Spectral and Sharp Sufficient Conditions for Graphs to Admit a Strong Star Factor

Abstract

LetGbe a graph. An odd $[1,k]$-factor of a graph G is a spanning subgraph H of G such that $de{g}_H(v)$ is odd and $1⩽de{g}_H(v)⩽k$ for every $v\in V(G)$ where k is a positive odd integer. We call a spanning subgraph H of a graph G a strong star factor if every component of H is isomorphic to an element of the stars $K_{1,1},K_{1,2},\cdots ,K_{1,r}$ and is an induced subgraph of G where $r⩾2$ is an integer. In a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor of G, each component is isomorphic to a member in $\{K_{1,1},K_{1,2},C_3,C_4\cdots ,C_m\}$. A graph G is a strong star factor deleted graph if $G-e$ has a strong star factor for each edge e of G. In this paper, through the typical spectral techniques, we obtain the respective necessary and sufficient conditions defining a strong star factor deleted graph, an odd $[1,k]$-factor deleted graph, and a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph. We determine a lower bound on the size to guarantee that G is a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph. We establish the upper bound of the signless Laplacian spectral radius (resp. the spectral radius) and the lower bound of the distance signless Laplacian spectral radius (resp. the distance spectral radius) to determine whether G admits a strong star factor. Furthermore, by constructing extremal graphs, we show that all the bounds obtained in this contribution are the best possible.

1. Introduction

In the framework of graph networks, vertices stand for sites, edges stand for the channels between sites, and the feasibility of network data transmission can be characterized by the existence of a fractional flow (the fractional factor works as a surrogate model) in the corresponding graph. From a theoretical perspective, a strong star factor deleted graph, an odd $[1,k]$-factor deleted graph, and a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph determine, after removing an edge from the original graph, whether there exists a fractional factor in the remaining subgraph, which allows us to efficiently calculate the possibility of data transfer during network attacks (see [1,2]). In this article, we only consider finite, simple, and undirected graphs. If there are graph theory symbols and terms not defined here, we refer to [3,4,5,6]. A factor of a graph is a spanning subgraph. An isolated edge is an edge in graph G that has no common endpoint with any other edges. The edge incident to the leaf vertex is a pendant edge of graph G. The complement of graph G, denoted as $\overline{G}$, is a graph with the vertex set $V(G)$ such that any two vertices in $\overline{G}$ are adjacent if and only if they are not adjacent in G. Let $K_{1,\nu -1}$, $K_{\nu }$, and $C_{\nu }$ be a star, a complete graph, and a cycle of order $\nu$, respectively.

Amahashi and Kano [7] gave the characterization of a $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor in 1982, which remains one of the cornerstone results in factor theory. As an extension of a $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor, the characterization of a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor was established by Egawa, Kano, and Kelmans [8]. In other words, a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor must be a $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor.

For the two graphs $G_1$ and $G_2$, let $G_1\cup G_2$ and $k{G}_1$ be the disjoint union and the disjoint union of k copies of $G_1$, respectively. The join$G_1\vee G_2$ is the graph obtained from $G_1\cup G_2$ by connecting every vertex of $G_1$ with every vertex of $G_2$ via an edge. The join graph of $G_1$ and $G_2$ is denoted as $G_1\vee G_2$. $V(G\vee H)=V(G)\cup V(H)$ and $E(G\vee H)=E(G)\cup E(H)\cup \{uv|\mathrm{for}\mathrm{all}u\in V(G)\mathrm{and}v\in V(H)\}$. Given a graph, let ${\eta }_1(G)$, $\kappa (G)$, $\rho (G)$, and ${\mu }_1(G)$ be the distance signless Laplacian spectral radius, signless Laplacian spectral radius, spectral radius, and distance spectral radius of G, respectively. As usual, we denote by $oddca(G)$, $o(G)$, and $iso(G)$ the number of those components that are odd complete cacti (i.e., each block of G is a complete odd order graph), odd components (i.e., components with odd order), and isolated vertices in a graph G, respectively. We are concerned about four types of factors with the given properties.

The first is a strong star factor deleted graph.

Theorem 1. 

A connected graph G has a strong star factor deleted graph if and only if, for any subset X of vertices of G, the following six statements hold:

(i) 

$oddca(G-X)⩽r|X|$;

(ii) 

If there exists an isolated edge in $G-X$, then $oddca(G-X)⩽r|X|-2$;

(iii) 

If there exists a pendant edge that is not an isolated edge and connects two odd complete cacti in $G-X$, then $oddca(G-X)⩽r|X|-2$;

(iv) 

If there exists a pendant edge that is not an isolated edge and connects an odd complete cactus in $G-X$, then $oddca(G-X)⩽r|X|-1$;

(v) 

If there exists neither a pendant edge nor an isolated edge e and e is a connected component in $G-X$, say C, where C is not an odd complete cactus, and there exists an odd complete cactus in $G-X-e$, then $oddca(G-X)⩽r|X|-1$;

(vi) 

If there exists neither a pendant edge nor an isolated edge e and e is a connected component in $G-X$, say C, where C is not an odd complete cactus, and there exists two odd complete cacti in $G-X-e$, then $oddca(G-X)⩽r|X|-2$.

The second is an odd $[1,k]$-factor deleted graph. As far as we know, there is no necessary and sufficient condition for the existence of an odd $[1,k]$-factor deleted graph. Our second main result gives a necessary and sufficient condition for the existence of an odd $[1,k]$-factor deleted graph.

Theorem 2. 

A connected graph G is an odd $[1,k]$-factor deleted graph if and only if, for any subset X of vertices of G, the following four statements hold:

(i) 

$o(G-X)⩽k|X|$;

(ii) 

If there exists an isolated edge in $G-X$, then $o(G-X)⩽k|X|-2$;

(iii) 

If there exists a pendant edge that is not an isolated edge and connects two odd components in $G-X$, then $o(G-X)⩽k|X|-2$;

(iv) 

If there exists neither a pendant edge nor an isolated edge that connect two odd components in $G-X$, then $o(G-X)⩽k|X|-2$.

The third is a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph. As far as we know, there is no necessary and sufficient condition for the existence of a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph. Our third main result gives a necessary and sufficient condition for the existence of a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph.

Theorem 3. 

A connected graph G is a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph if and only if, for any subset X of vertices of G, the following three statements hold:

(i) 

$iso(G-X)⩽2|X|$;

(ii) 

If there exists an isolated edge in $G-X$, then $iso(G-X)⩽2|X|-2$;

(iii) 

If there exists a pendant edge in $G-X$, then $iso(G-X)⩽2|X|-1$.

Our fourth main result gives a sufficient condition to ensure that a graph G admits a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph according to the size of G.

Theorem 4. 

Let G be a connected graph with $\nu ⩾4$ vertices. If $|E(G)|⩾\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1$ with $\nu ⩾4$, then G admits a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph unless $G\cong H_{\nu }$, where $H_{\nu }$ is a complete graph to which an edge has been attached and $|V(H_{\nu })|=\nu$.

The fourth is a strong star factor. As far as we know, there is no spectral sufficient condition for a graph that has a strong star factor. Our fifth main result gives a sufficient condition to ensure that a graph G admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor according to the distance signless Laplacian spectral radius of G.

Theorem 5. 

Let $r⩾3$ and $\nu ⩾r+3$ be two integers, where ν is an even number and r is an odd number, and let G be a connected graph with ν vertices. Assume that the largest root of $\alpha (x)=x^3-(5\nu +2r-7)x^2+(4r^2+8{\nu }^2+2\nu r-27\nu +4r+24)x-4\nu r^2-4{\nu }^3-8\nu r+6r^2+22{\nu }^2-42\nu +14r+28=0$ is $\beta (\nu ,r)$. For $r⩾3$, if ${\eta }_1(G)⩽\beta (\nu ,r)$ with $\nu ⩾r+3$, then G admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor, unless $G\cong K_1\vee ((r+1)K_1\cup K_{\nu -r-2})$.

Our sixth main result gives a sufficient condition to ensure that a graph G admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor according to the signless Laplacian spectral radius of G.

Theorem 6. 

Let $r⩾3$ and $\nu ⩾r+3$ be two integers, where ν is an even number and r is an odd number, and let G be a connected graph with ν vertices. Assume that the largest root of $\gamma (x)=x^3-(4\nu -3r-10)x^2+(2{\nu }^2-2\nu r-5\nu )x+4\nu r-2r^2-2{\nu }^2+10\nu -10r-12=0$ is $\phi (\nu ,r)$.

(i) 

If ${\kappa }_1(G)⩾r+3$ with $\nu =r+3$, then G admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor, unless $G\cong K_1\vee K_{r+2}$.

(ii) 

If ${\kappa }_1(G)⩾r+3+\sqrt{r^2+6r+5}$ with $\nu =2r+4$, then G admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor, unless $G\cong K_2\vee (2r+2)K_1$.

(iii) 

If ${\kappa }_1(G)⩾\phi (\nu ,r)$ with $\nu ⩾3r+5$, then G admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor, unless $G\cong K_1\vee (K_{\nu -r-2}\cup (r+1)K_1)$.

Our seventh main result gives a sufficient condition to ensure that a graph G admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor according to the spectral radius of G.

Theorem 7. 

Let $r⩾3$ and ν be two integers, where ν is an even number and r is an odd number, and let G be a connected graph with ν vertices. Assume that the largest root of $\chi (x)=x^3-(\nu -r-3)x^2-(\nu -1)x+\nu r-r^2+\nu -4r-3=0$ is $\theta (\nu )$. If $\rho (G)⩾\theta (\nu )$ with $\nu ⩾6$, then G admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor, unless $G\cong K_1\vee (K_{\nu -r-2}\cup (r+1)K_1)$.

Our eighth main result gives a sufficient condition to ensure that a graph G admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor according to the distance spectral radius of G.

Theorem 8. 

Let $r⩾3$ and $\nu ⩾r+3$ be two integers, where ν is an even number and r is an odd number, and let G be a connected graph with ν vertices.

(i) 

If ${\mu }_1(G)⩽{\mu }_1(K_1\vee (K_1\cup \overline{K_4}))$ with $\nu =6$, then G admits a strong $\{K_{1,1},K_{1,2},K_{1,3}\}$-factor, unless $G\cong K_1\vee (K_1\cup \overline{K_4})$.

(ii) 

If ${\mu }_1(G)⩽{\mu }_1(K_1\vee (K_3\cup \overline{K_4}))$ with $\nu =8$, then G admits a strong $\{K_{1,1},K_{1,2},K_{1,3}\}$-factor, unless $G\cong K_1\vee (K_3\cup \overline{K_4})$.

(iii) 

If ${\mu }_1(G)⩽{\mu }_1(K_1\vee (K_1\cup \overline{K_6}))$ with $\nu =8$, then G admits a strong $\{K_{1,1},K_{1,2},K_{1,3},K_{1,4},K_{1,5}\}$-factor, unless $G\cong K_1\vee (K_1\cup \overline{K_6})$.

(iv) 

If ${\mu }_1(G)⩽{\mu }_1(K_1\vee (K_{\nu -r-2}\cup \overline{K_{r+1}}))$ with $\nu ⩾10$, then G admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor, unless $G\cong K_1\vee (K_{\nu -r-2}\cup \overline{K_{r+1}})$.

2. Tools

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

Let $G=(V(G),E(G))$ be a graph where $V(G)=\{v_1,v_2,\cdots ,v_{\nu }\}$ is the vertex set and $E(G)$ is the edge set of the graph G. The order of G is the number $\nu =|V(G)|$ of its vertices, and the size of G is the number $|E(G)|$ of its edges. For $u\in V(G)$, the neighborhood $N_G(u)$ of u is the set of vertices adjacent to u in G, and the degree of u, denoted by ${deg}_G(u)$, is the number $|N_G(u)|$. For a given subset $X\subseteq V(G)$, the subgraph of G induced by X is denoted by $G[X]$. As usual, we let $G-v$ or $G-uv$ denote the graph obtained from G by deleting vertex $v\in V(G)$ or edge $uv\in E(G)$, respectively (this notation is naturally extended if more than one vertex or edge is deleted).

The spectral radius of a graph G is the largest eigenvalue of the adjacency matrix of G, which is defined as the symmetric matrix $\mathbf{A}(G)=(a_{ij})$, where $a_{ij}=1$ if $v_i$ and $v_j$ are adjacent, and $a_{ij}=0$ otherwise. The signless Laplacian spectral radius of a graph G is the largest eigenvalue of its signless Laplacian matrix $\mathbf{K}(G)=\mathbf{F}(G)+\mathbf{A}(G)$, where $\mathbf{F}(G)$ denotes the diagonal matrix of vertex degrees in a graph G.

The distance spectral radius of a graph G is the largest eigenvalue of its distance matrix$\mathbf{D}(G)=(d_{ij})$, where $d_{ij}$ is the distance between two vertices $v_i$ and $v_j$ in G. The distance signless Laplacian spectral radius of a graph G is the largest eigenvalue of its distance signless Laplacian matrix [9] $\mathbf{Q}(G)=\mathbf{Tr}(G)+\mathbf{D}(G)$, where $\mathbf{Tr}(G)=diag(Tr(v_1),Tr(v_2),\cdots ,Tr(v_{\nu }))$ is the diagonal matrix of the vertex transmissions in G, and the transmission $Tr(v_i)$ of a vertex $v_i\in V(G)$ is defined as the sum of distances from $v_i$ to all other vertices of G.

A $[b,b]$-factor is called a b-factor. Tutte [10] gave the characterization of 1-factors in 1947, which remains one of the cornerstone results in factor theory. As an extension of Tutte’s 1-Factor Theorem, the well-known sufficient and necessary condition for the existence of an odd $[1,k]$-factor was established by Amahashi. Lu, Wu, and Yang [11] gave a sufficient condition for the existence of an odd $[1,k]$-factor in a graph in terms of the third largest eigenvalue. Tutte [12] obtained the well-known b-Factor Theorem in 1952. More and more researchers are focusing on establishing the relationship between the eigenvalues of a graph and the existence of substructures of the graph. Liu, Pan, and Li [13] established a relationship between a perfect matching and a signless Laplacian spectral radius of a graph. Zhang and Lin [14] proved an upper bound for the distance spectral radius to guarantee the existence of perfect matching in a graph. In 1953, Tutte [15] gave a necessary and sufficient condition for the existence of a $\{K_2,C_n\}$-factor in a graph. In recent years, it has been of great interest for researchers to find spectral sufficient conditions such that a graph has a given factor (see the survey [16]). For more advances along this line, we may consult [17,18,19,20]. The existence of factors with given properties has received special attention. The following lemmas, following from the Perron–Frobenius theorem [4,21], are well known:

Lemma 1. 

Let G be a connected graph and let H be a proper subgraph of G. Then, ${\kappa }_1(H)<{\kappa }_1(G)$.

Lemma 2. 

Let G be a connected graph and let H be a proper subgraph of G. Then, $\rho (H)<\rho (G)$.

Lemma 3 

(Minc [22]). Let $\nu \ge 2$ be an integer, and $K_{\nu }$ be a complete graph of order ν. Then, ${\kappa }_1(K_{\nu })=2\nu -2$.

Lemma 4 

(Minc [23]). Let e be an edge of a graph G such that $G-e$ is still connected. Then, ${\eta }_1(G-e)>{\eta }_1(G)$.

Lemma 5 

([23]). Let G be a connected graph with two nonadjacent vertices $u,v\in V(G)$. Then, ${\mu }_1(G+uv)<{\mu }_1(G).$

Lemma 6 

([24]). Let $\nu ,p,|X|$, and ${\nu }_i(i=1,2,\cdots ,p)$ be positive integers and $\nu ={\displaystyle \sum _{i=1}^p}{\nu }_i+|X|$. If ${\nu }_1⩾{\nu }_2⩾\cdots ⩾{\nu }_p⩾1$, then ${\eta }_1(K_{|X|}\vee (K_{{\nu }_1}\cup K_{{\nu }_2}\cup \cdots \cup K_{{\nu }_p}))⩾{\eta }_1(K_{|X|}\vee (K_{\nu -|X|-(p-1)}\cup (p-1)K_1))$ with equality if and only if $K_{|X|}\vee (K_{{\nu }_1}\cup K_{{\nu }_2}\cup \cdots \cup K_{{\nu }_p})\cong K_{|X|}\vee (K_{\nu -|X|-(p-1)}\cup (p-1)K_1)$.

Lemma 7 

([25]). Let G be a ν-vertex connected graph and let $\mathbf{x}={(x_1,x_2,\cdots ,x_{\nu })}^{\top }$ be the Perron vector of $\mathbf{D}(G)$ corresponding to ${\mu }_1(G)$. If $v_i,v_j\in V(G)$ satisfy $N_G(v_i)\setminus \{v_j\}=N_G(v_j)\setminus \{v_i\}$, then $x_i=x_j$.

Let $\mathbf{M}$ be a real matrix of order $\nu$ and assume that $\mathbf{M}$ can be described as

$$\mathbf{M}=\left(\begin{array}{ccc}{\mathbf{M}}_{11}& \cdots & {\mathbf{M}}_{1s}\\ ⋮& \ddots & ⋮\\ {\mathbf{M}}_{s1}& \cdots & {\mathbf{M}}_{ss}\end{array}\right)$$

with respect to the partition $\pi :V=V_1\cup \cdots \cup V_s$, where ${\mathbf{M}}_{\mathbf{ij}}$ denotes the submatrix (block) of $\mathbf{M}$ formed by rows in $V_i$ and columns in $V_j$. Let $m_{ij}$ denote the average row sum of ${\mathbf{M}}_{\mathbf{ij}}$. Then, matrix ${\mathbf{M}}_{\pi }=(m_{ij})$ is called the quotient matrix of $\mathbf{M}$. If, for any i and j, ${\mathbf{M}}_{\mathbf{ij}}$ admits a constant row sum, then ${\mathbf{M}}_{\pi }$ is called the equitable matrix of $\mathbf{M}$ and the partition is called equitable.

Lemma 8 

([26,27]). Let $\mathbf{M}$ be a square matrix with an equitable partition π and let ${\mathbf{M}}_{\pi }$ be the corresponding quotient matrix. Then, every eigenvalue of ${\mathbf{M}}_{\pi }$ is an eigenvalue of $\mathbf{M}$. Furthermore, if $\mathbf{M}$ is nonnegative and ${\mathbf{M}}_{\pi }$ is irreducible, then the largest eigenvalues of $\mathbf{M}$ and ${\mathbf{M}}_{\pi }$ are equal.

The following lemma can be easily derived by the Rayleigh quotient [17].

Lemma 9. 

Let G be a connected graph with order ν. Then,

$${\mu }_1(G)=\underset{\mathbf{x}\ne \mathbf{0}}{max}{\displaystyle \frac{{\mathbf{x}}^{\top }\mathbf{D}(G)\mathbf{x}}{{\mathbf{x}}^{\top }\mathbf{x}}}⩾{\displaystyle \frac{{\mathbf{1}}^{\top }\mathbf{D}(G)\mathbf{1}}{{\mathbf{1}}^{\top }\mathbf{1}}}={\displaystyle \frac{2W(G)}{\nu }},$$

where $\mathbf{1}={(1,1,\cdots ,1)}^{\top }$ and $W(G)={\displaystyle \sum _{i<j}}{d}_{ij}$.

The rest of the paper is organized as follows: In Section 3, we present a necessary and sufficient condition for the existence of a strong star factor deleted graph. In Section 4, we present a necessary and sufficient condition for the existence of an odd $[1,k]$-factor deleted graph. In Section 5, we present a necessary and sufficient condition for the existence of a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph and obtain a bound for the size of G. In Section 6, we obtain some bounds for the distance signless Laplacian spectral radius of G, the signless Laplacian spectral radius of G, the spectral radius of G, and the distance spectral radius of G. In the last section, we give several extremal graphs to imply that all the bounds are the best possible.

3. A Necessary and Sufficient Condition for the Existence of a Strong Star Factor Deleted Graph

In this section, we give the proof of Theorem 1, which gives a necessary and sufficient condition for the existence of a strong star factor deleted graph.

A block of a graph G is a maximal connected subgraph of G that has no cut vertex. An isolated vertex is a vertex of degree 0. An isolated vertex in a graph is considered to be a block of a graph. Note that $K_1$ is an odd complete cactus. For $X,Y\subseteq V(G)$, we denote by $E(X,Y)$ the set of edges of G with one end in X and the other end in Y.

In 1997, Egawa, Kano, and Kelmans [8] gave a sufficient and necessary condition for the existence of a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor.

Lemma 10 

(Egawa, Kano, and Kelmans [8]). Let $r⩾2$ be an integer. Then, a simple graph G admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor if and only if

$$oddca(G-X)⩽r|X|$$

for all $X\subseteq V(G)$.

Now we will give a proof of Theorem 1.

Proof of Theorem 1. 

Proof of the Sufficiency. Assume that any subset X of vertices of G satisfies Theorem 1 (i)–(vi). We prove it by contradiction. Suppose that G admits no strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor deleted graph. Therefore, there exists $e\in E(G)$ such that $G_e:=G-e$ has no strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor. Then, by Lemma 10, there exists some subset X of vertices of G such that

$$oddca(G_e-X)=oddca(G-e-X)>r|X|.$$

Together with Lemma 10 and Theorem 1 (i), G admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor.

There are three cases for discussing the location of two endpoints of e: Case 1: Both endpoints of e belong to X. Case 2: One of the two endpoints of e belongs to X. Case 3: Neither of the endpoints of e belongs to X. By Case 1, we obtain $e\in E(G[X])$. By Case 2, we obtain $e\in E(X,V(G)-X)$. By Case 3, we obtain $e\in E(G-X)$.

In fact, we proceed by considering the following three cases (based on the position of e):

Case 1. $e\in E(G[X])$.

In this case, we obtain $oddca(G-X)=oddca(G_e-X)>r|X|$, which contradicts Theorem 1 (i).

Case 2. $e\in E(X,V(G)-X)$.

In this case, we obtain $oddca(G-X)=oddca(G_e-X)>r|X|$, which contradicts Theorem 1 (i).

Case 3. $e\in E(G-X)$.

Subcase 3.1. $e\in E(G-X)$ and e is an isolated edge.

In this subcase, we obtain $oddca(G-X)+2=oddca(G_e-X)>r|X|$. Hence, $oddca(G-X)>r|X|-2$, which contradicts Theorem 1 (ii).

Subcase 3.2. $e\in E(G-X)$ and e is a pendant edge that is not an isolated edge.

We proceed by considering the following two cases (based on the position of e):

$(a)$

e is not in an odd complete cactus but e connects two odd complete cacti. We obtain $oddca(G-X)+2=oddca(G_e-X)>r|X|$. Therefore, $oddca(G-X)>r|X|-2$, which contradicts Theorem 1 (iii).

$(b)$

e is not in an odd complete cactus and e connects an odd complete cactus. We have $oddca(G-X)+1=oddca(G_e-X)>r|X|$. Therefore, $oddca(G-X)>r|X|-1$, which contradicts Theorem 1 (iv).

Subcase 3.3. $e\in E(G-X)$ and e is neither a pendant edge nor an isolated edge.

We proceed by considering the following four cases (based on the position of e):

$(a)$

e is in an odd complete cactus, say C, but $C-e$ is not an odd complete cactus. We obtain $oddca(G-X)-1=oddca(G_e-X)>r|X|$. Hence, $oddca(G-X)>r|X|+1$, which contradicts Theorem 1 (i).

$(b)$

e is in a connected component, say C, where C is not an odd complete cactus, but $C-e$ is an odd complete cactus. We obtain $oddca(G-X)+1=oddca(G_e-X)>r|X|$. Hence, $oddca(G-X)>r|X|-1$, which contradicts Theorem 1 (v).

$(c)$

e is in a connected component, say C, where C is not an odd complete cactus, but $C-e$ are two odd complete cacti. We obtain $oddca(G-X)+2=oddca(G_e-X)>r|X|$. Hence, $oddca(G-X)>r|X|-2$, which contradicts Theorem 1 (vi).

$(d)$

e is in a connected component, say C, where C is not an odd complete cactus, and $C-e$ is not an odd complete cactus. We obtain $oddca(G-X)=oddca(G_e-X)>r|X|$, which contradicts Theorem 1 (i).

Proof of the Necessity. Assume that a connected graph G has a strong star factor deleted graph. Let $G_e:=G-e$ for each edge e of G. A graph G is a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor deleted graph if and only if $G_e$ admits a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor for each edge e of G. Together with Lemma 10, we obtain

$$oddca(G_e-X)⩽r|X|$$

for any subset X of vertices of G. Thus, $oddca(G-X)⩽oddca(G_e-X)⩽r|X|$. Therefore, Theorem 1 (i) holds.

If there exists an isolated edge e in $G-X$, then $oddca(G-X)+2=oddca(G_e-X)⩽r|X|$. Thus, $oddca(G-X)⩽r|X|-2$. Therefore, Theorem 1 (ii) holds.

If there exists a pendant edge e that is not isolated edge and connects two odd complete cacti in $G-X$, then $oddca(G-X)+2=oddca(G_e-X)⩽r|X|$. Thus, $oddca(G-X)⩽r|X|-2$. Therefore, Theorem 1 (iii) holds.

If there exists a pendant edge e that is not isolated edge and connects an odd complete cactus in $G-X$, then $oddca(G-X)+1=oddca(G_e-X)⩽r|X|$. Thus, $oddca(G-X)⩽r|X|-1$. Therefore, Theorem 1 (iv) holds.

If there exists neither a pendant edge nor an isolated edge e and e is a connected component in $G-X$, say C, where C is not an odd complete cactus and there exists an odd complete cactus in $G-X-e$, then $oddca(G-X)+1=oddca(G_e-X)⩽r|X|$. Thus, $oddca(G-X)⩽r|X|-1$. Therefore, Theorem 1 (v) holds.

If there exists neither a pendant edge nor an isolated edge e and e is a connected component in $G-X$, say C, where C is not an odd complete cactus, and there exists two odd complete cacti in $G-X-e$, then $oddca(G-X)+2=oddca(G_e-X)⩽r|X|$. Thus, $oddca(G-X)⩽r|X|-2$. Therefore, Theorem 1 (vi) holds. This completes the proof of Theorem 1. □

4. A Necessary and Sufficient Condition for the Existence of an Odd $[\mathbf{1},k]$-Factor Deleted Graph

In 1985, Amahashi [28] gave a sufficient and necessary condition for the existence of an odd $[1,k]$-factor.

Lemma 11 

(Amahashi [28]). Let k be a positive odd integer. Then, a graph G admits an odd $[1,k]$-factor if and only if

$$o(G-X)⩽k|X|$$

for all $X\subseteq V(G)$.

Now we will give a proof of Theorem 2.

Proof of Theorem 2. 

Proof of the Sufficiency. Assume that any subset X of vertices of G satisfies four statements (i)–(iv). We prove it by contradiction. Suppose that G admits no odd $[1,k]$-factor deleted graph. Therefore, there exists $e\in E(G)$ such that $G_e:=G-e$ has no odd $[1,k]$-factor. Then, by Lemma 11, there exists some subset X of vertices of G such that

$$o(G_e-X)>k|X|.$$

Together with Lemma 11 and Theorem 2 (i), G admits an odd $[1,k]$-factor.

There are three cases for discussing the location of two endpoints of e: Case 1: Both endpoints of e belong to X. Case 2: One of the two endpoints of e belongs to X. Case 3: Neither of the endpoints of e belongs to X. By Case 1, we obtain $e\in E(G[X])$. By Case 2, we obtain $e\in E(X,V(G)-X)$. By Case 3, we obtain $e\in E(G-X)$.

In fact, we proceed by considering the following three cases (based on the position of e):

Case 1. $e\in E(G[X])$.

In this case, we obtain $o(G-X)=o(G_e-X)>k|X|$, which contradicts Theorem 2 (i).

Case 2. $e\in E(X,V(G)-X)$.

In this case, we obtain $o(G-X)=o(G_e-X)>k|X|$, which contradicts Theorem 2 (i).

Case 3. $e\in E(G-X)$.

Subcase 3.1. $e\in E(G-X)$ and e is an isolated edge.

We obtain $o(G-X)+2=o(G_e-X)>k|X|$. Hence, $o(G-X)>k|X|-2$, which contradicts Theorem 2 (ii).

Subcase 3.2. $e\in E(G-X)$ and e is a pendant edge that is not an isolated edge.

We proceed by considering the following two cases (based on the position of e):

$(a)$

e is in an odd component, say C, and there is an odd component in $C-e$. Thus, we obtain $o(G-S)=o(G_e-X)>k|X|$, which contradicts Theorem 2 (i).

$(b)$

e is not in an odd component but e connects two odd components. We obtain $o(G-X)+2=o(G_e-X)>k|X|$. Thus, $o(G-X)>k|X|-2$, which contradicts Theorem 2 (iii).

Subcase 3.3. $e\in E(G-X)$ and e is neither a pendant edge nor an isolated edge.

We proceed by considering the following three cases (based on the position of e):

$(a)$

e is in an odd component, say C, and $C-e$ is still an odd component. We obtain $o(G-X)=o(G_e-X)>k|X|$, which contradicts Theorem 2 (i).

$(b)$

e is in a connected component, say C, where C is not an odd component, but $C-e$ is two odd components. We obtain $o(G-X)+2=o(G_e-X)>k|X|$. Hence, $o(G-X)>k|X|-2$, which contradicts Theorem 2 (iv).

$(c)$

e is in a connected component, say C, where C is not an odd component, and $C-e$ is not an odd component. We have $o(G-X)=o(G_e-X)>k|X|$, contrary to Theorem 2 (i).

By $(b)$ and $(c)$, we know that e is in a connected component, say C, where C is not an odd component. Hence, $|V(C)|$ is an even number. Then, $|V(C-e)|$ is also an even number. Note that an even number is equal to an odd number add an odd number. Hence, we know that $C-e$ is two odd components. Note that an even number is equal to an even number add an even number. Hence, we know that $C-e$ is not an odd component.

Proof of the Necessity. Assume that a connected graph G has an odd $[1,k]$-factor deleted graph. Let $G_e:=G-e$ for each edge e of G. A graph G is an odd $[1,k]$-factor deleted graph if and only if $G_e$ admits an odd $[1,k]$-factor for each edge e of G. Together with Lemma 11, we obtain

$$o(G_e-X)⩽k|X|$$

for any subset X of vertices of G. Thus, $o(G-X)⩽o(G_e-X)⩽k|X|$. Therefore, Theorem 2 (i) holds.

If there exists an isolated edge e in $G-X$, then $o(G-X)+2=o(G_e-X)⩽k|X|$. Thus, $o(G-X)⩽k|X|-2$. Therefore, Theorem 2 (ii) holds.

If there exists a pendant edge e that is not an isolated edge and connects two odd components in $G-X$, then $o(G-X)+2=o(G_e-X)⩽k|X|$. Thus, $o(G-X)⩽k|X|-2$. Therefore, Theorem 2 (iii) holds.

If there exists neither a pendant edge nor an isolated edge that connects two odd components in $G-X$, then $o(G-X)+2=o(G_e-X)⩽k|X|$. Thus, $o(G-X)⩽k|X|-2$. Therefore, Theorem 2 (iv) holds.

This completes the proof of Theorem 2. □

5. A $\{{\mathit{K}}_{\mathbf{1},\mathbf{1}},{\mathit{K}}_{\mathbf{1},\mathbf{2}},{\mathit{C}}_{\mathbf{m}}:\mathit{m}⩾\mathbf{3}\}$-Factor Deleted Graph

5.1. A Necessary and Sufficient Condition for the Existence of a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-Factor Deleted Graph

In 1980, Akiyama and Era [29] gave a sufficient and necessary condition for the existence of a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor.

Lemma 12. 

(Akiyama and Era [29]) A graph G has a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor if and only if

$$iso(G-X)⩽2|X|$$

for all $X\subseteq V(G)$.

Now we will give a proof of Theorem 3.

Proof of Theorem 3. 

Proof of the Sufficiency. Assume that any subset X of vertices of G satisfies Theorem 3 (i)–(iii). We prove it by contradiction. Suppose that G admits no $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph. Hence, there exists $e\in E(G)$ such that $G_e:=G-e$ has no $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor. Then, by Lemma 12, there exists some subset X of vertices of G such that

$$iso(G_e-X)>2|X|.$$

Together with Lemma 12 and Theorem 3 (i), G admits a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor.

In fact, we proceed by considering the following three cases (based on the position of e):

Case 1. $e\in E(G[X])$.

We obtain $iso(G-X)=iso(G_e-X)>2|X|$, which contradicts Theorem 3 (i).

Case 2. $e\in E(X,V(G)-X)$.

We obtain $iso(G-X)=iso(G_e-X)>2|X|$, which contradicts Theorem 3 (i).

Case 3. $e\in E(G-X)$.

Subcase 3.1. $e\in E(G-X)$ and e is an isolated edge.

We obtain $iso(G-X)+2=iso(G_e-X)>2|X|$. Hence, $iso(G-X)>2|X|-2$, which contradicts Theorem 3 (ii).

Subcase 3.2. $e\in E(G-X)$ and e is a pendant edge that is not an isolated edge.

We obtain $iso(G-X)+1=iso(G_e-X)>2|X|$. Hence, $iso(G-X)>2|X|-1$, which contradicts Theorem 3 (iii).

Subcase 3.3. $e\in E(G-X)$ and e is not a pendant edge.

We obtain $iso(G-X)=iso(G_e-X)>2|X|$, which contradicts Theorem 3 (i).

Proof of the Necessity. Assume that a connected graph G has a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph. Let $G_e:=G-e$ for each edge e of G. A graph G is a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph if and only if $G_e$ admits a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor for each edge e of G. Together with Lemma 12, we obtain

$$iso(G_e-X)⩽2|X|$$

for any subset X of vertices of G. Thus, $iso(G-X)⩽iso(G_e-X)⩽2|X|$. Therefore, Theorem 3 (i) holds.

If there exists an isolated edge e in $G-X$, then $iso(G-X)+2=iso(G_e-X)⩽2|X|$. Thus, $iso(G-X)⩽2|X|-2$. Therefore, Theorem 3 (ii) holds.

If there exists a pendant edge e that is an isolated edge in $G-S$, then $iso(G-X)\le 2|X|-2<2|X|-1$. If e is a pendant edge e which is not an isolated edge in $G-S$, then $iso(G-X)+1=iso(G_e-X)⩽2|X|$. Thus, $iso(G-X)⩽2|X|-1$. Therefore, Theorem 3 (iii) holds.

This completes the proof of Theorem 3. □

5.2. Size

Lemma 13 

(Ren, Zhang, and Wang [30]). Let $\nu ⩾4$ be a integer, and let G be a connected graph with ν vertices. If $|E(G)|⩾11$ with $\nu =7$, then G contains a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor, unless $G\cong K_2\vee 5K_1$. If $|E(G)|⩾\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -3}{2}}\right)+3$ with $\nu \ne 7$, then G contains a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor, unless $G\cong K_1\vee (K_{\nu -4}\cup 3K_1)$.

In this subsection, we give the proof of Theorem 4, which gives a sufficient condition via the size of a connected graph G to ensure that G is a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph.

Proof of Theorem 4. 

We prove it by contradiction. Suppose that G is not a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph. Choose such a connected graph G of order $\nu$ such that its size is as large as possible. We consider the following two cases:

Case 1.G has no $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor.

Together with Lemma 13, $\nu ⩾4$, and $\nu \ne 7$, we obtain $|E(G)|⩽\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -3}{2}}\right)+3$, which contradicts the condition that $|E(G)|>\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+3.$ Together with Lemma 13 and $\nu =7$, we obtain $|E(G)|⩽11$, which contradicts the condition that $|E(G)|>\left({\displaystyle \genfrac{}{}{0pt}{}{7-1}{2}}\right)+3=16.$

Case 2.G has a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor.

In view of Lemma 12, we obtain $iso(G-X)⩽2|X|$ for any subset X of vertices of G.

In view of Theorem 3, G is not $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted if and only if there exists a subset X of vertices of G that satisfies one of the following statements (i)–(ii):

(i)

$G-X$ has an isolated edge and $iso(G-X)⩾2|X|-1$;

(ii)

$G-X$ has a pendant edge that is not an isolated edge and $iso(G-X)⩾2|X|$.

Subcase 2.1. Statement (i) holds.

In view of $iso(G-X)⩽2|X|$ for any subset X of vertices of G and statement (i), we obtain $2|X|⩾iso(G-X)⩾2|X|-1$. Hence, $iso(G-X)=2|X|-1$ or $iso(G-X)=2|X|$. Since the size of G is as large as possible, we obtain the following:

(a) $G[X]$ is a complete graph;

(b) $G-X$ has at most two non-trivial connected components, one of which is $K_2$ and the other of which is a complete graph, say $G_1$;

(c) G is the join of $G[X]$ and $G-X$. Hence, $G=G[X]\vee (G-X)$.

Set $|V(G_1)|={\nu }_1$. Therefore, we obtain ${\nu }_1=0$ or ${\nu }_1⩾2$.

Claim 1.

1. 

If$iso(G-X)=2|X|-1$, then $|E(G)|⩽\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1$, where equality holds if and only if $G=K_1\vee (K_2\cup K_1)$.

2. 

If$iso(G-X)=2|X|$, then$|E(G)|<\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1.$

Proof. 

(1) Suppose that $X=\varnothing$. We have $iso(G)=1$, which contradicts the condition that G is a connected graph, and $\nu ⩾4$. Hence, $|X|⩾1$, $iso(G-X)=2|X|-1$, and $\nu =3|X|+{\nu }_1+1$. Therefore, $|E(G)|=\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -2|X|-1}{2}}\right)+|X|(2|X|+1)+1.$ Since $|X|⩾1$ and ${\nu }_1⩾0$, we obtain $\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1-|E(G)|=2|X|(|X|+{\nu }_1-1)⩾0$, where equality holds if and only if $|X|=1$ and ${\nu }_1=0$, i.e., $G=K_1\vee (K_2\cup K_1)$.

(2) Note that $iso(G-X)=2|X|$ and $\nu =3|X|+{\nu }_1+2.$ Hence, $|E(G)|=\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -2|X|-2}{2}}\right)+|X|(2|X|+2)+1.$ We have $|X|+{\nu }_1\ne 0$. Suppose that $|X|={\nu }_1=0$. We obtain $|V(G)|=2$, which contradicts the condition that $\nu ⩾4$. We obtain $\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1-|E(G)|={2|X|}^2+2{\nu }_1|X|+{\nu }_1>0.$ Therefore, $|E(G)|<\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1$ for $\nu ⩾4$. □

Subcase 2.2. Statement (ii) holds.

In view of $iso(G-X)⩽2|X|$ for any subset X of vertices of G and statement (ii), we obtain $2|X|⩾iso(G-X)⩾2|X|$. Thus, $iso(G-X)=2|X|$. Since the size of G is as large as possible, we obtain the following:

(a)

$G[X]$ is a complete graph;

(b)

$G-X$ has one non-trivial connected component, say $G_1$, where $G_1$ is a complete graph to which an edge has been attached;

(c)

G is the join of $G[X]$ and $G-X$. Hence, $G=G[X]\vee (G-X)$.

Set $|V(G_1)|={\nu }_1$. $H_{\nu }$ is a complete graph to which an edge has been attached and $|V(H_{\nu })|=\nu$.

Claim 2. $|E(G)|⩽\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1$. Equality holds if and only if G is $H_{{\nu }_1}$, where $H_{{\nu }_1}$ is a complete graph to which an edge has been attached.

Proof. 

We obtain $\nu =3|X|+{\nu }_1$ and ${\nu }_1⩾3$. Hence, $|E(G)|=\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -2|X|-1}{2}}\right)+|X|(2|X|+1)+1.$ Thus, $\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1-|E(G)|=2|X|(|X|+{\nu }_1-2)⩾0$, where equality holds if and only if $|X|=0$, i.e., G is $H_{{\nu }_1}$, where $H_{{\nu }_1}$ is a complete graph to which an edge has been attached. □

In view of Claim 1 and Claim 2, we have $|E(G)|⩽\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1$, which contradicts the condition that $|E(G)|>\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1$. This completes the proof of Theorem 4. □

6. A Strong $\{{\mathit{K}}_{\mathbf{1},\mathbf{1}},{\mathit{K}}_{\mathbf{1},\mathbf{2}},\cdots ,{\mathit{K}}_{\mathbf{1},\mathbf{r}}\}$-Factor

In this section, we will make use of typical spectral techniques to obtain two lower bounds for the distance signless Laplacian spectral radius and the distance signless Laplacian spectral radius and two upper bounds for the signless Laplacian spectral radius and the spectral radius of a graph.

6.1. Distance Signless Laplacian Spectral Radius

In this subsection, we give the proof of Theorem 5, which gives a sufficient condition via the distance signless Laplacian spectral radius of a connected graph G to ensure that G is a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor. Let $OddCa(G)$ denote the set of those components of a graph G that are odd complete cacti.

Proof of Theorem 5. 

Suppose that G has no strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor. Then, by Lemma 10, there exists some nonempty subset $X\subseteq V(G)$ such that $oddca(G-X)⩾r|X|+1$. Suppose that $X=\varnothing$. We have $oddca(G)⩾1$. Since G is a connected graph, we obtain $oddca(G)=1$. Therefore, $|V(G)|$ is an odd number, which contradicts the condition that $|V(G)|$ is an even number. Choose such a connected graph G of order $\nu$ such that its distance signless Laplacian spectral radius is as small as possible. Together with Lemma 4 and the choice of G, the induced subgraph $G[X]$ and each connected component of $G-X$ are complete graphs, respectively. Moreover, it holds that $G=G[X]\vee (G-X)$.

Claim 3.  $G-X$ has no even order connected component.

Proof. 

Otherwise, we can connect each vertex of an even order connected component to each vertex of a connected component of $OddCa(G-X)$ with an edge, which results in a larger order connected component, contradicting the choice of G. □

Let $\alpha (x)=x^3-(5\nu +2r-7)x^2+(4r^2+8{\nu }^2+2\nu r-27\nu +4r+24)x-4\nu r^2-4{\nu }^3-8\nu r+6r^2+22{\nu }^2-42\nu +14r+28$ be a real function in x and let $\beta (\nu ,r)$ be the largest root of $\alpha (x)=0.$ We aim to demonstrate that $oddca(G-X)⩽r|X|+2$. If $oddca(G-X)⩾r|X|+3$, then we create a new graph H obtained from G by connecting each vertex of a connected component of $OddCa(G-X)$ to each vertex of another connected component of $OddCa(G-X)$ with an edge and then connecting each vertex of the third connected component of $OddCa(G-X)$. Then, $oddca(H-X)⩾r|X|+1$ and G is a proper spanning subgraph of H. By Lemma 4, we find that ${\eta }_1(G)>{\eta }_1(H)$, which contradicts the initial choice of G. Therefore, $oddca(G-X)⩽r|X|+2$. Bear in mind that $oddca(G-X)⩾r|X|+1$, and it suffices to consider $oddca(G-X)=r|X|+1$ and $oddca(G-X)=r|X|+2$.

We consider two cases:

Case 1. $oddca(G-X)=r|X|+1$.

Assume that $G_1,G_2,\cdots ,G_{r|X|+1}$ are all the components of $G-X$ and let ${\nu }_i=|V(G_i)|$ with ${\nu }_1⩾{\nu }_2⩾\cdots ⩾{\nu }_{r|X|+1}$. Then, $G=K_{|X|}\vee (K_{{\nu }_1}\cup K_{{\nu }_2}\cup \cdots \cup K_{{\nu }_{r|X|+1}}).$

Claim 4.${\nu }_2={\nu }_3=\cdots ={\nu }_{r|X|+1}=1$.

Proof. 

Let ${\nu }_2={\nu }_3=\cdots ={\nu }_{r|X|+1}=1$. Hence, ${\nu }_1=\nu -|X|-r|X|$ is odd. Then, $oddca(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|})-X)=oddca(G-X)⩾r|X|+1.$ By Lemma 6, we have ${\eta }_1(G)⩾{\eta }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))$ with equality if and only if $G\cong K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|})$. Hence, $K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|})$ is a $\nu$-vertex connected graph with ${\eta }_1(G)>{\eta }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))$, a contradiction to the choice of G. Hence, we obtain $G=K_{|X|}\vee (K_{{\nu }_1}\cup K_{{\nu }_2}\cup \cdots \cup K_{{\nu }_{r|X|+1}})\cong K_{|X|}\vee (K_{\nu -|X|-r|X|}\cup (r|X|)K_1).$ Hence, we obtain ${\nu }_2={\nu }_3=\cdots ={\nu }_{r|X|+1}=1.$ This completes the proof of Claim 4. □

In view of Claim 4, we obtain $G=K_{|X|}\vee (K_{{\nu }_1}\cup (r|X|)K_1),$ where ${\nu }_1=\nu -|X|-r|X|$ is odd. As ${\nu }_1⩾1$, we obtain $\nu ⩾|X|(r+1)+1$.

Subcase 1.1. ${\nu }_1=1$.

In view of Claim 4, we obtain $G=K_{|X|}\vee (r|X|+1)K_1$ and $\nu =(r+1)|X|+1$. Since r is an odd number, $\nu$ is an odd number, which contradicts the condition that $|V(G)|$ is an even number.

Subcase 1.2. ${\nu }_1⩾3$.

We obtain $G=K_{|X|}\vee (K_{{\nu }_1}\cup (r|X|)K_1)$ and $\nu =(r+1)|X|+{\nu }_1$, where ${\nu }_1$ is odd. Since r is an odd number, $\nu$ is an odd number, which contradicts the condition that $|V(G)|$ is an even number.

Case 2. $oddca(G-X)=r|X|+2$.

Looking back at Case 1 and Claim 4, we have $G=K_{|X|}\vee (K_{{\nu }_1}\cup (r|X|+1)K_1)$, where ${\nu }_1=\nu -|X|-r|X|-1$ is odd.

We proceed by considering the following two subcases:

Subcase 2.1. ${\nu }_1=1$.

In this subcase, one has $G=K_{|X|}\vee (r|X|+2)K_1$ and $\nu =(r+1)|X|+2$. Consider the partition $V(G)=V(K_{|X|})\cup V((r|X|+2)K_1)$. The quotient matrix of $\mathbf{Q}(G)$ is

$${\mathbf{M}}_{\mathbf{1}}=\left(\begin{array}{cc}\nu +|X|-2& r|X|+2\\ |X|& |X|+4r|X|+4\end{array}\right).$$

Let $\mathbf{M}$ be a square matrix. Then, its characteristic polynomial is denoted by ${\Phi }_{\mathbf{M}}(x)=det(x\mathbf{I}-\mathbf{M})$, where $\mathbf{I}$ is the identity matrix, whose order is the same as that of $\mathbf{M}$. Then, the characteristic polynomial of ${\mathbf{M}}_{\mathbf{1}}$ is

$${\Phi }_{{\mathbf{M}}_{\mathbf{1}}}(x)=x^2{-(4r|X|+\nu +2|X|+2)x+4\nu r|X|+3r|X|}^2+\nu |X|+{|X|}^2-8r|X|+4\nu -8.$$

It is easy to see that the partition $V(G)=V(K_{|X|})\cup V((r|X|+2)K_1)$ is an equitable partition. Together with Lemma 8, we conclude that the largest root of ${\Phi }_{{\mathbf{M}}_{\mathbf{1}}}(x)=0$, say ${\beta }_1$, equals ${\eta }_1(G)$. Combining with $|X|={\displaystyle \frac{\nu -2}{r+1}}$, we have

$$\begin{array}{c}{\beta }_1={\eta }_1(G)\hfill \\ ={\displaystyle \frac{4r|X|+\nu +2|X|+2+\sqrt{16r^2{|X|}^2-8\nu r|X|+{4r|X|}^2+48r|X|+{\nu }^2-12\nu +8|X|+36}}{2}}\hfill \\ ={\displaystyle \frac{\nu (5r+3)-6r-2+\sqrt{{\nu }^2(9r^2-2r+1)-4\nu (3r^2-8r+1)+4(r^2-6r+5)}}{2r+2}}.\hfill \end{array}$$

Recall that $\alpha (x)=x^3-(5\nu +2r-7)x^2+(4r^2+8{\nu }^2+2\nu r-27\nu +4r+24)x-4\nu r^2-4{\nu }^3-8\nu r+6r^2+22{\nu }^2-42\nu +14r+28$ is a real function in x and $\beta (\nu ,r)$ is the largest root of $\alpha (x)=0.$

If $|X|=1$ and $\nu =r+3$, then ${\eta }_1(G)={\displaystyle \frac{5r^2+12r+7+(r+1)\sqrt{9r^2+22r+17}}{2r+2}}$ and $\alpha ({\eta }_1(G))=0$. Hence, ${\eta }_1(G)=\beta (\nu ,r)$, which contradicts ${\eta }_1(G)<\beta (\nu ,r)$ for $\nu =r+3$. If $|X|=2$ and $\nu =2r+4$, then ${\eta }_1(G)={\displaystyle \frac{5r^2+10r+5+(r+1)\sqrt{9r^2+10r+5}}{r+1}}$ and $\alpha ({\eta }_1(G))=12r^3-2r^2-24r-14+(4r^2-4r-6)\sqrt{9r^2+10r+5}>0.$ As usual, given a real function, say $f(x)$, in x, we denote by $f^{\prime }(x)$ the derivative function of $f(x)$. By ${\alpha }^{\prime }(x)=3x^2+2(7-5\nu -2r)x+24-27\nu +8{\nu }^2+4r+2\nu r+4r^2$, we have ${\displaystyle \frac{5\nu +2r-7}{3}}={\displaystyle \frac{5(2r+4)+2r-7}{3}}<{\displaystyle \frac{5r^2+10r+5+(r+1)\sqrt{9r^2+10r+5}}{r+1}}={\eta }_1(G)$. If $x>{\displaystyle \frac{5r^2+10r+5+(r+1)\sqrt{9r^2+10r+5}}{r+1}}={\eta }_1(G)$, then

$${\alpha }^{\prime }(x)>{\alpha }^{\prime }({\eta }_1(G))={\alpha }^{\prime }({\displaystyle \frac{5r^2+10r+5+(r+1)\sqrt{9r^2+10r+5}}{r+1}})=22r^2+16r+4+(6r+4)\sqrt{9r^2+10r+5}>0.$$

Hence, $\alpha (x)$ is a monotonically increasing function for $x>{\displaystyle \frac{5r^2+10r+5+(r+1)\sqrt{9r^2+10r+5}}{r+1}}={\eta }_1(G).$ Hence, ${\eta }_1(G)={\displaystyle \frac{5r^2+10r+5+(r+1)\sqrt{9r^2+10r+5}}{r+1}}>\beta (\nu ,r)$, which contradicts ${\eta }_1(G)<\beta (\nu ,r)$ for $\nu =2r+4$.

If $|X|=3$ and $\nu =3r+5$, then ${\eta }_1(G)={\displaystyle \frac{15r^2+28r+13+(r+1)\sqrt{81r^2+54r+25}}{2r+2}}$ and $\alpha ({\eta }_1(G))=72r^3-18r^2-68r-30+(8r^2-6r-6)\sqrt{81r^2+54r+25}>0.$ By ${\alpha }^{\prime }(x)=3x^2+2(7-5\nu -2r)x+24-27\nu +8{\nu }^2+4r+2\nu r+4r^2$, we have ${\displaystyle \frac{5\nu +2r-7}{3}}={\displaystyle \frac{5(3r+5)+2r-7}{3}}<{\displaystyle \frac{15r^2+28r+13+(r+1)\sqrt{81r^2+54r+25}}{2r+2}}={\eta }_1(G)$. If $x>{\displaystyle \frac{15r^2+28r+13+(r+1)\sqrt{81r^2+54r+25}}{2r+2}}={\eta }_1(G)$, then ${\alpha }^{\prime }(x)>{\alpha }^{\prime }({\eta }_1(G))={\alpha }^{\prime }({\displaystyle \frac{15r^2+28r+13+(r+1)\sqrt{81r^2+54r+25}}{2r+2}})={\displaystyle \frac{113r^2}{2}}+15r+{\displaystyle \frac{1}{2}}+({\displaystyle \frac{11}{2}}r+{\displaystyle \frac{3}{2}})\sqrt{81r^2+54r+25}>0.$ Hence, $\alpha (x)$ is a monotonically increasing function for $x>{\eta }_1(G)={\displaystyle \frac{15r^2+28r+13+(r+1)\sqrt{81r^2+54r+25}}{2r+2}}.$ Thus, ${\eta }_1(G)={\displaystyle \frac{15r^2+28r+13+(r+1)\sqrt{81r^2+54r+25}}{2r+2}}>\beta (\nu ,r)$, which contradicts ${\eta }_1(G)<\beta (\nu ,r)$ for $\nu =3r+5$.

If $|X|=4$ and $\nu =4r+6$, then ${\eta }_1(G)={\displaystyle \frac{2(5r^2+9r+4+(r+1)\sqrt{9r^2+4r+2})}{r+1}}$ and $\alpha ({\eta }_1(G))=6(36r^3-11r^2-21r-8)+12(6r^2-4r-3)\sqrt{9r^2+4r+2}-11)>0.$ By ${\alpha }^{\prime }(x)=3x^2+2(7-5\nu -2r)x+24-27\nu +8{\nu }^2+4r+2\nu r+4r^2$, we have ${\displaystyle \frac{5\nu +2r-7}{3}}={\displaystyle \frac{5(4r+6)+2r-7}{3}}<{\displaystyle \frac{2(5r^2+9r+4+(r+1)\sqrt{9r^2+4r+2})}{r+1}}={\eta }_1(G)$. If $x>{\displaystyle \frac{2(5r^2+9r+4+(r+1)\sqrt{9r^2+4r+2})}{r+1}}={\eta }_1(G)$, then ${\alpha }^{\prime }(x)>{\alpha }^{\prime }({\eta }_1(G))={\alpha }^{\prime }({\displaystyle \frac{2(5r^2+9r+4+(r+1)\sqrt{9r^2+4r+2})}{r+1}})=108r^2+8r-2+(32r+4)\sqrt{9r^2+4r+2}>0.$ Hence, $\alpha (x)$ is a monotonically increasing function for $x>{\eta }_1(G)={\displaystyle \frac{2(5r^2+9r+4+(r+1)\sqrt{9r^2+4r+2})}{r+1}}.$ Hence, ${\eta }_1(G)={\displaystyle \frac{2(5r^2+9r+4+(r+1)\sqrt{9r^2+4r+2})}{r+1}}>\beta (\nu ,r)$, which contradicts ${\eta }_1(G)<\beta (\nu ,r)$ for $\nu =4r+6$.

If $|X|=5$ and $\nu =5r+7$, then ${\eta }_1(G)={\displaystyle \frac{25r^2+44r+19+(r+1)\sqrt{225r^2+70r+41}}{2r+2}}$ and $\alpha ({\eta }_1(G))=4(120r^3-41r^2-48r-17)+4(8r^2-5r-3)\sqrt{225r^2+70r+41}>0.$ By ${\alpha }^{\prime }(x)=3x^2+2(7-5\nu -2r)x+24-27\nu +8{\nu }^2+4r+2\nu r+4r^2$, we have ${\displaystyle \frac{5\nu +2r-7}{3}}={\displaystyle \frac{5(5r+7)+2r-7}{3}}<{\displaystyle \frac{25r^2+44r+19+(r+1)\sqrt{225r^2+70r+41}}{2r+2}}={\eta }_1(G)$. If $x>{\displaystyle \frac{25r^2+44r+19+(r+1)\sqrt{225r^2+70r+41}}{2r+2}}={\eta }_1(G)$, then ${\alpha }^{\prime }(x)>{\alpha }^{\prime }({\eta }_1(G))={\alpha }^{\prime }({\displaystyle \frac{25r^2+44r+19+(r+1)\sqrt{225r^2+70r+41}}{2r+2}})={\displaystyle \frac{353r^2-10r-7+(21r+1)\sqrt{225r^2+70r+41}}{2}}>0.$ Hence, $\alpha (x)$ is a monotonically increasing function for $x>{\eta }_1(G)={\displaystyle \frac{25r^2+44r+19+(r+1)\sqrt{225r^2+70r+41}}{2r+2}}.$ Hence, ${\eta }_1(G)={\displaystyle \frac{25r^2+44r+19+(r+1)\sqrt{225r^2+70r+41}}{2r+2}}>\beta (\nu ,r)$, which contradicts ${\eta }_1(G)<\beta (\nu ,r)$ for $\nu =5r+7$.

If $|X|=6$ and $\nu =6r+8$, then ${\eta }_1(G)={\displaystyle \frac{15r^2+26r+11+(r+1)\sqrt{81r^2+18r+13}}{r+1}}$ and $\alpha ({\eta }_1(G))=10[90r^3+r^2(10\sqrt{81r^2+18r+13}-33)-2r(3\sqrt{81r^2+18r+13}+13)-3(\sqrt{81r^2+18r+13}+3)]>0.$ By ${\alpha }^{\prime }(x)=3x^2+2(7-5\nu -2r)x+24-27\nu +8{\nu }^2+4r+2\nu r+4r^2$, we have ${\displaystyle \frac{5\nu +2r-7}{3}}={\displaystyle \frac{5(6r+8)+2r-7}{3}}<{\displaystyle \frac{15r^2+26r+11+(r+1)\sqrt{81r^2+18r+13}}{r+1}}={\eta }_1(G)$. If $x>{\displaystyle \frac{15r^2+26r+11+(r+1)\sqrt{81r^2+18r+13}}{r+1}}={\eta }_1(G)$, then ${\alpha }^{\prime }(x)>{\alpha }^{\prime }({\eta }_1(G))={\alpha }^{\prime }({\displaystyle \frac{15r^2+26r+11+(r+1)\sqrt{81r^2+18r+13}}{r+1}})=262r^2-24r-4+26r\sqrt{81r^2+18r+13}>0.$ Hence, $\alpha (x)$ is a monotonically increasing function for $x>{\eta }_1(G)={\displaystyle \frac{15r^2+26r+11+(r+1)\sqrt{81r^2+18r+13}}{r+1}}.$ Thus, ${\eta }_1(G)={\displaystyle \frac{15r^2+26r+11+(r+1)\sqrt{81r^2+18r+13}}{r+1}}>\beta (\nu ,r)$, which contradicts ${\eta }_1(G)<\beta (\nu ,r)$ for $\nu =6r+8$.

Fact 1. If $\nu ⩾7r+8$, then $\beta (\nu ,r)<3\nu -r$.

Proof. 

By ${\alpha }^{\prime }(x)=3x^2+2(7-5\nu -2r)x+24-27\nu +8{\nu }^2+4r+2\nu r+4r^2$, we have ${\displaystyle \frac{5\nu +2r-7}{3}}<2\nu +3r+1$. If $x>2\nu +3r+1$, then

$${\alpha }^{\prime }(x)>{\alpha }^{\prime }(2\nu +3r+1)=19r^2+3\nu +60r+41>0.$$

By $\nu ⩾7r+8$ and ${\displaystyle \frac{-5r+8}{8}}<7r+8$, we obtain $\alpha (2\nu +3r+1)=-4(r+1){\nu }^2-(5r^2-3r-8)\nu +21r^3+100r^2+139r+60<-4(r+1){(7r+8)}^2-(5r^2-3r-8)(7r+8)+21r^3+100r^2+139r+60=-210r^3-563r^2-485r-132<0.$ Note that $\alpha (3\nu -r)=2{\nu }^3-(17r-4){\nu }^2+(22r^2-11r+30)\nu -7r^3+9r^2-10r+28.$ Let $h_1(x)=2x^3-(17r-4)x^2+(22r^2-11r+30)x-7r^3+9r^2-10r+28$ be a real function in x for $x\in [7r+8,+\infty )$. Note that ${\displaystyle \frac{17r-4+\sqrt{157r^2-70r-164}}{6}}-(7r+8)={\displaystyle \frac{-25r-52+\sqrt{157r^2-70r-164}}{6}}.$ Observe that $157r^2-70r-164-{(-25r-52)}^2=-6(78r^2+445r+478)<0$ and $\sqrt{157r^2-70r-164}<25r+52.$ Hence, ${\displaystyle \frac{-25r-52+\sqrt{157r^2-70r-164}}{6}}<{\displaystyle \frac{-25r-52+25r+52}{6}}=0.$ Together with ${\displaystyle \frac{17r-4+\sqrt{157r^2-70r-164}}{6}}<7r+8,$ we conclude that $h_1(x)$ is a monotonically increasing function for $x⩾7r+8.$ By $x⩾7r+8$, we obtain $h_1(x)⩾h_1(7r+8)=4(188r^2+540r+387)>0$. If $\nu ⩾7r+8$, then $\alpha (3\nu -r)>0$. Together with $\alpha (2\nu +3r+1)<0$ for $\nu ⩾7r+8$ and ${\alpha }^{\prime }(x)>0$, we find that $\beta (\nu ,r)<3\nu -r$ for $\nu ⩾7r+8$. Fact 1 is proved. □

Fact 2. If $|X|⩾4$ and $\nu ⩾4r+6$, then

$${\eta }_1(G)={\displaystyle \frac{\nu (5r+3)-6r-2+\sqrt{{\nu }^2(9r^2-2r+1)-4\nu (3r^2-8r+1)+4(r^2-6r+5)}}{2r+2}}>3\nu -r.$$

Proof. 

Note that

$$\begin{array}{cc}& {\nu }^2(9r^2-2r+1)-4\nu (3r^2-8r+1)+4(r^2-6r+5)\hfill \\ & -{[(3\nu -r)2(r+1)-(\nu (5r+3)-6r-2)]}^2\hfill \\ & =4(2r^2-2r-2){\nu }^2+4(r^3-2r^2+r-4)\nu -4(r^4-4r^3+r^2+10r-4).\hfill \end{array}$$

Let $h_2(x)=4(2r^2-2r-2)x^2+4(r^3-2r^2+r-4)x-4(r^4-4r^3+r^2+10r-4)$ be a real function in x for $x\in [4r+6,+\infty )$. Note that ${\displaystyle \frac{-r^3+2r^2-r+4}{4(r^2-r-1)}}-(4r+6)={\displaystyle \frac{-17r^3-6r^2+39r+28}{4(r^2-r-1)}}<0.$ Hence, $h_2(x)$ is increasing in the interval $[4r+6,+\infty )$. Thus, $h_2(x)⩾h_2(4r+6)=4{(r+1)}^2(35r^2-4r-92)⩾4{(r+1)}^2(315-12-92)>0.$ Hence, ${\nu }^2(9r^2-2r+1)-4\nu (3r^2-8r+1)+4(r^2-6r+5)>{[(3\nu -r)2(r+1)-(\nu (5r+3)-6r-2)]}^2.$ Thus, $\sqrt{{\nu }^2(9r^2-2r+1)-4\nu (3r^2-8r+1)+4(r^2-6r+5)}>(3\nu -r)2(r+1)-(\nu (5r+3)-6r-2).$ Therefore, ${\eta }_1(G)={\displaystyle \frac{\nu (5r+3)-6r-2+\sqrt{{\nu }^2(9r^2-2r+1)-4\nu (3r^2-8r+1)+4(r^2-6r+5)}}{2r+2}}>3\nu -r.$ This completes the proof of Fact 2. □

If $\nu ⩾7r+9$, then $\beta (\nu ,r)<3\nu -r$. If $|X|⩾4$ and $\nu ⩾7r+9$, then ${\eta }_1(G)>3\nu -r.$ Hence, ${\eta }_1(G)>3\nu -r>\beta (\nu ,r)$ for $\nu ⩾7r+9$, which contradicts ${\eta }_1(G)<\beta (\nu ,r)$ for $\nu ⩾7r+9$.

Subcase 2.2. ${\nu }_1⩾3$.

We obtain $\nu =(r+1)|X|+{\nu }_1+1⩾3(|X|+1)⩾6$ and $G=K_{|X|}\vee (K_{{\nu }_1}\cup (r|X|+1)K_1).$ Consider the partition $V(G)=V((r|X|+1)K_1)\cup V(K_{{\nu }_1})\cup V(K_{|X|})$. The quotient matrix of $\mathbf{Q}(G)$ is

$${\mathbf{M}}_{\mathbf{2}}=\left(\begin{array}{ccc}2\nu +(2r-1)|X|-2& 2(\nu -(r+1)|X|-1)& |X|\\ 2(r|X|+1)& 2\nu -|X|-2& |X|\\ r|X|+1& \nu -(r+1)|X|-1& \nu +|X|-2\end{array}\right).$$

Then, the characteristic polynomial of ${\mathbf{M}}_{\mathbf{2}}$ equals

$$\begin{array}{cc}\hfill {\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(x)=& x^3-(2r|X|+5\nu -|X|-6)x^2+(4r^2{|X|}^2+2\nu r|X|+{4r|X|}^2-3\nu |X|+8{\nu }^2\hfill \\ & -24\nu +8|X|+16)x-2r^2{|X|}^3-4\nu r^2{|X|}^2-{4\nu r|X|}^2+8r^2{|X|}^2-4\nu r|X|\hfill \\ & +{6r|X|}^2+2{\nu }^2|X|-4{\nu }^3-10\nu |X|+20{\nu }^2+8r|X|-32\nu +12|X|+16.\hfill \end{array}$$

Since $V(G)=V((r|X|+1)K_1)\cup V(K_{{\nu }_1})\cup V(K_{|X|})$ is an equitable partition, by Lemma 8, the largest root, say ${\beta }_2$, of ${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(x)=0$ equals ${\eta }_1(G)$.

To show $\beta (\nu ,r)⩽{\eta }_1(G)$, it suffices to prove that ${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(\beta (\nu ,r))⩽0.$ Together with $\alpha (\beta (\nu ,r))=0$, we obtain

$${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(\beta (\nu ,r))={\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(\beta (\nu ,r))-\alpha (\beta (\nu ,r))=(|X|-1)f,$$

where ${f=[-(2r-1)\beta (\nu ,r)}^2+(4r^2|X|+2\nu r+4r^2+4r|X|-3\nu +4r+8)\beta (\nu ,r)-4\nu r^2(|X|+1)-2r^2{|X|}^2-4\nu r|X|+6r^2|X|-8\nu r+2{\nu }^2+6r^2+6r|X|-10\nu +14r+12]$. Clearly, ${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(\beta (\nu ,r))=0$ for $|X|=1$. If $|X|=1$, then the polynomial ${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(x)$ becomes $\alpha (x)$. Hence, ${\eta }_1(G)={\beta }_2=\beta (\nu ,r)$, which contradicts ${\eta }_1(G)<\beta (\nu ,r)$ for $\nu ⩾r+5$.

Next, we will prove that $f<0$ for $|X|⩾2$. By $\nu ⩾2(r+1)+4$, we have

$$\begin{array}{cc}\hfill \alpha (2\nu +3r+1)=& -4(r+1){\nu }^2-(5r^2-3r-8)\nu +21r^3+100r^2+139r+60\hfill \\ \hfill ⩽& -4(r+1){[2(r+1)+4]}^2-(5r^2-3r-8)[2(r+1)+4]\hfill \\ & +21r^3+100r^2+139r+60\hfill \\ \hfill =& -5r^3-36r^2-67r-36<0.\hfill \end{array}$$

Hence, $\beta (\nu ,r)>2\nu +3r+1.$ Let $h_3(x)=-(2r-1)x^2+(4r^2|X|+2\nu r+4r^2+4r|X|-3\nu +4r+8)x-4\nu r^2(|X|+1)-2r^2{|X|}^2-4\nu r|X|+6r^2|X|-8\nu r+2{\nu }^2+6r^2+6r|X|-10\nu +14r+12$ be a real function in x for $x\in (2\nu +3r+1,+\infty )$. By $\nu ⩾(r+1)|X|+3$, we have ${\displaystyle \frac{4r^2|X|+2\nu r+4r^2+4r|X|-3\nu +4r+8}{2(2r-1)}}<2\nu +3r+1.$ Hence, $h_3(x)$ is a monotonically decreasing function for $x>2\nu +3r+1.$ Thus, $h_3(x)<h_3(2\nu +3r+1)=-4r{\nu }^2+[2r^2(2|X|-7)+r(4|X|-3)+7]\nu +6r^3(2|X|-1)-r^2{(2|X|}^2-22|X|-19)+2r(5|X|+23)+21$ when $x>2\nu +3r+1$. Let $h_4(x)=-4r{x}^2+[2r^2(2|X|-7)+r(4|X|-3)+7]x+6r^3(2|X|-1)-r^2{(2|X|}^2-22|X|-19)+2r(5|X|+23)+21$ be a real function in x for $x\in [(r+1)|X|+4,+\infty )$. By $\nu ⩾(r+1)|X|+4$, we have ${\displaystyle \frac{2r^2(2|X|-7)+r(4|X|-3)+7}{8r}}<(r+1)|X|+4.$ Thus, $h_4(x)$ is a monotonically decreasing function for $x⩾(r+1)|X|+4.$ Hence, $h_4(x)⩽h_4((r+1)|X|+4)=-2(|X|+3)r^3-{(2|X|}^2+11|X|+37)r^2-2(|X|+15)r+7(|X|+7)<0$ when $x⩾(r+1)|X|+4$. Thus, $f<0$ and ${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(\beta (\nu ,r))={\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(\beta (\nu ,r))-\alpha (\beta (\nu ,r))<0$ when $|X|⩾2$. Therefore, $\beta (\nu ,r)<{\beta }_2={\eta }_1(G)$ when $\nu ⩾(r+1)|X|+4$, a contradiction to the condition. This completes the proof of Theorem 5. □

6.2. Signless Laplacian Spectral Radius

In this subsection, we give the proof of Theorem 6, which gives a sufficient condition via the signless Laplacian spectral radius of a connected graph G to ensure that G is a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor.

Proof of Theorem 6. 

Suppose that G has no strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor. Then, by Lemma 10, there exists some nonempty subset $X\subseteq V(G)$ such that $oddca(G-X)⩾r|X|+1$. Suppose that $X=\varnothing$. We have $oddca(G)⩾1$. Since G is a connected graph, we obtain $oddca(G)=1$. Therefore, $|V(G)|$ is an odd number, which contradicts the condition that $|V(G)|$ is an even number. Choose such a connected graph G of order $\nu$ such that its signless Laplacian spectral radius is as large as possible. Together with Lemma 1 and the choice of G, the induced subgraph $G[X]$ and each connected component of $G-X$ are complete graphs, respectively. Moreover, it holds that $G=G[X]\vee (G-X)$.

Claim 5. $G-X$ has no even order connected component.

Proof. 

Otherwise, we can connect each vertex of an even order connected component to each vertex of a connected component of $OddCa(G-X)$ with an edge, which results in a larger order connected component, contradicting the choice of G. □

Let $\gamma (x)=x^3-(4\nu -3r-10)x^2+(2{\nu }^2-2\nu r-5\nu )x+4\nu r-2r^2-2{\nu }^2+10\nu -10r-12$ be a real function in x and let $\phi (\nu ,r)$ be the largest root of $\gamma (x)=0.$ We aim to demonstrate that $oddca(G-X)⩽r|X|+2$. If $oddca(G-X)⩾r|X|+3$, then we create a new graph H obtained from G by connecting each vertex of a connected component of $OddCa(G-X)$ to each vertex of another connected component of $OddCa(G-X)$ with an edge and then connecting each vertex of the third connected component of $OddCa(G-X)$. Then, $oddca(H-X⩾r|X|+1$ and G is a proper spanning subgraph of H. By Lemma 1, we find that ${\kappa }_1(G)<{\kappa }_1(H)$, which contradicts the initial choice of G. Therefore, $oddca(G-X)⩽r|X|+2$. Bear in mind that $oddca(G-X)⩾r|X|+1$. It suffices to consider $oddca(G-X)=r|X|+1$ and $oddca(G-X)=r|X|+2$.

In what follows, we proceed by considering the two possible cases:

Case 1. $oddca(G-X)=r|X|+1$. Assume that $G_1,G_2,\cdots ,G_{r|X|+1}$ are all the components of $G-X$ and let ${\nu }_i=|V(G_i)|$ with ${\nu }_1⩾{\nu }_2⩾\cdots ⩾{\nu }_{r|X|+1}$. Then, $G=K_{|X|}\vee (K_{{\nu }_1}\cup K_{{\nu }_2}\cup \cdots \cup K_{{\nu }_{r|X|+1}}).$

Claim 6.${\nu }_2={\nu }_3=\cdots ={\nu }_{r|X|+1}=1$.

Proof. 

If ${\nu }_2⩾3$, then $oddca(K_{|X|}\vee (K_{{\nu }_1+2}\cup K_{{\nu }_2-2}\cup \cdots K_{{\nu }_{r|X|+1}})-X)=oddca(G-X)⩾r|X|+1.$ Let $\mathbf{x}={(x_1,\cdots ,x_{\nu })}^{\top }$ be the Perron vector of $\mathbf{K}(G)$, and let $x_i$ denote the entry of $\mathbf{x}$ corresponding to the vertex $v_i\in V(G)$. By Lemma 7, one has $x_j=x_k$ for all $v_j,v_k$ in X (resp. $V(G_i)$), and $i\in \{1,2,\cdots ,r|X|+1\}$). For convenience, let $x_0=x_r$ for all $v_r\in X$ and $x_i=x_r$ for all $v_r\in V(G_i)$, $i\in \{1,2,\cdots ,r|X|+1\}$. Then,

$$\left\{\begin{array}{cc}{\kappa }_1(G)x_1=|X|x_0+(|X|+2{\nu }_1-2)x_1,\hfill & \\ {\kappa }_1(G)x_2=|X|x_0+(|X|+2{\nu }_2-2)x_2.\hfill \end{array}\right.$$

Consequently, $({\kappa }_1(G)-|X|-2{\nu }_1+2)x_1=({\kappa }_1(G)-|X|-2{\nu }_2+2)x_2.$ Observe Lemmas 1 and 3. Since $K_{|X|+{\nu }_1}$ and $K_{|X|+{\nu }_2}$ are two proper subgraphs of G, ${\kappa }_1(G)>{\kappa }_1(K_{|X|+{\nu }_1})=2(|X|+{\nu }_1)-2>|X|+2{\nu }_1-2$ and ${\kappa }_1(G)>{\kappa }_1(K_{|X|+{\nu }_2})=2(|X|+{\nu }_2)-2>|X|+2{\nu }_2-2.$ By $({\kappa }_1(G)-|X|-2{\nu }_1+2)x_1=({\kappa }_1(G)-|X|-2{\nu }_2+2)x_2$, ${\kappa }_1(G)>|X|+2{\nu }_1-2$, and ${\nu }_1⩾{\nu }_2$, we have $x_1⩾x_2$. Because of the Rayleigh quotient,

$$\begin{array}{cc}& {\kappa }_1(K_{|X|}\vee (K_{{\nu }_1+2}\cup K_{{\nu }_2-2}\cup \cdots K_{{\nu }_{r|X|+1}}))-{\kappa }_1(G))\hfill \\ \hfill ⩾& {\mathbf{x}}^{\top }(\mathbf{K}(K_{|X|}\vee (K_{{\nu }_1+2}\cup K_{{\nu }_2-2}\cup \cdots K_{{\nu }_{r|X|+1}}))-\mathbf{K}(G))\mathbf{x}\hfill \\ \hfill =& 2({\nu }_1{x}_1(x_1+x_2)-4({\nu }_2-2)x_2^2+{\nu }_1{x}_2(x_1+x_2))\hfill \\ \hfill ⩾& 8{\nu }_1{x}_2^2-8({\nu }_2-2)x_2^2\hfill \\ \hfill =& 8x_2^2({\nu }_1-{\nu }_2+2)\hfill \\ \hfill >& 0.\hfill \end{array}$$

Hence, $K_{|X|}\vee (K_{{\nu }_1+2}\cup K_{{\nu }_2-2}\cup \cdots K_{{\nu }_{r|X|+1}})$ is a $\nu$-vertex connected graph with ${\eta }_1(G)>{\eta }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))$, a contradiction to the choice of G. Therefore, we obtain ${\nu }_2={\nu }_3=\cdots ={\nu }_{r|X|+1}=1.$ □

In view of Claim 6, we obtain $G=K_{|X|}\vee (K_{{\nu }_1}\cup (r|X|)K_1),$ where ${\nu }_1$ is odd. As ${\nu }_1⩾1$, we obtain $\nu ⩾|X|(r+1)+1$.

We proceed by considering the following two subcases:

Subcase 1.1. ${\nu }_1=1$.

In view of Claim 6, we obtain $G=K_{|X|}\vee (r|X|+1)K_1$ and $\nu =(r+1)|X|+1$. Since r is an odd number, $\nu$ is an odd number, which contradicts the condition that $|V(G)|$ is an even number.

Subcase 1.2. ${\nu }_1⩾3$.

We obtain $G=K_{|X|}\vee (K_{{\nu }_1}\cup (r|X|)K_1)$ and $\nu =(r+1)|X|+{\nu }_1$, where ${\nu }_1$ is odd. Since r is an odd number, $\nu$ is an odd number, which contradicts the condition that $|V(G)|$ is an even number.

Case 2. $oddca(G-X)=r|X|+2$.

Looking back at Case 1 and Claim 6, we have $G=K_{|X|}\vee (K_{{\nu }_1}\cup (r|X|+1)K_1)$, where ${\nu }_1$ is odd.

We proceed by considering the following two subcases:

Subcase 2.1. ${\nu }_1=1$.

In this subcase, one has $G=K_{|X|}\vee (r|X|+2)K_1$ and $\nu =(r+1)|X|+2$. Consider the partition $V(G)=V(K_{|X|})\cup V((r|X|+2)K_1)$. The quotient matrix of $\mathbf{K}(G)$ is

$${\mathbf{M}}_{\mathbf{1}}=\left(\begin{array}{cc}\nu +|X|-2& r|X|+2\\ |X|& |X|\end{array}\right).$$

Then, the characteristic polynomial of ${\mathbf{M}}_{\mathbf{1}}$ is

$${\Phi }_{{\mathbf{M}}_{\mathbf{1}}}(x)=x^2-(\nu +2|X|-2)x-r{s}^2+\nu |X|+{|X|}^2-4|X|.$$

Since $V(G)=V(K_{|X|})\cup V((r|X|+2)K_1)$ is an equitable partition, together with Lemma 8, we conclude that the largest root of ${\Phi }_{{\mathbf{M}}_{\mathbf{1}}}(x)=0$, say ${\theta }_1$, equals ${\kappa }_1(G)$. Together with $\nu =(r+1)|X|+2$, we obtain

$${\theta }_1={\kappa }_1(G)={\displaystyle \frac{\nu +2|X|-2+\sqrt{{4r|X|}^2+{\nu }^2-4\nu +8|X|+4}}{2}}={\displaystyle \frac{(r+3)|X|+\sqrt{|X|((r^2+6r+1)|X|+8)}}{2}}>(r+2)|X|.$$

For $|X|=1$ and $\nu =r+3$, we have ${\kappa }_1(G)={\theta }_1=r+3$, which contradicts ${\kappa }_1(G)>r+3$ for $\nu =r+3$. For $|X|=2$ and $\nu =2r+4$, we have ${\kappa }_1(G)={\theta }_1=r+3+\sqrt{(r+5)(r+1)}$, which contradicts ${\kappa }_1(G)>r+3+\sqrt{(r+5)(r+1)}$ for $\nu =2r+4$.

To show $\phi (\nu ,r)>{\kappa }_1(G)$, it suffices to prove $\gamma ({\theta }_1)<0.$ Note that ${\Phi }_{{\mathbf{M}}_{\mathbf{1}}}({\theta }_1)=0$. For $|X|⩾3$ and $\nu =(r+1)|X|+2⩾3r+5$, we have

$$\begin{array}{cc}\hfill \gamma ({\theta }_1)=& \gamma ({\theta }_1)-{\theta }_1{\Phi }_{{\mathbf{M}}_{\mathbf{1}}}({\theta }_1)\hfill \\ \hfill =& -(3\nu -2|X|-3r-8){\theta }_1^2+{[r|X|}^2+2{\nu }^2-2\nu r-{|X|}^2-\nu (|X|+5)+4|X|]{\theta }_1\hfill \\ & +4\nu r-2r^2-2{\nu }^2+10\nu -10r-12\hfill \\ \hfill =& -(3((r+1)|X|+2)-2|X|-3r-8){\theta }_1^2+{[r|X|}^2+2{((r+1)|X|+2)}^2\hfill \\ & {-2((r+1)|X|+2)r-|X|}^2-((r+1)|X|+2)(|X|+5)+4|X|]{\theta }_1\hfill \\ & +4((r+1)|X|+2)r-2r^2-2{((r+1)|X|+2)}^2\hfill \\ & +10((r+1)|X|+2)-10r-12\hfill \\ \hfill =& -(3r|X|-3r+|X|-2){\theta }_1^2+{[4r|X|}^2+2r^2(|X|-1)|X|+r|X|-4r+5|X|\hfill \\ & -2]{\theta }_1-{4r|X|}^2-2r^2(|X|-1)|X|\hfill \\ & +2r^2{(|X|-1)-2|X|}^2+6r|X|-2r+2|X|\hfill \\ \hfill <& -(3r|X|-3r+|X|-2)(r+2)|X|{\theta }_1+{[4r|X|}^2+2r^2(|X|-1)|X|+r|X|\hfill \\ & -4r+5|X|-2]{\theta }_1-{4r|X|}^2-2r^2(|X|-1)|X|\hfill \\ & +2r^2{(|X|-1)-2|X|}^2+6r|X|-2r+2|X|\hfill \\ \hfill =& -(r^2(|X|-1)|X|+{2|X|}^2+{3r|X|}^2-9r|X|-9|X|+4r+2){\theta }_1\hfill \\ & -{4r|X|}^2-2r^2(|X|-1)|X|+2r^2{(|X|-1)+6r|X|-2|X|}^2+2|X|-2r\hfill \\ \hfill =& (-(r^2+3r+2){|X|}^2+(r^2+9r+9)|X|-4r-2){\theta }_1\hfill \\ & -{4r|X|}^2-2r^2(|X|-1)|X|+2r^2{(|X|-1)+6r|X|-2|X|}^2+2|X|-2r\hfill \\ \hfill ⩽& (-3(r^2+3r+2)|X|+(r^2+9r+9)|X|-4r-2){\theta }_1\hfill \\ & -{4r|X|}^2-2r^2(|X|-1)|X|+2r^2{(|X|-1)+6r|X|-2|X|}^2+2|X|-2r\hfill \\ \hfill =& -(2r^2|X|+4r-3|X|+2)x\hfill \\ & -2r^2{|X|}^2-{4r|X|}^2+4r^2|X|-2r^2+6r|X|-{2|X|}^2-2r+2|X|\hfill \\ \hfill <& 0,\hfill \end{array}$$

which contradicts ${\kappa }_1(G)>\phi (\nu ,r)$ for $\nu ⩾3r+5$.

Subcase 2.2. ${\nu }_1⩾3$.

We obtain $\nu =(r+1)|X|+{\nu }_1+1$ and $G=K_{|X|}\vee (K_{{\nu }_1}\cup (r|X|+1)K_1).$ Consider the partition $V(G)=V((r|X|+1)K_1)\cup V(K_{{\nu }_1})\cup V(K_{|X|})$. The quotient matrix of $\mathbf{K}(G)$ is

$${\mathbf{M}}_{\mathbf{2}}=\left(\begin{array}{ccc}|X|& 0& |X|\\ 0& 2\nu -(2r+1)|X|-4& |X|\\ r|X|+1& \nu -(r+1)|X|-1& \nu +|X|-2\end{array}\right).$$

Then, the characteristic polynomial of ${\mathbf{M}}_{\mathbf{2}}$ equals

$$\begin{array}{cc}\hfill {\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(x)=& x^3-(3\nu -2r|X|+|X|-6)x^2-(2\nu r|X|+{4r|X|}^2-2{\nu }^2-3\nu |X|-4r|X|\hfill \\ & {+8\nu +8|X|-8)x+4\nu r|X|}^2-2r^2{|X|}^3-2{\nu }^2|X|-{10r|X|}^2+10\nu |X|-12|X|.\hfill \end{array}$$

Since $V(G)=V((r|X|+1)K_1)\cup V(K_{{\nu }_1})\cup V(K_{|X|})$ is an equitable partition, together with Lemma 8, we find that the largest root of ${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(x)=0$, denoted as ${\theta }_2$, equals ${\kappa }_1(G)$.

Observe that $K_{|X|}\vee (\nu -|X|)K_1$ is a proper subgraph of G. Consider the partition $V(K_{|X|}\vee (\nu -|X|)K_1)=V(K_{|X|})\cup V((\nu -|X|)K_1)$. Its corresponding quotient matrix of $\mathbf{K}(K_{|X|}\vee (\nu -|X|)K_1)$ is represented by

$${\mathbf{M}}_{\mathbf{3}}=\left(\begin{array}{cc}\nu +|X|-2& \nu -|X|\\ |X|& |X|\end{array}\right).$$

Then, the characteristic polynomial of ${\mathbf{M}}_{\mathbf{3}}$ is given by

$${\Phi }_{{\mathbf{M}}_{\mathbf{3}}}(x)=x^2{-(\nu +2|X|-2)x+2|X|}^2-2|X|.$$

It is evident that the partition $V(K_{|X|}\vee (\nu -|X|)K_1)=V(K_{|X|})\cup V((\nu -|X|)K_1)$ is equitable. Referring to Lemma 8, the largest root, say ${\theta }_3$, of ${\Phi }_{{\mathbf{M}}_{\mathbf{3}}}(x)=0$ equals ${\kappa }_1(K_{|X|}\vee (\nu -|X|)K_1)$. As ${\Phi }_{{\mathbf{M}}_{\mathbf{3}}}(x)=0$ is a quadratic equation with respect to x, by the root formula, we have ${\kappa }_1(K_{|X|}\vee (\nu -|X|)K_1)={\theta }_3$, where ${\theta }_3={\displaystyle \frac{\nu +2|X|-2+\sqrt{{\nu }^2+4\nu |X|-{4|X|}^2-4\nu +4}}{2}}$. According to Lemma 1, it follows that

$${\theta }_2>{\theta }_3={\displaystyle \frac{\nu +2|X|-2+\sqrt{{(\nu -2)}^2+4\nu |X|-{4|X|}^2}}{2}}>{\displaystyle \frac{\nu +2|X|-2+\nu -2}{2}}=\nu +|X|-2.$$

To demonstrate that $\phi (\nu ,r)>{\kappa }_1(G)$, it suffices to prove $\gamma ({\theta }_2)<0$. Notice that ${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}({\theta }_2)=0$. Hence,

$$\begin{array}{cc}\hfill \gamma ({\theta }_2)=& \gamma ({\theta }_2)-{\Phi }_{{\mathbf{M}}_{\mathbf{2}}}({\theta }_2)\hfill \\ \hfill =& -(\nu +2r|X|-3r-|X|-4){\theta }_2^2+{[4r|X|}^2+2\nu r(|X|-1)-3\nu |X|+3\nu \hfill \\ & -4r|X|+8|X|-8]{\theta }_2+2r^2{(|X|}^3{-1)-4\nu r(|X|-1)|X|+10r|X|}^2\hfill \\ & +2{\nu }^2(|X|-1)-4\nu r(|X|-1)+10\nu -10r+12|X|-10\nu |X|-12\hfill \\ \hfill ⩽& -((r+1)|X|+4)+2r|X|-3r-|X|-4){\theta }_2^2+{[4r|X|}^2+2\nu r(|X|-1)\hfill \\ & -3\nu |X|+3\nu -4r|X|+8|X|-8]{\theta }_2+2r^2{(|X|}^3-1)-4\nu r(|X|-1)|X|\hfill \\ & +{10r|X|}^2+2{\nu }^2(|X|-1)-4\nu r(|X|-1)+10\nu -10r+12|X|-10\nu |X|-12\hfill \\ \hfill =& (|X|-1)(-3r{\theta }_2^2+(4r|X|+2\nu r-3\nu +8){\theta }_2+2r^2{(|X|}^2+|X|+1)-4\nu r|X|\hfill \\ & -4\nu r+10r|X|+2{\nu }^2+10r-10\nu +12)\hfill \end{array}$$

$$\begin{array}{cc}\hfill ⩽& (|X|-1)(-3r(\nu +|X|-2){\theta }_2+(4r|X|+2\nu r-3\nu +8){\theta }_2\hfill \\ & +2r^2{(|X|}^2+|X|+1)-4\nu r|X|-4\nu r+10r|X|+2{\nu }^2+10r-10\nu +12)\hfill \\ \hfill =& (|X|-1)(-(3\nu +\nu r-r|X|-6r-8){\theta }_2+2r^2{(|X|}^2+|X|+1)\hfill \\ & -4\nu r|X|+2{\nu }^2-4\nu r+10r|X|+10r-10\nu +12)\hfill \\ \hfill ⩽& (|X|-1)(-(3\nu +\nu r-r|X|-6r-8)(\nu +|X|-2)+2r^2{(|X|}^2+|X|+1)\hfill \\ & -4\nu r|X|+2{\nu }^2-4\nu r+10r|X|+10r-10\nu +12)\hfill \\ \hfill =& (|X|-1)(-(r+1){\nu }^2-(4r(|X|-1)+3|X|-4)\nu +8|X|\hfill \\ & +2r^2{(|X|}^2{+|X|+1)+r(|X|}^2+14|X|-2)-4)\hfill \\ \hfill ⩽& (|X|-1)(-(r+1)((r+1)|X|+4)\nu -(4r(|X|-1)+3|X|-4)\nu +8|X|\hfill \\ & +2r^2{(|X|}^2{+|X|+1)+r(|X|}^2+14|X|-2)-4)\hfill \\ \hfill =& (|X|-1)(-(4|X|+6r|X|+r^2|X|)\nu +2r^2{|X|}^2+2r^2|X|\hfill \\ & +{r|X|}^2+2r^2+14r|X|-2r+8|X|-4)\hfill \\ \hfill ⩽& (|X|-1)(-(4|X|+6r|X|+r^2|X|)((r+1)|X|+4)+2r^2{|X|}^2+2r^2|X|\hfill \\ & +{r|X|}^2+2r^2+14r|X|-2r+8|X|-4)\hfill \\ \hfill =& (|X|-1)(-(2r^2+23r+16){|X|}^2-2(r^2+5r+4)|X|+2r^2-2r-4)\hfill \\ \hfill ⩽& 0,\hfill \end{array}$$

which contradicts ${\kappa }_1(G)>\phi (\nu ,r)$ for $\nu ⩾(r+1)|X|+4$. This completes the proof of Theorem 6. □

6.3. Spectral Radius

In this subsection, we give the proof of Theorem 7, which gives a sufficient condition via the spectral radius of a connected graph G to ensure that G is a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor.

Proof of Theorem 7. 

Suppose that G has no strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor. Then, by Lemma 10, there exists some nonempty subset $X\subseteq V(G)$ such that $oddca(G-X)⩾r|X|+1$. Suppose that $X=\varnothing$. We have $oddca(G)⩾1$. Since G is a connected graph, we obtain $oddca(G)=1$. Therefore, $|V(G)|$ is an odd number, which contradicts the condition that $|V(G)|$ is an even number. Choose a connected graph G of order $\nu$ such that its adjacency spectral radius is as large as possible. Together with Lemma 2 and the choice of G, the induced subgraph $G[X]$ and each connected component of $G-X$ are complete graphs, respectively. Moreover, it holds that $G=G[X]\vee (G-X)$.

Claim 7. $G-X$ has no even order connected component.

Proof. 

Otherwise, we can connect each vertex of an even order connected component to each vertex of a connected component of $OddCa(G-X)$ with an edge, which results in a larger order connected component, contradicting the choice of G. □

Let $\chi (x)=x^3-(\nu -r-3)x^2-(\nu -1)x+\nu r-r^2+\nu -4r-3$ be a real function in x and let $\theta (\nu )$ be the largest root of $\chi (x)=0$.

We aim to demonstrate that $oddca(G-X)⩽r|X|+2$. If $oddca(G-X)⩾r|X|+3$, then we create a new graph H obtained from G by connecting each vertex of a connected component of $OddCa(G-X)$ to each vertex of another connected component of $OddCa(G-X)$ with an edge and then connecting each vertex of the third connected component of $OddCa(G-X)$. Then, $oddca(H-X)⩾r|X|+1$ and G is a proper spanning subgraph of H. By Lemma 2, we find that $\rho (G)<\rho (H)$, which contradicts the initial choice of G. Therefore, $oddca(G-X)⩽r|X|+2$. Bear in mind that $oddca(G-X)⩾r|X|+1$. It will be sufficient to show that $oddca(G-X)=r|X|+1$ and $oddca(G-X)=r|X|+2$.

We consider the following two cases:

Case 1. $oddca(G-X)=r|X|+1$.

Suppose that $G_1,G_2,\cdots ,G_{r|X|+1}$ are all the components in $G-X$ such that $|V(G_1)|⩾|V(G_2)|⩾\cdots ⩾|V(G_{r|X|+1})|$. Hence, $G=K_{|X|}\vee (K_{|V(G_1)|}\cup K_{|V(G_2)|}\cup \cdots \cup K_{|V(G_{r|X|+1})|})$. We would like to prove that $|V(G_2)|=|V(G_3)|=\cdots =|V(G_{r|X|+1})|=1$. Note that $|V(G_2)|⩾3$ and $oddca(K_{|X|}\vee (K_{|V(G_1)|+2}\cup K_{|V(G_2)|-2}\cup \cdots \cup K_{|V(G_{r|X|+1})|})-X)=oddca(G-X)⩾r|X|+1.$

Let $\mathbf{x}={(x_1,\cdots ,x_{\nu })}^{\top }$ be the Perron vector of $A(G)$ and $x_i$ denote the entry of x corresponding to the vertex $v_i\in V(G)$. In view of Lemma 7, we refer to $x_r=x_s$ for all $v_r,v_s$ in X (resp. $V(G_i),i\in \{1,2,\cdots ,r|X|+1\})$. Let $x_i=x_r$ for all $v_r\in V(G_i)$, $i\in \{1,2,\cdots ,r|X|+1\}$ and $x_0=x_r$ for all $v_r\in X$. Thus,

$$\left\{\begin{array}{cc}\rho (G)x_1=|X|x_0+(|V(G_1)|-1)x_1,\hfill & \\ \rho (G)x_2=|X|x_0+(|V(G_2)|-1)x_2.\hfill \end{array}\right.$$

Therefore, $x_1⩾x_2$. Using the Rayleigh quotient, we obtain

$$\begin{array}{cc}& \rho (K_{|X|}\vee (K_{|V(G_1)|+2}\cup K_{|V(G_2)|-2}\cup \cdots \cup K_{|V(G_{r|X|+1})|}))-\rho (G)\hfill \\ \hfill ⩾& {\mathbf{x}}^{\top }(\mathbf{A}(K_{|X|}\vee (K_{|V(G_1)|+2}\cup K_{|V(G_2)|-2}\cup \cdots \cup K_{|V(G_{r|X|+1})|}))-\mathbf{A}(G))\mathbf{x}\hfill \\ \hfill =& 4(|V(G_1)|x_1{x}_2-(|V(G_2)|-2)x_2^2)\hfill \\ \hfill ⩾& 4(|V(G_1)|-|V(G_2)|+2)x_2^2>0,\hfill \end{array}$$

which implies that $K_{|X|}\vee (K_{|V(G_1)|+2}\cup K_{|V(G_2)|-2}\cup \cdots \cup K_{|V(G_{r|X|+1})|})$ is a $\nu$-vertex connected graph with $\rho (K_{|X|}\vee (K_{|V(G_1)|+2}\cup K_{|V(G_2)|-2}\cup \cdots \cup K_{|V(G_{r|X|+1})|}))>\rho (G)$, which is a contradiction to our assumption of G. Hence, $|V(G_2)|=|V(G_3)|=\cdots =|V(G_{r|X|+1})|=1$.

We obtain $G=K_{|X|}\vee (K_{|V(G_1)|}\cup (r|X|)K_1)$.

We proceed by considering the following two subcases:

Subcase 1.1. $|V(G_1)|=1$.

In this subcase, one obtains $G=K_{|X|}\vee (r|X|+1)K_1$ and $\nu =(r+1)|X|+1$. Since r is an odd number, $\nu$ is an odd number, which contradicts the condition that $|V(G)|$ is an even number.

Subcase 1.2. $|V(G_1)|⩾3$.

In this subcase, one has $G=K_{|X|}\vee (K_{|V(G_1)|}\cup (r|X|)K_1)$ and $\nu =(r+1)|X|+|V(G_1)|$, where ${\nu }_1$ is odd. Since r is an odd number, $\nu$ is an odd number, which contradicts the condition that $|V(G)|$ is an even number.

Case 2. $oddca(G-X)=r|X|+2$.

Looking back at Case 1, we have $G=K_{|X|}\vee (K_{|V(G_1)|}\cup (r|X|+1)K_1)$, where $|V(G_1)|$ is odd.

We proceed by considering the following two subcases:

Subcase 2.1. $|V(G_1)|=1$.

In this subcase, one has $G=K_{|X|}\vee (r|X|+2)K_1$ and $\nu =(r+1)|X|+2$. Consider the partition $V(K_{|X|}\vee (r|X|+2)K_1)=V(K_{|X|})\cup V((r|X|+2)K_1)$. Its corresponding quotient matrix of $\mathbf{A}(G)$ is

$${\mathbf{M}}_{\mathbf{1}}=\left(\begin{array}{cc}|X|-1& r|X|+2\\ |X|& 0\end{array}\right).$$

Then, the characteristic polynomial of ${\mathbf{M}}_{\mathbf{1}}$ equals

$${\Phi }_{{\mathbf{M}}_{\mathbf{1}}}(x)=x^2{-(|X|-1)x-r|X|}^2-2|X|.$$

It is easy to see that the partition $V(G)=V(K_{|X|})\cup V((r|X|+2)K_1)$ is equitable. By Lemma 8, the largest root, say ${\rho }_1$, of ${\Phi }_{{\mathbf{M}}_{\mathbf{1}}}(x)=0$ equals $\rho (G)$. As ${\Phi }_{{\mathbf{M}}_{\mathbf{1}}}(x)=0$ is a quadratic equation with respect to x, by the root formula, we obtain $\rho (G)={\rho }_1$, where ${\rho }_1={\displaystyle \frac{|X|-1+\sqrt{{(4r+1)|X|}^2+6|X|+1}}{2}}$.

• $r⩾3,\nu =(r+1)|X|+2⩾4|X|+2⩾6.$

Our goal is to prove that $\chi (\rho (G))<0$. By a direct calculation, we obtain

$$\begin{array}{cc}\hfill \chi (\rho (G))& ={\displaystyle \frac{1}{2}}(|X|-1)g,\hfill \end{array}$$

where $g=(-2r^2|X{|}^2-(3r-1)|X|+2r^2+3r+1+(r+1)\sqrt{{4r|X|}^2+{|X|}^2+6|X|+1})$. If $|X|=1$, then $\chi (\rho (G))=0$. Hence, $\rho (G)={\rho }_1⩽\theta (\nu )$ for $|X|=1$, which is a contradiction. We now proceed to show that $\chi (\rho (G))<0$ for $|X|⩾2$. Since $\sqrt{{4r|X|}^2+{|X|}^2+6|X|+1}<\sqrt{{(4r+4)|X|}^2+6|X|+1}=\sqrt{(4r+4)(|X|+{\displaystyle \frac{3}{4r+4}}{)}^2+{\displaystyle \frac{4r-5}{4r+4}}}<\sqrt{2}(r+1)(|X|+{\displaystyle \frac{3}{4r+4}})+1.$, we obtain

$$\begin{array}{cc}\hfill g& <-2r^2{|X|}^2-(3r-1)|X|+2r^2+3r+1+\sqrt{2}(r+1)(|X|+{\displaystyle \frac{3}{4r+4}})+1\hfill \\ & =-2r^2{|X|}^2+(\sqrt{2}{r}^2+2\sqrt{2}r-3r+\sqrt{2}+1)|X|+(r+1){\displaystyle \frac{3\sqrt{2}}{4}}+2r^2+4r+2\hfill \\ & ⩽-4r^2|X|+(\sqrt{2}{r}^2+2\sqrt{2}r-3r+\sqrt{2}+1)|X|+(r+1){\displaystyle \frac{3\sqrt{2}}{4}}+2r^2+4r+2\hfill \\ & ={\displaystyle \frac{1}{4}}(r+1)((4(\sqrt{2}+1)+4(\sqrt{2}-4)r)|X|+8r+3\sqrt{2}+8)\hfill \\ & ⩽{\displaystyle \frac{1}{4}}(r+1)(3(4(\sqrt{2}+1)+4(\sqrt{2}-4)r)+8r+3\sqrt{2}+8)\hfill \\ & ={\displaystyle \frac{1}{4}}(r+1)(-8(3-\sqrt{2})r+11\sqrt{2}+16)\hfill \\ & ⩽{\displaystyle \frac{1}{4}}(r+1)(-24(3-\sqrt{2})+11\sqrt{2}+16)\hfill \\ & =-{\displaystyle \frac{1}{4}}(r+1)(7(8-5\sqrt{2}))\hfill \\ & <0.\hfill \end{array}$$

Therefore, $\chi (\rho (G))<0$, and so $\rho (G)={\rho }_1<\theta (\nu )$ for $|X|⩾2$, which is a contradiction.

Subcase 2.2. $|V(G_1)|⩾3$.

We obtain $G=K_{|X|}\vee (K_{|V(G_1)|}\cup (r|X|+1)K_1)$ and $\nu =|X|(r+1)+|V(G_1)|+1⩾(r+1)|X|+4⩾8$. Consider the partition $V(G)=V(K_{|X|})\cup V((r|X|+1)K_1)\cup V(K_{|V(G_1)|})$. The quotient matrix of $\mathbf{A}(G)$ is denoted as

$${\mathbf{M}}_{\mathbf{2}}=\left(\begin{array}{ccc}|X|-1& r|X|+1& \nu -|X|(r+1)-1\\ |X|& 0& 0\\ |X|& 0& \nu -|X|(r+1)-2\end{array}\right).$$

Then, the characteristic polynomial of ${\mathbf{M}}_{\mathbf{2}}$ is given by

$${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(x)=x^3-(\nu -r|X|-3)x^2-{(r|X|}^2-r|X|+\nu +|X|-2)x-r^2{|X|}^3+{r\nu |X|}^2-{r|X|}^3-{3r|X|}^2+\nu |X|-{|X|}^2-2|X|.$$

Since $V(G)=V(K_{|X|})\cup V((r|X|+1)K_1)\cup V(K_{|V(G_1)|})$ is an equitable partition, according to Lemma 8, we find that the largest root of ${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(x)=0$, say ${\rho }_2$, equals $\rho (G)$.

To demonstrate that $\theta (\nu ,r)⩾\rho (G)$, it suffices to prove $\chi ({\rho }_2)⩽0$. Observe that ${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}({\rho }_2)=0$. Obviously,

$$\begin{array}{cc}\hfill \chi ({\rho }_2)=& \chi ({\rho }_2)-{\Phi }_{{\mathbf{M}}_{\mathbf{2}}}({\rho }_2)\hfill \\ \hfill =& -(|X|-1)[r{\rho }_2^2-(r|X|+1){\rho }_2\hfill \\ & -r^2{(|X|}^2{+|X|+1)+\nu (r|X|+r+1)-r|X|}^2-4r|X|-4r-|X|-3].\hfill \end{array}$$

If $|X|=1$, then $\chi (x)={\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(x)$ and ${\rho }_2=\theta (\nu ,r)$. We now proceed to show that $\chi ({\rho }_2)<0$ for $|X|⩾2$.

• $r⩾3,\nu =(r+1)|X|+1+{\nu }_1⩾(r+1)|X|+4⩾4|X|+4.$

Notice that $\nu ⩾(r+1)|X|+4$. Observe that $K_{|X|}\vee (\nu -|X|)K_1$ is a proper subgraph of G. Consider the partition $V(K_{|X|}\vee (\nu -|X|)K_1)=V(K_{|X|})\cup V((\nu -|X|)K_1)$. Its corresponding quotient matrix of $\mathbf{A}(K_{|X|}\vee (\nu -|X|)K_1)$ is represented by

$${\mathbf{M}}_{\mathbf{3}}=\left(\begin{array}{cc}|X|-1& \nu -|X|\\ |X|& 0\end{array}\right).$$

Then, the characteristic polynomial of ${\mathbf{M}}_{\mathbf{3}}$ is given by

$${\Phi }_{{\mathbf{M}}_{\mathbf{3}}}(x)=x^2{-(|X|-1)x+|X|}^2-\nu |X|.$$

It is evident that the partition $V(K_{|X|}\vee (\nu -|X|)K_1)=V(K_{|X|})\cup V((\nu -|X|)K_1)$ is equitable. Referring to Lemma 8, the largest root, say ${\rho }_3$, of ${\Phi }_{{\mathbf{M}}_{\mathbf{3}}}(x)=0$ equals $\rho (K_{|X|}\vee (\nu -|X|)K_1)$. As ${\Phi }_{{\mathbf{M}}_{\mathbf{3}}}(x)=0$ is a quadratic equation with respect to x, by the root formula, we have $\rho (K_{|X|}\vee (\nu -|X|)K_1)={\rho }_3$, where ${\rho }_3={\displaystyle \frac{|X|-1+\sqrt{-{3|X|}^2+4|X|\nu -2|X|+1}}{2}}$. According to Lemma 2, it follows that ${\rho }_2>{\rho }_3$. Observe that ${(1-|X|)}^2-4|X|(|X|-\nu )={-3|X|}^2+4|X|\nu -2|X|+1⩾-{3|X|}^2+4|X|((r+1)|X|+4)-2|X|+1={(4(r+1)-3)|X|}^2+14|X|+{1>9|X|}^2>0.$

Thus, ${\rho }_3={\displaystyle \frac{|X|-1+\sqrt{-{3|X|}^2+4|X|\nu -2|X|+1}}{2}}>{\displaystyle \frac{|X|-1+\sqrt{{9|X|}^2}}{2}}={\displaystyle \frac{4|X|-1}{2}}.$ Let $h(x)=r{x}^2-(r|X|+1)x-r^2{(|X|}^2{+|X|+1)+\nu (r|X|+r+1)-r|X|}^2-4r|X|-4r-|X|-3$ be a real function in x. Clearly, ${\displaystyle \frac{r|X|+1}{2r}}<{\displaystyle \frac{4|X|-1}{2}}<{\rho }_3.$ This implies that $h(x)$ is increasing in the interval $[{\rho }_3,+\infty )$. Hence, $h({\rho }_2)>h({\rho }_3)=(2(r+1)|X|+r+1-2|X|)\nu -{\displaystyle \frac{(r+1)\sqrt{-{3|X|}^2+4\nu |X|-2|X|+1}}{2}}-r^2{(|X|}^2+|X|+1)-{\displaystyle \frac{r}{2}}{(4|X|}^2+9|X|+7)-{\displaystyle \frac{3|X|}{2}}-{\displaystyle \frac{5}{2}}.$ Observe that ${\displaystyle \frac{(r+1)|X|\nu }{2}}>{\displaystyle \frac{(r+1)\sqrt{4|X|\nu }}{2}}>{\displaystyle \frac{(r+1)\sqrt{-{3|X|}^2+4|X|\nu -2|X|+1}}{2}}.$ Then, $h({\rho }_2)>({\displaystyle \frac{3(r+1)|X|}{2}}+r+1-2|X|)\nu -r^2{(|X|}^2+|X|+1)-{\displaystyle \frac{r}{2}}{(4|X|}^2+9|X|+7)-{\displaystyle \frac{3|X|}{2}}-{\displaystyle \frac{5}{2}}.$ Note that $\nu ⩾(r+1)|X|+4.$ Thus, $h({\rho }_2)>({\displaystyle \frac{3(r+1)|X|}{2}}+r+1-2|X|)((r+1)|X|+4)-r^2{(|X|}^2+|X|+1)-{\displaystyle \frac{r}{2}}{(4|X|}^2+9|X|+7)-{\displaystyle \frac{3|X|}{2}}-{\displaystyle \frac{5}{2}}={\displaystyle \frac{1}{2}}((r^2-2r-1){|X|}^2+(7r-5)|X|-2r^2+r+3).$ Observe that $|X|⩾2$ and $r⩾3$. Then, $h({\rho }_2)>{\displaystyle \frac{1}{2}}(4(r^2-2r-1)+2(7r-5)-2r^2+r+3)=-{\displaystyle \frac{11}{2}}+{\displaystyle \frac{7r}{2}}+r^2>0.$ Therefore, $\chi ({\rho }_2)<0$, and so $\rho (G)<\theta (\nu ,r)$ when $|X|⩾2$, which is a contradiction. This completes the proof of Theorem 7. □

6.4. Distance Spectral Radius

In this subsection, we give the proof of Theorem 8, which gives a sufficient condition via the distance spectral radius of a connected graph G with even order to ensure that G is a strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor.

Proof of Theorem 8. 

We prove it by contradiction. Suppose that G has no strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor. Then, by Lemma 10, there exists some nonempty subset $X\subseteq V(G)$ such that $oddca(G-X)⩾r|X|+1.$ Suppose that $X=\varnothing$. We have $oddca(G)⩾1$. Since G is a connected graph, we obtain $oddca(G)=1$. Therefore, $|V(G)|$ is an odd number, which contradicts the condition that $|V(G)|$ is an even number. Choose a connected graph G of order $\nu$ such that its distance spectral radius is as small as possible. Together with Lemma 5 and the choice of G, the induced subgraph $G[X]$ and each connected component of $G-X$ are complete graphs, respectively. Moreover, it holds that $G=G[X]\vee (G-X).$

Claim 8. $G-X$ has no even order connected component.

Proof. 

Otherwise, we can connect each vertex of an even order connected component to each vertex of a connected component of $OddCa(G-X)$ with an edge, which results in a larger order connected component, contradicting the choice of G. This completes the proof of Claim 8. □

We aim to demonstrate that $oddca(G-X)⩽r|X|+2$. If $oddca(G-X)⩾r|X|+3$, then we create a new graph H obtained from G by connecting each vertex of a connected component of $OddCa(G-X)$ to each vertex of another connected component of $OddCa(G-X)$ with an edge and then connecting each vertex of the third connected component of $OddCa(G-X)$. Then, $oddca(H-X)⩾r|X|+1$ and G is a proper spanning subgraph of H. By Lemma 5, we find that ${\mu }_1(G)>{\mu }_1(H)$, which contradicts the initial choice of G. Therefore, $oddca(G-X)⩽r|X|+2$. Bear in mind that $oddca(G-X)⩾r|X|+1$. It suffices to consider $oddca(G-X)=r|X|+1$ and $oddca(G-X)=r|X|+2$.

Claim 9.${\mu }_1(G)⩾{\mu }_1(K_1\vee (K_{\nu -r-2}\cup \overline{K_{r+1}})$ if $\nu ⩾10$.

Proof. 

We consider the following two cases:

Case 1. $oddca(G-X)=r|X|+1$.

Let $G_1,G_2,\cdots ,G_{r|X|+1}$ be all the components in $G-X$ such that $|V(G_1)|⩾|V(G_2)|⩾\cdots ⩾|V(G_{r|X|+1})|$. Thus, $G=K_{|X|}\vee (K_{|V(G_1)|}\cup K_{|V(G_2)|}\cup \cdots \cup K_{|V(G_{r|X|+1})|}).$

Claim 10.$|V(G_2)|=|V(G_3)|=\cdots =|V(G_{r|X|+1})|=1$.

Proof. 

If $|V(G_2)|⩾3$, then $oddca(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|})-X)=oddca(G-X)⩾r|X|+1.$ Let $\mathbf{x}={(x_1,\cdots ,x_{\nu })}^{\top }$ be the Perron vector of $\mathbf{D}(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))$ and $x_i$ denote the entry of $\mathbf{x}$ corresponding to the vertex $v_i\in V(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))$. In view of Lemma 7, we obtain $x_j=x_k$ for all $v_j,v_k$ in $V(K_{|X|})$ (resp. $V((r|X|)K_1)$ and $V(K_{\nu -|X|-r|X|}$)).

Let $x_0=x_j$ for all $v_j\in V(K_{|X|})$, $x_1=x_j$ for all $v_j\in V((r|X|)K_1)$, and $x_2=x_j$ for all $v_j\in V(K_{\nu -|X|-r|X|})$. Hence,

$$\left\{\begin{array}{cc}{\mu }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -(r+1)|X|}))x_1=|X|x_0+2(\nu -(r+1)|X|)x_2+2(r|X|-1)x_1,\hfill & \\ {\mu }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))x_2=|X|x_0+(\nu -|X|-r|X|-1)x_2+2r|X|x_1.\hfill \end{array}\right.$$

Consequently, $x_1=(1+{\displaystyle \frac{\nu -r|X|-1-|X|}{{\mu }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))+2}})x_2.$

Because of the Rayleigh quotient, we obtain

$$\begin{array}{cc}\hfill & {\mu }_1(G)-{\mu }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))\hfill \\ \hfill ⩾& {\mathbf{x}}^{\top }(\mathbf{D}(G)-\mathbf{D}(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|})))\mathbf{x}\hfill \\ \hfill =& |V(G_1)|\sum _{i=2}^{r|X|+1}(|V(G_i)|-1)x_2^2\hfill \\ \hfill & +(|V(G_2)|-1)[(\nu -|X|-|V(G_2)|-(r|X|-1))x_2^2-2x_1{x}_2]\hfill \\ \hfill & +(|V(G_3)|-1)[(\nu -|X|-|V(G_3)|-(r|X|-1))x_2^2-2x_1{x}_2]\hfill \\ \hfill & +(|V(G_4)|-1)[(\nu -|X|-|V(G_4)|-(r|X|-1))x_2^2-2x_1{x}_2]+\cdots \hfill \\ \hfill & +(|V(G_{r|X|+1})|-1)[(\nu -|X|-|V(G_{r|X|+1})|-(r|X|-1))x_2^2-2x_1{x}_2].\hfill \end{array}$$

(1)

We consider the following two cases:

(a) $|V(G_3)|=1$.

We consider the following two cases:

(i) $|V(G_1)|=3.$

(ii) $|V(G_1)|⩾5$.

(b) $|V(G_3)|⩾3$.

If $|V(G_3)|=1$ and $|V(G_1)|=3$, then $|V(G_2)|=3$. Thus,

$$\begin{array}{cc}& {\mu }_1(G)-{\mu }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))\hfill \\ \hfill ⩾& {\mathbf{x}}^{\top }(\mathbf{D}(G)-\mathbf{D}(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|})))\mathbf{x}\hfill \\ \hfill =& |V(G_1)|(|V(G_2)|-1)x_2^2+(|V(G_2)|-1)(|V(G_1)|x_2^2-2x_1{x}_2)\hfill \\ \hfill =& 4(3x_2-x_1)x_2\hfill \\ \hfill =& 4(2-{\displaystyle \frac{\nu -|X|-r|X|-1}{{\mu }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))+2}})x_2^2\hfill \\ \hfill =& 4(2-{\displaystyle \frac{4}{{\mu }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))+2}})x_2^2\hfill \\ \hfill >& 0.\hfill \end{array}$$

Hence, $K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|})$ is a $\nu$-vertex connected graph with ${\mu }_1(G)>{\mu }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))$, which is a contradiction to our assumption of G. For $|V(G_3)|=1$, $|V(G_1)|⩾5$, and $|V(G_3)|⩾3$, we would like to prove that each term in (1) is positive. In view of (1), it suffices to show that $(\nu -|X|-|V(G_2)|-(r|X|-1))x_2^2-2x_1{x}_2>0.$ Observe that $K_{\nu -r|X|}$ is a subgraph of $K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|})$. Together with Lemma 5, we may conclude that ${\mu }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))⩾{\mu }_1(K_{\nu -r|X|})=\nu -r|X|-1.$

Hence,

$$\begin{array}{cc}\hfill & (\nu -|X|-|V(G_2)|-(r|X|-1))x_2^2-2x_1{x}_2\hfill \\ \hfill =& (\nu -|X|-|V(G_2)|-r|X|-1-{\displaystyle \frac{2\nu -2|X|-2(r|X|+1)}{{\mu }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))+2}})x_2^2\hfill \\ \hfill ⩾& (\nu -|X|-|V(G_2)|-r|X|-1-{\displaystyle \frac{2\nu -2|X|-2(r|X|+1)}{\nu -r|X|+1}})x_2^2\hfill \\ \hfill =& (\nu -|X|-|V(G_2)|-r|X|-3+{\displaystyle \frac{2|X|+4}{\nu -r|X|+1}})x_2^2\hfill \\ \hfill >& (\nu -|X|-|V(G_2)|-r|X|-3)x_2^2\hfill \\ \hfill =& [\sum _{j=1,j\ne 2}^{r|X|+1}(|V(G_j)|-1)-3]x_2^2.\hfill \end{array}$$

(2)

If $|V(G_1)|⩾5$, $|V(G_2)|⩾3$, and $|V(G_3)|=|V(G_4)|=\cdots =|V(G_{r|X|+1})|=1$, we obtain ${\displaystyle \sum _{j=1,j\ne 2}^{r|X|+1}}(|V(G_j)|-1)-3=|V(G_1)|-4>0.$ If $|V(G_3)|⩾3$, then we obtain $|V(G_1)|⩾|V(G_2)|⩾|V(G_3)|⩾3$. Consequently, ${\displaystyle \sum _{j=1,j\ne 2}^{r|X|+1}}(|V(G_j)|-1)-3⩾2(3-1)-3>0.$ Combining with (1) and (2), we may conclude that ${\mu }_1(G)>{\mu }_1(K_{|X|}\vee ((r|X|)K_1\cup K_{\nu -|X|-r|X|}))$, which is a contradiction to our assumption of G. Hence, we obtain $|V(G_2)|=|V(G_3)|=\cdots =|V(G_{r|X|+1})|=1.$ This completes the proof of Claim 10. □

In view of Claim 10, we obtain $G=K_{|X|}\vee (K_{|V(G_1)|}\cup (r|X|)K_1),$ where $|V(G_1)|$ is odd. As $|V(G_1)|⩾1$, we obtain $\nu ⩾|X|(r+1)+1$.

We proceed by considering the following two subcases:

Subcase 1.1. $|V(G_1)|=1$.

In view of Claim 10, we obtain $G=K_{|X|}\vee (r|X|+1)K_1$ and $\nu =|X|(r+1)+1$. Since r is an odd number, $\nu$ is an odd number, which contradicts the condition that $|V(G)|$ is an even number.

Subcase 1.2. $|V(G_1)|⩾3$.

By Claim 10, we obtain $G=K_{|X|}\vee (K_{|V(G_1)|}\cup (r|X|)K_1).$ We have $\nu =|X|(r+1)+|V(G_1)|$. Since r is an odd number, $\nu$ is an odd number, which contradicts the condition that $|V(G)|$ is an even number.

Case 2. $oddca(G-X)=r|X|+2$.

Looking back at Case 1 and Claim 10, we have $G=K_{|X|}\vee (K_{|V(G_1)|}\cup (r|X|+1)K_1)$, where $|V(G_1)|$ is odd.

Hence, we have $G=K_{|X|}\vee (K_{|V(G_1)|}\cup \overline{K_{r+1}}),$ where $|V(G_1)|$ is odd. As $|V(G_1)|=\nu -|X|(r+1)-1⩾1$, one has $\nu ⩾|X|(r+1)+2$.

Consider the equitable partition $V(K_1\vee (K_{\nu -r-2}\cup \overline{K_{r+1}})=V(K_{\nu -r-2})\cup V(K_1)\cup V(\overline{K_{r+1}})$. Then, one may obtain the quotient matrix of $\mathbf{D}(K_1\vee (K_{\nu -r-2}\cup \overline{K_{r+1}}))$ corresponding to the partition as

$$\mathbf{B}=\left(\begin{array}{ccc}\nu -r-3& 1& 2(r+1)\\ \nu -r-2& 0& r+1\\ 2(\nu -r-2)& 1& 2r\end{array}\right).$$

Therefore, its characteristic polynomial is

$${\Phi }_{\mathbf{B}}(x)=x^3-(\nu +r-3)x^2+(2r^2-2\nu r-5\nu +6r+9)x-\nu r+r^2-3\nu +4r+5.$$

By Lemma 8, the largest root, say $\sigma$, of ${\Phi }_{\mathbf{B}}(x)=0$ satisfies $\sigma ={\mu }_1(K_1\vee (K_{\nu -r-2}\cup \overline{K_{r+1}})$. In view of Lemma 9, we have

$$\sigma ={\mu }_1(K_1\vee (K_{\nu -r-2}\cup \overline{K_{r+1}})⩾{\displaystyle \frac{2W(K_1\vee (K_{\nu -r-2}\cup (r+1)K_1)}{\nu }}={\displaystyle \frac{{\nu }^2+(2r+1)\nu -(r+1)(r+4)}{\nu }}.$$

Since $r+1⩽\nu -2$, we obtain $\sigma ⩾\nu +(2r+1)-{\displaystyle \frac{(\nu -2)(r+4)}{\nu }}>\nu +r-3.$

We proceed by considering the following two subcases:

Subcase 2.1. $|V(G_1)|=1$.

By Claim 10, one has $G=K_{|X|}\vee (r|X|+2)K_1=K_{|X|}\vee ((r|X|+1)K_1\cup K_1)$ and $\nu =|X|(r+1)+2$.

• $r=3$.

Since $\nu =4|X|+2⩾10$, we obtain $|X|⩾2$. By Claim 10, one has $G=K_{|X|}\vee (3|X|+2)K_1$. Consider the partition $V(G)=V(K_{|X|})\cup V((3|X|+2)K_1)$. Then, the quotient matrix of $\mathbf{D}(G)$ corresponding to the partition is given as

$${\mathbf{M}}_{\mathbf{1}}=\left(\begin{array}{cc}|X|-1& 3|X|+2\\ |X|& 2(3|X|+1)\end{array}\right).$$

Then, the characteristic polynomial of ${\mathbf{M}}_{\mathbf{1}}$ equals

$${\Phi }_{{\mathbf{M}}_{\mathbf{1}}}(x)=x^2{-(7|X|+1)x+3|X|}^2-6|X|-2.$$

It is easy to see that the partition $V(G)=V(K_{|X|})\cup V((3|X|+2)K_1)$ is equitable. By Lemma 8, the largest root, say ${\sigma }_1$, of ${\Phi }_{{\mathbf{M}}_{\mathbf{1}}}(x)=0$ satisfies ${\sigma }_1={\mu }_1(G)$. We would like to prove that ${\sigma }_1>\sigma$. Let

$$h_1(x)=x{\Phi }_{{\mathbf{M}}_{\mathbf{1}}}(x)-{\Phi }_{\mathbf{B}}(x)$$

be a real function in x. Hence, $h_1(x)=(1-3|X|)x^2+{(3|X|}^2+38|X|-25)x+24|X|-14$. We would like to prove that $h_1(\sigma )<0$. Since ${\displaystyle \frac{{3|X|}^2+38|X|-25}{6|X|-2}}<4|X|+6,$$h_1(x)$ is monotonically decreasing in the interval $[4|X|+6,+\infty )$. Hence, we obtain $\sigma ={\mu }_1(K_1\vee (K_{\nu -r-2}\cup \overline{K_{r+1}})⩾{\displaystyle \frac{2W(K_1\vee (K_{\nu -r-2}\cup (r+1)K_1)}{\nu }}⩾4|X|+9-{\displaystyle \frac{28}{4s+2}}>4|X|+6.$ Then, $h_1(x)<h_1(4|X|+6)={-36|X|}^3+{42|X|}^2+92|X|-128.$ Let $h_2(x)=-36x^3+42x^2+92x-128$ be a real function in x for $x\in [2,+\infty )$. Together with ${\displaystyle \frac{7-5\sqrt{13}}{18}}<{\displaystyle \frac{7+5\sqrt{13}}{18}}<2,$ we find that $h_2(x)$ is monotonically decreasing in the interval $[2,+\infty )$. Then, $h_1(x)<h_2(x)⩽h_2(4)=-64<0.$ Thus, ${\sigma }_1>\sigma$ and ${\mu }_1(G)>{\mu }_1(K_1\vee (K_{\nu -r-2}\cup \overline{K_{r+1}})$.

• $r⩾5$.

In this case, $G=K_{|X|}\vee ((r|X|+1)K_1\cup K_1)$. Consider the partition $V(G)=V(K_1)\cup V(K_{|X|})\cup V((r|X|+1)K_1)$. Then, one may obtain the quotient matrix of $\mathbf{D}(G)$ corresponding to this partition as

$${\mathbf{M}}_{\mathbf{2}}=\left(\begin{array}{ccc}0& |X|& 2r|X|+2\\ 1& |X|-1& r|X|+1\\ 2& |X|& 2r|X|\end{array}\right).$$

Then, the characteristic polynomial of ${\mathbf{M}}_{\mathbf{2}}$ is given by

$$\begin{array}{cc}\hfill {\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(x)=& x^3-(2rs+s-1)x^2+(r{s}^2-6rs-2s-4)x+2r{s}^2-4rs-4.\hfill \end{array}$$

Since $V(G)=V(K_1)\cup V(K_{|X|})\cup V((r|X|+1)K_1)$ is an equitable partition, by Lemma 8, the largest root, say ${\sigma }_2$, of ${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(x)=0$, satisfies ${\sigma }_2={\mu }_1(G)$. Next, we would like to prove that ${\sigma }_2⩾\sigma$. It is straightforward to check that

$${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(x)-{\Phi }_{\mathbf{B}}(x)=(|X|-1)[-r{x}^2+(2r^2+r|X|+2r+3)x+2r|X|+r^2+2r+3].$$

If $|X|=1$, then ${\Phi }_{{\mathbf{M}}_{\mathbf{2}}}(\sigma )={\Phi }_{\mathbf{B}}(\sigma )=0$ and ${\sigma }_2=\sigma$.

Next, we would like to show that ${\sigma }_2>\sigma$ for $|X|⩾2$. Let $h_3(x)=-r{x}^2+(2r^2+r|X|+2r+3)x+2r|X|+r^2+2r+3$ be a real function in x. Obviously, ${\displaystyle \frac{2r^2+r|X|+2r+3}{2r}}<|X|(r+1)+r-1.$ Therefore, $h_3(x)$ is decreasing for $x\in [|X|(r+1)+r-1,+\infty )$. Since $\sigma >\nu +r-3=|X|(r+1)+r-1,$ we obtain $h_3(\sigma )<h_3(|X|(r+1)+r-1)=-r^2{(r+1)|X|}^2+(r+1)(5r+3)|X|+(r+1)(r^2+2r).$ Let $h_4(x)=-r^2(r+1)x^2+(r+1)(5r+3)x+(r+1)(r^2+2r)$ be a real function in x for $x\in [2,+\infty )$. As ${\displaystyle \frac{(r+1)(5r+3)}{2r^2(r+1)}}<2,$ one finds that $h_4(x)$ is decreasing for $x\in [2,+\infty )$. Therefore, we have $h_4(x)⩽h_4(2)=-3r^3+9r^2+18r+6=-3(r+1)(r-(2+\sqrt{6}))(r-(2-\sqrt{6})).$ Since $r⩾5$ and $2-\sqrt{6}<2+\sqrt{6}<5$, $h_3(\sigma )<h_4(x)<0$, ${\sigma }_2>\sigma$, and ${\mu }_1(G)>{\mu }_1(K_1\vee (K_{\nu -r-2}\cup (r+1)K_1)$.

Subcase 2.2. $|V(G_1)|⩾3$.

By Claim 10, we obtain $G=K_{|X|}\vee (K_{|V(G_1)|}\cup (r|X|+1)K_1)$, where $|V(G_1)|=\nu -|X|(r+1)-1$. We have $\nu =|X|(r+1)+1+|V(G_1)|⩾|X|(r+1)+4$.

If $|X|=1$, then $G\cong K_1\vee (K_{\nu -r-2}\cup \overline{K_{r+1}})$ and ${\mu }_1(G)={\mu }_1(K_1\vee (K_{\nu -r-2}\cup \overline{K_{r+1}})$. Next, we consider $|X|⩾2$. Consider the partition $V(G)=V(K_{\nu -(r+1)|X|-1})\cup V(K_{|X|})\cup V((r|X|+1)K_1)$. Then, the quotient matrix of $\mathbf{D}(G)$ with respect to this partition equals

$${\mathbf{M}}_{\mathbf{3}}=\left(\begin{array}{ccc}\nu -(r+1)|X|-2& |X|& 2r|X|+2\\ \nu -(r+1)|X|-1& |X|-1& r|X|+1\\ 2(\nu -(r+1)|X|-1)& |X|& 2r|X|\end{array}\right).$$

Then, the characteristic polynomial of ${\mathbf{M}}_{\mathbf{3}}$ equals

$$\begin{array}{cc}\hfill {\Phi }_{{\mathbf{M}}_{\mathbf{3}}}(x)=& x^3-(r|X|+\nu -3)x^2+(2r^2{|X|}^2+{3r|X|}^2-2\nu r|X|+3r|X|-5\nu +3|X|\hfill \\ & +6)x-r^2{|X|}^3-{r|X|}^3+2r^2{s}^2+{\nu r|X|}^2+{r|X|}^2-2\nu r|X|+\nu |X|\hfill \\ & +4r|X|-{|X|}^2-4\nu +2|X|+4.\hfill \end{array}$$

Since $V(G)=V(K_{\nu -(r+1)|X|-1})\cup V(K_{|X|})\cup V((r|X|+1)K_1)$ is an equitable partition, by Lemma 8, the largest root, say ${\sigma }_3$, of ${\Phi }_{{\mathbf{M}}_{\mathbf{3}}}(x)=0$, satisfies ${\sigma }_3={\mu }_1(G)$. Hence, we obtain

$$\begin{array}{cc}\hfill {\Phi }_{{\mathbf{M}}_{\mathbf{3}}}(x)-{\Phi }_{\mathbf{B}}(x)=& (|X|-1)(-r{x}^2+(2r^2|X|-2\nu r+2r^2+3r|X|+6r+3)x\hfill \\ & -r^2{|X|}^2+\nu r|X|+r^2|X|-{r|X|}^2+4r-\nu r+r^2+\nu -|X|+1).\hfill \end{array}$$

Let $h_5(x)=-r{x}^2+(2r^2|X|-2\nu r+2r^2+3r|X|+6r+3)x-r^2{|X|}^2+\nu r|X|+r^2|X|-{r|X|}^2+4r-\nu r+r^2+\nu -|X|+1$ be a real function in $x\in [\nu +r-3,+\infty )$. We would like to prove that $h_5(\sigma )<0$.

• $r⩾3.$

Obviously, ${\displaystyle \frac{2r^2|X|-2\nu r+2r^2+3r|X|+6r+3}{2r}}<\nu +r-3.$ Therefore, $h_5(x)$ is decreasing in the interval $[\nu +r-3,+\infty )$. Together with $\sigma >\nu +r-3$,

$$\begin{array}{cc}\hfill h_5(\sigma )<h_5(\nu +r-3)=& -3r{\nu }^2+[r(4|X|+17)+2r^2(|X|-1)+4]\nu +r^3(2|X|+1)\hfill \\ & -r^2{(|X|}^2{+2|X|-7)-r(|X|}^2+9|X|+20)-|X|-8.\hfill \end{array}$$

Let $h_6(x)=-3r{x}^2+[r(4|X|+17)+2r^2(|X|-1)+4]x+r^3(2|X|+1)-r^2{(|X|}^2{+2|X|-7)-r(|X|}^2+9|X|+20)-|X|-8$ be a real function in x for $x\in [(r+1)|X|+4,+\infty )$. Together with ${\displaystyle \frac{r(4|X|+17)+2r^2(|X|-1)+4}{6r}}<(r+1)|X|+4,$ we find that $h_6(x)$ is a decreasing function in the interval $[(r+1)|X|+4,+\infty )$. Hence, $h_6(x)⩽h_6((r+1)|X|+4)=-(r^3+r^2){|X|}^2-(3r^2-4r-3)|X|+r^3-r^2+8.$ Let $h_7(x)=-(r^3+r^2)x^2-(3r^2-4r-3)x+r^3-r^2+8$ be a real function in x. Together with ${\displaystyle \frac{-(3r^2-4r-3)}{2(r^3+r^2)}}<2,$ we find that $h_7(x)$ is a decreasing function in the interval $[2,+\infty )$. Hence, $h_7(x)<h_7(2)=-3r^3-11r^2+8r+14.$ Let $h_8(x)=-3x^3-11x^2+8x+14$ be a real function in x for $x\in [3,+\infty )$. Obviously, ${\displaystyle \frac{-11-\sqrt{193}}{9}}<{\displaystyle \frac{-11+\sqrt{193}}{9}}<3.$ Thus, $h_8(x)$ is a decreasing function in the interval $[3,+\infty )$. Hence, $h_8(x)⩽h_8(3)=-142<0$. Hence, $h_5(\sigma )<0$ and ${\mu }_1(G)>{\mu }_1(K_1\vee (K_{\nu -r-2}\cup (r+1)K_1)$. This completes the proof of Claim 9. □

Note for Claim 9 that, if $\nu ⩾10$, then ${\mu }_1(G)>{\mu }_1(K_1\vee (K_{\nu -r-2}\cup (r+1)K_1)$, which contradicts the assumption that ${\mu }_1(G)<{\mu }_1(K_1\vee (K_{\nu -r-2}\cup (r+1)K_1)$. Next, we consider $\nu =6,8$. Recall that $G=K_{|X|}\vee ((r|X|+1)K_1\cup K_{\nu -(r+1)|X|-1})$.

• $\nu =6.$

In view of $\nu ⩾|X|(r+1)+2$, we obtain $r=3$ and $|X|=1.$ Therefore, we obtain $G=K_1\vee (K_1\cup \overline{K_4})$ and ${\mu }_1(G)={\mu }_1(K_1\vee (K_1\cup \overline{K_4}))$, which contradicts the assumption that ${\mu }_1(G)<{\mu }_1(K_1\vee (K_1\cup \overline{K_4}))$.

• $\nu =8.$

In view of $\nu ⩾|X|(r+1)+2$, we obtain $r=3$ and $|X|=1.$ Therefore, we obtain $G=K_1\vee (K_3\cup \overline{K_4})$ and ${\mu }_1(G)={\mu }_1(K_1\vee (K_3\cup \overline{K_4}))$, which contradicts the assumption that ${\mu }_1(G)<{\mu }_1(K_1\vee (K_3\cup \overline{K_4}))$. In view of $\nu ⩾|X|(r+1)+2$, we have $r=5$ and $|X|=1.$ Therefore, we obtain $G=K_1\vee (K_1\cup \overline{K_6})$ and ${\mu }_1(G)={\mu }_1(K_1\vee (K_1\cup \overline{K_6}))$, which contradicts the assumption that ${\mu }_1(G)<{\mu }_1(K_1\vee (K_1\cup \overline{K_6}))$. This completes the proof of Theorem 8. □

7. Concluding Remarks

At last, we prove that the bounds obtained in Theorems 4–8 are the best possible.

Next, we prove that the bound obtained in Theorem 4 is the best possible. $H_{\nu }$ is a complete graph to which an edge has been attached, and $|V(H_{\nu })|=\nu$. $H_{{\nu }_1}$ is a complete graph to which an edge has been attached. Let e be the pendant edge in $H_{{\nu }_1}$. Since, obviously, $H_{{\nu }_1}-e$ does not have a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor, $H_{{\nu }_1}$ is not a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted graph.

Theorem 9. 

For some positive integer $\nu ⩾3$, $H_{\nu }$ is not $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted and $E(H_{\nu })=\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1$. Furthermore, let G be a connected graph with $G\ne H_{\nu }$ of order $\nu ⩾4$. If G is not $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor deleted, then $|E(G)|<\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1.$

Proof. 

It suffices to prove the second half of Theorem 9. We consider the following two cases:

Case 1. G has no $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor.

Together with Lemma 13, $\nu ⩾5$, and $\nu \ne 7$, we obtain $|E(G)|⩽\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -3}{2}}\right)+3<\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1.$ Together with Lemma 13 and $\nu =7$, we obtain $|E(G)|⩽11<\left({\displaystyle \genfrac{}{}{0pt}{}{7-1}{2}}\right)+1=16$.

Case 2. G has a $\{K_{1,1},K_{1,2},C_m:m⩾3\}$-factor.

Choose such a connected graph G of order $\nu$ such that its size is as large as possible. By Case 2 in the proof of Theorem 4, we find that G is one of $K_{|X|}\vee ((2|X|-1)K_1\cup K_2)$, $K_{|X|}\vee ((2|X|-1)K_1\cup K_2\cup K_{{\nu }_1})$ (${\nu }_1⩾2$), $K_{|X|}\vee ((2|X|)K_1\cup K_2)$, $K_{|X|}\vee ((2|X|)K_1\cup K_2\cup K_{{\nu }_1})$$({\nu }_1⩾2)$, or $K_{|X|}\vee ((2|X|)K_1\cup H_{{\nu }_1})$$({\nu }_1⩾3)$. For any $G\in \{K_{|X|}\vee ((2|X|-1)K_1\cup K_2),K_{|X|}\vee ((2|X|-1)K_1\cup K_2\cup K_{{\nu }_1})({\nu }_1⩾2),K_{|X|}\vee ((2|X|)K_1\cup K_2),K_{|X|}\vee ((2|X|)K_1\cup K_2\cup K_{{\nu }_1})({\nu }_1⩾2)\}$, by Claim 1, we obtain $|E(G)|⩽\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{2}}\right)+1$, where equality holds if and only if $G=K_1\vee (K_2\cup K_1)=H_4$. Since $G\ne H_4$, we have $|E(G)|<\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{2}}\right)+1$. By ${\nu }_1⩾3$ and Claim 2, we obtain $|E(K_{|X|}\vee ((2|X|)K_1\cup H_{{\nu }_1}))|⩽\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1$, where equality holds if and only if $K_{|X|}\vee ((2|X|)K_1\cup H_{{\nu }_1})=H_{{\nu }_1^{\prime }}$. As $G\ne H_{{\nu }_1^{\prime }},$ $|E(K_{|X|}\vee ((2|X|)K_1\cup H_{{\nu }_1}))|<\left({\displaystyle \genfrac{}{}{0pt}{}{\nu -1}{2}}\right)+1$ follows. □

Next, we prove that the bound obtained in Theorem 5 is the best possible.

Theorem 10. 

Let $r⩾3$ and $\nu ⩾r+3$ be two integers, where ν is an even number and r is an odd number, and let G be a connected graph with ν vertices. Then,

(i) 

$K_1\vee ((r+1)K_1\cup K_{\nu -r-2})$ has no strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor;

(ii) 

${\eta }_1(K_1\vee ((r+1)K_1\cup K_{\nu -r-2}))$ is equal to the largest root of $x^3-(5\nu +2r-7)x^2+(4r^2+8{\nu }^2+2\nu r-27\nu +4r+24)x-4\nu r^2-4{\nu }^3-8\nu r+6r^2+22{\nu }^2-42\nu +14r+28=0.$

Proof of Theorem 10. 

(i) Let u be the vertex with the maximum degree of $K_1\vee ((r+1)K_1\cup K_{\nu -r-2})$. Let $X=\{u\}$. Then, $oddca(K_1\vee ((r+1)K_1\cup K_{\nu -r-2})-X)=r+2>r=r|X|$. According to Lemma 10, we find that $K_1\vee ((r+1)K_1\cup K_{\nu -r-2})$ has no strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor.

(ii) Consider the partition $V(K_1\vee ((r+1)K_1\cup K_{\nu -r-2}))=V((r+1)K_1)\cup V(K_{\nu -r-2})\cup V(K_1)$. Correspondingly, we obtain the quotient matrix of $\mathbf{Q}(K_1\vee ((r+1)K_1\cup K_{\nu -r-2}))$ as

$${\mathbf{M}}_{\mathbf{1}}=\left(\begin{array}{ccc}2\nu +2r-3& 2(\nu -r-2)& 1\\ 2(r+1)& 2\nu -3& 1\\ r+1& \nu -r-2& \nu -1\end{array}\right).$$

The corresponding characteristic polynomial is $x^3-(5\nu +2r-7)x^2+(4r^2+8{\nu }^2+2\nu r-27\nu +4r+24)x-4\nu r^2-4{\nu }^3-8\nu r+6r^2+22{\nu }^2-42\nu +14r+28$. By Lemma 8, one has that ${\eta }_1(K_1\vee ((r+1)K_1\cup K_{\nu -r-2}))$ is equal to the largest root of $x^3-(5\nu +2r-7)x^2+(4r^2+8{\nu }^2+2\nu r-27\nu +4r+24)x-4\nu r^2-4{\nu }^3-8\nu r+6r^2+22{\nu }^2-42\nu +14r+28=0.$ This completes the proof of Theorem 10. □

Next, we claim that the signless Laplacian spectral radius conditions in Theorem 6 are sharp.

Theorem 11. 

Let $r⩾3$ and $\nu ⩾r+3$ be two integers, where ν is an even number and r is an odd number, and let G be a connected graph with ν vertices. Then,

(i) 

$K_1\vee K_{r+2}$ and $K_2\vee (2r+2)K_1$ have no strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor.

(ii) 

${\kappa }_1(K_1\vee (K_{\nu -r-2}\cup (r+1)K_1))$ is equal to the largest root of $x^3-(4\nu -3r-10)x^2+(2{\nu }^2-2\nu r-5\nu )x+4\nu r-2r^2-2{\nu }^2+10\nu -10r-12=0.$${\kappa }_1(K_1\vee K_{r+2})=r+3$, ${\kappa }_1(K_2\vee (2r+2)K_1)=r+3+\sqrt{r^2+6r+5}$.

Proof of Theorem 11. 

(i) Let u be the vertex with the maximum degree of $K_1\vee K_{r+2}$. Let $X=\{u\}$. Then, $oddca(K_1\vee K_{r+2}-X)=r+2>r=r|X|$. According to Lemma 10, we obtain $K_1\vee K_{r+2}$, which has no strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor. Let $X=\{V(K_2)\}$. Then, $oddca(K_2\vee (2r+2)K_1-X)=2r+2>2r=r|X|$. According to Lemma 10, we find that $K_2\vee (2r+2)K_1$ has no strong $\{K_{1,1},K_{1,2},\cdots ,K_{1,r}\}$-factor.

(ii) According to the partition $V(K_1\vee (r+2)K_1)=V(K_1)\cup V((r+2)K_1)$, the quotient matrix of $\mathbf{K}(K_1\vee K_{r+2})$ is

$${\mathbf{M}}_{\mathbf{2}}=\left(\begin{array}{cc}r+2& r+2\\ 1& 1\end{array}\right).$$

The corresponding characteristic polynomial is $x^2-(r+3)x$. By Lemma 8, one has that ${\kappa }_1(K_1\vee (r+2)K_1)$ is equal to the largest root of $x^2-(r+3)x=0.$ Hence, ${\kappa }_1(K_1\vee K_{r+2})=r+3$.

According to the partition $V(K_2\vee (2r+2)K_1)=V(K_2)\cup V((2r+2)K_1)$, the quotient matrix of $\mathbf{K}(K_2\vee (2r+2)K_1)$ is

$${\mathbf{M}}_{\mathbf{3}}=\left(\begin{array}{cc}2r+4& 2r+2\\ 2& 2\end{array}\right).$$

The corresponding characteristic polynomial is $x^2-(2b+6)x+4$. By Lemma 8, one finds that ${\kappa }_1(K_1\vee (r+2)K_1)$ is equal to the largest root of $x^2-(2b+6)x+4=0.$ Hence, ${\kappa }_1(K_2\vee (2r+2)K_1)=r+3+\sqrt{r^2+6r+5}$.

According to the partition $V(K_1\vee (K_{\nu -r-2}\cup (r+1)K_1))=V((r+1)K_1)\cup V(K_{\nu -r-2})\cup V(K_1)$, the quotient matrix of $\mathbf{K}(K_1\vee (K_{\nu -r-2}\cup (r+1)K_1))$ equals

$${\mathbf{M}}_{\mathbf{4}}=\left(\begin{array}{ccc}1& 0& 1\\ 0& 2\nu -2r-5& 1\\ r+1& \nu -r-2& \nu -1\end{array}\right).$$

The corresponding characteristic polynomial is $x^3-(4\nu -3r-10)x^2+(2{\nu }^2-2\nu r-5\nu )x+4\nu r-2r^2-2{\nu }^2+10\nu -10r-12$. By Lemma 8, one has that ${\eta }_1(K_1\vee ((r+1)K_1\cup K_{\nu -r-2}))$ is equal to the largest root of $x^3-(4\nu -3r-10)x^2+(2{\nu }^2-2\nu r-5\nu )x+4\nu r-2r^2-2{\nu }^2+10\nu -10r-12=0.$ This completes the proof of Theorem 11. □

Next, we claim that the spectral radius conditions in Theorem 7 are sharp.

Theorem 12. 

Let $r⩾3$ and ν be two integers, where ν is an even number and r is an odd number, and let G be a connected graph with ν vertices. Then, $\rho (K_1\vee (K_{\nu -r-2}\cup (r+1)K_1))$ is equal to the largest root of $x^3-(\nu -r-3)x^2-(\nu -1)x+\nu r-r^2+\nu -4r-3=0.$

Proof of Theorem 12. 

According to the partition $V(G)=V(K_1)\cup V((r+1)K_1)\cup V(K_{\nu -r-2})$, the quotient matrix of $\mathbf{A}(G)$ is denoted as

$${\mathbf{M}}_{\mathbf{5}}=\left(\begin{array}{ccc}0& r+1& \nu -r-2\\ 1& 0& 0\\ 1& 0& \nu -r-3\end{array}\right).$$

The corresponding characteristic polynomial is $x^3-(\nu -r-3)x^2-(\nu -1)x+\nu r-r^2+\nu -4r-3$. By Lemma 8, one has that $\rho (K_1\vee ((r+1)K_1\cup K_{\nu -r-2}))$ is equal to the largest root of $x^3-(\nu -r-3)x^2-(\nu -1)x+\nu r-r^2+\nu -4r-3=0.$ This completes the proof of Theorem 12. □

Next, we claim that the distance spectral radius conditions in Theorem 8 are sharp.

Theorem 13. 

Let $r⩾3$ and $\nu ⩾r+3$ be two integers, where ν is an even number and r is an odd number, and let G be a connected graph with ν vertices.

(i) 

$K_1\vee (K_1\cup \overline{K_4})$ and $K_1\vee (K_3\cup \overline{K_4})$ have no strong $\{K_{1,1},K_{1,2},K_{1,3}\}$-factor.

(ii) 

$K_1\vee (K_1\cup \overline{K_6})$ has no strong $\{K_{1,1},K_{1,2},K_{1,3},K_{1,4},K_{1,5}\}$-factor.

Proof of Theorem 13. 

(i) Let u be the vertex with the maximum degree of $K_1\vee (K_1\cup \overline{K_4})$. Let $X=\{u\}$. Then, $oddca(K_1\vee (K_1\cup \overline{K_4})-X)=5>3=3|X|$. According to Lemma 10, we find that $K_1\vee (K_1\cup \overline{K_4})$ has no strong $\{K_{1,1},K_{1,2},K_{1,3}\}$-factor. Let u be the vertex with the maximum degree of $K_1\vee (K_3\cup \overline{K_4})$. Let $X=\{u\}$. Then, $oddca(K_1\vee (K_3\cup \overline{K_4})-X)=5>3=3|X|$. According to Lemma 10, we find that $K_1\vee (K_3\cup \overline{K_4})$ has no strong $\{K_{1,1},K_{1,2},K_{1,3}\}$-factor.

(ii) Let u be the vertex with the maximum degree of $K_1\vee (K_1\cup \overline{K_6})$. Let $X=\{u\}$. Then, $oddca(K_1\vee (K_1\cup \overline{K_6})-X)=7>5=5|X|$. According to Lemma 10, we find that $K_1\vee (K_1\cup \overline{K_6})$ has no strong $\{K_{1,1},K_{1,2},K_{1,3},K_{1,4},K_{1,5}\}$-factor. □

Hence, the condition in Theorem 8 cannot be replaced by the condition that ${\mu }_1(G)⩽{\mu }_1(K_1\vee (K_1\cup \overline{K_4}))$, ${\mu }_1(G)⩽{\mu }_1(K_1\vee (K_3\cup \overline{K_4}))$, ${\mu }_1(G)⩽{\mu }_1(K_1\vee (K_1\cup \overline{K_6}))$, and ${\kappa }_1(G)⩽{\mu }_1(K_1\vee (K_{\nu -r-2}\cup \overline{K_{r+1}}))$, which implies that the bounds established in Theorem 8 are the best possible.