}, 1].$ The first derivative of $\psi _4(\alpha )$ can be written as

$$\begin{aligned} \psi '_4(\alpha )=3k(8-3k)\alpha ^2+ 4(3k^2-3k-4)\alpha +(2k-3)(6k-7). \end{aligned}$$

We are to show $\psi '_4(\alpha )>0.$ In fact, when $k=2$, one has $\psi '_4(\alpha )=12\alpha ^2+8\alpha +5>0$. Now we consider $k\geqslant 3$. Clearly, $\psi '_4(\frac{2}{3})=\frac{1}{3}(48k^2-88k+31)>0$ and $\psi '_4(1)=5(3k^2-4k+1)>0.$ Hence, one has $\psi '_4(\alpha )\geqslant \min \{\psi '_4(\frac{2}{3}),\psi '_4(1)\}>0$. Consequently, $\psi _4(\alpha )$ is monotonically increasing for $\frac{2}{3}\leqslant \alpha \leqslant 1$. Therefore, we may easily deduce that $\psi _4(\alpha )<\psi _4(1)=-9k^2+6k-1<0$ for $\frac{2}{3}<\alpha <1$. Note that $\psi _3(n)\leqslant \psi _3(\frac{3k-1}{1-\alpha })=\frac{1}{4(1-\alpha )}\psi _4(\alpha )$, then we obtain $\psi _3(n)<0$ for $\frac{2}{3}<\alpha <1$. That is to say, $\psi '(\frac{n-k}{2})<0.$

Therefore, together with $\psi '(2)>0, \psi '(\frac{n-k}{2})<0$ and (3.19), we obtain

$$\begin{aligned} \Phi _{B_5^s}(n-k-1)=\psi (s)\geqslant \min \left\{ \psi (2),\psi ((n-k)/2)\right\} \end{aligned}$$

(3.24)

for $2\leqslant s\leqslant \frac{n-k}{2}$. In order to complete the proof, it suffices to show that $\psi (2)>0$ and $\psi ((n-k)/2)>0$.

In fact, by (3.18) we obtain $\psi (2)=\psi _5(n),$ where $\psi _5(n)=(1-\alpha )n^2+[(2k+4)\alpha ^2-(k+3)\alpha -2k-1]n-2(k+1)(k+6)\alpha ^2 +2(k^2+11k+9)\alpha +(k+1)(k-6),$ with $\frac{2}{3}<\alpha <1$ and $n\geqslant \frac{3k-1}{1-\alpha }$. In what follows, we show the real function $\psi _5(x)$ is monotonically increasing for $x\geqslant \frac{3k-1}{1-\alpha }$ with $\frac{2}{3}< \alpha < 1$. This is equivalent to show the first derivative $\psi _5'(x)$ is positive. Indeed, by some calculations, we see that

$$\begin{aligned} \psi '_5(x)&=2(1-\alpha )x+(2k+4)\alpha ^2-(k+3)\alpha -2k-1\\&\geqslant 2(1-\alpha )\cdot \frac{3k-1}{1-\alpha }+(2k+4)\alpha ^2-(k+3)\alpha -2k-1\qquad \qquad \qquad \qquad (\text {As } \alpha< 1 )\\&=2(k+2)\alpha ^2-(k+3)\alpha +4k-3\\&> 2(k+2)\cdot \left( \frac{2}{3}\right) ^2-(k+3)\cdot \frac{2}{3}+4k-3\qquad \qquad \qquad \left( \text {As } \frac{k+3}{4(k+2)}<\frac{2}{3}< \alpha \right) \\&=\frac{1}{9}(38k-29)>0, \end{aligned}$$

as desired. Consequently,

$$\begin{aligned} \psi _5(n) \geqslant&(1-\alpha )\cdot \left( \frac{3k-1}{1-\alpha }\right) ^{2}+[(2k+4)\alpha ^2-(k+3)\alpha -2k-1]\cdot \frac{3k-1}{1-\alpha }\nonumber \\&-2(k+1)(k+6)\alpha ^2+2(k^2+11k+9)\alpha +(k+1)(k-6)\nonumber \\ =&\frac{1}{1-\alpha }[2(k+1)(k+6)\alpha ^3+2(k^2-13k-17)\alpha ^2\nonumber \\&-(2k^2-19k-27)\alpha +4k^2-12k-4]. \end{aligned}$$

(3.25)

Let $\psi _6(\alpha )=2(k+1)(k+6)\alpha ^3+2(k^2-13k-17)\alpha ^2- (2k^2-19k-27)\alpha +4k^2-12k-4$ be a real function in $\alpha$ with $\frac{2}{3}\leqslant \alpha \leqslant 1$. Consider the second derivative of $\psi _6(\alpha )$. Then, we have

$$\begin{aligned} \psi ''_6(\alpha )&=12(k+1)(k+6)\alpha +4(k^2-13k-17)\\&\geqslant 12(k+1)(k+6)\cdot \frac{2}{3}+4(k^2-13k-17)\\&=4(3k^2+k-5)>0. \end{aligned}$$

This implies the first derivative of $\psi _6(\alpha )$ is a monotonically increasing function for $\frac{2}{3}\leqslant \alpha \leqslant 1$. Consequently,

$$\begin{aligned} \psi '_6(\alpha )&=6(k+1)(k+6)\alpha ^2+4(k^2-13k-17)\alpha -(2k^2-19k-27)\\&\geqslant 6(k+1)(k+6)\cdot (\frac{2}{3})^2+4(k^2-13k-17)\cdot \frac{2}{3} -(2k^2-19k-27)\\&=\frac{1}{3}(10k^2+9k-7)>0. \end{aligned}$$

Hence, $\psi _6(\alpha )$ is a monotonically increasing function in $\alpha$ for $\frac{2}{3}\leqslant \alpha \leqslant 1$. Therefore,

$$\begin{aligned} \psi _6(\alpha )&\geqslant 2(k+1)(k+6)\cdot (\frac{2}{3})^3\\&\quad +2(k^2-13k-17)\cdot (\frac{2}{3})^2-(2k^2-19k-27)\cdot \frac{2}{3}+4k^2-12k-4\\&=\frac{2}{27}(56k^2-91k+33) >0 \end{aligned}$$

for $\frac{2}{3}\leqslant \alpha \leqslant 1$. Recall that $\psi (2)=\psi _5(n)$. Then, together with (3.25), we obtain $\psi (2)\geqslant \frac{1}{1-\alpha }\psi _6(\alpha )>0$ for $\frac{2}{3}<\alpha <1$.

Now we show $\psi ((n-k)/2)>0$ in what follows. In view of (3.18) we obtain $\psi ((n-k)/2)=\frac{1}{8}[(2-\alpha )n+\alpha k-2k-2][(1-\alpha )n^2+((2k+2)\alpha -4k-2)n-(k^2-2k)\alpha +3k^2+2k]$. Let

$$\begin{aligned} \psi _7(x)=(1-\alpha )x^2+((2k+2)\alpha -4k-2)x-(k^2-2k)\alpha +3k^2+2k \end{aligned}$$

be a real function in x with $x\geqslant \frac{3k-1}{1-\alpha }$. Then, consider the first derivative of $\psi _7(x)$. Thereby,

$$\begin{aligned} \psi '_7(x)&=2(1-\alpha )x+(2k+2)\alpha -4k-2\\&\geqslant 2(1-\alpha )\cdot \frac{3k-1}{1-\alpha }+(2k+2)\alpha -4k-2\\&=2(k+1)\alpha +2k-4\\&>2(k+1)\cdot \frac{2}{3}+2k-4>0. \end{aligned}$$

Hence, $\psi _7(x)$ is a monotonically increasing function for $x\geqslant \frac{3k-1}{1-\alpha }$. Consequently,

$$\begin{aligned} \psi _7(n)&\geqslant (1-\alpha )\cdot \left( \frac{3k-1}{1-\alpha }\right) ^2+[(2k+2)\alpha -4k-2]\cdot \frac{3k-1}{1-\alpha }\\&\quad -(k^2-2k)\alpha +3k^2+2k\\&=\frac{1}{1-\alpha }[(k-2)k\alpha ^2+2(k^2+2k-1)\alpha -6k+3]\\&>\frac{1}{1-\alpha }\left[ (k-2)k\cdot \left( \frac{2}{3}\right) ^2+2(k^2+2k-1)\cdot \frac{2}{3}-6k+3\right] \\&=\frac{1}{9(1-\alpha )}(16k^2-38k+15)>0. \end{aligned}$$

On the other hand, for $\alpha \in (\frac{2}{3},1)$ and $n\geqslant \frac{3k-1}{1-\alpha }$, one has

$$\begin{aligned} (2-\alpha )n+\alpha k-2k-2&\geqslant (2-\alpha )\cdot \frac{3k-1}{1-\alpha }+\alpha k-2k-2\\&=\frac{1}{(1-\alpha )}(-k\alpha ^2+3\alpha +4k-4)\\&>\frac{1}{(1-\alpha )}(-k+3\alpha +4k-4)\\&>0. \end{aligned}$$

Together with $\psi _7(n)>0$, one has

$$\begin{aligned} \psi ((n-k)/2) =\frac{1}{8}[(2-\alpha )n+\alpha k-2k-2]\psi _7(n)>0. \end{aligned}$$

Note that $\psi (2)>0$. Together with (3.24), we obtain that $\Phi _{B_5^s}(n-k-1)=\psi (s)>0$ for $2\leqslant s\leqslant \frac{n-k}{2}$, as desired.

Combining with Subcases 1.1–1.4, we obtain that Claim 3 holds if n, k have the same parity.

Case 2. n and k have different parity. In this case, $\lceil \frac{n-k}{2}\rceil =\frac{n-k+1}{2}.$ Thus, $K_s\vee (K_{2\lceil \frac{n-k}{2}\rceil -2s+1}\cup (s+n-2\lceil \frac{n-k}{2}\rceil -1)K_1)=K_s\vee (K_{n-2s-k+2}\cup (s+k-2)K_1).$ Let

$$\begin{aligned} G_6^s=K_s\vee (K_{n-2s-k+2}\cup (s+k-2)K_1). \end{aligned}$$

It may be rewritten as $G_6^s=K_s\vee (K_{n-2s-l+1}\cup (s+l-1)K_1)$ with $l=k-1$.

To prove Claim 3, it suffices to show $\rho _\alpha (G_6^s)<\rho _\alpha (G_6^1)$ for $n\geqslant f(\alpha ,l+1)$ and $2\leqslant s\leqslant \frac{n-l}{2}$. It is easy to see that n, l have the same parity. By Case 1, we know that Claim 3 holds for $n\geqslant f(\alpha ,l)$ and $2\leqslant s\leqslant \frac{n-l}{2}$. Note that $f(\alpha ,l+1)>f(\alpha ,l)$. Hence, Claim 3 holds for $n\geqslant f(\alpha ,l+1)$ and $2\leqslant s\leqslant \frac{n-l}{2}$ obviously.

Combining Cases 1 and 2, we conclude that Claim 3 holds. $\square$

Now we come back to show Theorem 1.1. For all n-vertex graphs G and positive integer $k\geqslant 2$, if $\mu (G)\leqslant \lceil \frac{n-k}{2}\rceil$ satisfying $n\geqslant f(\alpha ,k)$ defined in (1.1), then by Claims 1-3, one has

$$\begin{aligned} \rho _\alpha (G)\leqslant \rho _\alpha (K_1\vee (K_{2\lceil (n-k)/2\rceil -1}\cup (n-2\lceil (n-k)/2\rceil )K_1)) \end{aligned}$$

with equality if and only if $G\cong K_1\vee (K_{2\lceil (n-k)/2\rceil -1}\cup (n-2\lceil (n-k)/2\rceil )K_1)$.

In fact, $K_1\vee (K_{2\lceil (n-k)/2\rceil -1}\cup (n-2\lceil (n-k)/2\rceil )K_1)$ is an n-vertex graph obtained from $K_{2\lceil (n-k)/2\rceil }$ by attaching $(n-2\lceil (n-k)/2\rceil )$ pendant edges to one vertex, say v,  of $K_{2\lceil (n-k)/2\rceil }$. Then, we have a vertex partition of $K_1\vee (K_{2\lceil (n-k)/2\rceil -1}\cup (n-2\lceil (n-k)/2\rceil )K_1)$ as $\{v\}\cup W\cup U,$ where v is a vertex of degree $n-1, U$ is the set of all pendant vertices and W is the set of the remaining vertices. Clearly, this partition is equitable. The quotient matrix of $A_\alpha (K_1\vee (K_{2\lceil (n-k)/2\rceil -1}\cup (n-2\lceil (n-k)/2\rceil )K_1)$ with respect to this partition is

$$\begin{aligned} M=\left( \begin{array}{ccc} (n-1)\alpha &{} (2\lceil \frac{n-k}{2}\rceil -1)(1-\alpha ) &{} (n-2\lceil \frac{n-k}{2}\rceil )(1-\alpha )\\ 1-\alpha &{} 2\lceil \frac{n-k}{2}\rceil +\alpha -2 &{} 0\\ 1-\alpha &{} 0 &{} \alpha \end{array} \right) . \end{aligned}$$

By a direct calculation, we may obtain characteristic polynomial of M as

$$\begin{aligned} \Phi _M(x)&=x^3-(\alpha n+\alpha -2+2\lceil (n-k)/2\rceil ) x^2\\&\quad +[\alpha ^2n-n-2\alpha +1+2\alpha n\lceil (n-k)/2\rceil ]x-(\alpha -2)(2\alpha -1)n\\&\quad -\alpha -4(\alpha -1)^2\lceil (n-k)/2\rceil ^2+2(n-2\alpha n+3\alpha ^2-4\alpha +2)\lceil (n-k)/2\rceil . \end{aligned}$$

By Lemma 2.3, we obtain that $\rho _\alpha (K_1\vee (K_{2\lceil (n-k)/2\rceil -1}\cup (n-2\lceil (n-k)/2\rceil )K_1))$ is equal to the largest root, say $\theta _0$, of $\Phi _M(x)=0$. This completes the proof. $\square$

4 Proof of Theorem 1.2

In this section, we give the proof of Theorem 1.2, which gives an upper bound on the $A_\alpha$-spectral radius for a graph G with order $n\leqslant 3k$ ensuring $\mu (G)\leqslant \frac{n-k}{2},$ where $k\geqslant 2$ is an integer having the same parity with that of n.

Proof of Theorem 1.2

Assume that G is connected and $\mu (G)\leqslant \frac{n-k}{2}$. By Lemma 2.1, we have

$$\begin{aligned} \mu (G)=\min _{S\subseteq V(G)}\frac{n-o(G-S)+|S|}{2}\leqslant \frac{n-k}{2}, \end{aligned}$$

which deduces that there exists some nonempty subset S of V(G) such that $\ell -s\geqslant k$, where $s=|S|, \ell =o(G-S)$ and all components of $G-S$ are odd. Let $H_1, H_2,\ldots ,H_\ell$ be all the odd components of $G-S$ satisfying $|V(H_1)|=n_1\geqslant |V(H_2)|=n_2\geqslant \cdots \geqslant |V(H_\ell )|=n_\ell$. By a similar discussion as that of Theorem 1.1, we have $\rho _\alpha (G)\leqslant \rho _{\alpha }(K_s\vee (K_{n-2s-k+1}\cup (s+k-1)K_1)),$ where $s\in \{1,2,\ldots , \frac{n-k}{2}\}$.

Let $G_5^s=K_s\vee (K_{n-2s-k+1}\cup (s+k-1)K_1)$. In what follows, we show that $G_5^{\frac{n-k}{2}}$ is the unique graph among $\{G_5^s:\, 1\leqslant s\leqslant \frac{n-k}{2}\}$ having the largest $A_\alpha$-spectral radius.

Partition the vertex set of $G_5^s$ as $V(G_5^s)=V(K_s)\cup V(K_{n-2s-k+1})\cup V((s+k-1)K_1)$, where $V(K_s)=\{u_1,u_2,\ldots ,u_s\}$, $V((s+k-1)K_1)=\{v_1,v_2,\ldots ,v_{s+k-1}\}$ and $V(K_{n-2s-k+1})=\{w_1,w_2,\ldots ,w_{n-2s-k+1}\}$. Let $\varvec{y}$ be the Perron vector of $A_\alpha (G_5^s)$ with respect to $\rho _\alpha (G_5^s)$. By symmetry (see also [27, Proposition 16]), $\varvec{y}$ takes the same value (say $y_1$, $y_2$ and $y_3$) on the vertices of $V(K_s)$, $V((s+k-1)K_1)$ and $V(K_{n-2s-k+1})$, respectively.

For convenience, let $r=n-2s-k.$ Clearly, r is even. Then, we construct a new graph $G_6$ as

$$\begin{aligned} G_6=G_5^s+\sum _{i=1}^{\frac{r}{2}}\sum _{j=1}^{s+k-1}w_iv_j- \sum _{i=\frac{r}{2}+1}^{r}\sum _{j=i+1}^{r+1}w_iw_j. \end{aligned}$$

(4.1)

Then, $G_6\cong K_{\frac{n-k}{2}}\vee \frac{n+k}{2}K_1$, i.e., $G_6\cong G_5^{\frac{n-k}{2}}$. We are to show that $\rho _\alpha (G_5^s)<\rho _\alpha (G_5^{\frac{n-k}{2}})$ for $1\leqslant s <\frac{n-k}{2}$.

Similarly, partition the vertex set $V(G_6)$ as $V(K_{\frac{n-k}{2}})\cup V(\frac{n+k}{2}K_1)$, where $V(K_{\frac{n-k}{2}})=\{u_1,u_2,\ldots , u_{\frac{n-k}{2}}\}$, $V(\frac{n+k}{2}K_1)=\{v_1,v_2,\ldots ,v_{\frac{n+k}{2}}\}$. This partition is equitable. The corresponding quotient matrix is

$$\begin{aligned} W=\left( \begin{array}{cc} \frac{n+k}{2}\alpha +\frac{n-k}{2}-1 &{} \frac{n+k}{2}(1-\alpha ) \\ \frac{n-k}{2}(1-\alpha )&{} \frac{n-k}{2}\alpha \end{array} \right) . \end{aligned}$$

(4.2)

On the other hand, let $\varvec{z}$ be the Perron vector of $A_\alpha (G_6)$ with respect to $\rho _\alpha (G_6)$. By symmetry (see also [27, Proposition 16]), $\varvec{z}$ takes the same value (say $z_1$ and $z_2$) on the vertices of $V(K_{\frac{n-k}{2}})$ and $V(\frac{n+k}{2}K_1)$, respectively. Then, by $A_\alpha (G_6)\varvec{z}=\rho _\alpha (G_6)\varvec{z}$, we have

$$\begin{aligned} \rho _\alpha (G_6) z_2=\frac{n-k}{2}(1-\alpha )z_1+\frac{n-k}{2}\alpha z_2. \end{aligned}$$

This gives

$$\begin{aligned} z_2=\frac{(1-\alpha )(n-k)}{2\rho _\alpha (G_6)-\alpha (n-k)}z_1. \end{aligned}$$

(4.3)

Note that $K_{\frac{n-k}{2}+1}$ is a proper subgraph of $G_6$. By Lemmas 2.4 and 2.5, one has $\frac{n-k}{2}=\rho _\alpha (K_{\frac{n-k}{2}+1})<\rho _\alpha (G_6)$. Together with (4.3), we obtain $z_2<z_1$.

In what follows, we compare the $A_\alpha$-spectral radius of $G_5^s$ with that of $G_6$ for $1\leqslant s\leqslant \frac{n-k}{2}-1$. Recall that $r=n-2s-k.$ Together with (4.1) and $z_2<z_1,$ one has

$$\begin{aligned}&\varvec{z}^T(\rho _\alpha (G_6)-\rho _\alpha (G_5^s))\varvec{y}\nonumber \\ =&\ \varvec{z}^T(A_\alpha (G_6)-A_\alpha (G_5^s))\varvec{y}\nonumber \\ =&\sum _{i=1}^{\frac{r}{2}}\sum _{j=1}^{s+k-1}(1-\alpha )(y_3z_2+y_2z_1)\nonumber \\&+\sum _{i=1}^{\frac{r}{2}}\alpha (s+k-1)y_3z_1 +\sum _{i=1}^{s+k-1}\frac{r}{2}\alpha y_2z_2\nonumber \\&-\sum _{i=\frac{r}{2}+1}^{r}\sum _{j=i+1}^{r+1}(1-\alpha )(y_3z_2+y_3z_2)-\sum _{i=\frac{r}{2}+1}^{r+1}\frac{r}{2}\alpha y_3z_2\nonumber \\ =&\ \frac{r}{2}\big [((s+k-1)(1-\alpha )-(\frac{r}{2}+1))y_3z_2+(s+k-1)(1-\alpha )y_2z_1\nonumber \\&\ +(s+k-1)\alpha y_3z_1+(s+k-1)\alpha y_2z_2\big ]\nonumber \\>&\ \frac{r}{2}[((s+k-1)(1-\alpha )-(\frac{r}{2}+1))y_3z_2+(s+k-1)\alpha y_3z_1]\nonumber \\ \geqslant&\ \frac{r}{2}[-\alpha (s+k-1)y_3z_2+(s+k-1)\alpha y_3z_1]\\ =&\ \frac{r}{2}\left[ \alpha (s+k-1)(z_1-z_2)y_3\right] \nonumber \\>&\ 0,\qquad \qquad \qquad \qquad (\text {As } z_1>z_2)\nonumber \end{aligned}$$

(4.4)

where inequality in (4.4) follows from $s+k-1\geqslant \frac{n-2s-k}{2}+1=\frac{r}{2}+1$ for $n\leqslant 3k$. So, for $1\leqslant s<\frac{n-k}{2}$, we get $\rho _\alpha (G_6)>\rho _\alpha (G_5^s)$ with $\alpha \in [0,1)$ and $n\leqslant 3k$. That is to say, $G_6=G_5^{\frac{n-k}{2}}$ is the unique graph among $\{G_5^s:\, 1\leqslant s\leqslant \frac{n-k}{2}\}$ having the largest $A_\alpha$-spectral radius.

Now we determine the $A_\alpha$-spectral radius of $G_5^{\frac{n-k}{2}}.$ By Lemma 2.3, it is equivalent to determine the largest eigenvalue of its quotient matrix W in (4.2). By an elementary calculation, we obtain that the largest eigenvalue, say $\theta _1,$ of W is given as $\theta _1=\frac{1}{4}[(2\alpha +1)n-k-2+g(n,k,\alpha )]$, where

$$\begin{aligned} g(n,k,\alpha )=\sqrt{(4\alpha ^2-8\alpha +5)n^2+(4\alpha k-2k-4)n+(k-2)(4\alpha k-3k-2)}. \end{aligned}$$

This completes the proof. $\square$

5 Concluding Remarks

In this paper, we provide some $A_\alpha$-spectral conditions to ensure that a connected graph has no large matchings. In fact, we use a unified approach to deal with these problems. Consequently, by our results we may deduce some known or new results on these problems.

In Theorems 1.1 and 1.2, if n is even and let $k=2$ and $\alpha =0$, then we may deduce the following results obtained by O [28], which provides an adjacency spectral condition to guarantee that there exists a perfect matching in graph G.

Theorem 5.1

[28] Let G be an n-vertex connected graph with even n and let $\bar{\theta }(n)$ is the largest root of $x^3-(n-4)x^2-(n-1)x+2(n-4)=0$. For $n=4$ or $n\geqslant 8,$ if $\rho _0(G)>\bar{\theta }(n),$ then G has a perfect matching. For $n=6$, if $\rho _0(G)>\frac{1+\sqrt{33}}{2}$, then G has a perfect matching.

In Theorem 1.1, let n be even and $k=2$, then we deduce the following results obtained by Zhao, Huang and Wang [34]. It is an $A_\alpha$-spectral condition which guarantees that there exists a perfect matching in graph G.

Theorem 5.2

[34] Let $\alpha \in [0,1),$ and let G be a connected graph of even order n with $n\geqslant f(\alpha )$, where

$$\begin{aligned} f(\alpha )={\left\{ \begin{array}{ll} 10,&{}\text {if } 0\leqslant \alpha \leqslant \frac{1}{2};\\ 14,&{}\text {if } \frac{1}{2}<\alpha \leqslant \frac{2}{3};\\ \frac{5}{1-\alpha },&{}\text {if } \frac{2}{3}<\alpha <1. \end{array}\right. } \end{aligned}$$

If $\rho _\alpha (G)\geqslant \beta _0$, then G has a perfect matching unless $G=K_1\vee (K_{n-3}\cup 2K_1)$, where $\beta _0$ is the largest root of $x^3-((\alpha +1)n+\alpha -4)x^2+(\alpha n^2+(\alpha ^2-2\alpha -1)n-2\alpha +1)x-\alpha ^2n^2 +(5\alpha ^2-3\alpha +2)n-10\alpha ^2+15\alpha -8=0$.

In Theorems 1.1 and 1.2, assume that n, k have the same parity and let $\alpha =0$. Then, we deduce the following results obtained by Kim, O, Sim and Shin [10]. It gives an adjacency spectral condition which guarantees that there exists a maximum matching of size greater than $\frac{n-k}{2}$ in graph G.

Theorem 5.3

[10] Let n and k be positive integers with the same parity. Let $\theta '(n,k)$ be the largest root of $x^3-(n-k-2)x^2-(n-1)x+k(n-k-2)=0$ and let

$$\begin{aligned} \theta (n,k)={\left\{ \begin{array}{ll} \theta '(n,k),&{}\text {if } n\geqslant 3k+2;\\ \frac{1}{4}[n-k-2+\sqrt{(n-k-2)^2+4(n^2-k^2)}],&{}\text {if } n\leqslant 3k. \end{array}\right. } \end{aligned}$$

If G is an n-vertex connected graph satisfying $\rho _0(G)>\theta (n,k)$ with $n\geqslant k+2$, then $\mu (G)>\frac{n-k}{2}$.

In Theorems 1.1 and 1.2, if n is even and let $k=2$ and $\alpha =\frac{1}{2}$, then we may deduce the following results obtained by Liu, Pan and Li [23]. It gives a signless Laplacian spectral condition which guarantees that there exists a perfect matching in graph G.

Theorem 5.4

[23] Let G be an n-vertex connected graph with even n, and let $\tilde{\theta }(n)$ be the largest root of equation $x^3-(3n-7)x^2+n(2n-7)x-2(n^2-7n+12)=0$. For $n\geqslant 10$, if $q(G)>\tilde{\theta }(n),$ then G has a perfect matching. For $n=4$ (resp. $n=6$), if $q(G)>4$ (resp. $q(G)>4+2\sqrt{3}$), then G has a perfect matching.

Similarly, in Theorems 1.1 and 1.2, assume that n, k have the same parity and let $\alpha =\frac{1}{2}$, then we obtain the following signless Laplacian spectral condition which guarantees that there exists a maximum matching of size at least $\frac{n-k}{2}$ in graph G.

Theorem 5.5

Let n and k be positive integers with the same parity. Let $\sigma (n,k)$ be the largest root of $4x^3-(6n-4k-6)x^2+n(2n-2k-3)x-n^2+(2k+3)n-k^2-3k-2=0$ and let

$$\begin{aligned} \hat{\theta }(n,k)={\left\{ \begin{array}{ll} \sigma (n,k),&{}\text {if } n\geqslant 5k;\\ \frac{1}{4}[2n-k-2+\sqrt{2n(n-2)-(k+2)(k-2)}],&{}\text {if } n\leqslant 3k. \end{array}\right. } \end{aligned}$$

(5.1)

If G is an n-vertex connected graph satisfying $q(G)>2\hat{\theta }(n,k)$ with $n\geqslant k+2$, then $\mu (G)>\frac{n-k}{2}$.

In view of (1.1), Theorem 1.1 does not completely characterize all the n-vertex connected graph with $\mu (G)\leqslant \frac{n-k}{2}$ having the largest $A_\alpha$-spectral radius. We should make up the gap in Theorem 1.1. On the other hand, in Theorem 1.2 we only consider the case that n, k have the same parity. When n, k have different parity, it is still open. We will do it in the near future.