#  Spectral radius and $[a,b]$-factors in graphs 111This work is supported by
the National Natural Science Foundation of China (Grant Nos. 11771141,
12011530064 and 11871391)000 _Email address:_<EMAIL_ADDRESS>(D. Fan),
<EMAIL_ADDRESS>(H. Lin), luhongliang@mail.xjtu.edu.cn(H.Lu).

Dandan Fana, Huiqiu Lina and Hongliang Lub
a School of Mathematics, East China University of Science and Technology,
Shanghai 200237, P.R. China
b School of Mathematics and Statistics, Xi’an Jiaotong University
Xi’an, Shanxi 710049, China
Corresponding author.

Abstract

An $[a,b]$-factor of a graph $G$ is a spanning subgraph $H$ such that $a\leq
d_{H}(v)\leq b$ for each $v\in V(G)$. In this paper, we provide spectral
conditions for the existence of an odd $[1,b]$-factor in a connected graph
with minimum degree $\delta$ and the existence of an $[a,b]$-factor in a
graph, respectively. Our results generalize and improve some previous results
on perfect matchings of graphs. For $a=1$, we extend the result of O[30] to
obtain an odd $[1,b]$-factor and further improve the result of Liu, Liu and
Feng[27] for $a=b=1$. For $n\geq 3a+b-1$, we confirm the conjecture of Cho,
Hyun, O and Park[6]. We conclude some open problems in the end.

Keywords: $[a,b]$-factor; unique perfect matching; spectral radius.

AMS Classification: 05C50

##  1 Introduction

The study of factors from graph eigenvalues has a rich history. In 2005,
Brouwer and Haemers [3] described a regular graph to contain a perfect
matching in terms of the third largest eigenvalue. Their result was improved
in [9, 8, 10] and extended in [22, 23] to obtain a regular factor. Very
recently, O [30] provided a spectral condition to guarantee the existence of a
perfect matching in a graph. One problem concerning perfect matching which has
attracted considerable interest is that of determining the structure of graphs
with a unique perfect matching[35, 5, 32]. Yang and Ye[37] characterized all
bipartite graphs with a unique perfect matching whose adjacency matrices have
inverses diagonally similar to non-negative matrices, which settles an open
problem of Godsil on inverses of bipartite graphs in [14]. Lovász[26] proved
that a graph of order $n$ with a unique perfect matching cannot have more than
$\frac{n^{2}}{4}$ edges. Based on the structural properties of graphs with a
unique perfect matching, in this paper, we first determine the graph attaining
the maximum spectral radius among all graphs of order $2n$ with a unique
perfect matching.

Let $G$ be a graph with adjacency matrix $A(G)$. The largest eigenvalue of
$A(G)$, denoted by $\rho(G)$, is called the spectral radius of $G$. Denote by
${}^{\prime}\nabla^{\prime}$ and ${}^{\prime}\cup^{\prime}$ the join and union
products, respectively. Suppose that $G_{1}$ is an empty graph with vertex set
$U=\\{u_{1},u_{2},\ldots,u_{n}\\}$, and $G_{2}$ is a complete graph with
vertex set $W=\\{w_{1},w_{2},\ldots,w_{n}\\}$. Let $G(2n,1)$ be the graph of
order $2n$ obtained from $G_{1}\cup G_{2}$ by letting
$N_{G_{2}}(u_{i})=\\{w_{1},w_{2},\ldots,w_{i}\\}$ for $1\leq i\leq n$.
Clearly, $G(2n,1)$ contains a unique perfect matching.

######  Theorem 1.1.

If $G$ is a connected graph of order $2n$ with a unique perfect matching, then
$\rho(G)\leq\rho(G(2n,1))$, with equality if and only if $G\cong G(2n,1)$.

Another problem concerning perfect matching is extending it to a general
factor. An odd [1,b]-factor of a graph $G$ is a spanning subgraph $H$ such
that $d_{H}(v)$ is odd and $1\leq d_{H}(v)\leq b$ for each $v\in V(G)$. Lu, Wu
and Yang [20] gave a sufficient condition for the existence of an odd
$[1,b]$-factor in a graph in terms of the third largest eigenvalue. Recently,
Kim, O, Park and Ree [19] improved the spectral condition of Lu, Wu and Yang.
In this paper, we generalize their result by giving a spectral condition to
guarantee the existence of an odd $[1,b]$-factor with minimum degree $\delta$.
Let $T(n,b,\delta)=K_{\delta}\nabla(K_{n-(b+1)\delta-1}\cup(b\delta+1)K_{1})$
and $F(b,\delta)=\max\\{4(b+1)\delta+4,b\delta^{3}+\delta\\}$, where $b$ is a
positive odd integer.

######  Theorem 1.2.

Suppose that $G$ is a connected graph of even order $n\geq F(b,\delta)$ with
minimum degree $\delta$. If $\rho(G)\geq\rho(T(n,b,\delta))$, then $G$
contains an odd $[1,b]$-factor, unless $G\cong T(n,b,\delta)$.

An $[a,b]$-factor of a graph $G$ is a spanning subgraph $H$ such that $a\leq
d_{H}(v)\leq b$ for each $v\in V(G)$. In the past decades, some researchers
provided various parameter conditions for a graph to have an $[a,b]$-factor,
such as the degree condition[28], the neighborhood condition [25], the
stability number[17], and the binding number[7]. In addition, the conditions
for a graph to have a fractional $[a,b]$-factor (see [34] for the definition)
was also investigated by several researchers[11, 21, 18]. Only very recently,
Cho, Hyun, O and Park[6] posed a conjecture regarding the spectral condition
for the existence of an $[a,b]$-factor as follows.

######  Conjecture 1.1.

(See [6]) Let $a\cdot n$ be an even integer at least 2, where $n\geq a+1$. If
$G$ is a graph of order $n$ with $\rho(G)>\rho(H_{n,a})$ where
$H_{n,a}=K_{a-1}\nabla(K_{1}\cup K_{n-a})$, then $G$ contains an
$[a,b]$-factor.

In this paper, we confirm Conjecture 1.1 for $n\geq 3a+b-1$.

######  Theorem 1.3.

Let $a\cdot n$ be an even integer, where $n\geq 3a+b-1$ and $b\geq a\geq 1$.
If $\rho(G)>\rho(H_{n,a})$, then $G$ contains an $[a,b]$-factor.

A $[k,k]$-factor is called a $k$-factor. As a corollary of Theorem 1.3, we
have the following result.

######  Corollary 1.1.

Let $k\cdot n$ be an even integer, where $n\geq 4k-1$. If
$\rho(G)>\rho(H_{n,k})$, then $G$ contains a $k$-factor.

##  2 Proof of Theorem 1.1

If $G$ is connected, then $A(G)$ is irreducible. By the Perron-Frobenius
theorem(cf. [15, Section 8.8]), the Perron vector $x$ is a positive
eigenvector of $A(G)$ respect to $\rho(G)$. For any $v\in V(G)$, let
$N_{G}(v)$ and $d_{G}(v)$ be the neighborhood and degree of $v$, respectively.
In this section, we give the proof of Theorem 1.1. Before proceeding, the
following lemmas are needed.

######  Lemma 2.1.

(See [24]) Let $G$ be a connected graph, and let $u,v$ be two vertices of $G$.
Suppose that $v_{1},v_{2},\ldots,v_{s}\in N_{G}(v)\backslash N_{G}(u)$ with
$s\geq 1$, and $G^{*}$ is the graph obtained from $G$ by deleting the edges
$vv_{i}$ and adding the edges $uv_{i}$ for $1\leq i\leq s$. Let $x$ be the
Perron vector of $A(G)$. If $x_{u}\geq x_{v}$, then $\rho(G)<\rho(G^{*})$.

An edge is said to be a _cut edge_ if its removal increases the number of
components of a graph.

######  Lemma 2.2.

(See [16]) Let $G$ be a connected graph with a unique perfect matching. Then
$G$ contains a cut edge $uv$ that is an edge of the perfect matching of $G$.

######  Lemma 2.3.

(See [2]) If $u$ and $v$ are two nonadjacent vertices of graph $G$, then
$\rho(G+uv)>\rho(G)$.

For any $S\subseteq V(G)$, let $G[S]$ be the subgraph of $G$ induced by $S$.
An edge of a graph is called _pendant edge_ if exactly one of its ends has the
degree 1. We now give a short proof of Theorem 1.1.

###### Proof of Theorem 1.1.

Suppose that $G$ is a connected graph of order $2n$ with a unique perfect
matching $M$. By Lemma 2.2, there exists a cut edge, say $u_{0}v_{0}$, that is
contained in $M$. Observe that $G-u_{0}v_{0}$ consists of two odd components.
Let $x^{(0)}$ be the Perron vector of $A(G)$. Without loss of generality, we
assume that $x^{(0)}_{u_{0}}\geq x^{(0)}_{v_{0}}$. Let

$G_{1}=G-\sum_{w\in N_{G}(v_{0})\setminus\\{u_{0}\\}}v_{0}w+\sum_{w\in
V(G)\setminus(N(u_{0})\cup\\{u_{0}\\})}u_{0}w.$

We see that $G_{1}$ also has a unique perfect matching, say $M_{1}$. By Lemmas
2.1 and 2.3, we have $\rho(G)\leq\rho(G_{1})$, with equality if and only if
$G\cong G_{1}$. Let $S_{1}=V(G_{1})-\\{u_{0},v_{0}\\}$. Since $u_{0}v_{0}$ is
a pendant edge of $G_{1}$ that is contained in $M_{1}$, the induced subgraph
$G_{1}[S_{1}]$ also contains a unique perfect matching, i.e.,
$M_{1}\setminus\\{u_{0}v_{0}\\}$. Again by Lemma 2.2, there exists a cut edge
$u_{1}v_{1}$ in $G_{1}[S_{1}]$ that is contained in
$M_{1}\setminus\\{u_{0}v_{0}\\}$. Let $x^{(1)}$ be the Perron vector of
$A(G_{1})$. Assume that $x^{(1)}_{u_{1}}\geq x^{(1)}_{v_{1}}$. Let

$G_{2}=G_{1}-\sum_{w\in
N_{G_{1}[S_{1}]}(v_{1})\setminus\\{u_{1}\\}}v_{1}w+\sum_{w\in
S_{1}\setminus(N(u_{1})\cup\\{u_{1}\\})}u_{1}w.$

Clearly, $G_{2}$ also has a unique perfect matching. Again by Lemmas 2.1 and
2.3, $\rho(G_{1})\leq\rho(G_{2})$, where the equality holds if and only if
$G_{1}\cong G_{2}$. By repeating this procedure, we can construct a sequence
of graphs $G_{0},G_{1},G_{2},\cdots,G_{n-1}$:

  * (i)

$G_{0}=G$;

  * (ii)

for $i\in[0,n-2]$, let
$S_{i}=V(G_{i})-\\{v_{0},v_{1},\ldots,v_{i-1},u_{0},u_{1},\ldots,u_{i-1}\\}$
and

$G_{i+1}=G_{i}-\sum_{w\in
N_{G_{i}[S_{i}]}(v_{i})\setminus\\{u_{i}\\}}v_{i}w+\sum_{w\in
S_{i}\setminus(N(u_{i})\cup\\{u_{i}\\})}u_{i}w,$

where $u_{i}v_{i}$ is a cut edge of $G_{i}[S_{i}]$ that is contained in the
unique perfect matching of $G_{i}[S_{i}]$ and $x^{(i)}_{v_{i}}\leq
x^{(i)}_{u_{i}}$, where $x^{(i)}$ is the Perron vector of $A(G_{i})$.

As above, we see that $G_{i}$ has a unique perfect matching for each $i$, and
$\rho(G_{i})\leq\rho(G_{i+1})$ with equality if and only if $G_{i}\cong
G_{i+1}$ ($0\leq i\leq n-2$). Note that $G_{n-1}\cong G(2n,1)$. Thus we
conclude that $\rho(G)\leq\rho(G(2n,1))$, where the equality holds if and only
if $G\cong G(2n,1)$.

We complete the proof. ∎

##  3 Proof of Theorem 1.2

The well-known sufficient and necessary condition for the existence of an odd
$[1,b]$-factor established by Amahashi[1].

######  Lemma 3.1.

(See [1]) Let $G$ be a graph and let $b$ be a positive odd integer. Then $G$
contains an odd $[1,b]$-factor if and only if for every subset $S\subseteq
V(G)$,

$o(G-S)\leq b|S|,$

where $o(H)$ is the number of odd components in a graph $H$.

######  Lemma 3.2.

(See [12]) Let $n=\sum_{i=1}^{t}n_{i}+s$. If $n_{1}\geq n_{2}\geq\cdots\geq
n_{t}\geq p$ and $n_{1}<n-s-p(t-1)$, then

$\rho(K_{s}\nabla(K_{n_{1}}\cup K_{n_{2}}\cup\cdots\cup
K_{n_{t}}))<\rho(K_{s}\nabla(K_{n-s-p(t-1)}\cup(t-1)K_{p})).$

Now we shall give the proof of Theorem 1.2.

###### Proof of Theorem 1.2.

Suppose that $G$ contains no odd $[1,b]$-factor, by Lemma 3.1, there exists
some nonempty subset $S$ of $V(G)$ such that $q=o(G-S)>b|S|$. Since $n$ is
even, $q$ and $b|S|$ have the same parity, we have $q\geq b|S|+2$. Let $|S|=s$
and $t=bs+2$. Then $G$ is a spanning subgraph of
$G_{s}^{1}=K_{s}\nabla(K_{n_{1}}\cup K_{n_{2}}\cup\cdots\cup K_{n_{t}})$ for
some positive odd integers $n_{1}\geq n_{2}\geq\cdots\geq n_{t}$ with
$\sum_{i=1}^{t}n_{i}=n-s$. Thus,

$\rho(G)\leq\rho(G_{s}^{1}),$ (1)

where the equality holds if and only if $G\cong G_{s}^{1}$. Note that the
graph $G_{s}^{1}$ contains no odd $[1,b]$-factor. Then we shall derive the
proof into the following three cases.

Case 1.

$s\geq\delta+1.$

Let $G_{s}^{2}=K_{s}\nabla(K_{n-s-t+1}\cup(t-1)K_{1})$. Note that
$s\geq\delta+1$ and $t=bs+2$. Then by Lemma 3.2, we have

$\rho(G_{s}^{1})\leq\rho(G_{s}^{2}),$ (2)

with equality if and only if $(n_{1},\ldots,n_{t})=(n-s-t+1,1,\ldots,1)$. The
vertex set of $G_{s}^{2}$ can be partitioned as $V(G_{s}^{2})=V(K_{s})\cup
V((bs+1)K_{1})\cup V(K_{n-(b+1)s-1})$, where
$V(K_{s})=\\{v_{1},\ldots,v_{s}\\}$,
$V((bs+1)K_{1})=\\{u_{1},\ldots,u_{bs+1}\\}$ and
$V(K_{n-(b+1)s-1})=\\{w_{1},\ldots,w_{n-(b+1)s-1}\\}$. Let

$G^{\prime}_{s}=G_{s}^{2}+\sum_{i=b\delta+2}^{bs+1}\sum_{j=1}^{n-(b+1)s-1}u_{i}w_{j}+\sum_{i=b\delta+2}^{bs}\sum_{j=i+1}^{bs+1}u_{i}u_{j}-\sum_{i=\delta+1}^{s}\sum_{j=1}^{b\delta+1}v_{i}u_{j}.$

Clearly, $G^{\prime}_{s}\cong
K_{\delta}\nabla(K_{n-(b+1)\delta-1}\cup(b\delta+1)K_{1})$. Let $x$ be the
Perron vector of $A(G_{s}^{2})$ with respect to $\rho=\rho(G_{s}^{2})$. By
symmetry, $x$ takes the same value (say $x_{1}$, $x_{2}$ and $x_{3}$) on the
vertices of $V(K_{s})$, $V((bs+1)K_{1})$ and $V(K_{n-(b+1)s-1)})$,
respectively. Then, by $A(G_{s}^{2})x=\rho x$, we have

$\displaystyle\rho x_{2}$ $\displaystyle=sx_{1},$ $\displaystyle\rho x_{3}$
$\displaystyle=sx_{1}+(n-(b+1)s-2)x_{3}.$

Note that $n\geq s+t=(b+1)s+2$. Then $x_{3}\geq x_{2}$ and

$\displaystyle x_{2}=\frac{sx_{1}}{\rho}.$ (3)

Similarly, let $y$ be the Perron vector of $A(G^{\prime}_{s})$ corresponding
to $\rho(G^{\prime}_{s})=\rho^{\prime}$. By symmetry, $y$ takes the same
values $y_{1}$, $y_{2}$ and $y_{3}$ on the vertices of $V(K_{\delta})$,
$V((b\delta+1)K_{1})$ and $V(K_{n-(b+1)\delta-1})$, respectively. Then, by
$A(G^{\prime}_{s})y=\rho^{\prime}y$, we obtain

$\displaystyle\rho^{\prime}y_{2}$ $\displaystyle=\delta y_{1},$ (4)
$\displaystyle\rho^{\prime}y_{3}$ $\displaystyle=\delta
y_{1}+(n-(b+1)\delta-2)y_{3}.$ (5)

Note that $G^{\prime}_{s}$ contains $K_{n-(b+1)\delta-1}\cup
K_{\delta}\cup(b\delta+1)K_{1}$ as a proper spanning subgraph. Then
$\rho^{\prime}>\rho(K_{n-(b+1)\delta-1}\cup
K_{\delta}\cup(b\delta+1)K_{1})=n-(b+1)\delta-2$. Putting (4) into (5), and
considering that $\rho^{\prime}>n-(b+1)\delta-2$, we have

$y_{3}=\frac{\rho^{\prime}y_{2}}{\rho^{\prime}-(n-(b+1)\delta-2)}.$ (6)

Recall that $n\geq(b+1)s+2$. Then $\delta+1\leq s\leq(n-2)/(b+1)$. Since
$G_{s}^{2}$ is not a regular graph, it follows that $\rho<n-1$. Suppose to the
contrary that $\rho\geq\rho^{\prime}$. Combining this with (3) and (6) yields
that

$\displaystyle y^{T}(\rho^{\prime}-\rho)x$
$\displaystyle=y^{T}(A(G^{\prime}_{s})-A(G_{s}^{2}))x$
$\displaystyle=\sum_{i=b\delta\\!+\\!2}^{bs+1}\sum_{j=1}^{n\\!-\\!(b\\!+\\!1)s\\!-\\!1}\\!(x_{u_{i}}y_{w_{j}}\\!+\\!x_{w_{j}}y_{u_{i}})\\!+\sum_{i=b\delta\\!+\\!2}^{bs}\sum_{j=i\\!+\\!1}^{bs\\!+\\!1}\\!(x_{u_{i}}y_{u_{j}}\\!+\\!x_{u_{j}}y_{u_{i}})\\!\\!-\\!\sum_{i=\delta\\!+\\!1}^{s}\sum_{j=1}^{b\delta\\!+\\!1}\\!(x_{v_{i}}y_{u_{j}}\\!+\\!x_{u_{j}}y_{v_{i}})\\!$
$\displaystyle=(s\\!-\\!\delta)[b(n\\!-\\!(b\\!+\\!1)s\\!-\\!1)(x_{2}y_{3}\\!+\\!x_{3}y_{3})\\!+\\!b(bs\\!-\\!b\delta\\!-\\!1)x_{2}y_{3}\\!-\\!(b\delta\\!+\\!1)(x_{1}y_{2}\\!+\\!x_{2}y_{3})]$
$\displaystyle=(s\\!-\\!\delta)[(b(n\\!-\\!(b\\!+\\!1)\delta\\!-\\!s\\!-\\!2)\\!-\\!1)x_{2}y_{3}\\!+\\!b(n\\!-\\!(b\\!+\\!1)s\\!-\\!1)x_{3}y_{3}\\!-\\!(b\delta\\!+\\!1)x_{1}y_{2}]$
$\displaystyle\geq(s\\!-\\!\delta)[b(n\\!-\\!(b\\!+\\!1)\delta\\!-\\!s\\!-\\!2)x_{2}y_{3}\\!-\\!(b\delta\\!+\\!1)x_{1}y_{2}]~{}(\mbox{since
$x_{3}\\!\geq\\!x_{2}$ and $n\geq(b\\!+\\!1)s\\!+\\!2$})$
$\displaystyle\geq(s\\!-\\!\delta)[b(n\\!-\\!(b\\!+\\!1)\delta\\!-\\!\frac{n\\!-\\!2}{b\\!+\\!1}\\!-\\!2)x_{2}y_{3}\\!-\\!(b\delta\\!+\\!1)x_{1}y_{2}]~{}~{}(\mbox{since
$s\leq\frac{n-2}{b+1}$})$
$\displaystyle=(s\\!-\\!\delta)x_{1}y_{2}\left(\frac{bs\rho^{\prime}(n-(b+1)\delta-\frac{n-2}{b+1}-2)}{\rho(\rho^{\prime}-(n-(b+1)\delta-2))}-(b\delta+1)\right)$
$\displaystyle\geq\frac{(b\delta\\!+\\!1)(s\\!-\\!\delta)x_{1}y_{2}}{\rho(\rho^{\prime}\\!-\\!(n\\!-\\!(b\\!+\\!1)\delta\\!-\\!2))}[\rho^{\prime}(n\\!-\\!(b\\!+\\!1)\delta\\!-\\!\frac{n\\!-\\!2}{b\\!+\\!1}\\!-\\!2)\\!-\\!\rho(\rho^{\prime}\\!-\\!(n\\!-\\!(b\\!+\\!1)\delta\\!-\\!2))]$
$\displaystyle(\mbox{since $s\geq\delta+1$ and $b\geq 1$})$
$\displaystyle\geq\frac{(b\delta\\!+\\!1)(s\\!-\\!\delta)x_{1}y_{2}}{\rho(\rho^{\prime}\\!-\\!(n\\!-\\!(b\\!+\\!1)\delta\\!-\\!2))}(\rho^{\prime}(2n\\!-\\!2(b\\!+\\!1)\delta\\!-\\!\frac{n\\!-\\!2}{b\\!+\\!1}\\!-\\!4)\\!-\\!\rho\rho^{\prime})~{}~{}(\mbox{since
$\rho\geq\rho^{\prime}$})$
$\displaystyle\geq\frac{(b\delta\\!+\\!1)(s\\!-\\!\delta)\rho^{\prime}(\frac{3n}{2}\\!-\\!2(b\\!+\\!1)\delta\\!-\\!3\\!-\\!\rho)x_{1}y_{2}}{\rho(\rho^{\prime}-(n-(b+1)\delta-2))}~{}~{}(\mbox{since
$b\geq 1$})$
$\displaystyle\geq\frac{(b\delta\\!+\\!1)(s\\!-\\!\delta)\rho^{\prime}(n\\!-\\!1\\!-\\!\rho)x_{1}y_{2}}{\rho(\rho^{\prime}\\!-\\!(n\\!-\\!(b\\!+\\!1)\delta\\!-\\!2))}~{}~{}(\mbox{since
$n\geq 4(b\\!+\\!1)\delta\\!+\\!4$})$ $\displaystyle>0~{}~{}(\mbox{since
$\rho<n\\!-\\!1$, $\rho^{\prime}>n\\!-\\!(b\\!+\\!1)\delta\\!-\\!2$ and
$s\geq\delta\\!+\\!1$}).$

This implies that $\rho^{\prime}>\rho$, which contradicts the assumption
$\rho^{\prime}\leq\rho$. It follows that $\rho^{\prime}>\rho$. Combining this
with (1) and (2), we may conclude that
$\rho(G)\leq\rho(G_{s}^{1})\leq\rho(G_{s}^{2})<\rho(T(n,b,\delta))$.

Case 2.

$s<\delta$.

Let $G_{s}^{3}=K_{s}\nabla(K_{n-s-(\delta+1-s)(t-1)}\cup(t-1)K_{\delta+1-s})$.
Recall that $G$ is a spanning subgraph of $G_{s}^{1}=K_{s}\nabla(K_{n_{1}}\cup
K_{n_{2}}\cup\cdots\cup K_{n_{t}})$, where $t=bs+2$, $n_{1}\geq
n_{2}\geq\cdots\geq n_{t}$ and $\sum_{i=1}^{t}n_{i}=n-s$. Clearly,
$n_{t}\geq\delta+1-s$ because the minimum degree of $G_{s}^{1}$ is at least
$\delta$. By Lemma 3.2, we have

$\rho(G_{s}^{1})\leq\rho(G_{s}^{3}),$ (7)

where the equality holds if and only if
$(n_{1},\ldots,n_{t})=(n-s-(\delta+1-s)(t-1),\delta+1-s,\ldots,\delta+1-s)$.

Assume that $\rho(G^{3}_{s})=\rho^{*}\geq n-(\delta+1-s)(t-1)$. Let $x$ be the
Perron vector of $A(G^{3}_{s})$ corresponding to $\rho^{*}$. By symmetry, $x$
takes the same values $x_{1}$, $x_{2}$, and $x_{3}$ on the vertices of
$K_{n-s-(\delta+1-s)(t-1)}$, $K_{\delta+1-s}$ and $K_{s}$, respectively. Then,
by $A(G^{3}_{s})x=\rho^{*}x$, we have

$\displaystyle\rho^{*}x_{1}$ $\displaystyle\\!\\!=\\!\\!$
$\displaystyle(n-1-(\delta+1-s)(t-1)-s)x_{1}+sx_{3},$ (8)
$\displaystyle\rho^{*}x_{2}$ $\displaystyle\\!\\!=\\!\\!$
$\displaystyle(\delta-s)x_{2}+sx_{3},$ (9) $\displaystyle\rho^{*}x_{3}$
$\displaystyle\\!\\!=\\!\\!$
$\displaystyle(n\\!-\\!(\delta\\!+\\!1\\!-\\!s)(t\\!-\\!1)\\!-\\!s)x_{1}\\!+\\!(\delta\\!+\\!1\\!-\\!s)(t\\!-\\!1)x_{2}\\!+\\!(s\\!-\\!1)x_{3}.$
(10)

From (8) and (9), we obtain

$\left\\{\begin{aligned}
&x_{1}=\frac{sx_{3}}{\rho^{*}-(n-1-(\delta+1-s)(t-1))+s},\\\
&x_{2}=\frac{sx_{3}}{\rho^{*}-\delta+s}.\end{aligned}\right.$ (11)

Note that $n\geq F(b,\delta)\geq b\delta^{3}+\delta$, $b\geq 1$ and
$\delta\geq s+1\geq 2$. Then $\rho^{*}\geq n-(\delta+1-s)(t-1)>\delta+1$.
Putting (11) into (10), and considering that $\rho^{*}\geq
n-(\delta+1-s)(t-1)>\delta+1$, we get

$\displaystyle~{}~{}~{}~{}\rho^{*}+1$
$\displaystyle=s+\frac{s(n-(\delta+1-s)(t-1)-s)}{\rho^{*}-(n-1-(\delta+1-s)(t-1))+s}+\frac{s(\delta+1-s)(t-1)}{\rho^{*}-(\delta-s)}$
$\displaystyle<s+\frac{s(n-(\delta+1-s)(t-1)-s)}{s+1}+\frac{s(\delta+1-s)(t-1)}{s+1}$
$\displaystyle=s+\frac{s(n-s)}{s+1}$
$\displaystyle=n\\!-\\!(\delta\\!+\\!1\\!-\\!s)(\\!t-\\!1)\\!-\\!\frac{n\\!-\\!s\\!-\\!(\delta\\!+\\!1\\!-\\!s)(t\\!-\\!1)(s\\!+\\!1)}{s\\!+\\!1}$
$\displaystyle\leq
n\\!-\\!(\delta\\!+\\!1\\!-\\!s)(t\\!-\\!1)\\!-\\!\frac{b\delta^{3}\\!+\\!\delta\\!-\\!s\\!-\\!(\delta\\!+\\!1\\!-\\!s)(t\\!-\\!1)(s\\!+\\!1)}{s\\!+\\!1}~{}~{}(\mbox{since
~{}$n\geq b\delta^{3}\\!+\\!\delta$})$
$\displaystyle<n\\!-\\!(\delta\\!+\\!1\\!-\\!s)(t\\!-\\!1)~{}~{}(\mbox{since
~{}$\delta\geq s+1$, $b\geq 1$ and $s\geq 1$})$ $\displaystyle\leq\rho^{*},$

which is impossible. Thus we have

$\displaystyle\rho^{*}$
$\displaystyle<n\\!-\\!(\delta\\!+\\!1\\!-\\!s)(t\\!-\\!1)$ (12)
$\displaystyle=n\\!-\\!b\delta\\!-\\!2\\!-\\!((s\\!-\\!1)(\delta\\!-\\!s)b\\!+\\!\delta\\!-\\!(s\\!+\\!1))~{}~{}(\mbox{since
~{}$\delta\geq s\\!+\\!1$, $b\\!\geq\\!1$ and $s\\!\geq\\!1$})$
$\displaystyle\leq n\\!-\\!b\delta\\!-\\!2.$

Since $T(n,b,\delta)$ contains $K_{n-b\delta-1}\cup(b\delta+1)K_{1}$ as a
proper spanning subgraph, it follows that
$\rho(T(n,b,\delta))>\rho(K_{n-b\delta-1}\cup(b\delta+1)K_{1})=n-b\delta-2$.
Combining this with (1), (7) and (12), we may conclude that
$\rho(G)\leq\rho(G_{s}^{1})\leq\rho(G_{s}^{3})<\rho(T(n,b,\delta))$.

Case 3.

$s=\delta.$

Note that $G_{s}^{1}$ is a spanning subgraph of $T(n,b,\delta)$. Then

$\rho(G_{s}^{1})\leq\rho(T(n,b,\delta)),$

with equality holding if and only if $G_{s}^{1}\cong T(n,b,\delta)$. Combining
this with (1), we may conclude that

$\rho(G)\leq\rho(T(n,b,\delta)),$

where the equality holds if and only if $G\cong T(n,b,\delta)$. This completes
the proof. ∎

From Theorem 1.2, we can get the result of Liu, Liu and Feng[27] immediately.

######  Corollary 3.1.

(See [27]) Let $G$ be a connected graph of even order $n$ with minimum degree
$\delta(G)\geq 2$. If
$n\geq\max\\{7+7\delta+2\delta^{2},\delta^{3}+3\delta^{2}+2\delta\\}$ and
$\rho(G)\geq\rho(T(n,1,\delta))$, then $G$ has a perfect matching, unless
$G\cong T(n,1,\delta)$.

##  4 Proof of Theorem 1.3

The following two structural lemmas will play an essential role in the proof
of Theorem 1.3.

######  Lemma 4.1.

(See [28]) Let $G$ be a graph of order $n$ with minimum degree $\delta(G)$ and
let $a$ and $b$ be integers such that $1\leq a<b$. Then if $\delta(G)\geq a$,
$n\geq 2a+b+\frac{a^{2}-a}{b}$ and

$\max\\{d_{G}(u),d_{G}(w)\\}\geq\frac{an}{a+b},$

for any two nonadjacent vertices $u$ and $w$ of $G$, then $G$ contains an
$[a,b]$-factor.

######  Lemma 4.2.

(See [29]) Suppose that $k\geq 3$. Let $G$ be a connected graph of order
$n\geq 4k-3$ with minimum degree $\delta(G)$, where $k\cdot n$ is even and
$\delta(G)\geq k$. If

$\max\\{d_{G}(u),d_{G}(w)\\}\geq\frac{n}{2},$

for any two nonadjacent vertices $u$ and $w$ of $G$, then $G$ contains a
$k$-factor.

######  Lemma 4.3.

(See [13]) Let $G$ be a connected graph of order $n$. If

$\displaystyle\rho(G)\geq n-2,$ (13)

then $G$ contains a Hamiltonian path unless $G\cong H_{n,1}$. If the
inequality (13) is strict, then $G$ contains a Hamiltonian cycle unless
$G\cong H_{n,2}$.

######  Lemma 4.4.

(See [2]) Let $M$ be a real symmetric matrix, and let $\lambda_{1}(M)$ be the
largest eigenvalue of $M$. If $B_{\Pi}$ is an equitable quotient matrix of
$M$, then the eigenvalues of $B_{\Pi}$ are also eigenvalues of $M$.
Furthermore, if $M$ is nonnegative and irreducible, then
$\lambda_{1}(M)=\lambda_{1}(B_{\Pi}).$

######  Lemma 4.5.

Suppose that $G$ is a connected graph of order $n$. Let $u,w$ be two
nonadjacent vertices of $G$ such that $\max\\{d_{G}(u),d_{G}(w)\\}\leq t$,
where $t\geq 1$. Then $\rho(G)\leq\rho(K_{t}\nabla(2K_{1}\cup K_{n-t-2}))$,
with equality if and only if $G\cong K_{t}\nabla(2K_{1}\cup K_{n-t-2})$.

###### Proof.

Let $x$ be the Perron vector of $A(G)$ with respect to $\rho(G)$. Assume that
$V(G)\setminus\\{u,w\\}=\\{v_{1},v_{2},\ldots,v_{n-2}\\}$ with $x_{v_{1}}\geq
x_{v_{2}}\geq\cdots\geq x_{v_{n-2}}$. Let

$G^{\prime}=G-\sum_{v\in N_{G}(u)}uv-\sum_{v\in
N_{G}(w)}wv+\sum_{i=1}^{t}(uv_{i}+wv_{i}).$

Since $d_{G}(u)\leq t$ and $d_{G}(w)\leq t$, by Lemma 2.1, it follows that
$\rho(G)\leq\rho(G^{\prime})$, where equality holds if and only if $G\cong
G^{\prime}$. Obviously, $G^{\prime}$ is a spanning subgraph of
$K_{t}\nabla(2K_{1}\cup K_{n-t-2})$. Thus

$\displaystyle\rho(G)\leq\rho(K_{t}\nabla(2K_{1}\cup K_{n-t-2})),$

with equality if and only if $G\cong K_{t}\nabla(2K_{1}\cup K_{n-t-2})$. ∎

Now we shall give the proof of Theorem 1.3.

###### Proof of Theorem 1.3.

Let $G$ be a graph of order $n\geq 3a+b-1$ with $\rho(G)>\rho(H_{n,a})$, where
$a\cdot n$ is an even integer and $b\geq a\geq 1$. We first assert that $G$ is
connected. If not, suppose that $G_{1},\ldots,G_{p}$ ($p\geq 2$) are the
components of $G$. Then
$\rho(G)=\max\\{\rho(G_{1}),\ldots,\rho(G_{p})\\}\leq\rho(K_{n-1})=n-2$, which
contradicts that $\rho(G)>\rho(H_{n,a})\geq n-2$, and so we assume that $G$ is
connected in the following. We next assert that $\delta(G)\geq a$. Otherwise,
let $1\leq\delta(G)\leq a-1$. Then $G$ is a spanning subgraph of $H_{n,a}$ for
$a\geq 2$, and so $\rho(G)\leq\rho(H_{n,a})$, a contradiction.

If $a=b=1$. Since $n$ is even and $\rho(G)>\rho(H_{n,1})$, by Lemma 4.3, it
follows that $G$ contains a Hamiltonian path of even number, and so $G$
contains a 1-factor. If $a=b=2$, then by using the same analysis as above, we
deduce that $G$ contains a 2-factor. Next, we assume that both $a$ and $b$ are
not equal to 1 or 2 simultaneously. Suppose that $G$ contains no
$[a,b]$-factor of order $n\geq 3a+b-1$. By Lemmas 4.1 and 4.2, there exist two
nonadjacent vertices $u$ and $w$ such that
$\max\\{d_{G}(u),d_{G}(w)\\}\leq\frac{an}{a+b}-1$. Let $t=\frac{an}{a+b}-1$.
Then $t\geq\delta(G)\geq a\geq 1$. Suppose that $G_{1}\cong
K_{t}\nabla(2K_{1}\cup K_{n-t-2})$. By Lemma 4.5, we obtain

$\displaystyle\rho(G)\leq\rho(G_{1}).$ (14)

The vertex set of $G_{1}$ can be partitioned as $V(G_{1})=V(2K_{1})\cup
V(K_{t})\cup V(K_{n-t-2})$, where $V(2K_{1})=\\{u,w\\}$,
$V(K_{t})=\\{v_{1},\ldots,v_{t}\\}$ and
$V(K_{n-t-2})=\\{v_{t+1},\ldots,v_{n-2}\\}$. Let
$G^{\prime}_{1}=G_{1}-\\{uv_{a},uv_{a+1},\ldots,uv_{t}\\}+\\{wv_{t+1},wv_{t+2},\ldots,wv_{n-2}\\}$.
Clearly, $G^{\prime}_{1}\cong H_{n,a}$. If $a=1$ and $b\geq 2$, then
$G^{\prime}_{1}\cong K_{1}\cup K_{n-1}$. Note that $t=\frac{an}{a+b}-1$. Then
$n\geq 3t+3$. Observe that $A(G_{1})$ has the equitable quotient matrix

$B_{1}=\begin{bmatrix}0&t&0\\\ 2&t-1&n-t-2\\\ 0&t&n-t-3\\\ \end{bmatrix}.$

The characteristic polynomial of $B_{1}$ is

$f(\lambda)=\lambda^{3}-(n-4)\lambda^{2}-(n+2t-3)\lambda+2tn-2t^{2}-6t.$

By a simple computation, we have

$\displaystyle
f(-\infty)=-\infty,~{}~{}f(0)=2t(n-t-3)>0,~{}~{}f(n-3)=-2t^{2}<0,$

and

$\displaystyle f(n-2)=(n-\frac{3}{2})^{2}-2t^{2}-2t-\frac{1}{4}\geq
7t^{2}+7t+2>0.$

By Lemma 4.4, we have $\lambda_{1}(B_{1})=\rho(G_{1})$. Notice that
$\rho(G^{\prime}_{1})=\rho(K_{1}\cup K_{n-1})=n-2$ and
$\lambda_{1}(B_{1})<n-2$. Then $\rho(G_{1})<\rho(G^{\prime}_{1})$. Combining
this with (14), we obtain that $\rho(G)\leq\rho(G_{1})<\rho(H_{n,1})$, a
contradiction. Next, we assume that $a\geq 2$ and $b\geq 3$. Recall that
$t=\frac{an}{a+b}-1$. Then $n\geq 2t+2$. Let $x$ be the Perron vector of
$A(G_{1})$ with respect to $\rho(G_{1})$, and let $\rho=\rho(G_{1})$. By
symmetry, $x$ takes the same value (say $x_{1}$, $x_{2}$ and $x_{3}$) on the
vertices of $V(2K_{1})$, $V(K_{t})$ and $V(K_{n-t-2})$, respectively. Then, by
$A(G_{1})x=\rho x$, we have

$x_{2}=\frac{(\rho-(n-t-3))x_{3}}{t}.$ (15)

Similarly, let $y$ be the Perron vector of $A(G^{\prime}_{1})$ corresponding
to $\rho(G^{\prime}_{1})=\rho^{\prime}$. By symmetry, $y$ takes the same
values $y_{1}$, $y_{2}$ and $y_{3}$ on the vertices of $V(K_{1})$,
$V(K_{a-1})$ and $V(K_{n-a})$, respectively. Then, by
$A(G^{\prime}_{1})y=\rho^{\prime}y$, we obtain

$\displaystyle\rho^{\prime}y_{1}$ $\displaystyle=(a-1)y_{2},$ (16)
$\displaystyle\rho^{\prime}y_{3}$ $\displaystyle=(a-1)y_{2}+(n-a-1)y_{3}.$
(17)

Since $G^{\prime}_{1}$ contains $K_{1}\cup K_{n-1}$ as a proper spanning
subgraph, we have $\rho^{\prime}>n-2>n-a-1$. Putting (16) into (17), and
considering that $\rho^{\prime}>n-a-1$, we have

$y_{3}=\frac{\rho^{\prime}y_{1}}{\rho^{\prime}-(n-a-1)}.$ (18)

Obviously, neither $G_{1}$ nor $G^{\prime}_{1}$ is a regular graph. Thus
$\rho<n-1$ and $\rho^{\prime}<n-1$. Combining this with (15) and (18) yields
that

$\displaystyle y^{T}(\rho^{\prime}-\rho)x$
$\displaystyle=y^{T}(A(G^{\prime}_{1})-A(G_{1}))x$
$\displaystyle=\sum_{i=t+1}^{n-2}(x_{w}y_{v_{i}}+x_{v_{i}}y_{w})-\sum_{i=a}^{t}(x_{u}y_{v_{i}}+x_{v_{i}}y_{u})$
$\displaystyle=(n\\!-\\!t\\!-\\!2)(x_{1}y_{3}+x_{3}y_{3})\\!-\\!(t\\!-\\!a+1)(x_{1}y_{3}+x_{2}y_{1})$
$\displaystyle=(n-2t+a-3)x_{1}y_{3}+(n-t-2)x_{3}y_{3}-(t-a+1)x_{2}y_{1}$
$\displaystyle>(n-t-2)x_{3}y_{3}-(t-a+1)x_{2}y_{1}~{}~{}(\mbox{since $n\geq
2t+2$ and $a\geq 2$})$
$\displaystyle=x_{3}y_{1}\left(\frac{\rho^{\prime}(n-t-2)}{\rho^{\prime}-(n-a-1)}-\frac{(\rho-(n-t-3))(t-a+1)}{t}\right)$
$\displaystyle>x_{3}y_{1}\left(\frac{\rho^{\prime}(n\\!-\\!t\\!-\\!2)}{a}\\!-\\!\frac{(t\\!-\\!a\\!+\\!1)(t+2)}{t}\right)~{}~{}(\mbox{since
$\rho\\!<\\!n\\!-\\!1$ and $\rho^{\prime}\\!<\\!n\\!-\\!1$})$
$\displaystyle>x_{3}y_{1}\left(\frac{(n-2)(n-t-2)}{a}-\frac{(t-a+1)(t+2)}{t}\right)~{}~{}(\mbox{since
$\rho^{\prime}>n-2$})$
$\displaystyle\geq\frac{((n-2)(n-t-2)-(t-a+1)(t+2))x_{3}y_{1}}{t}~{}(\mbox{since
$t\geq a$})$ $\displaystyle>0~{}~{}(\mbox{since $t\geq a\geq 2$ and $n\geq
2t+2$}).$

Therefore, we have $\rho<\rho^{\prime}$. Combining this with (14), we may
conclude that $\rho(G)\leq\rho(G_{1})<\rho(H_{n,a})$, a contradiction.
Therefore, the proof is complete. ∎

##  5 Concluding remarks

A fractional matching of a graph $G$ is a function $f$ giving each edge a
number in $[0,1]$ such that $\sum_{e\in\Gamma(v)}f(e)\leq 1$ for each $v\in
V(G)$, where $\Gamma(v)$ is the set of edges incident to $v$. A fractional
perfect matching of a graph $G$ is a fractional matching $f$ with $\sum_{e\in
E(G)}f(e)=\frac{n}{2}$. The relations between the eigenvalues and the
fractional matching number of a graph were studied by several researchers [31,
33, 36].

The following fundamental lemmas provide some sufficient and necessary
conditions for the existence of a $[1,b]$-factor for $b\geq 2$ and a
fractional perfect matching in a graph, respectively. Let $i(H)$ be the number
of isolated vertices of a graph $H$.

######  Lemma 5.1.

(See [4]) Let $G$ be a graph and let $b\geq 2$ be a positive integer. Then $G$
contains a $[1,b]$-factor if and only if for every subset $S\subseteq V(G)$,

$i(G-S)\leq b|S|.$

######  Lemma 5.2.

(See [34]) A graph $G$ has a fractional perfect matching if and only if for
every subset $S\subseteq V(G)$,

$i(G-S)\leq|S|.$

By Lemmas 5.1 and 5.2 and using the same analysis as the proof of Theorem 1.2.
We easily obtain the following result.

######  Theorem 5.1.

Suppose that $G$ is a connected graph of order $n$ with minimum degree
$\delta$.

  1. (i)

If $n\geq 4(b+1)\delta+4$ and
$\rho(G)\geq\rho(K_{\delta}\nabla(K_{n-(b+1)\delta-1}\cup(b\delta+1)K_{1}))$
where $b\geq 2$, then $G$ contains a $[1,b]$-factor, unless $G\cong
K_{\delta}\nabla(K_{n-(b+1)\delta-1}\cup(b\delta+1)K_{1})$.

  2. (ii)

If $n\geq 8\delta+4$ and
$\rho(G)\geq\rho(K_{\delta}\nabla(K_{n-2\delta-1}\cup(\delta+1)K_{1}))$, then
$G$ contains a fractional perfect matching, unless $G\cong
K_{\delta}\nabla(K_{n-2\delta-1}\cup(\delta+1)K_{1})$.

Corollary 1.1 provides a spectral condition to guarantee the existence a
$k$-factor in a graph. It is natural to ask the following question for $k\geq
1$.

######  Problem 5.1.

What is the maximum spectral radius and what is the corresponding extremal
graph among all graphs with a unique $k$-factor for $k\geq 1$?

In this paper, we give the answer to Problem 5.1 for $k=1$. However, the
structure of graphs with a unique $k$-factor is more complicated for $k\geq
2$, and it seems difficult to determine the extremal graphs about the problem.

If $k\leq n$. Let $n=sk+t$ with $s\geq 1$ and $0\leq t\leq k-1$. We give the
process to construct the graph $G(2n,k)$. First define a graph $F_{1}$ on
$2(k+t)$ vertices as follows. Let $H_{1}\cong K_{t}\nabla tK_{1}$, and let
$A_{11}=V(tK_{1})$ and $A_{12}=V(K_{t})$. Denote by $H_{2}$ the graph obtained
from $kK_{1}\cup K_{k}$ by adding edges between $V(kK_{1})$ and $V(K_{k})$
such that $d_{H_{2}}(v)=k-t$ for $v\in V(kK_{1})$ and $d_{H_{2}}(u)=2k-t-1$
for $u\in V(K_{k})$. Suppose that $A_{21}=V(kK_{1})$ and $A_{22}=V(K_{k})$.
Let $F_{1}$ be the graph of order $2(k+t)$ obtained from $H_{1}\cup H_{2}$ by
connecting all vertices of $A_{1j}$ with $A_{2(3-j)}$ for $1\leq j\leq 2$ and
adding all edges between $A_{12}$ and $A_{22}$. The resulting graph $F_{1}$
has exactly one $k$-factor. Suppose that $U_{1}=A_{11}\cup A_{21}$ and
$W_{1}=A_{12}\cup A_{22}$. Next take $s-1$ copies of $K_{k}\nabla kK_{1}$
labeled $F_{2},\ldots,F_{s}$. For $2\leq i\leq s$, let $U_{i}$ and $W_{i}$ be
the vertices set of $V(kK_{1})$ and $V(K_{k})$ in each $F_{i}$, respectively.
Then the graph $G(2n,k)$ is obtained by adding edges connecting all vertices
of $W_{i}$ in $F_{i}$ to all vertices in $F_{j}$ for each $i,j$ with $1\leq
i<j\leq s$. The resulting graph $G(2n,k)$ has a unique $k$-factor. We end the
paper by proposing the following problem for further research.

######  Problem 5.2.

For $k\geq 1$. Suppose that $G$ is a graph of order $2n$ with a unique
$k$-factor.

(I) Does $\rho(G)\leq\rho(K_{2n-k}\nabla H)$, where $H$ is a $2(k-n)$-regular
graph if $k>n$?

(II) Does $\rho(G)\leq\rho(G(2n,k))$, if $k\leq n$?

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