Sufficient conditions for component factors in a graph

Abstract

Let G be a graph and $\mathcal {H}$ be a set of connected graphs. A spanning subgraph H of G is called an $\mathcal {H}$–factor if each component of H is isomorphic to a member of $\mathcal {H}$. In this paper, we first present a lower bound on the size (resp. the spectral radius) of G to guarantee that G has a $\{P_2,\, C_n: n\ge 3\}$–factor (or a perfect k–matching for even k) and construct extremal graphs to show all this bounds are best possible. We then provide a lower bound on the signless laplacian spectral radius of G to ensure that G has a $\{K_{1,j}:1\le j\le k\}$–factor, where $k\ge 2$ is an integer. Moreover, we also provide some Laplacian eigenvalue (resp. toughness) conditions for the existence of $\{P_2,\, C_{n}:n\ge 3\}$–factor, $P_{\ge 3}$–factor and $\{K_{1,j}: 1\le j\le k\}$–factor in G, respectively. Some of our results extend or improve the related existing results.

1 Introduction

All graphs considered in this paper are undirected, connected and simple. Let $G=(V(G),E(G))$ be a graph of order n and size m. For a subset $S\subseteq V(G)$, we use G[S] and $G-S$ to denote the subgraphs of G induced by S and $V(G)\backslash S$, respectively. We denote by $G_1\cup G_2$ the disjoint union of $G_1$ and $G_2$. The join $G_1\vee G_2$ is the graph obtained from $G_1\cup G_2$ by adding all possible edges between $V(G_1)$ and $V(G_2)$. As usual, the star and the cycle of order n are denoted by $K_{1,n-1}$ and $C_{n}$, respectively. Let $G_1$ and $G_2$ be two vertex–disjoint graphs. Other undefined notations can be found in [4].

The Laplacian and the signless Laplacian matrices of a graph G are defined as $L(G)=D(G)-A(G)$ and $Q(G)=D(G)+A(G)$, where A(G) and D(G) are the adjacency matrix and the diagonal degree matrix of G, respectively. The largest eigenvalues of A(G) and Q(G) are called the spectral radius and the signless Laplacian spectral radius of G, denoted by $\rho (G)$ and q(G), respectively. Let $\mu _{i}(G)$ (or $\mu _{i}$ for short) be the i-th smallest Laplacian eigenvalues of G. In particular, the second smallest Laplacian eigenvalue $\mu _{2}(G)$ is also known as the algebraic connectivity of G [8].

For a set of connected graphs $\mathcal {H}$, a spanning subgraph H of G is called an $\mathcal {H}$–factor if each component of H is isomorphic to a member of $\mathcal {H}$. In particular, an $\mathcal {H}$–factor is a ${{\mathcal {P}} }_{\ge \ell }$–factor if $\mathcal {H}=\{P_{\ell }, P_{\ell +1}, \ldots \}$, where $P_{\ell }$ is a path with $\ell$ vertices. Up to now, there have been lots of research work to seek the conditions for the existence of $\mathcal {H}$–factor in a graph, such as $\{P_2,\, C_n: n\ge 3\}$–factor [9, 17], $\{K_{1,j}:1\le j\le k\}$–factor [1], $P_{\ge 3}$–factor [13], etc. We won’t list them all here, but we’ll focus primarily on those related to the spectral conditions for the existence of $\mathcal {H}$–factor in a graph. Zhang [20] characterize the extremal graphs with maximum spectral radius among all connected graphs of given order with prescribed minimum degree and without a ${{\mathcal {P}} }_{\ge 2}$–factor or a ${{\mathcal {P}} }_{\ge 3}$–factor, which generalizes the result in Li and Miao [14]. In [16], Miao and Li give some sufficient conditions (size, the spectral radius, or the distance spectral radius) to ensure that a graph contains a $\{K_{1,j}:1\le j\le k\}$–factor, where $k\ge 2$ be an integer. We refer the readers to [18] for more details on graph factors.

Motivated by [17], it is natural and interesting to ask whether or not there is a spectral condition to guarantee the existence of $\{P_2,\, C_n: n\ge 3\}$–factor in a graph? Note that every even cycle can be decomposed into a combination of $P_2$. Then the above problem has another statement as follows: whether or not there is a spectral condition to guarantee the existence of $\{P_2,\, C_{2i+1}: i\ge 1\}$–factor in a graph?

Inspired by the ideas from Miao and Li [16] and using the typical spectral techniques, we provide some sufficient conditions to ensure a graph G contains a $\{P_2,\, C_{2i+1}: i\ge 1\}$–factor in terms of the size and the spectral radius of G, respectively.

Theorem 1.1

Let G be a graph of order n. Then we have

• (i) for $n\in \{3,4,10\}$ or $n\ge 12$, if $|E(G)|> \left( {\begin{array}{c}n-2\\ 2\end{array}}\right) +2,$ then G has a $\{P_2,\, C_{2i+1}: i\ge 1\}$–factor;

• (ii) for $n\in \{5,7,9,11\}$, if $|E(G)|> \frac{(3n-1)(n-1)}{8},$ then G has a $\{P_2,\, C_{2i+1}: i\ge 1\}$–factor;

• (iii) for $n\in \{6,8\}$, if $|E(G)|> \frac{3n(n-2)}{8},$ then G has a $\{P_2,\, C_{2i+1}: i\ge 1\}$–factor.

Theorem 1.2

Let G be a graph of order n. Then we have

• (i) for $n\in \{3,4,8\}$ or $n\ge 10$, if $\rho (G)> \theta (n)$, then G has a $\{P_2,\, C_{2i+1}: i\ge 1\}$–factor, where $\theta (n)$ is the largest root of $x^3-(n-4) x^2-(n-1)x+2(n-4)=0$;

• (ii) for $n\in \{5,7,9\}$, if $\rho (G)>\frac{n-3+\sqrt{5n^2-6n+5}}{4}$, then G has a $\{P_2,\, C_{2i+1}: i\ge 1\}$–factor;

• (iii) for $n=6$, if $\rho (G)>\frac{1+\sqrt{33}}{2}$, then G has a $\{P_2,\, C_{2i+1}: i\ge 1\}$–factor.

Recently, Miao and Li [16] provided a sufficient condition to ensure a graph G has a $\{K_{1,j}:1\le j\le k\}$–factor in terms of $\rho (G)$. As a continuance of their work, we further deduce a sharp lower bound on q(G) to ensure that G has a $\{K_{1,j}:1\le j\le k\}$–factor.

Theorem 1.3

Let $k\ge 2$ be an integer and G be a graph of order $n\ge 2k+12$. If

$$\begin{aligned} q(G)\ge q(K_1\vee (K_{n-k-2}\cup (k+1) K_1)), \end{aligned}$$

then G has a $\{K_{1,j}:1\le j\le k\}$–factor unless $K_1\vee (K_{n-k-2}\cup (k+1) K_1)$.

For a graph G of order n, the Laplacian eigenratio of G is defined as $\frac{\mu _{2}(G)}{\mu _{n}(G)}$, and it has attracted great concern about the relations between the Laplacian eigenratio and other graph properties [10, 11]. For example, it is known that for a graph G of order n, if n is even and $\frac{\mu _{2}(G)}{\mu _{n}(G)} \ge \frac{1}{2}$, then G has a perfect matching [5]. Furthermore, we also present the following Laplacian eigenratio conditions for the existence of $\{P_2, C_{2i+1}:i\ge 1\}$–factor, $P_{\ge 3}$–factor and $\{K_{1,j}:1\le j\le k\}$–factor in a graph, respectively.

Theorem 1.4

Let G be a graph of order n. If $\frac{\mu _{2}(G)}{\mu _{n}(G)} \ge \frac{1}{2}$, then G has a $\{P_2,\, C_{2i+1}:i\ge 1\}$–factor.

Theorem 1.5

Let G be a graph of order n. If $\frac{\mu _{2}(G)}{\mu _{n}(G)} \ge \frac{1}{3}$, then G has a $P_{\ge 3}$–factor.

Theorem 1.6

Let $k\ge 2$ be an integer and G be a graph of order n. If $\frac{\mu _{2}(G)}{\mu _{n}(G)} \ge \frac{1}{k+1}$, then G has a $\{K_{1,j}:1\le j\le k\}$–factor.

For $S \subseteq V(G)$, let $c(G-S)$ and $i(G-S)$ denote the numbers of connected components and isolated vertices in $G-S$, respectively. Recall that the toughness t(G) of a connected graph G is defined as

$$\begin{aligned} t(G)=\min \left\{ \frac{\vert S \vert }{c(G-S)}:\,\,S \subseteq V(G), \,\,c(G-S)\ge 2 \right\} . \end{aligned}$$

By convention, a complete graph has infinite toughness. This parameter was introduced by Chvátal [6] in 1973 and is closely related to many graph properties, including Hamiltonicity, pancyclicity and spanning trees, see [3]. Very recently, Gu and Haemers [11] derived a lower bound of t(G) by building the relationship between the toughness of G and its Laplacian eigenvalues as follows: $t(G) \ge \frac{\mu _{2}(G)}{\mu _{n}(G)-\mu _{2}(G)}$ for any connected graph G of order n. This implies that $t(G) \ge \frac{\mu _{2}(G)}{\mu _{n}(G)-\mu _{2}(G)}\ge 1$ when $\frac{\mu _{2}(G)}{\mu _{n}(G)} \ge \frac{1}{2}$. Thus, by Theorem 1.4, we have the following result immediately.

Corollary 1.7

Let G be a graph of order n. If $t(G)\ge 1$, then G has a $\{P_2,\, C_{2i+1}:i\ge 1\}$–factor.

Remark 1.1

Recall that Guan, Ma and Shi [12] showed that for a graph G, if $t(G)\ge 1$, then G has a $\{P_2, C_{2i+1}:i\ge 2\}$–factor. So Theorem 1.4 can be viewed as a generalization of their result.

Similarly, we have the following results for the existence of $P_{\ge 3}$–factor and $\{K_{1,j}:1\le j\le k\}$–factor in a graph, respectively.

Corollary 1.8

Let G be a graph of order n. If $t(G)\ge \frac{1}{2}$, then G has a $P_{\ge 3}$–factor.

Remark 1.2

Recall that Zhou et al. [21] showed that for a graph G, if $t(G)\ge \frac{2}{3}$, then G has a ${{\mathcal {P}} }_{\ge 3}$–factor. So Corollary 1.8 can be viewed as a slight improvement of the result due to Zhou et.al.

Corollary 1.9

Let $k\ge 2$ be an integer and G be a graph of order n. If $t(G)\ge \frac{1}{k}$, then G has a $\{K_{1,j}:1\le j\le k\}$–factor.

A matching M in a graph is a set of pairwise non-adjacent edges. A perfect matching is a matching which matches all vertices of a graph. Let $f: E(G) \rightarrow \{0,1, \ldots , k\}$ be an assignment such that the sum of weights of edges incident with any vertex is at most k, i.e., $\sum _{e \sim v} f(e) \le k$ for any vertex $v \in V(G)$. A k--matching is a subgraph induced by the edges with weight among $1, \ldots , k$ such that $\sum _{e \sim v} f(e) \le k$. The sum of all weights, i.e., $\sum _{e \in E(G)} f(e)$, is called the size of a k-matching f. A k–matching is perfect if $\sum _{e \sim v} f(e)=k$ for every vertex $v \in V(G)$. Clearly, a k–matching is perfect if and only if its size is k|V(G)| /2. For $k=1$, the perfect 1–matching is also known as the the perfect matching. For $k=2$, Tutte [17] showed that a connected graph G has a perfect 2–matching is equivalent to G has a $\{P_2,\, C_{2i+1}: i\ge 1\}$–factor. Moreover, for $k\ge 4$ be an even integer, it was shown by Lu and Wang [15] that a graph G contains a perfect k–matching if and only if G contains a perfect 2–matching. Thus, by Theorems 1.1, 1.2 and 1.4 and Corollary 1.7, we then have the following conditions for a graph G has a perfect k–matching.

Corollary 1.10

Let $k\ge 2$ be an even integer and G be a graph of order n. Then we have

• (i) for $n\in \{3,4,10\}$ or $n\ge 12$, if $|E(G)|> \left( {\begin{array}{c}n-2\\ 2\end{array}}\right) +2,$ then G has a perfect k–matching;

• (ii) for $n\in \{5,7,9,11\}$, if $|E(G)|> \frac{(3n-1)(n-1)}{8},$ then G has a perfect k–matching;

• (iii) for $n\in \{6,8\}$, if $|E(G)|> \frac{3n(n-2)}{8},$ then G has a perfect k–matching.

Corollary 1.11

Let $k\ge 2$ be an even integer and G be a graph of order n. Then we have

• (i) for $n\in \{3,4,8\}$ or $n\ge 10$, if $\rho (G)> \theta (n)$, then G has a perfect k–matching, where $\theta (n)$ is the largest root of $x^3-(n-4) x^2-(n-1)x+2(n-4)=0$;

• (ii) for $n\in \{5,7,9\}$, if $\rho (G)>\frac{n-3+\sqrt{5n^2-6n+5}}{4}$, then G has a perfect k–matching;

• (iii) for $n=6$, if $\rho (G)>\frac{1+\sqrt{33}}{2}$, then G has a perfect k–matching.

Corollary 1.12

Let $k\ge 2$ be an even integer and G be a graph of order n. If $\frac{\mu _{2}(G)}{\mu _{n}(G)} \ge \frac{1}{2}$, then G has a perfect k–matching.

Corollary 1.13

Let $k\ge 2$ be an even integer and G be a graph of order n. If $t(G) \ge 1$, then G has a perfect k–matching.

The remainder of the paper is organized as follows. In Section 2, we present some preliminary results, which will be used in the subsequent section. In Section 3, we will give the proofs of Theorems 1.1, 1.2 and 1.3, respectively. In section 4, we present the proofs of Theorems 1.4, 1.5 and 1.6, respectively. In the last section we construct extremal graph to show the bound obtained in Theorems 1.1 and 1.2 are best possible.

2 Preliminary

In this section, we present some preliminary results and lemmas which are useful.

Let $\phi _{M}(x):=det(xI-M)$ be the characteristic polynomial of a square matrix M, where I is the identity matrix, whose order is the same as that of M. Consider an $n\times n$ real symmetric matrix

$$\begin{aligned} M=\left( \begin{array}{cccc} M_{1,1} &{} M_{1,2} &{} \cdots &{} M_{1, m} \\ M_{2,1} &{} M_{2,2} &{} \cdots &{} M_{2, m} \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ M_{m, 1} &{} M_{m, 2} &{} \cdots &{} M_{m, m} \end{array}\right) , \end{aligned}$$

whose rows and columns are partitioned according to a partitioning $X_{1}, X_{2}, \ldots , X_{m}$ of $\{1,2, \ldots , n\}$. The quotient matrix $\mathcal {B}$ of the matrix M is the $m \times m$ matrix whose entries are the average row sums of the blocks $M_{i, j}$ of M. The partition is equitable if each block $M_{i, j}$ of M has constant row (and column) sum.

Lemma 2.1

([19]) Let M be a square 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 and $M_{\pi }$ is irreducible, then the largest eigenvalues of M and $M_{\pi }$ are equal.

Lemma 2.2

([17]) A graph G has a $\{P_2,\, C_{2i+1}: i\ge 1\}$–factor if and only if $i(G-S)\le |S|$ for any $S \subseteq V(G)$.

Lemma 2.3

([1]) For any integer $k\ge 2$, a graph G has a $\{K_{1,j}:1\le j\le k\}$–factor if and only if $i(G-S)\le k |S|$ for any $S \subseteq V(G)$.

In order to prove the main results for $\rho (G)$ and q(G) simultaneously, we introduce the matrix $A_{a}(G) = aD(G) + A(G)$ and denote by $\rho _{a}(G)$ the largest eigenvalue of $A_{a}(G)$, where $a\ge 0$. Clearly, $A_0(G) = A(G)$ (resp. $A_1(G) = Q(G)$) and $\rho _{0}(G)=\rho (G)$ (resp. $\rho _{1}(G) = q(G)$).

Lemma 2.4

([2]) If H is a spanning subgraph of a graph G, then $\rho _{a}(H)\le \rho _{a}(G)$, with equality if and only if $G\cong H$; Moreover, if H is a proper subgraph of G, then $\rho _{a}(H)< \rho _{a}(G)$.

Lemma 2.5

([7]) Let G be a graph of order n with m edges. Then

$$\begin{aligned} q(G)\le \frac{2m}{n-1}+n-2. \end{aligned}$$

3 Proofs of Theorems 1.1, 1.2 and 1.3

We now give the proofs of Theorems 1.1, 1.2 and 1.3, respectively.

Proof of Theorem 1.1:

Suppose to the contrary that G has no $\left\{ P_{2},\, C_{2i+1}: i \ge 1\right\}$–factor. Then Lemma 2.2 implies that there exists a non-empty subset $S \subseteq V(G)$ satisfying $i(G-S) \geqslant |S|+1$. We choose such a connected graph G of order n so that its size is as large as possible. According to the choice of G, we see that the induced subgraph G[S] and each connected component of $G-S$ are complete graphs, respectively, and $G \cong G[S] \vee (G-S)$.

First, we claim that there is at most one non-trivial connected component in $G-S$. Otherwise, we can add edges among all nontrivial connected components to get a bigger non-trivial connected component, which is a contradiction to the choice of G. For convenience, let $i(G-S)=i$ and $|S|=s$. We now consider the following two possible cases.

Case 1   $G-S$ has only one non-trivial connected component, say $G_1$.

In this case, let $\left| V\left( G_1\right) \right| =n_1 \ge 2$. We are to show $i=s+1$. If $i \ge s+2$, let $H_1$ be a new graph obtained from G by joining each vertex of $G_1$ with one vertex in $I(G-S)$ by an edge, where $I(G-S)$ is a set of isolated vertices in $G-S$. Then we have $\left| E\left( H_1\right) \right| =|E(G)|+n_1>|E(G)|$ and $i\left( H_1-S\right) \ge s+1$, a contradiction to the choice of G. Hence $i \le s+1$. Recall that $i \ge s+1$. Therefore, we have $i=s+1$ and $G=K_s \vee \left( K_{n-2s-1} \cup (s+1) K_1\right)$.

Bear in mind that $n=s+s+1+n_1 \ge 2s+3 \ge 5$ and $|E(G)|=s(s+1)+\left( {\begin{array}{c}n-s-1\\ 2\end{array}}\right)$. By a directed calculation, we have

$$\begin{aligned} \begin{aligned} \left( {\begin{array}{c}n-2\\ 2\end{array}}\right) +2-|E(G)|&=\frac{1}{2}(s-1)(2n-3s-8)\\&\ge \frac{1}{2}(s-1)(4s+6-3s-8)\\&=\frac{1}{2}(s-1)(s-2)\ge 0. \end{aligned} \end{aligned}$$

Thus, $|E(G)| \le \left( {\begin{array}{c}n-2\\ 2\end{array}}\right) +2$ for $n \ge 5$. By a direct calculation, we have $\left( {\begin{array}{c}n-2\\ 2\end{array}}\right) +2<\frac{(3n-1)(n-1)}{8}$ for $n\in \{5,7,9,11\}$ and $\left( {\begin{array}{c}n-2\\ 2\end{array}}\right) +2<\frac{3n(n-2)}{8}$ for $n\in \{6, 8\}$, a contradiction.

Case 2   $G-S$ has no non-trivial connected component.

In this case, we are to prove $i \le s+2$. If $i\ge s+3$, let $H_2$ be a new graph obtained from G by adding an edge in $I(G-S)$. Clearly, $i(H_2-S) \ge s+1$, $H_2-S$ has exactly one non-trivial connected component and $|E(G)|<|E(H_2)|$, contradicting to the choice of G. Bear in mind that $i \ge s+1$, it suffices to consider $i=s+1$ (i.e., $n=2s+1$) and $i=s+2$ (i.e., $n=2s+2$).

For $i=s+1$, we have $G\cong K_s\vee (s+1)K_1$. Therefore, $n=2s+1$ and $|E(G)|=s(s+1)+\left( {\begin{array}{c}s\\ 2\end{array}}\right)$. By a directed calculation, we have

$$\begin{aligned} \left( {\begin{array}{c}n-2\\ 2\end{array}}\right) +2-|E(G)|=\left( {\begin{array}{c}2s-1\\ 2\end{array}}\right) +2-s(s+1)-\left( {\begin{array}{c}s\\ 2\end{array}}\right) =\frac{1}{2}(s-1)(s-6). \end{aligned}$$

Thus, $|E(G)| \le \left( {\begin{array}{c}n-2\\ 2\end{array}}\right) +2$ for $s=1$ or $s\ge 6$, which is a contradiction for $n=3$ or $n\ge 13$. For $s\in \{2,3,4,5\}$ (or $n\in \{5,7,9,11\}$), we have

$$\begin{aligned} |E(G)|=s(s+1)+\left( {\begin{array}{c}s\\ 2\end{array}}\right) =\frac{3s^2+s}{2}=\frac{3(n-1)^2+2(n-1)}{8} =\frac{(3n-1)(n-1)}{8}, \end{aligned}$$

a contradiction.

For $i=s+2$, we have $G\cong K_s\vee (s+2)K_1$. Therefore, $n=2s+2$ and $|E(G)|=s(s+2)+\left( {\begin{array}{c}s\\ 2\end{array}}\right)$. By a directed calculation, we have

$$\begin{aligned} \left( {\begin{array}{c}n-2\\ 2\end{array}}\right) +2-|E(G)|=\left( {\begin{array}{c}2s\\ 2\end{array}}\right) +2-s(s+2)-\left( {\begin{array}{c}s\\ 2\end{array}}\right) =\frac{1}{2}(s-1)(s-4). \end{aligned}$$

Thus, $|E(G)| \le \left( {\begin{array}{c}n-2\\ 2\end{array}}\right) +2$ for $s=1$ or $s\ge 4$, which is a contradiction for $n=4$ or $n\ge 10$. For $s\in \{2,3\}$ (or $n\in \{6,8\}$), we have

$$\begin{aligned} |E(G)|=s(s+2)+\left( {\begin{array}{c}s\\ 2\end{array}}\right) =\frac{3s^2+3s}{2} =\frac{3}{2}\cdot \frac{n-1}{2}\cdot \left( \frac{n-1}{2}+1\right) =\frac{3n(n-2)}{8}, \end{aligned}$$

a contradiction.

In view of Cases 1 and 2, the proof of Theorem 1.1 is completed.  $\square$

Proof of Theorem 1.2:

Suppose to the contrary that G has no $\left\{ P_{2},\, C_{2i+1}: i\ge 1\right\}$–factor. Then Lemma 2.1 implies that there exists a non-empty subset $S \subseteq V(G)$ satisfying $i(G-S) \ge |S|+1$. We choose such a connected graph G of order n so that its adjacency spectral radius is as large as possible. By Lemma 2.4 and the choice of G, it follows that the induced subgraph G[S] and each connected component of $G-S$ are complete graphs, respectively, and $G\cong G[S] \vee (G-S)$.

For convenience, let $i(G-S)=i$ and $|S|=s$. We claim that there exists at most one non–trivial connected component in $G-S$. Otherwise, we can add edges among all non–trivial connected components to get a non–trivial connected component of larger size, which gives a contradiction (based on Lemma 2.4). Let $\phi (x)=x^3-(n-4) x^2-(n-1)x+2(n-4)$ be a real function in x and $\theta (n)$ be the largest root of $\phi (x)=0$. We now consider the following two possible cases.

Case 1   $G-S$ has just one non–trivial connected component, say $G_1$.

In this case, let $|V(G_1)|=n_1 \ge 2$. We are to show $i=s+1$. If $i \ge s+2$, let $H_1$ be a new graph obtained from G by joining each vertex of $G_1$ with one vertex in $I(G-S)$ by an edge. Then $i(H_1-S) \ge s+1$ and G is a proper spanning subgraph of $H_1$. By Lemma 2.4, we have $\rho (G)<\rho (H_1)$, a contradiction to the choice of G. Therefore $i \le s+1$. Recall that $i \ge s+1$. Hence $i=s+1$ $(n=2s+1+n_1 \ge 2s+3)$ and $G=K_s \vee (K_{n-2s-1} \cup (s+1) K_1)$. We now consider the partition $V(G)=V(K_s) \cup V((s+1) K_1) \cup V(K_{n-2s-1})$. Then the corresponding quotient matrix of A(G) is

$$\begin{aligned} B_1=\left( \begin{array}{ccc} s-1 &{} s+1 &{} n-2s-1 \\ s &{} 0 &{} 0 \\ s &{} 0 &{} n-2s-2 \end{array}\right) , \end{aligned}$$

and the characteristic polynomial of $B_1$ is

$$\begin{aligned} \begin{aligned} \phi _{B_1}(x)&=x^3-(n-s-3) x^2-\left( s^2+n-2\right) x-s^3 +n s^2-s^3-3s^2+n s-s^2-2s\\&=x^3-(n-s-3) x^2-\left( s^2+n-2\right) x-2s^3 +(n-4)s^2+(n-2)s. \end{aligned} \end{aligned}$$

Note that the partition $V(G)=V(K_s) \cup V((s+1) K_1) \cup V(K_{n-2s-1})$ is equitable. Then Lemma 2.1 implies that the largest root $\theta _1$ of $\phi _{B_1}(x)=0$ is $\rho (G)$. In order to prove $\theta (n)\ge \rho (G)$, it suffices to show $\phi (\theta _1)<0$. Note that

$$\begin{aligned} \begin{aligned} \phi (\theta _1)&=\phi (\theta _1)-\phi _{B_1}(\theta _1)\\&=(s-1)(-\theta _{1}^2+(s+1)\theta _1+2s^2+s-ns+1-n+4s+4-n+s+3)\\&=(s-1)(-\theta _{1}^2+(s+1)\theta _1+2s^2-(n-6)s-2n+8)\\&\le (s-1)(-\theta _{1}^2+(s+1)\theta _1+2-s)\\&=(s-1)(-\theta _{1}(\theta _1-s-1)+2-s) \end{aligned} \end{aligned}$$

and $K_{s+2}$ is a proper subgraph of $K_s \vee (K_{n-2s-1} \cup (s+1) K_1)$. Then by Lemma 2.4, we have $\rho (G)=\theta _1>s+1$. It follows that

$$\begin{aligned} \begin{aligned} \phi (\theta _1)&=\phi (\theta _1)-\phi _{B_1}(\theta _1)\\&\le (s-1)(-\theta _{1}(\theta _1-s-1)+2-s)\\&<(s-1)(-\theta _1 (s+1-s-1)+2-s)\\&=(s-1)(2-s)\le 0. \end{aligned} \end{aligned}$$

Thus, $\phi (\theta _1)=\phi (\theta _1)-\phi _{B_1}(\theta _1)<0$ for $s\ge 1$, which implies that $\rho (G)=\theta _1<\theta (n)$, a contradiction for $n\notin \{5,6,7,9\}$. By a direct calculation, we have $\theta (5)\thickapprox 2.34< \frac{5-3+\sqrt{5\times 5^2-6\times 5+5}}{4}=3$, $\theta (6)\thickapprox 3.18< \frac{1+\sqrt{33}}{2}$, $\theta (7)\thickapprox 4.11< \frac{7-3+\sqrt{5\times 7^2-6\times 7+5}}{4}=1+\sqrt{33}$, $\theta (9)\thickapprox 6.05< \frac{9-3+\sqrt{5\times 9^2-6\times 9+5}}{4}=\frac{3+\sqrt{89}}{2}$.

Case 2   $G-S$ has no non–trivial connected component.

If $i\ge s + 3$, let $H_2$ be a graph obtained form G by adding an edge in $I(G-S)$. Then $i(H_2-S)\ge s+1$ and $H_2-S$ has one non–trivial connected component. From Case 1, we have $\rho (G)\le \theta (n)$, contradicting to the choice of G. Thus, we only consider $i=s+1$ (i.e., $n=2s+1$) and $i=s+2$ (i.e., $n=2s+2$).

For $i=s+1$, we have $G\cong K_s\vee (s+1)K_1$ and $n=2s+1$. One may consider the partition $V(K_s \vee (s+1) K_1)=V(K_s) \cup V((s+ 1) K_1)$. Its corresponding quotient matrix of $A(K_s \vee (k s+1) K_1)$ is

$$\begin{aligned} B_2=\left( \begin{array}{cc} s-1 &{} s+1 \\ s &{} 0 \end{array}\right) . \end{aligned}$$

And the characteristic polynomial of $B_2$ is

$$\begin{aligned} \phi _{B_2}(x)=x^2-(s-1) x-s(s+1). \end{aligned}$$

Note that the partition $V(K_s \vee (s+1) K_1)=V(K_s) \cup V((s+ 1) K_1)$ is equitable. Then Lemma 2.1 implies that $\rho (K_s \vee (s+1) K_1)$ is the largest root $\theta _2$ of $\phi _{B_2}(x)=0$. As $\phi _{B_2}(x)=0$ is a quadratic equation with respect to x, by the root formula, we have

$$\begin{aligned} \rho \left( K_s \vee (s+1) K_1\right) =\theta _2=\frac{s-1+\sqrt{5 s^2+2 s+1}}{2}. \end{aligned}$$

If $s=1$, then we have $n=3$, $\phi _{B_2}(x)=x^2-2$ and $\phi (x)=(x+1)(x^2-1)$. That is $\theta _2=\sqrt{2}=\theta (3)$, a contradiction to the condition. If $s\in \{2,3,4\}$ (or $n\in \{5,7,9\}$), then $\theta _2=\frac{s-1+\sqrt{5 s^2+2 s+1}}{2}=\frac{n-3+\sqrt{5 n^2-6n+5}}{4}$, a contradiction. Next, we consider $s\ge 5$. Note that

$$\begin{aligned} \begin{aligned} \phi (\theta _2)=\phi (\theta _2)-\theta _2 \phi _{B_2}(\theta _2)&= -(s-2)\theta _{2}^{2}+(s^2-s)\theta _2+4s-6\\&=-\theta _2[(s-2)\theta _2-s^2+s]+4s-6, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \theta _2=\frac{s-1+\sqrt{5 s^2+2 s+1}}{2} =\frac{s-1+\sqrt{(2s+1)^2+s^2-2s}}{2}>\frac{3s}{2}. \end{aligned}$$

We then have

$$\begin{aligned} \begin{aligned} \phi (\theta _2)&=\phi (\theta _2)-\theta _2 \phi _{B_2}(\theta _2)\\&=-\theta _2[(s-2)\theta _2-s^2+s]+4s-6\\&<-\theta _2[(s-2)\cdot \frac{3s}{2}-s^2+s]+4s-6\\&=-\theta _2 (\frac{s^2}{2}-2s)+4s-6\\&<-\frac{3s^3}{4}+3s^2+4s-6. \end{aligned} \end{aligned}$$

Let $f(s)=-\frac{3s^3}{4}+3s^2+4s-6$. Note that $f'(s)=-\frac{9}{4}s^2+6s+4<0$ when $s\ge 5$. Hence f(s) is a monotonically decreasing functions with respect to s. Therefore,

$$\begin{aligned} \begin{aligned} \phi (\theta _2)&=\phi (\theta _2)-\theta _2 \phi _{B_2}(\theta _2)<-\frac{3s^3}{4}+3s^2+4s-6\\&\le f(5)=-\frac{19}{4}<0. \end{aligned} \end{aligned}$$

Thus, $\phi (\theta _2)<0$ for $s\ge 5$ and $\rho (G)=\theta _2<\theta (n)$, a contradiction.

For $i=s+2$. In this subcase one has $G=K_s \vee (s+2) K_1$ and $n=2s+2$. We consider the partition $V(G)=V(K_s) \cup V((s+2) K_1)$. Then the corresponding quotient matrix of A(G) is

$$\begin{aligned} B_3=\left( \begin{array}{cc} s-1 &{} s+2 \\ s &{} 0 \end{array}\right) , \end{aligned}$$

and its characteristic polynomial is

$$\begin{aligned} \phi _{B_3}(x)=x^2-(s-1) x-s(s+2). \end{aligned}$$

Note that the partition $V(G)=V(K_s) \cup V((s+2) K_1)$ is equitable. Then Lemma 2.1 implies that the largest root $\theta _3$ of $\phi _{B_3}(x)=0$ is $\rho (G)$. As $\phi _{B_3}(x)=0$ is a quadratic equation with respect to x, we may easily obtain that

$$\begin{aligned} \theta _3=\rho (G)=\frac{s-1+\sqrt{5s^2+6 s+1}}{2} . \end{aligned}$$

If $s=1$, then $n=4$, we have $\phi _{B_3}(x)=x^2-3$ and $\phi (x)=x(x^2-3)$, which implies that $\theta _3=\sqrt{3}=\theta (4)$, a contradiction to the condition. If $s=2$ (or $n=6$), then $\theta _3=\frac{1+\sqrt{33}}{2}$, a contradiction. Next, we consider $s\ge 3$.

Bear in mind that $n=2s+2$ and so

$$\begin{aligned} \phi (x)=x^3-(2s-2)x^2-(2s+1)x-4(s-1). \end{aligned}$$

In what follows, it suffices to prove $\phi (\theta _3)<0$. Note that

$$\begin{aligned} \begin{aligned} \phi (\theta _3)&=\phi (\theta _3)-\theta _3 \phi _{B_3}(\theta _3)\\&= -\theta _{3}^{2}(s-1)+\theta _3 (s^2-1)+4(s-1)\\&=-(s-1)(\theta _{3}^{2}-(s+1)\theta _{3}-4), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \theta _3 =\frac{s-1+\sqrt{5 s^2+6s+1}}{2} =\frac{s-1+\sqrt{(s+5)^2+4(s^2-s-6)}}{2}> s+2. \end{aligned}$$

Then we have

$$\begin{aligned} \begin{aligned} \phi (\theta _3)&=\phi (\theta _3)-\theta _3 \phi _{B_3}(\theta _3)\\&=-(s-1)(\theta _{3}^{2}-(s+1)\theta _{3}-4)\\&< -(s-1)[(s+2)^{2}-(s+1)(s+2)-4]\\&=-(s-1)(s-2)<0. \end{aligned} \end{aligned}$$

Thus, $\phi (\theta _3)<0$ for $s\ge 3$ and $\rho (G)=\theta _3<\theta (n)$, a contradiction.

In view of Cases 1 and 2, the proof of Theorem 1.2 is completed.  $\square$

Proof of Theorem 1.3:

Assume that G has no $\{K_{1,j}:1\le j\le k\}$–factor. Then Lemma 2.3 implies that there exists some nonempty subset $S\subseteq V(G)$ such that $i(G-S)\ge k |S|+1$. Let $|S|=s$. Then G is a spanning subgraph of $G_1=K_{s}\vee (K_{n-(k+1)s-1}\cup (ks+1) K_{1})$. Hence Lemma 2.4 implies that

$$\begin{aligned} q(G)\le q(G_1), \end{aligned}$$

(1)

with equality if and only if $G\cong G_1$.

If $s=1$, then $G_1\cong K_{1}\vee (K_{n-k-2} \cup (k+1) K_{1})$. Combining this with (1), we conclude that

$$\begin{aligned} q(G)\le q(K_{1}\vee (K_{n-k-2} \cup (k+1) K_{1})), \end{aligned}$$

where the equality holds if and only if $G\cong K_{1}\vee (K_{n-k-2}\cup (k+1) K_{1})$.

For $s\ge 2$, note that Lemma 2.5 implies that

$$\begin{aligned} \begin{aligned} q(G_1)&\le \frac{2m(G_1)}{n-1}+n-2\\&=\frac{(n-ks-1)(n-ks-2)+2s(ks+1)}{n-1}+n-2\\&=\frac{(k^2+2k)s^2-(2kn-3k-2)s+2n^2-6n+4}{n-1}. \end{aligned} \end{aligned}$$

(2)

Let $f(s)=(k^2+2k)s^2-(2kn-3k-2)s+2n^2-6n+4$. Since $n\ge (k+1)s+1$, then $2\le s\le \frac{n-1}{k+1}$. By a simple calculation, we have

$$\begin{aligned} \begin{aligned} f(2)-f\left( \frac{n-1}{k+1}\right)&=(k^2+2k)\left( 2^2-\left( \frac{n-1}{k+1}\right) ^2\right) -(2kn-3k-2)\left( 2-\frac{n-1}{k+1}\right) \\&=\frac{(n-2k-3)(-2k^3+(n-8)k^2-7k-2)}{(k+1)^2}\\&\ge \frac{9(k-2)(4k+1)}{(k+1)^2} \ge 0, \end{aligned} \end{aligned}$$

(3)

where the inequality follows from the fact that $n\ge 2k+12$ and $k\ge 2$. This implies that, for $2\le s\le \frac{n-1}{k+1}$, the maximum value of f(s) is attained at $s=2$. This together with (2) and (3) imply that

$$\begin{aligned} \begin{aligned} q(G_1) \le \frac{f(2)}{n-1}&= \frac{2n^2-(4k+6)n+4k^2+14k+8}{n-1}\\&= 2(n-k-2)-\frac{2(n-2k-6)k-4}{n-1}\\&\le 2(n-k-2)-\frac{12k-4}{n-1}\\&<2(n-k-2), \end{aligned} \end{aligned}$$

(4)

where the penultimate inequality follows from $n\ge 2k+12$. Note that $K_{n-k-1}$ is a proper subgraph of $K_1\vee (K_{n-k-2}\cup (k+1) K_1)$. Then we have

$$\begin{aligned} q(K_1\vee (K_{n-k-2}\cup (k+1) K_1))>q(K_{n-k-1})=2(n-k-2). \end{aligned}$$

Combining this with (1) and (4), we have

$$\begin{aligned} q(G)\le q(G_1)<q(K_1\vee (K_{n-k-2}\cup (k+1) K_1)). \end{aligned}$$

Concluding the above results, we obtain

$$\begin{aligned} q(G)\le q(K_1\vee (K_{n-k-2}\cup (k+1) K_1)), \end{aligned}$$

where the equality holds if and only if $G\cong K_1\vee (K_{n-k-2}\cup (k+1) K_1)$. Let $S=\{u\}$ denote the unique vertex of degree $n-1$ in $K_1\vee (K_{n-k-2}\cup (k+1) K_1)$. Then $i(K_1\vee (K_{n-k-2}\cup (k+1) K_1)-S)\ge k+1>|S|$ since u is adjacent to $k+1$ pendant vertices. Thus by Lemma 2.3, we have $K_1\vee (K_{n-k-2}\cup (k+1) K_1)$ has no $\{K_{1,j}:1\le j\le k\}$–factor, and so the result follows.   $\square$

4 Proofs of Theorems 1.4, 1.5 and 1.6

In this section, we will present the following Laplacian eigenvalue conditions for the existence of $\{P_2,\, C_{2i+1}:i\ge 1\}$–factor, $P_{\ge 3}$–factor and $\{K_{1,j}: 1\le j\le k\}$–factor, respectively. In order to prove our results, the following lemma is needed.

Lemma 4.1

([10]) Let G be a graph of order n. Suppose that $S\subseteq V(G)$ such that $G-S$ is disconnected. Let X and Y be disjoint vertex subsets of $G-S$ such that $X\cup Y = V (G)-S$ with $|X|\le |Y|$. Then

$$\begin{aligned} |X|\le \frac{\mu _n-\mu _2}{2\mu _n}n, \end{aligned}$$

and

$$\begin{aligned} |S|\ge \frac{2\mu _2}{\mu _n - \mu _2}|X|. \end{aligned}$$

Now we are in a position to give the proofs of our results. Our strategy for proving the following results is employed the similar argument as that was used in [10].

Proof of Theorem 1.4:

Suppose to the contrary that G has no a $\{P_2,C_{2i+1}:i\ge 1\}$–factor. Then Lemma 2.2 implies that there exists a nonempty subset $S\subseteq V(G)$ such that

$$\begin{aligned} i(G-S)\ge |S|+1\ge 2. \end{aligned}$$

Let $v_1, v_2,\ldots , v_c$ be the isolated vertices of $G-S$, where $c=i(G-S)$. Defined $X=\cup _{1\le i\le \lfloor c/2 \rfloor }v_i$ and $Y=V(G)-S-X$. Then $\frac{c-1}{2}\le |X|\le |Y|$. By Lemma 4.1, we have

$$\begin{aligned} |S|\ge \frac{2\mu _2}{\mu _n - \mu _2}\cdot \frac{c-1}{2}, \end{aligned}$$

with equality holding only when $|X|= |Y|= \frac{c-1}{2}$.

However, note that $2\mu _2\ge \mu _n$ and $X \ge \frac{c}{2}>\frac{c-1}{2}$ when c is even. Then we have

$$\begin{aligned} |S|\ge \frac{2\mu _2}{\mu _n - \mu _2}\cdot \frac{c}{2}\ge c=i(G-S). \end{aligned}$$

When c is odd, by the definitions of X and Y, we have $|X|<|Y|$. Thus the equality never hold. Then

$$\begin{aligned} |S|> \frac{2\mu _2}{\mu _n - \mu _2}\cdot \frac{c-1}{2}\ge c-1=i(G-S)-1, \end{aligned}$$

a contradiction.   $\square$

Let $sun(G-S)$ denote the number of sun components in $G-S$. It is proved in [13] that every graph G admits a $P_{\ge 3}$–factor if and only if $sun(G-X)\le 2|X|$ for every $X \subseteq V(G)$. Then we have the following result.

Proof of Theorem 1.5:

Suppose to the contrary that G has no a $P_{\ge 3}$–factor. Then there exists a nonempty subset $S\subseteq V(G)$ such that

$$\begin{aligned} sun(G-S)\ge 2 |S|+1\ge 3. \end{aligned}$$

Let $V_1, V_2,\ldots , V_c$ be the sun components of $G-S$, where $c=sun(G-S)$. Without loss of generality, suppose that $|V_1|\le |V_2|\le \cdots \le |V_c|$. Defined $X=\cup _{1\le i\le \lfloor c/2 \rfloor }V_i$ and $Y=V(G)-S-X$. Then $\frac{c-1}{2}\le |X|\le |Y|$. By Lemma 4.1, we have

$$\begin{aligned} |S|\ge \frac{2\mu _2}{\mu _n - \mu _2}\cdot \frac{c-1}{2}, \end{aligned}$$

with equality holding only when $|X|= |Y|= \frac{c-1}{2}$.

However, note that $3\mu _2\ge \mu _n$ and $X \ge \frac{c}{2}>\frac{c-1}{2}$ when c is even. Then

$$\begin{aligned} |S|\ge \frac{2\mu _2}{\mu _n - \mu _2}\cdot \frac{c}{2}\ge \frac{c}{2}=\frac{1}{2}\cdot sun(G-S). \end{aligned}$$

When c is odd, by the definitions of X and Y, we have $|X|<|Y|$. Thus the equality never hold. Then

$$\begin{aligned} |S|> \frac{2\mu _2}{\mu _n - \mu _2}\cdot \frac{c-1}{2}\ge \frac{c-1}{2}=\frac{1}{2}\cdot (sun(G-S)-1), \end{aligned}$$

a contradiction.   $\square$

Proof of Theorem 1.6:

Suppose to the contrary that G has no a $\{K_{1,j}:1\le j\le k\}$–factor. Then Lemma 2.3 implies that there exists a nonempty subset $S\subseteq V(G)$ such that

$$\begin{aligned} i(G-S)\ge k |S|+1\ge 3. \end{aligned}$$

Let $v_1, v_2,\ldots , v_c$ be the isolated vertices of $G-S$, where $c=i(G-S)$. Defined $X=\cup _{1\le i\le \lfloor c/2 \rfloor }v_i$ and $Y=V(G)-S-X$. Then $\frac{c-1}{2}\le |X|\le |Y|$. By Lemma 4.1, we have

$$\begin{aligned} |S|\ge \frac{2\mu _2}{\mu _n - \mu _2}\cdot \frac{c-1}{2}, \end{aligned}$$

with equality holding only when $|X|= |Y|= \frac{c-1}{2}$.

However, note that $(k+1)\mu _2\ge \mu _n$ and $X \ge \frac{c}{2}>\frac{c-1}{2}$ when c is even. Then we have

$$\begin{aligned} |S|\ge \frac{2\mu _2}{\mu _n - \mu _2}\cdot \frac{c}{2}\ge \frac{c}{k}=\frac{1}{k}\cdot i(G-S). \end{aligned}$$

When c is odd, by the definitions of X and Y, we have $|X|<|Y|$. Thus the equality never hold. Then

$$\begin{aligned} |S|> \frac{2\mu _2}{\mu _n - \mu _2}\cdot \frac{c-1}{2}\ge \frac{c-1}{k}=\frac{1}{k}\cdot (i(G-S)-1), \end{aligned}$$

a contradiction.   $\square$

5 Extremal graphs

In this section, we construct the extremal graphs to show that the bounds established in Theorems 1.1 and 1.2 are best possible, respectively.

Note that for a positive integer n, we have

• (i) $|E(K_1\vee (K_{n-3}\cup 2 K_1))|=\left( {\begin{array}{c}n-2\\ 2\end{array}}\right) +2$ for $n\in \{3,4,10\}$ or $n\ge 12$;

• (ii) $|E\left( K_{\frac{n-1}{2}}\vee \frac{n+1}{2} K_1\right) |=\frac{3(n-1)(n-1)}{8}$ for $n\in \{5,7,9,11\}$;

• (iii) $|E\left( K_{\frac{n-2}{2}}\vee \frac{n+2}{2} K_1\right) |=\frac{3n(n-2)}{8}$ for $n\in \{6,8\}$.

Clearly, the graphs $K_1\vee (K_{n-3}\cup 2 K_1)$, $K_{\frac{n-1}{2}}\vee \frac{n+1}{2} K_1$ and $K_{\frac{n-2}{2}}\vee \frac{n+2}{2} K_1$ have no $\{P_2,\, C_{2i+1}:i\ge 1\}$–factor, respectively. This shows that the bound in Theorem 1.1 is best possible.

Theorem 5.1

Let n be a positive integer and $\theta (n)$ be the largest root of $x^3-(n-4) x^2-(n-1)x+2(n-4)=0$. Then we have

• (i) $\rho (K_1\vee (K_{n-3}\cup 2 K_1))=\theta (n)$ for $n\in \{3,4,8\}$ or $n\ge 10$;

• (ii) $\rho \left( K_{\frac{n-1}{2}}\vee \frac{n+1}{2} K_1\right) =\frac{n-3+\sqrt{5n^2-6n+5}}{4}$ for $n\in \{5,7,9\}$;

• (iii) $\rho (P_2\vee 4 K_1)=\frac{1+\sqrt{33}}{2}$ for $n=6$.

Proof

For $n\in \{3,4,8\}$ or $n\ge 10$, we consider the partition $V(K_1\vee (K_{n-3}\cup 2 K_1)) =V(K_1) \cup V(2 K_1) \cup V(K_{n-3})$. The quotient matrix of $A(K_1\vee (K_{n-3}\cup 2 K_1))$ corresponding to the above partition is

$$\begin{aligned} B_1=\left( \begin{array}{ccc} 0 &{} 2 &{} n-3 \\ 1 &{} 0 &{} 0 \\ 1 &{} 0 &{} n-4 \end{array}\right) . \end{aligned}$$

Then we have

$$\begin{aligned} \phi _{B_{1}}(x)=x^3-(n-4)x^2-(n-1)x+2(n-4). \end{aligned}$$

Note that the partition is equitable. Then the spectral radius of graph $K_1\vee (K_{n-3}\cup 2 K_1)$ is the largest root of $x^3-(n-4)x^2-(n-1)x+2(n-4)=0$. It follows that $\rho (K_1\vee (K_{n-3}\cup 2 K_1))=\theta (n)$.

For $n\in \{5,7,9\}$, we consider the partition $V(K_{\frac{n-1}{2}}\vee \frac{n+1}{2} K_1)=V(K_{\frac{n-1}{2}}) \cup V(\frac{n+1}{2} K_1)$. Then the corresponding quotient matrix of $A(K_{\frac{n-1}{2}}\vee \frac{n+1}{2} K_1)$ is

$$\begin{aligned} B_2=\left( \begin{array}{cc} \frac{n-3}{2} &{} \frac{n+1}{2} \\ \frac{n-1}{2} &{} 0 \end{array}\right) . \end{aligned}$$

Hence we have

$$\begin{aligned} \phi _{B_2}(x)=x^2-\frac{n-3}{2} x-\frac{n^2-1}{4}. \end{aligned}$$

Note that the spectral radius of graph $K_{\frac{n-1}{2}}\vee \frac{n+1}{2} K_1$ is the largest root of $x^2-\frac{n-3}{2} x-\frac{n^2-1}{4}=0$ since the above partition is equitable. It follows that $\rho \left( K_{\frac{n-1}{2}}\vee \frac{n+1}{2} K_1\right) =\frac{n-3+\sqrt{5n^2-6n+5}}{4}$.

For $n=6$, consider the partition $V(K_{2}\vee 4 K_1)=V(K_{2}) \cup V(4 K_1)$. The quotient matrix of $A(K_{2}\vee 4 K_1)$ corresponding to the above partition equals

$$\begin{aligned} B_3=\left( \begin{array}{cc} 1 &{} 4 \\ 2 &{} 0 \end{array}\right) . \end{aligned}$$

Hence we have $\phi _{B_3}(x)=x^2-x-8$. Note that the spectral radius of graph $K_{2}\vee 4 K_1$ is the largest root of $x^2-x-8=0$ since the above partition is equitable. It follows that $\rho \left( K_{2}\vee 4 K_1\right) =\frac{1+\sqrt{33}}{2}$. This completes the proof.$\square$

In view of Theorem 5.1, we know that the bound obtained in Theorem 1.2 is best possible.