© Wuhan University 2025

0 Introduction

All graphs considered are undirected, simple, and connected throughout this paper. Let $G$ be a graph with vertex set $V(G)$ and edge set $E(G)$. We denote by $|V(G)|=n$ and $|E(G)|=e\left(G\right)$ the order and the size of $G$, respectively. Moreover, we write $d_G(u)$ for the degree of vertex $u$ in $G$, and $N_G(v)$ for the neighbor set of vertex $v$ in $G$. For any $S\subseteq V(G)$, let $G[S]$ be the subgraph of $G$ induced by $S$, and let $G-S$ be the subgraph induced by $V(G)\backslash S$. The set of all edges between two vertex sets $X$ and $Y$ is denoted by $e\left(X,Y\right)$. For any two graphs $G$ and $H$, we denote by $G\bigcup H$ the disjoint union of $G$ and $H$. The join $G\vee H$ is the graph obtained from $G\bigcup H$ by adding all possible edges between $V(G)$ and $V(H)$.

The adjacency matrix $\mathit{A}(G)=(a_{i,j})$ of $G$ is a $n\times n$ symmetric matrix, whose entries $a_{i,j}$ are given by $a_{i,j}=\mathrm{1}$ if $v_i$ and $v_j$ are adjacent in $G$, and $a_{i,j}=\mathrm{0}$ otherwise. The degree diagonal matrix of $G$ is denoted by $\mathit{D}(G)=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(d_G(v_{\mathrm{1}}),d_G(v_{\mathrm{2}}),\cdots ,d_G(v_n))$, where $d_G(v_i)$ is the degree of $v_i$ in $G$. For any real $\alpha \in [\mathrm{0,1}]$, Nikiforov^[1] defined the ${\mathit{A}}_{\alpha }$-matrix of $G$ as ${\mathit{A}}_{\alpha }(G)=\alpha \mathit{D}(G)+(\mathrm{1}-$$\alpha )\mathit{A}(G)$. The largest eigenvalue of ${\mathit{A}}_{\alpha }(G)$ is called the ${\mathit{A}}_{\alpha }$-spectral radius of $G$, and denoted by ${\rho }_{\alpha }(G)$. Obviously, ${\mathit{A}}_{\mathrm{0}}(G)=\mathit{A}(G)$, ${\mathit{A}}_{\mathrm{1}}(G)=\mathit{D}(G)$ and $\mathrm{2}{\mathit{A}}_{\mathrm{1}/\mathrm{2}}(G)=\mathit{Q}(G)$$=\mathit{D}(G)+\mathit{A}(G)$, where $\mathit{Q}(G)$ is the signless Laplacian matrix of $G$. Note that ${\mathit{A}}_{\alpha }(G)$ is a real symmetric non-negative matrix. By the Perron-Frobenius Theorem, ${\rho }_{\alpha }(G)$ is a positive number and there exists a unique positive unit eigenvector corresponding to ${\rho }_{\alpha }(G)$, which is called the Perron vector of $G$.

A graph is $k$-extendable if each matching consisting of $k$ edges can be extended to a perfect matching. Clearly, the existence of a perfect matching can be regarded as 0-extendability. The concept of $k$-extendability gradually evolved from the concept of elementary (that is, 1-extendable) bipartite graphs. In 1980, Plummer^[2] gave the first results on $k$-extendable graphs for arbitrary $k$, he studied the properties of $k$-extendable graphs and proved that nearly all $k$-extendable graphs $(k\ge \mathrm{2})$ are $(k-\mathrm{1})$-extendable and $(k+\mathrm{1})$-connected. Then Plummer^[3-4] provided the relationships between the toughness and genus of a graph and the $k$-extendability of the graph. Moreover, many researchers further studied the relationship between $k$-extendability and other graph parameters, such as minimum degree, connectivity, binding number, and independence number etc^[5-10].

In the past few years, much attention has been paid to the connections of the spectral radius and the $k$-extendable graphs. Fan and Lin^[11] investigated the adjacency spectral conditions for a graph (as well as a balanced bipartite graph) with a minimum degree $\delta$ to be $k$-extendable. In Ref. [12], Zhou and Zhang gave a lower bound on the signless Laplacian spectral radius to ensure that $G$ is $k$-extendable. Then, Zhang and Dam^[13] studied the $k$-extendability of graphs from a distance spectral perspective. The main goal of this paper is to study the relationship between $k$-extendability of graphs and the ${\mathit{A}}_{\alpha }$-spectral radius.

In the following, we first provide a condition for the $k$-extendability of matchings in a graph in terms of the ${\mathit{A}}_{\alpha }$-spectral radius.

Theorem 1   Let $k\ge \mathrm{1}$ be an integer, and let $G$ be a connected graph of even order $n$ with $n>\mathrm{2}k+\mathrm{4}$. For $\alpha \in [\mathrm{0,1})$, if

${\rho }_{\alpha }(G)\ge {\rho }_{\alpha }(K_{\mathrm{2}k}\vee (K_{n-\mathrm{2}k-\mathrm{1}}\bigcup K_{\mathrm{1}}))$

then $G$ is $k$-extendable, unless $G\cong K_{\mathrm{2}k}\vee (K_{n-\mathrm{2}k-\mathrm{1}}\bigcup K_{\mathrm{1}})$.

A bipartite graph is called balanced if both parts of the bipartition have equal size. Note that every bipartite graph with a perfect matching must be balanced. Hence a $k$-extendable bipartite graph must be balanced. Furthermore, we denote by ${\mathrm{ℬ}}_{n-k,n-\mathrm{1}}$ the bipartite graph obtained from $K_{n,n-\mathrm{1}}$ by attaching a vertex to $k$ vertices in $n$-vertex part (see Fig. 1). In Ref. [13], Zhang and Dam gave a sufficient condition in terms of the distance spectral radius for the $k$-extendability of a bipartite graph. Along this line, we have the result below.

Fig. 1 The extremal graph ${\mathcal{B}}_{\mathit{n}-\mathit{k},\mathit{n}-\mathrm{1}}$

Theorem 2   Let $k\ge \mathrm{1}$ be an integer, and let $G$ be a connected balanced bipartite graph of order $\mathrm{2}n$ with $n\ge k+\mathrm{1}$. For $\alpha \in [\mathrm{0,1})$, if

${\rho }_{\alpha }(G)\ge {\rho }_{\alpha }({\mathrm{ℬ}}_{n-k,n-\mathrm{1}})$

then $G$ is $k$-extendable, unless $G\cong {\mathrm{ℬ}}_{n-k,n-\mathrm{1}}$, see Fig. 1 for instance.

1 Preliminaries

In this section, we give several preliminary lemmas, which are useful to the proof of our main results.

For any $S\subseteq V(G)$, let $o\left(G-S\right)$ denote the number of odd components in $G-S$. In 1995, Chen^[7] provided a necessary and sufficient condition for the existence of $k$-extendable graphs.

Lemma 1^[7] Let $k\ge \mathrm{1}$. A graph $G$ is $k$-extendable if and only if

$o\left(G-S\right)\le \left|S\right|-\mathrm{2}k$

for any $S\subseteq V(G)$ such that $G[S]$ contains $k$ independent edges.

Lemma 2^[1] If $G$ is connected, and $H$ is a proper subgraph of $G$, then ${\rho }_{\alpha }(H)<{\rho }_{\alpha }(G)$ for any $\alpha \in [\mathrm{0,1})$.

Given any $X\subseteq V(G)$, let $N(X)$ be the set of all neighbors of the vertices in $X$. In Ref. [14], Plummer gave the necessary and sufficient conditions for bipartite graphs to be $k$-extendable.

Lemma 3^[14] Let $k\ge \mathrm{1}$ and let $G$ be a connected bipartite graph with parts $U$ and $W$. Then the following are equivalent:

(i) $G$ is $k$-extendable;

(ii) $\left|U\right|=\left|W\right|$ and for all nonempty subsets $X$ of $U$, if $\left|X\right|\le \left|U\right|-k$, then $\left|N(X)\right|\ge \left|X\right|+k$;

(iii) For all $u_{\mathrm{1}},u_{\mathrm{2}},\cdots ,u_k\in U$ and $w_{\mathrm{1}},w_{\mathrm{2}},\cdots ,w_k\in W$, the graph $G^{\text{'}}=G-\{u_{\mathrm{1}},\cdots ,u_k,w_{\mathrm{1}},\cdots ,w_k\}$ has a perfect matching.

Finally, we need a known result mentioned as Claim 1 of the proof of Theorem 1.1 in Ref. [15].

Lemma 4^[15] Let $n_{\mathrm{1}}\ge n_{\mathrm{2}}\ge \cdots \ge n_q\ge \mathrm{1}$ be positive integers and $n={\displaystyle \sum _{i=\mathrm{1}}^q}{n}_i+s$. For $\alpha \in [\mathrm{0,1})$,we have ${\rho }_{\alpha }(K_s\vee (K_{n_{\mathrm{1}}}\bigcup K_{n_{\mathrm{2}}}\bigcup \cdots \bigcup K_{n_q}))\le {\rho }_{\alpha }(K_s\vee (K_{n-s-q+\mathrm{1}}\bigcup (q-\mathrm{1})K_{\mathrm{1}}))$, where the equality holds if and only if $(n_{\mathrm{1}},n_{\mathrm{2}},\cdots ,n_q)=(n-s-q+\mathrm{1,1},\cdots ,\mathrm{1})$.

2 Proof of Theorem 1

In this section, we elaborate on the proof of Theorem 1, which provides a sufficient condition via the ${\mathit{A}}_{\alpha }$-spectral radius of a connected graph to ensure that the graph is $k$-extendable.

Proof of Theorem 1   Suppose to the contrary that $G$ is not $k$-extendable, we will prove that if $G$ is not $k$-extendable, then ${\rho }_{\alpha }(G)\le {\rho }_{\alpha }(K_{\mathrm{2}k}\vee (K_{n-\mathrm{2}k-\mathrm{1}}\bigcup K_{\mathrm{1}}))$ with equality only if $G\cong K_{\mathrm{2}k}\vee$$(K_{n-\mathrm{2}k-\mathrm{1}}\bigcup K_{\mathrm{1}})$. By Lemma 1, there exists some nonempty subset $S\subseteq V(G)$, such that $\left|S\right|\ge \mathrm{2}k$ and $o(G-S)>\left|S\right|-\mathrm{2}k$. For convenience, let $o(G-S)=q$ and $|S|=s$. We assert that $q$ and $s$ have the same parity. If $s$ is odd, then $n-s$ is an odd number since $n$ is even, and so, the number of odd components in $G-S$ is odd, i.e., $q$ is odd. Hence, $q$ and $s$ are odd. Similarly, we obtain that $q$ is even if $s$ is even. Thus, it concludes that $q\ge s-\mathrm{2}k+\mathrm{2}$. To promote the proof, we have the following three claims.

Claim 1 For some odd integers $n_{\mathrm{1}}\ge n_{\mathrm{2}}\ge \cdots \ge n_q\ge \mathrm{1}$ and ${\displaystyle \sum _{i=\mathrm{1}}^q}{n}_i=n-s$, $G$ is a spanning subgraph of $K_s\vee (K_{n_{\mathrm{1}}}\bigcup K_{n_{\mathrm{2}}}\bigcup \cdots \bigcup K_{n_q})$.

Proof   Let $G_{\mathrm{1}},G_{\mathrm{2}},\cdots ,G_q$ be the orders of the $q$ odd components of $G-S$ with $\left|V(G_{\mathrm{1}})\right|\ge$$\left|V(G_{\mathrm{2}})\right|\ge \cdots \ge \left|V(G_q)\right|$, and let $H_{\mathrm{1}},H_{\mathrm{2}},\cdots ,H_p$ be the $p$ even components of $G-S$. Without loss of generality, we assume that all even components join to component $G_{\mathrm{1}}$, i.e.,$G_{\mathrm{1}}\vee (H_{\mathrm{1}}\bigcup H_{\mathrm{2}}\bigcup$$\cdots \bigcup H_p)$. So, $G_{\mathrm{1}}\vee (H_{\mathrm{1}}\bigcup H_{\mathrm{2}}\bigcup \cdots \bigcup H_p)$ is the spanning subgraph of $K_{n_{\mathrm{1}}}$ for some integer $n_{\mathrm{1}}=$$\left|V(G_{\mathrm{1}}\vee (H_{\mathrm{1}}\bigcup H_{\mathrm{2}}\bigcup \cdots \bigcup H_p))\right|$, and $\left|V(G_{\mathrm{1}}\vee (H_{\mathrm{1}}\bigcup H_{\mathrm{2}}\bigcup \cdots \bigcup H_p))\right|$ is an odd integer. Let $G_i$ be the spanning subgraph of $K_{n_i}$ for some $\mathrm{2}\le i\le q$ and $n_i=\left|V(G_i)\right|$. Then all components of $G-S$ are odd and $G$ is the spanning subgraph of $G[S]\vee (G-S)$, and thus, $G$ is a spanning subgraph of $K_s\vee (K_{n_{\mathrm{1}}}\bigcup K_{n_{\mathrm{2}}}\bigcup \cdots \bigcup K_{n_q})$ for some odd integers $n_{\mathrm{1}}\ge n_{\mathrm{2}}\ge \cdots \ge n_q\ge \mathrm{1}$ and ${\displaystyle \sum _{i=\mathrm{1}}^q}{n}_i=n-s$.

Furthermore, by Lemma 2 we have ${\rho }_{\alpha }(G)\le {\rho }_{\alpha }(K_s\vee (K_{n_{\mathrm{1}}}\bigcup K_{n_{\mathrm{2}}}\bigcup \cdots \bigcup K_{n_q}))$ with equality holds if and only if $G\cong K_s\vee (K_{n_{\mathrm{1}}}\bigcup K_{n_{\mathrm{2}}}\bigcup \cdots \bigcup K_{n_q})$.

If $n_{\mathrm{2}}=n_{\mathrm{3}}=\cdots =n_q=\mathrm{1}$, then $K_s\vee (K_{n_{\mathrm{1}}}\bigcup K_{\mathrm{1}}\bigcup \cdots \bigcup K_{\mathrm{1}})\cong K_s\vee (K_{n-s-(q-\mathrm{1})}\bigcup (q-\mathrm{1})K_{\mathrm{1}})$, say $G_{\mathrm{1}}$.

Hence, by Lemma 4 we have that ${\rho }_{\alpha }(K_s\vee (K_{n_{\mathrm{1}}}\bigcup K_{n_{\mathrm{2}}}\bigcup \cdots \bigcup K_{n_q}))\le {\rho }_{\alpha }(K_s\vee (K_{n-s-q+\mathrm{1}}\bigcup (q-\mathrm{1})K_{\mathrm{1}}))$, where the equality holds if and only if $(n_{\mathrm{1}},n_{\mathrm{2}},\cdots ,n_q)=(n-s-q+\mathrm{1,1},\cdots ,\mathrm{1})$.

Note that $q\ge s-\mathrm{2}k+\mathrm{2}$. Let $G_{\mathrm{2}}\cong K_s\vee (K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}\bigcup (s-\mathrm{2}k+\mathrm{1})K_{\mathrm{1}})$. We next compare the ${\mathit{A}}_{\alpha }$-spectral radius of $G_{\mathrm{1}}$ and $G_{\mathrm{2}}$ in the following.

Claim 2 For $\alpha \in [\mathrm{0,1})$, we have ${\rho }_{\alpha }(K_s\vee (K_{n-s-q+\mathrm{1}}\bigcup (q-\mathrm{1})K_{\mathrm{1}}))\le {\rho }_{\alpha }(K_s\vee (K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}\bigcup (s-\mathrm{2}k+\mathrm{1})K_{\mathrm{1}}))$ with equality holds if and only if $q=s-\mathrm{2}k+\mathrm{2}$.

Proof   If $q=s-\mathrm{2}k+\mathrm{2}$, we have $K_s\vee (K_{n-s-q+\mathrm{1}}\bigcup (q-\mathrm{1})K_{\mathrm{1}})\cong K_s\vee (K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}\bigcup (s-\mathrm{2}k+\mathrm{1})$$K_{\mathrm{1}})$. Hence, ${\rho }_{\alpha }(K_s\vee$

$(K_{n-s-q+\mathrm{1}}\bigcup (q-\mathrm{1})K_{\mathrm{1}}))={\rho }_{\alpha }(K_s\vee (K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}\bigcup (s-\mathrm{2}k+\mathrm{1})K_{\mathrm{1}}))$

Now we consider $q>s-\mathrm{2}k+\mathrm{2}$. Clearly, $q\ge s-\mathrm{2}k+\mathrm{4}$ since $q$ and $s$ has the same parity, here it is sufficient to prove ${\rho }_{\alpha }(K_s\vee (K_{n-s-q+\mathrm{1}}\bigcup (q-\mathrm{1})K_{\mathrm{1}}))<{\rho }_{\alpha }(K_s\vee (K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}\bigcup (s-\mathrm{2}k+\mathrm{1})K_{\mathrm{1}}))$ for $q\ge s-\mathrm{2}k+\mathrm{4}$. We notice that $n=s+{\displaystyle \sum _{i=\mathrm{1}}^q}{n}_i\ge \mathrm{2}s-\mathrm{2}k+\mathrm{2}$, $G_{\mathrm{1}}\cong K_s\vee (K_{n-s-(q-\mathrm{1})}\bigcup (q-\mathrm{1})K_{\mathrm{1}})$ and $G_{\mathrm{2}}$$\cong K_s\vee (K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}\bigcup (s-\mathrm{2}k+\mathrm{1})K_{\mathrm{1}}))$.

Suppose that $\mathit{X}$ is the unit Perron eigenvector of ${\mathit{A}}_{\alpha }(G_{\mathrm{1}})$, and ${\mathit{X}}_u$ denotes the entry of $\mathit{X}$ corresponding to the vertex $u\in V(G_{\mathrm{1}})$. From the equitable partition $V(G_{\mathrm{1}})=V(K_s)\bigcup V(K_{n-s-q+\mathrm{1}})$$\bigcup \left(V((q-s+\mathrm{2}k-\mathrm{2})K_{\mathrm{1}}\right)\bigcup V\left(\left(s-\mathrm{2}k+\mathrm{1}\right)K_{\mathrm{1}}\right)$, we have that $V(K_s)$, $V(K_{n-s-q+\mathrm{1}})$ and $V((q-\mathrm{1})K_{\mathrm{1}})$ has the same entries in $\mathit{X}$. Let ${\mathit{X}}_u=x_{\mathrm{1}}$ for $u\in V(K_s)$, ${\mathit{X}}_v=x_{\mathrm{2}}$ for $v\in V(K_{n-s-q+\mathrm{1}})$ and ${\mathit{X}}_w=x_{\mathrm{3}}$ for $w\in V((q-\mathrm{1})K_{\mathrm{1}})$, respectively. For convenience, we write $\mathit{X}$ as follows:

$\mathit{X}={(\underset{s}{\underbrace{x_{\mathrm{1}},\dots ,x_{\mathrm{1}}}},\underset{n-s-q+\mathrm{1}}{\underbrace{x_{\mathrm{2}},\cdots ,x_{\mathrm{2}}}},\underset{q-s+\mathrm{2}k-\mathrm{2}}{\underbrace{x_{\mathrm{3}},\dots ,x_{\mathrm{3}}}},\underset{s-\mathrm{2}k+\mathrm{1}}{\underbrace{x_{\mathrm{3}},\dots ,x_{\mathrm{3}}}})}^{\mathrm{T}}$

For the graph $G_{\mathrm{2}}$, we partition $V(G_{\mathrm{2}})$ as $V(K_s)\bigcup V(K_{n-s-q+\mathrm{1}})\bigcup V(K_{q-s+\mathrm{2}k-\mathrm{2}})\bigcup V((s-\mathrm{2}k+\mathrm{1})$$K_{\mathrm{1}})$. Thus, ${\mathit{A}}_{\alpha }(G_{\mathrm{2}})-{\mathit{A}}_{\alpha }(G_{\mathrm{1}})$ has the form of a block matrix below:

$\left(\begin{array}{cccc}{\mathit{O}}_{n_{\mathrm{1}}\times n_{\mathrm{1}}}& {\mathit{O}}_{n_{\mathrm{1}}\times n_{\mathrm{2}}}& {\mathit{O}}_{n_{\mathrm{1}}\times n_{\mathrm{3}}}& {\mathit{O}}_{n_{\mathrm{1}}\times n_{\mathrm{4}}}\\ {\mathit{O}}_{n_{\mathrm{2}}\times n_{\mathrm{1}}}& \alpha (q-s+\mathrm{2}k-\mathrm{2}){\mathit{I}}_{n_{\mathrm{2}}\times n_{\mathrm{2}}}& (\mathrm{1}-\alpha ){\mathit{I}}_{n_{\mathrm{2}}\times n_{\mathrm{3}}}& {\mathit{O}}_{n_{\mathrm{2}}\times n_{\mathrm{4}}}\\ {\mathit{O}}_{n_{\mathrm{3}}\times n_{\mathrm{1}}}& (\mathrm{1}-\alpha ){\mathit{I}}_{n_{\mathrm{3}}\times n_{\mathrm{2}}}& \alpha (n-\mathrm{2}s+\mathrm{2}k-\mathrm{2}){\mathit{I}}_{n_{\mathrm{3}}\times n_{\mathrm{3}}}+(\mathrm{1}-\alpha ){(\mathit{J}-\mathit{I})}_{n_{\mathrm{3}}\times n_{\mathrm{3}}}& {\mathit{O}}_{n_{\mathrm{3}}\times n_{\mathrm{4}}}\\ {\mathit{O}}_{n_{\mathrm{4}}\times n_{\mathrm{1}}}& {\mathit{O}}_{n_{\mathrm{4}}\times n_{\mathrm{2}}}& {\mathit{O}}_{n_{\mathrm{4}}\times n_{\mathrm{3}}}& {\mathit{O}}_{n_{\mathrm{4}}\times n_{\mathrm{4}}}\end{array}\right)$

where $n_{\mathrm{1}}=s$,$n_{\mathrm{2}}=n-s-q+\mathrm{1}$,$n_{\mathrm{3}}=q-s+\mathrm{2}k-\mathrm{2}$ and $n_{\mathrm{4}}=s-\mathrm{2}k+\mathrm{1}$, meanwhile, $\mathit{J}$ is an all-ones matrix and $\mathit{I}$ is an identity matrix.

Furthermore, by Rayleigh's principle, we have

$\begin{array}{l}{\rho }_{\alpha }(G_{\mathrm{2}})-{\rho }_{\alpha }(G_{\mathrm{1}})\\ \ge {\mathit{X}}^{\mathrm{T}}({\mathit{A}}_{\alpha }(G_{\mathrm{2}})-{\mathit{A}}_{\alpha }(G_{\mathrm{1}}))\mathit{X}\\ =\alpha (q-s+\mathrm{2}k-\mathrm{2})((n-s-q+\mathrm{1})x_{\mathrm{2}}^{\mathrm{2}}+(n-\mathrm{2}s+\mathrm{2}k-\mathrm{2})x_{\mathrm{3}}^{\mathrm{2}})+(\mathrm{1}-\alpha )(q-s+\mathrm{2}k-\mathrm{2})\\ \times (\mathrm{2}(n-s-q+\mathrm{1})x_{\mathrm{2}}{x}_{\mathrm{3}}+(q-s+\mathrm{2}k-\mathrm{3})x_{\mathrm{3}}^{\mathrm{2}})\\ =(q-s+\mathrm{2}k-\mathrm{2})(\alpha ((n-s-q+\mathrm{1})x_{\mathrm{2}}^{\mathrm{2}}+(n-\mathrm{2}s+\mathrm{2}k-\mathrm{2})x_{\mathrm{3}}^{\mathrm{2}})+(\mathrm{1}-\alpha )(\mathrm{2}(n-s-q+\mathrm{1})\\ x_{\mathrm{2}}{x}_{\mathrm{3}}+(q-s+\mathrm{2}k-\mathrm{3})x_{\mathrm{3}}^{\mathrm{2}}))\\ \ge \mathrm{2}(\alpha ((n-s-q+\mathrm{1})x_{\mathrm{2}}^{\mathrm{2}}+(n-\mathrm{2}s+\mathrm{2}k-\mathrm{2})x_{\mathrm{3}}^{\mathrm{2}})+(\mathrm{1}-\alpha )((n-s-q+\mathrm{1})\mathrm{2}{x}_{\mathrm{2}}{x}_{\mathrm{3}}+x_{\mathrm{3}}^{\mathrm{2}}))\\ (\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e}q\ge s-\mathrm{2}k+\mathrm{4})\\ \ge \mathrm{2}(\alpha (n-s-q+\mathrm{1})x_{\mathrm{2}}^{\mathrm{2}}+(\mathrm{1}-\alpha )((n-s-q+\mathrm{1})\mathrm{2}{x}_{\mathrm{2}}{x}_{\mathrm{3}}+x_{\mathrm{3}}^{\mathrm{2}}))(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e}n\ge \mathrm{2}s-\mathrm{2}k+\mathrm{2})\\ >\mathrm{0}.\end{array}$

Therefore, ${\rho }_{\alpha }(K_s\vee (K_{n-s-q+\mathrm{1}}\bigcup (q-\mathrm{1})K_{\mathrm{1}}))<{\rho }_{\alpha }(K_s\vee (K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}\bigcup (s-\mathrm{2}k+\mathrm{1})K_{\mathrm{1}}))$.

Recall that $s\ge \mathrm{2}k$. Let $G_{\mathrm{3}}\cong K_{\mathrm{2}k}\vee (K_{n-\mathrm{2}k-\mathrm{1}}\bigcup K_{\mathrm{1}})$. Next, we compare the ${\mathit{A}}_{\alpha }$-spectral radius of $G_{\mathrm{2}}$ and $G_{\mathrm{3}}$ and give the following claim.

Claim 3 Let $k\ge \mathrm{1}$ and $n>\mathrm{2}k+\mathrm{4}$. For $\alpha \in [\mathrm{0,1})$, we have ${\rho }_{\alpha }(K_s\vee (K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}\bigcup (s-\mathrm{2}k+\mathrm{1})K_{\mathrm{1}}))\le {\rho }_{\alpha }(K_{\mathrm{2}k}\vee (K_{n-\mathrm{2}k-\mathrm{1}}\bigcup K_{\mathrm{1}}))$

with equality holds if and only if $s=\mathrm{2}k$.

Proof   If $s=\mathrm{2}k$, we have $K_s\vee (K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}\bigcup (s-\mathrm{2}k+\mathrm{1})K_{\mathrm{1}})\cong K_{\mathrm{2}k}\vee (K_{n-\mathrm{2}k-\mathrm{1}}\bigcup K_{\mathrm{1}})$. Hence, ${\rho }_{\alpha }(K_s\vee (K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}\bigcup (s-\mathrm{2}k+\mathrm{1})K_{\mathrm{1}}))={\rho }_{\alpha }(K_{\mathrm{2}k}\vee (K_{n-\mathrm{2}k-\mathrm{1}}\bigcup K_{\mathrm{1}}))$.

If $s\ge \mathrm{2}k+\mathrm{1}$, we should verify that ${\rho }_{\alpha }(K_s\vee (K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}\bigcup (s-\mathrm{2}k+\mathrm{1})K_{\mathrm{1}}))<{\rho }_{\alpha }(K_{\mathrm{2}k}\vee (K_{n-\mathrm{2}k-\mathrm{1}}$$\bigcup K_{\mathrm{1}}))$. Since $n\ge \mathrm{2}s-\mathrm{2}k+\mathrm{2}$ we have $s\le \frac{n}{\mathrm{2}}+k-\mathrm{1}$. Thus, it concludes that $\mathrm{2}k+\mathrm{1}\le s\le \frac{n}{\mathrm{2}}+k-\mathrm{1}$. Note that $G_{\mathrm{2}}\cong K_s\vee (K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}\bigcup (s-\mathrm{2}k+\mathrm{1})K_{\mathrm{1}})$ and $G_{\mathrm{3}}\cong K_{\mathrm{2}k}\vee (K_{n-\mathrm{2}k-\mathrm{1}}\bigcup K_{\mathrm{1}})$.

Let $\mathit{Y}$ be the unit Perron eigenvector of ${\mathit{A}}_{\alpha }(G_{\mathrm{3}})$, and ${\mathit{Y}}_v$ be the entry of $\mathit{Y}$ corresponding to the vertex $v\in V(G_{\mathrm{3}})$. Since $V(G_{\mathrm{3}})=V(K_{\mathrm{2}k})\bigcup (V(K_{s-\mathrm{2}k})\bigcup V(K_{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}})\bigcup V(K_{s-\mathrm{2}k}))\bigcup$$V(K_{\mathrm{1}})$ is an equitable partition of $G_{\mathrm{3}}$, $\mathit{Y}$ takes the same value on the vertices of $V(K_{\mathrm{2}k})$ (respectively $V(K_{n-\mathrm{2}k-\mathrm{1}})$ and $V(K_{\mathrm{1}})$). Let ${\mathit{Y}}_u=y_{\mathrm{1}}$ for $u\in V(K_{\mathrm{2}k})$, ${\mathit{Y}}_v=y_{\mathrm{2}}$ for $v\in V(K_{n-\mathrm{2}k-\mathrm{1}})$ and ${\mathit{Y}}_w=y_{\mathrm{3}}$ for $w\in V(K_{\mathrm{1}})$, then $\mathit{Y}$ can be written below:

$\mathit{Y}=(\underset{\mathrm{2}k}{\underbrace{y_{\mathrm{1}},\cdots ,y_{\mathrm{1}}}},\underset{s-\mathrm{2}k}{\underbrace{y_{\mathrm{2}},\cdots ,y_{\mathrm{2}}}},\underset{n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}}{\underbrace{y_{\mathrm{2}},\cdots ,y_{\mathrm{2}}}},\underset{s-\mathrm{2}k}{\underbrace{y_{\mathrm{2}},\cdots ,y_{\mathrm{2}}}},y_{\mathrm{3}}{)}^{\mathrm{T}}$

Let $V(G_{\mathrm{2}})=V(K_s)\bigcup V(K_{n-s-q+\mathrm{1}})\bigcup V(K_{q-s+\mathrm{2}k-\mathrm{2}})\bigcup V((s-\mathrm{2}k+\mathrm{1})K_{\mathrm{1}})$. Then ${\mathit{A}}_{\alpha }(G_{\mathrm{3}})-{\mathit{A}}_{\alpha }(G_{\mathrm{2}})$ can be partitioned as

$\left(\begin{array}{ccccc}{\mathit{O}}_{m_{\mathrm{1}}\times m_{\mathrm{1}}}& {\mathit{O}}_{m_{\mathrm{1}}\times m_{\mathrm{2}}}& {\mathit{O}}_{m_{\mathrm{1}}\times m_{\mathrm{3}}}& {\mathit{O}}_{m_{\mathrm{1}}\times m_{\mathrm{2}}}& {\mathit{O}}_{m_{\mathrm{1}}\times \mathrm{1}}\\ {\mathit{O}}_{m_{\mathrm{2}}\times m_{\mathrm{1}}}& -\alpha {\mathit{I}}_{m_{\mathrm{2}}\times m_{\mathrm{2}}}& {\mathit{O}}_{m_{\mathrm{2}}\times m_{\mathrm{3}}}& {\mathit{O}}_{m_{\mathrm{2}}\times m_{\mathrm{2}}}& -(\mathrm{1}-\alpha ){\mathit{I}}_{m_{\mathrm{2}}\times \mathrm{1}}\\ {\mathit{O}}_{m_{\mathrm{3}}\times m_{\mathrm{1}}}& {\mathit{O}}_{m_{\mathrm{3}}\times m_{\mathrm{2}}}& \alpha (s-\mathrm{2}k){\mathit{I}}_{m_{\mathrm{3}}\times m_{\mathrm{3}}}& (\mathrm{1}-\alpha ){\mathit{I}}_{m_{\mathrm{3}}\times m_{\mathrm{2}}}& {\mathit{O}}_{m_{\mathrm{3}}\times \mathrm{1}}\\ {\mathit{O}}_{m_{\mathrm{2}}\times m_{\mathrm{1}}}& {\mathit{O}}_{m_{\mathrm{2}}\times m_{\mathrm{2}}}& (\mathrm{1}-\alpha ){\mathit{I}}_{m_{\mathrm{2}}\times m_{\mathrm{3}}}& \alpha (n-s-\mathrm{2}){\mathit{I}}_{m_{\mathrm{2}}\times m_{\mathrm{2}}}+(\mathrm{1}-\alpha ){(\mathit{J}-\mathit{I})}_{m_{\mathrm{2}}\times m_{\mathrm{2}}}& {\mathit{O}}_{m_{\mathrm{2}}\times \mathrm{1}}\\ {\mathit{O}}_{\mathrm{1}\times m_{\mathrm{1}}}& -(\mathrm{1}-\alpha ){\mathit{I}}_{\mathrm{1}\times m_{\mathrm{2}}}& {\mathit{O}}_{\mathrm{1}\times m_{\mathrm{3}}}& {\mathit{O}}_{\mathrm{1}\times m_{\mathrm{2}}}& \alpha (\mathrm{2}k-s){\mathit{I}}_{\mathrm{1}\times \mathrm{1}}\end{array}\right)$

where $m_{\mathrm{1}}=\mathrm{2}k$, $m_{\mathrm{2}}=s-\mathrm{2}k$ and $m_{\mathrm{3}}=n-\mathrm{2}s+\mathrm{2}k-\mathrm{1}$.

Thus, by ${\mathit{A}}_{\alpha }(G_{\mathrm{3}})\mathit{Y}={\rho }_{\alpha }(G_{\mathrm{3}})\mathit{Y}$, we can get that

${\rho }_{\alpha }(G_{\mathrm{3}})y_{\mathrm{2}}=\alpha (n-\mathrm{2})y_{\mathrm{2}}+(\mathrm{1}-\alpha )(\mathrm{2}k{y}_{\mathrm{1}}+(n-\mathrm{2}k-\mathrm{2})y_{\mathrm{2}})$(1)

${\rho }_{\alpha }(G_{\mathrm{3}})y_{\mathrm{3}}=\alpha (\mathrm{2}k)y_{\mathrm{3}}+(\mathrm{1}-\alpha )\mathrm{2}k{y}_{\mathrm{1}}$(2)

Hence, it follows from Eqs. (1) and (2) that

$({\rho }_{\alpha }(G_{\mathrm{3}})-\mathrm{2}\alpha k-(n-\mathrm{2}k-\mathrm{2}))y_{\mathrm{2}}=({\rho }_{\alpha }(G_{\mathrm{3}})-\mathrm{2}\alpha k)y_{\mathrm{3}}$

which implies $y_{\mathrm{2}}>y_{\mathrm{3}}$.

According to the Rayleigh quotient, we derive that

$\begin{array}{l}{\rho }_{\alpha }(G_{\mathrm{3}})-{\rho }_{\alpha }(G_{\mathrm{2}})\\ \ge {\mathit{Y}}^{\mathrm{T}}({\mathit{A}}_{\alpha }(G_{\mathrm{3}})-{\mathit{A}}_{\alpha }(G_{\mathrm{2}}))\mathit{Y}\\ =\alpha ((s-\mathrm{2}k)(\mathrm{2}n-\mathrm{3}s+\mathrm{2}k-\mathrm{4})y_{\mathrm{2}}^{\mathrm{2}}+(\mathrm{2}k-s)y_{\mathrm{3}}^{\mathrm{2}})+(\mathrm{1}-\alpha )((s-\mathrm{2}k)(\mathrm{2}n-\mathrm{3}s+\mathrm{2}k-\mathrm{3})y_{\mathrm{2}}^{\mathrm{2}}-\mathrm{2}(s-\mathrm{2}k)y_{\mathrm{2}}{y}_{\mathrm{3}})\\ =(s-\mathrm{2}k)(\alpha ((\mathrm{2}n-\mathrm{3}s+\mathrm{2}k-\mathrm{4})y_{\mathrm{2}}^{\mathrm{2}}-y_{\mathrm{3}}^{\mathrm{2}})+(\mathrm{1}-\alpha )((\mathrm{2}n-\mathrm{3}s+\mathrm{2}k-\mathrm{3})y_{\mathrm{2}}^{\mathrm{2}}-\mathrm{2}{y}_{\mathrm{2}}{y}_{\mathrm{3}}))\\ >(s-\mathrm{2}k)(\alpha ((\mathrm{2}n-\mathrm{3}s+\mathrm{2}k-\mathrm{4})y_{\mathrm{2}}^{\mathrm{2}}-y_{\mathrm{2}}^{\mathrm{2}})+(\mathrm{1}-\alpha )((\mathrm{2}n-\mathrm{3}s+\mathrm{2}k-\mathrm{3})y_{\mathrm{2}}^{\mathrm{2}}-\mathrm{2}{y}_{\mathrm{2}}^{\mathrm{2}}))(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e}{y}_{\mathrm{2}}>y_{\mathrm{3}})\\ =(s-\mathrm{2}k)(\alpha (\mathrm{2}n-\mathrm{3}s+\mathrm{2}k-\mathrm{5})y_{\mathrm{2}}^{\mathrm{2}}+(\mathrm{1}-\alpha )(\mathrm{2}n-\mathrm{3}s+\mathrm{2}k-\mathrm{5})y_{\mathrm{2}}^{\mathrm{2}})\\ =(s-\mathrm{2}k)(\mathrm{2}n-\mathrm{3}s+\mathrm{2}k-\mathrm{5})y_{\mathrm{2}}^{\mathrm{2}}\\ >\mathrm{0}(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{2}k\le s\le \frac{n}{\mathrm{2}}+k-\mathrm{1}\mathrm{a}\mathrm{n}\mathrm{d}n>\mathrm{2}k+\mathrm{4}).\end{array}$