For a positive integer $n$, $N$ denotes the set $\{1,2,\cdots,n\}$. The set of all $n\times n$ complex matrices is denoted by $C^{n\times n}$ and $R^{n\times n}$ denotes the set of all $n\times n$ real matrices.

We let $Z_{n}$ denote the class of all $n\times n$ real matrices all of whose off-diagonal entries are nonpositive. An $n\times n$ matrix $A$ is called an $M$-matrix if there exists an $n\times n$ nonnegative matrix $B$ and a nonnegative real number $\lambda$ such that $A=\lambda I-B$ and $\lambda\geq\rho(B)$, $I$ is the identity matrix; if $\lambda>\rho(B)$, we call $A$ a nonsingular $M$-matrix; if $\lambda=\rho(B)$, we call $A$ a singular $M$-matrix. Denote by $M_{n}$ the set of nonsingular $M$-matrices.

Let $A\in Z_{n}$ and $\tau(A)=$min$\{Re(\lambda):\lambda\in\sigma(A)\}$. Basic for our purpose are the following simple facts (see Problems $16,19 $ and $28$ in Section $2.5$ of [1] ):

(1) $\tau(A)\in\sigma(A)$; $\tau(A)$ is called the minimum eigenvalue of $A$.

(2) If $A,B\in M_{n}$, and $A\geq B$, then $\tau(A)\geq\tau(B)$.

(3) If $A\in M_{n}$, then $\rho(A^{-1})$ is the Perron eigenvalue of the nonnegative matrix $A^{-1}$, and $\tau(A)=\frac{1}{\rho(A^{-1})}$ is a positive real eigenvalue of $A$.

Let $A,B\in C^{n\times n}$. The Fan product of $A$ and $B$ is denoted by $A\bigstar B\equiv C=(c_{ij})\in C^{n\times n}$ and is defined by

$\begin{equation*} c_{ij}=\left\{ \begin{aligned} -a_{ij}b_{ij}, \quad &\mbox{if $i\neq j$},\\ a_{ii}b_{ii}, \quad &\mbox{if $i=j$}. \end{aligned} \right. \end{equation*}$

If $A,B\in M_{n}$, then so is $A\bigstar B$. In [1-3], the following bounds for $\tau(A\bigstar B)$ are given for two nonsingular $M$-matrices $A$ and $B$, respectively.

$\begin{eqnarray*} &&\tau(A\bigstar B)\geq\tau(A)\tau(B), \\ &&\tau(A\bigstar B)\geq(1-\rho(J_{A})\rho(J_{B}))\min\limits_{1\leq i\leq n}(a_{ii}b_{ii}), \\ &&\tau(A\bigstar B)\geq\min\limits_{1\leq i\leq n}\{a_{ii}\tau(B)+b_{ii}\tau(A)-\tau(A)\tau(B)\}. \end{eqnarray*}$

Recently, Li [4] gave a sharper lower bound for $\tau(A\bigstar B)$, that is

$\begin{eqnarray*} \tau(A\bigstar B)\geq\min\limits_{1\leq i\leq n}\{a_{ii}b_{ii}-m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}}\}. \end{eqnarray*}$

Let $A=(a_{ij})\in C^{n\times n}$ and $B=(b_{ij})\in C^{n\times n}$. We write $A\geq B(>B)$ if $a_{ij}\geq b_{ij}(>b_{ij})$ for all $i,j\in\{1,2,\cdots,n\}$. If $0$ is the null matrix and $A\geq 0(>0)$, we say that $A$ is a nonnegative (positive) matrix. The spectral radius of $A$ is denoted by $\rho (A)$. If $A$ is a nonnegative matrix, the Perron-Frobenius theorem guarantees that $\rho(A)$ is an eigenvalue of $A$.

An $n\times n$ matrix $A$ is reducible if there exists a permutation matrix $P$ such that

$ P^{T}AP= [ $\begin{array}{cc} B&0\\ C&D \end{array}$ ], $

where $B,D$ are square matrices of order at least one. If $A$ is not reducible, then it is called irreducible. Note that any $1\times 1$ matrix is irreducible.

For two real matrices $A=(a_{ij})$ and $B=(b_{ij})$ of the same size, the Hadamard product of $A$ and $B$ is $A\circ B=(a_{ij}b_{ij})$.

In [1-3], the following bounds for $\rho(A\circ B)$ are given for $A,B\geq0$, respectively.

$\begin{eqnarray*} &&\rho(A\circ B)\leq\rho(A)\rho(B), \\ &&\rho(A\circ B)\leq(1+\rho(J^{\prime}_{A}\rho(J^{\prime}_{B})))\max\limits_{1\leq i\leq n}a_{ii}b_{ii}, \\ &&\rho(A\circ B)\leq\max\limits_{1\leq i\leq n}\{2a_{ii}b_{ii}+\rho(A)\rho(B)-a_{ii}\rho(B)-b_{ii}\rho(A)\}. \end{eqnarray*}$

Recently, Li [4] gave a sharper upper bound for $\rho(A\circ B)$, that is

$\begin{eqnarray*} \rho(A\circ B)\leq\max\limits_{1\leq i\leq n}\{a_{ii}b_{ii}+m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}}\}. \end{eqnarray*}$

In this paper, we give a new lower bound on $\tau(A\bigstar B)$ for two matrices $A,B\in M_{n}$ in Section 2 and a new upper bound on $\rho(A\circ B)$ for two nonnegative matrices $A$ and $B$ in Section 3. Some examples are given to illustrate our results.

In the following, we will need the notations:

$\begin{eqnarray*} & & R_{i}=\sum\limits_{k\neq i}|a_{ik}|,\quad\quad d_{i}=\frac{R_{i}}{|a_{ii}|}, i\in N;\\ & & m_{ji}=|a_{ji}|h_{j},\,\,\,\,\,\,m_{i}=\max\limits_{j\neq i}\{m_{ji}\},\quad i,j\in N,\,\,\,\, h_{j}=\left\{ \begin{aligned} d_{j}, \quad &d_{j}\neq0,\\ 1, \quad &\mbox{$d_{j}=0$}. \end{aligned} \right. \end{eqnarray*}$

In this section, we will give a lower bound for $\tau(A\bigstar B)$. In order to prove our results, we first give some lemmas.

Lemma2.1 [5] Let $A=(a_{ij})\in C^{n\times n}$. Then all the eigenvalues of $A$ lie in the region:

$ \bigcup\limits_{i,j=1\atop i\neq j}^{n}\{z\in C:|z-a_{ii}||z-a_{jj}|\leq \sum\limits_{k\neq i}|a_{ki}|\sum\limits_{l\neq j}|a_{lj}|\}. $

Lemma 2.2 Let $A=(a_{ij})\in C^{n\times n}$ and let $x_{1},x_{2},\cdots,x_{n}$ be positive real numbers. Then all the eigenvalues of $A$ lie in the region:

$ \bigcup\limits_{i,j=1\atop i\neq j}^{n}\{z\in C:|z-a_{ii}||z-a_{jj}|\leq (x_{i}\sum\limits_{k\neq i}\frac{1}{x_{k}}|a_{ki}|)(x_{j}\sum\limits_{l\neq j}\frac{1}{x_{l}}|a_{lj}|)\}. $

Proof Let $x_{1},x_{2},\cdots,x_{n}$ be positive real numbers, and define $X=$diag$(x_{1},x_{2},\cdots,x_{n}), B=(b_{ij})=X^{-1}AX$. Then we have

$ B=(b_{ij})=X^{-1}AX= \left[ \begin{array}{cccc} a_{11}&\frac{x_{2}}{x_{1}}a_{12}&\cdots&\frac{x_{n}}{x_{1}}a_{1n}\\ \frac{x_{1}}{x_{2}}a_{21}&a_{22}&\cdots&\frac{x_{n}}{x_{2}}a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{x_{1}}{x_{n}}a_{n1}&\frac{x_{2}}{x_{n}}a_{n2}&\cdots&a_{nn} \end{array} \right ]. $

Since $B=X^{-1}AX$, we have that $\sigma(A)=\sigma(B)$. By Lemma $2.1,$ there exists a pair$(i,j)$ for positive integers with $i\neq j$ such that

$ \bigcup\limits_{i,j=1\atop i\neq j}^{n}\{z\in C:|z-b_{ii}||z-b_{jj}|\leq \sum\limits_{k\neq i}|b_{ki}|\sum\limits_{l\neq j}|b_{lj}|\}. $

Observe that

$ a_{ii}=b_{ii},\quad a_{jj}=b_{jj},\quad b_{ki}=\frac{x_{i}}{x_{k}}a_{ki},\quad b_{lj}=\frac{x_{j}}{x_{l}}a_{lj}. $

Thus, we have

$ \bigcup\limits_{i,j=1\atop i\neq j}^{n}\{z\in C:|z-a_{ii}||z-a_{jj}|\leq (\sum\limits_{k\neq i}\frac{x_{i}}{x_{k}}|a_{ki}|)(\sum\limits_{l\neq j}\frac{x_{j}}{x_{l}}|a_{lj}|)\}. $

That is

$ \bigcup\limits_{i,j=1\atop i\neq j}^{n}\{z\in C:|z-a_{ii}||z-a_{jj}|\leq (x_{i}\sum\limits_{k\neq i}\frac{1}{x_{k}}|a_{ki}|)(x_{j}\sum\limits_{l\neq j}\frac{1}{x_{l}}|a_{lj}|)\}. $

Theorem 2.1 Let $A,B\in R^{n\times n}$ be two nonsingular $M$-matrices. Then

$\begin{eqnarray} \tau(A\bigstar B)\geq\min\limits_{i\neq j}\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}-[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} +4(m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{|b_{lj}|}{h_{l}})]^{\frac{1}{2}}\}. \end{eqnarray}$

(2.1)

Proof It is evident that (2.1) is an equality for $n=1$.

We next assume that $n\geq 2$.

If $A\bigstar B$ is irreducible, then $A$ and $B$ are irreducible. Let $\lambda$ be an eigenvalue of $A\bigstar B$ and satisfy $\tau(A\bigstar B)=\lambda$, so that $0<\lambda<a_{ii}b_{ii}$, $\forall i\in N$. Thus, by Lemma $2.2$, there is a pair $(i,j)$ of positive integers with $i\neq j$ such that

$\begin{eqnarray*} |\lambda-a_{ii}b_{ii}||\lambda-a_{jj}b_{jj}|&\leq& (m_{i}\sum\limits_{k\neq i}\frac{1}{m_{k}}|a_{ki}b_{ki}|)(m_{j}\sum\limits_{l\neq j}\frac{1}{m_{l}}|a_{lj}b_{lj}|)\\ & \leq& (m_{i}\sum\limits_{k\neq i}\frac{|a_{ki}b_{ki}|}{|a_{ki}|h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{|a_{lj}b_{lj}|}{|a_{lj}|h_{l}})\\ &=& (m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{|b_{lj}|}{h_{l}}). \end{eqnarray*}$

Thus, we have

$\begin{eqnarray*} \lambda&\geq&\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}-[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} +4(m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{|b_{lj}|}{h_{l}})]^{\frac{1}{2}}\}. \end{eqnarray*}$

That is

$\begin{eqnarray*} &&\tau(A\bigstar B)\geq\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}-[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} +4(m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{|b_{lj}|} {h_{l}})]^{\frac{1}{2}}\} \\ &&\geq\min\limits_{i\neq j}\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}-[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} +4(m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{|b_{lj}|}{h_{l}})]^{\frac{1}{2}}\}. \end{eqnarray*}$

Now, assume that $A\bigstar B$ is reducible. It is well known that a matrix in $Z_{n}$ is a nonsingular $M$-matrix if and only if all its leading principal minors are positive(see condition (E17) of Theorem 6.2.3 of [6] ). If we denote by $D=(d_{ij})$ the $n\times n$ permutation matrix with $d_{12}=d_{23}=\cdots=d_{n-1,n}=d_{n1}=1$, the remaining $d_{ij}$ zero, then both $A-tD$ and $B-tD$ are irreducible nonsingular $M$-matrices for any chosen positive real number $t$, sufficiently small such that all the leading principal minors of both $A-tD$ and $B-tD$ are positive. Now we substitute $A-tD$ and $B-tD$ for $A$ and $B$, respectively in the previous case, and then letting $t\longrightarrow 0$, the result follows by continuity.

Theorem 2.2 Let $A,B\in R^{n\times n}$ be two nonsingular $M$-matrices. Then

$\begin{eqnarray*} &&\min\limits_{i\neq j}\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}-[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} +4(m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{|b_{lj}|}{h_{l}})]^{\frac{1}{2}}\}\\ & &\geq \min\limits_{1\leq i\leq n}\{a_{ii}b_{ii}-m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}}\}. \end{eqnarray*}$

Proof Without loss of generality, for $i\neq j$, assume that

$\begin{eqnarray} a_{ii}b_{ii}-m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}}{h_{k}}\leq a_{jj}b_{jj}-m_{j}\sum\limits_{l\neq j}\frac{|b_{lj}|}{h_{l}}. \end{eqnarray}$

(2.2)

Thus, (2.2) is equivalent to

$\begin{eqnarray} m_{j}\sum\limits_{l\neq j}\frac{|b_{lj}|}{h_{l}}\leq m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}}+a_{jj}b_{jj}-a_{ii}b_{ii}. \end{eqnarray}$

(2.3)

From (2.1) and (2.3), we have

$\begin{eqnarray*} & & \frac{1}{2} \{a_{ii}b_{ii}+a_{jj}b_{jj}- [(a_{ii}b_{ii}-a_{jj}b_{jj})^{2}+4(m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{|b_{lj}|} {h_{l}})]^{\frac{1}{2}}\\ &\geq&\frac{1}{2} \{a_{ii}b_{ii}+a_{jj}b_{jj}- [(a_{ii}b_{ii}-a_{jj}b_{jj})^{2}\\ & & +4(m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})(m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}}+a_{jj}b_{jj}-a_{ii} b_{ii})]^{\frac{1}{2}}\}\\ & =&\frac{1}{2} \{a_{ii}b_{ii}+a_{jj}b_{jj}- [(a_{ii}b_{ii}-a_{jj}b_{jj})^{2}\\ & & +4(m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})^{2}+4(m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})(a_{jj} b_{jj}-a_{ii}b_{ii})]^{\frac{1}{2}}\} \\ &=&\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}-[(a_{jj}b_{jj}- a_{ii}b_{ii}+2m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})^{2}]^{\frac{1}{2}}\}\\ &=&\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}-(a_{jj}b_{jj}- a_{ii}b_{ii}+2m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})\}\\ & =&a_{ii}b_{ii}-m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}}. \end{eqnarray*}$

Thus, we have

$\begin{eqnarray*} \tau(A\bigstar B)&\geq&\min\limits_{i\neq j}\frac{1}{2} \{a_{ii}b_{ii}+a_{jj}b_{jj}- [(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} +4(m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{|b_{lj}|}{h_{l}})]^{\frac{1}{2}}\}\\ & \geq& \min\limits_{1\leq i\leq n}\{a_{ii}b_{ii}-m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}}\}. \end{eqnarray*}$

Example 2.1 Let

$ A= \left[ \begin{array}{cccc} 4&-1&-1&-1\\ -2&5&-1&-1\\ 0&-2&4&-1\\ -1&-1&-1&4 \end{array} \right],\quad\quad B=\left[ \begin{array}{cccc} 1&-0.5&0&0\\ -0.5&1&-0.5&0\\ 0&-0.5&1&-0.5\\ 0&0&-0.5&1 \end{array} \right]. $

Then

$ A\bigstar B=\left [ \begin{array}{cccc} 4&-0.5&0&0\\ -1&5&-0.5&0\\ 0&-1&4&-0.5\\ 0&0&-0.5&4 \end{array} \right]. $

By calculating with Matlab 7.0, we have $\tau(A\bigstar B)=3.2296$. By Theorem $3.1$ in [4], we have

$\begin{eqnarray*} \tau(A\bigstar B)\geq\min\limits_{i}\{a_{ii}b_{ii}-m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}}\}=2.4333. \end{eqnarray*}$

By Theorem $2.1$ in this paper, we have

$\begin{eqnarray*} \tau(A\bigstar B)&\geq&\min\limits_{i\neq j}\frac{1}{2} \{a_{ii}b_{ii}+a_{jj}b_{jj}- [(a_{ii}b_{ii}-a_{jj}b_{jj})^{2}\nonumber\\ & &+(m_{i}\sum\limits_{k\neq i}\frac{|b_{ki}|}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{|b_{lj}|}{h_{l}})]^{\frac{1}{2}}\}=2.9779. \end{eqnarray*}$

This numerical example shows that the result in Theorem 2.1 is better than that in Theorem 3.1 in [4].

In this section, we will give an upper bound for $\rho(A\circ B)$.

Theorem 3.1 Let $A,B\in R^{n\times n}$, $A\geq0$ and $B\geq0$. Then

$\begin{eqnarray} \rho(A\circ B)&\leq&\max\limits_{i\neq j}\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}+[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2}\nonumber\\ & &+(m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{b_{lj}}{h_{l}})]^{\frac{1}{2}}\}. \end{eqnarray}$

(3.1)

Proof It is evident that (3.1) is an equality for $n=1$.

We next assume that $n\geq 2$.

If $A\circ B$ is irreducible, then $A$ and $B$ are irreducible. Let $\lambda$ be an eigenvalue of $A\circ B$ and satisfy $\rho(A\circ B)=\lambda$, so that $\rho(A\circ B)\geq a_{ii}b_{ii}$, $\forall i\in N$. Thus, by Lemma $2.2$, there is a pair $(i,j)$ of positive integers with $i\neq j$ such that

$\begin{eqnarray*} |\lambda-a_{ii}b_{ii}||\lambda-a_{jj}b_{jj}|&\leq& (m_{i}\sum\limits_{k\neq i}\frac{1}{m_{k}}|a_{ki}b_{ki}|)(m_{j}\sum\limits_{l\neq j}\frac{1}{m_{l}}|a_{lj}b_{lj}|)\\ & \leq& (m_{i}\sum\limits_{k\neq i}\frac{|a_{ki}b_{ki}|}{a_{ki}h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{|a_{lj}b_{lj}|}{a_{lj}h_{l}})\\ &=& (m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{b_{lj}}{h_{l}}). \end{eqnarray*}$

Thus, we have

$\begin{eqnarray*} \lambda&\leq&\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}+[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2}+4(m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{b_{lj}}{h_{l}})]^{\frac{1}{2}}\}. \end{eqnarray*}$

That is

$\begin{eqnarray*} &&\rho(A\circ B)\leq\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}+[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} +4(m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{b_{lj}}{h_{l}})]^{\frac{1}{2}}\} \\ &\leq& \max\limits_{i\neq j}\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}+[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} +4(m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{b_{lj}}{h_{l}})]^{\frac{1}{2}}\}. \end{eqnarray*}$

Now, assume that $A\circ B$ is reducible. If we denote by $D=(d_{ij})$ the $n\times n$ permutation matrix with $d_{12}=d_{23}=\cdots=d_{n-1,n}=d_{n1}=1$, the remaining $d_{ij}$ zero, then both $A+tD$ and $B+tD$ are nonsingular irreducible matrices for any chosen positive real number $t$. Now we substitute $A+tD$ and $B+tD$ for $A$ and $B$, respectively in the previous case, and then let $t\longrightarrow 0$, the result follows by continuity.

Theorem 3.2 Let $A,B\in R^{n\times n}$, $A\geq0$ and $B\geq0$. Then

$\begin{eqnarray*} &&\max\limits_{i\neq j}\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}+[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} +4(m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{b_{lj}}{h_{l}})]^{\frac{1}{2}}\}\\ &&\leq\max\limits_{1\leq i\leq n}\{a_{ii}b_{ii}+m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}}\}. \end{eqnarray*}$

Proof Without loss of generality, for $i\neq j$, assume that

$\begin{eqnarray} a_{ii}b_{ii}+m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}}\geq a_{jj}b_{jj}+m_{j}\sum\limits_{l\neq j}\frac{b_{lj}}{h_{l}}. \end{eqnarray}$

(3.2)

Thus, (3.2) is equivalent to

$\begin{eqnarray} m_{j}\sum\limits_{l\neq j}\frac{b_{lj}}{h_{l}}\leq m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}}+a_{ii}b_{ii}-a_{jj}b_{jj}. \end{eqnarray}$

(3.3)

From (3.1) and (3.3), we have

$\begin{eqnarray*} & & \frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}+ [(a_{ii}b_{ii}-a_{jj}b_{jj})^{2}+4(m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})(m_{j}\sum\limits_{l\neq j} \frac{b_{lj}}{h_{l}})]^{\frac{1}{2}}\}\\ & \leq&\frac{1}{2} \{a_{ii}b_{ii}+a_{jj}b_{jj}+ [(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} \\ & & +4(m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})(m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}} +a_{ii}b_{ii}-a_{jj}b_{jj})]^{\frac{1}{2}}\}\\ &=&\frac{1}{2} \{a_{ii}b_{ii}+a_{jj}b_{jj}+ [(a_{ii}b_{ii}-a_{jj}b_{jj})^{2}\\ & & +4(m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})^{2}+4(m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}}) (a_{ii}b_{ii}-a_{jj}b_{jj})]^{\frac{1}{2}}\}\\ &=&\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}+[(a_{ii}b_{ii}- a_{jj}b_{jj}+2m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})^{2}]^{\frac{1}{2}}\}\\ &=&\frac{1}{2}\{a_{ii}b_{ii}+a_{jj}b_{jj}+(a_{ii}b_{ii}- a_{jj}b_{jj}+2m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})\} \\ &=& a_{ii}b_{ii}+m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}}. \end{eqnarray*}$

Thus, we have

$\begin{eqnarray*} \rho(A\circ B)&\leq&\max\limits_{i\neq j}\frac{1}{2} \{a_{ii}b_{ii}+a_{jj}b_{jj}+ [(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} +4(m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{b_{lj}}{h_{l}})]^{\frac{1}{2}}\}\\ &\leq& \max\limits_{1\leq i\leq n}\{a_{ii}b_{ii}+m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}}\}. \end{eqnarray*}$

Example 3.1 Let

$ A=\left [ \begin{array}{cccc} 4&1&1&1\\ 2&5&1&1\\ 0&2&4&1\\ 1&1&1&4 \end{array} \right],\quad\quad B=\left[ \begin{array}{cccc} 1&1&0&0\\ 1&3&2&0\\ 0&1&4&3\\ 0&0&1&5 \end{array} \right]. $

Then

$ A\circ B= \left[ \begin{array}{cccc} 4&1&0&0\\ 2&15&2&0\\ 0&2&16&3\\ 0&0&1&20 \end{array} \right]. $

By calculating with Matlab 7.0, we have $\rho(A\circ B)=20.7439$. By Theorem $4.1$ in [4], we have

$ \rho(A\circ B)\leq\max\limits_{i}\{a_{ii}b_{ii}+m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}}\}=23.2. $

By Theorem $3.1$ in this paper, we have

$\begin{eqnarray*} \rho(A\circ B)&\leq&\max\limits_{i\neq j}\frac{1}{2} \{a_{ii}b_{ii}+a_{jj}b_{jj}+ [(a_{ii}b_{ii}-a_{jj}b_{jj})^{2}\nonumber\\ & &+4 (m_{i}\sum\limits_{k\neq i}\frac{b_{ki}}{h_{k}})(m_{j}\sum\limits_{l\neq j}\frac{b_{lj}}{h_{l}})]^{\frac{1}{2}}\}=21.865. \end{eqnarray*}$

This numerical example shows that the result in Theorem 3.1 is better than that in Theorem 4.1 in [4].