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# Some new bounds for the minimum eigenvalue of the Hadamard product of an M-matrix and an inverse M-matrix

Journal of Inequalities and Applications20132013:480

https://doi.org/10.1186/1029-242X-2013-480

• Accepted: 27 September 2013
• Published:

## Abstract

Let A and B be nonsingular M-matrices. Several new bounds on the minimum eigenvalue for the Hadamard product of B and the inverse matrix of A are given. These bounds can improve considerably some previous results.

MSC:15A42, 15B34.

## Keywords

• M-matrix
• minimum eigenvalue

## 1 Introduction

Let ${\mathbb{C}}^{n×n}$ (${\mathbb{R}}^{n×n}$) denote the set of all $n×n$ complex (real) matrices, $A=\left({a}_{ij}\right)\in {\mathbb{C}}^{n×n}$, $N=\left\{1,2,\dots ,n\right\}$. We write $A\ge 0$ if ${a}_{ij}\ge 0$ for any $i,j\in N$. If $A\ge 0$, A is called a nonnegative matrix. The spectral radius of A is denoted by $\rho \left(A\right)$.

We denote by ${Z}_{n}$ the class of all $n×n$ real matrices, whose off-diagonal entries are nonpositive. A matrix $A=\left({a}_{ij}\right)\in {Z}_{n}$ is called a nonsingular M-matrix if there exist a nonnegative matrix B and a nonnegative real number s such that $A=sI-B$ with $s>\rho \left(B\right)$, where I is the identity matrix. ${M}_{n}$ will be used to denote the set of all $n×n$ nonsingular M-matrices. Let us denote $\tau \left(A\right)=min\left\{Re\left(\lambda \right):\lambda \in \sigma \left(A\right)\right\}$, where $\sigma \left(A\right)$ denotes the spectrum of A.

The Hadamard product of two matrices $A=\left({a}_{ij}\right)\in {\mathbb{C}}^{n×n}$ and $B=\left({b}_{ij}\right)\in {\mathbb{C}}^{n×n}$ is the matrix $A\circ B=\left({a}_{ij}{b}_{ij}\right)\in {\mathbb{C}}^{n×n}$. If $A,B\in {M}_{n}$, then $B\circ {A}^{-1}$ is also an M-matrix (see ).

Let $A=\left({a}_{ij}\right)$ be an $n×n$ matrix with all diagonal entries being nonzero throughout. For $i,j,k\in N$, $i\ne j$, denote
$\begin{array}{c}{R}_{i}=\sum _{j\ne i}|{a}_{ij}|,\phantom{\rule{2em}{0ex}}{d}_{i}=\frac{{R}_{i}}{|{a}_{ii}|};\hfill \\ {r}_{ji}=\frac{|{a}_{ji}|}{|{a}_{jj}|-{\sum }_{k\ne j,i}|{a}_{jk}|},\phantom{\rule{2em}{0ex}}{r}_{i}=\underset{j\ne i}{max}\left\{{r}_{ji}\right\};\hfill \\ {m}_{ji}=\frac{|{a}_{ji}|+{\sum }_{k\ne j,i}|{a}_{jk}|{r}_{i}}{|{a}_{jj}|},\phantom{\rule{2em}{0ex}}{m}_{i}=\underset{j\ne i}{max}\left\{{m}_{ij}\right\};\hfill \\ {u}_{ji}=\frac{|{a}_{ji}|+{\sum }_{k\ne j,i}|{a}_{jk}|{m}_{ki}}{|{a}_{jj}|},\phantom{\rule{2em}{0ex}}{u}_{i}=\underset{j\ne i}{max}\left\{{u}_{ij}\right\}.\hfill \end{array}$
In 2013, Zhou et al.  obtained the following result: If $A=\left({a}_{ij}\right)\in {M}_{n}$ is a strictly row diagonally dominant matrix, $B=\left({b}_{ij}\right)\in {M}_{n}$ and ${A}^{-1}=\left({\alpha }_{ij}\right)$, then
$\tau \left(B\circ {A}^{-1}\right)\ge \underset{i\in N}{min}\left\{\frac{{b}_{ii}-{m}_{i}{\sum }_{j\ne i}|{b}_{ji}|}{{a}_{ii}}\right\}.$
(1)
In 2013, Cheng et al.  presented the following result: If $A=\left({a}_{ij}\right)\in {M}_{n}$ and ${A}^{-1}=\left({\alpha }_{ij}\right)$ is a doubly stochastic matrix, then
$\tau \left(A\circ {A}^{-1}\right)\ge \underset{1\le i\le n}{min}\left\{\frac{{a}_{ii}-{u}_{i}{\sum }_{j\ne i}|{a}_{ji}|}{1+{\sum }_{j\ne i}{u}_{ji}}\right\}.$
(2)

In this paper, we present some new lower bounds of $\tau \left(B\circ {A}^{-1}\right)$ and $\tau \left(A\circ {A}^{-1}\right)$, which improve (1) and (2).

## 2 Main results

In this section, we present our main results. Firstly, we give some lemmas.

Lemma 1 

Let $A=\left({a}_{ij}\right)\in {\mathbb{R}}^{n×n}$. If A is a strictly row diagonally dominant matrix, then ${A}^{-1}=\left({\alpha }_{ij}\right)$ satisfies
$|{\alpha }_{ji}|\le {d}_{j}|{\alpha }_{ii}|,\phantom{\rule{1em}{0ex}}j,i\in N,j\ne i.$
Lemma 2 Let $A=\left({a}_{ij}\right)\in {\mathbb{R}}^{n×n}$. If A is a strictly row diagonally dominant M-matrix, then ${A}^{-1}=\left({\alpha }_{ij}\right)$ satisfies
${\alpha }_{ji}\le {w}_{ji}{\alpha }_{ii},\phantom{\rule{1em}{0ex}}j,i\in N,j\ne i,$
where
${w}_{ji}=\frac{|{a}_{ji}|+{\sum }_{k\ne j,i}|{a}_{jk}|{m}_{ki}{h}_{i}}{|{a}_{jj}|},\phantom{\rule{2em}{0ex}}{h}_{i}=\underset{j\ne i}{max}\left\{\frac{|{a}_{ji}|}{|{a}_{jj}|{m}_{ji}-{\sum }_{k\ne j,i}|{a}_{jk}|{m}_{ki}}\right\}.$

Proof This proof is similar to the one of Lemma 2.2 in . □

Lemma 3 If $A=\left({a}_{ij}\right)\in {M}_{n}$ and ${A}^{-1}=\left({\alpha }_{ij}\right)$ is a doubly stochastic matrix, then
${\alpha }_{ii}\ge \frac{1}{1+{\sum }_{j\ne i}{w}_{ji}},\phantom{\rule{1em}{0ex}}i\in N,$

where ${w}_{ji}$ is defined as in Lemma  2.

Proof This proof is similar to the one of Lemma 3.1 in . □

Lemma 4 

If $A=\left({a}_{ij}\right)\in {\mathbb{R}}^{n×n}$ is a strictly row diagonally dominant M-matrix, then, for ${A}^{-1}=\left({\alpha }_{ij}\right)$,
${\alpha }_{ii}\ge \frac{1}{{a}_{ii}},\phantom{\rule{1em}{0ex}}i\in N.$

Lemma 5 

If $A=\left({a}_{ij}\right)\in {\mathbb{C}}^{n×n}$ and ${x}_{1},{x}_{2},\dots ,{x}_{n}$ are positive real numbers, then all the eigenvalues of A lie in the region
$\bigcup _{i\ne j}\left\{z\in \mathbb{C}:|z-{a}_{ii}|\le {x}_{i}\sum _{k\ne i}\frac{1}{{x}_{k}}|{a}_{ki}|,i\in N\right\}.$

Lemma 6 

If $A=\left({a}_{ij}\right)\in {\mathbb{C}}^{n×n}$ and ${x}_{1},{x}_{2},\dots ,{x}_{n}$ are positive real numbers, then all the eigenvalues of A lie in the region
$\bigcup _{i\ne j}\left\{z\in \mathbb{C}:|z-{a}_{ii}||z-{a}_{jj}|\le \left({x}_{i}\sum _{k\ne i}\frac{1}{{x}_{k}}|{a}_{ki}|\right)\left({x}_{j}\sum _{k\ne j}\frac{1}{{x}_{k}}|{a}_{kj}|\right),i,j\in N\right\}.$
Theorem 1 If $A=\left({a}_{ij}\right)$, $B=\left({b}_{ij}\right)\in {M}_{n}$ and ${A}^{-1}=\left({\alpha }_{ij}\right)$, then
$\tau \left(B\circ {A}^{-1}\right)\ge \underset{1\le i\le n}{min}\left\{\frac{{b}_{ii}-{w}_{i}{\sum }_{j\ne i}|{b}_{ji}|}{{a}_{ii}}\right\},$
(3)

where ${w}_{i}={max}_{j\ne i}\left\{{w}_{ij}\right\}$ and ${w}_{ij}$ is defined as in Lemma  2.

Proof It is evident that the result holds with equality for $n=1$.

We next assume that $n\ge 2$.

Since A is an M-matrix, there exists a positive diagonal matrix D such that ${D}^{-1}AD$ is a strictly row diagonally dominant M-matrix, and
$\tau \left(B\circ {A}^{-1}\right)=\tau \left({D}^{-1}\left(B\circ {A}^{-1}\right)D\right)=\tau \left(B\circ {\left({D}^{-1}AD\right)}^{-1}\right).$

Therefore, for convenience and without loss of generality, we assume that A is a strictly row diagonally dominant matrix.

(i) First, we assume that A and B are irreducible matrices. Then, for any $i\in N$, we have $0<{w}_{i}<1$. Since $\tau \left(B\circ {A}^{-1}\right)$ is an eigenvalue of $B\circ {A}^{-1}$, then by Lemma 2 and Lemma 5, there exists an i such that
$\begin{array}{rcl}|\tau \left(B\circ {A}^{-1}\right)-{b}_{ii}{\alpha }_{ii}|& \le & {w}_{i}\sum _{j\ne i}\frac{1}{{w}_{j}}|{b}_{ji}{\alpha }_{ji}|\le {w}_{i}\sum _{j\ne i}\frac{1}{{w}_{j}}|{b}_{ji}|{w}_{ji}|{\alpha }_{ii}|\\ \le & {w}_{i}\sum _{j\ne i}\frac{1}{{w}_{j}}|{b}_{ji}|{w}_{j}|{\alpha }_{ii}|={w}_{i}|{\alpha }_{ii}|\sum _{j\ne i}|{b}_{ji}|.\end{array}$
By Lemma 4, the above inequality and $0\le \tau \left(B\circ {A}^{-1}\right)\le {b}_{ii}{\alpha }_{ii}$, for any $i\in N$, we obtain
$|\tau \left(B\circ {A}^{-1}\right)|\ge {b}_{ii}{\alpha }_{ii}-{w}_{i}|{\alpha }_{ii}|\sum _{j\ne i}|{b}_{ji}|\ge \frac{{b}_{ii}-{w}_{i}{\sum }_{j\ne i}|{b}_{ji}|}{{a}_{ii}}\ge \underset{1\le i\le n}{min}\left\{\frac{{b}_{ii}-{w}_{i}{\sum }_{j\ne i}|{b}_{ji}|}{{a}_{ii}}\right\}.$

(ii) Now, assume that one of A and 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 ). If we denote by $T=\left({t}_{ij}\right)$ the $n×n$ permutation matrix with ${t}_{12}={t}_{23}=\cdots ={t}_{n-1,n}={t}_{n1}=1$, the remaining ${t}_{ij}$ zero, then both $A-ϵT$ and $B-ϵT$ are irreducible nonsingular M-matrices for any chosen positive real number ϵ sufficiently small such that all the leading principal minors of both $A-ϵT$ and $B-ϵT$ are positive. Now, we substitute $A-ϵT$ and $B-ϵT$ for A and B, respectively, in the previous case, and then letting $ϵ\to 0$, the result follows by continuity. □

From Lemma 3 and Theorem 1, we can easily obtain the following corollaries.

Corollary 1 If $A=\left({a}_{ij}\right),B=\left({b}_{ij}\right)\in {M}_{n}$ and ${A}^{-1}=\left({\alpha }_{ij}\right)$ is a doubly stochastic matrix, then
$\tau \left(B\circ {A}^{-1}\right)\ge \underset{1\le i\le n}{min}\left\{\frac{{b}_{ii}-{w}_{i}{\sum }_{j\ne i}|{b}_{ji}|}{1+{\sum }_{j\ne i}{w}_{ji}}\right\}.$
Corollary 2 If $A=\left({a}_{ij}\right)\in {M}_{n}$ and ${A}^{-1}=\left({\alpha }_{ij}\right)$ is a doubly stochastic matrix, then
$\tau \left(A\circ {A}^{-1}\right)\ge \underset{1\le i\le n}{min}\left\{\frac{{a}_{ii}-{w}_{i}{\sum }_{j\ne i}|{a}_{ji}|}{1+{\sum }_{j\ne i}{w}_{ji}}\right\}.$
(4)
Remark 1 We next give a simple comparison between (3) and (1), (4) and (2), respectively. Since ${m}_{ji}{h}_{i}\le {r}_{i}$, $0\le {h}_{i}\le 1$, $j,i\in N$, $j\ne i$, then ${w}_{ji}\le {m}_{ji}$, ${w}_{i}\le {m}_{i}$ and ${w}_{ji}\le {u}_{ji}$, ${w}_{i}\le {u}_{i}$ for any $j,i\in N$, $j\ne i$. Therefore,
$\begin{array}{c}\tau \left(B\circ {A}^{-1}\right)\ge \underset{1\le i\le n}{min}\left\{\frac{{b}_{ii}-{w}_{i}{\sum }_{j\ne i}|{b}_{ji}|}{{a}_{ii}}\right\}\ge \underset{1\le i\le n}{min}\left\{\frac{{b}_{ii}-{m}_{i}{\sum }_{j\ne i}|{b}_{ji}|}{{a}_{ii}}\right\},\hfill \\ \tau \left(A\circ {A}^{-1}\right)\ge \underset{1\le i\le n}{min}\left\{\frac{{a}_{ii}-{w}_{i}{\sum }_{j\ne i}|{a}_{ji}|}{1+{\sum }_{j\ne i}{w}_{ji}}\right\}\ge \underset{1\le i\le n}{min}\left\{\frac{{a}_{ii}-{u}_{i}{\sum }_{j\ne i}|{a}_{ji}|}{1+{\sum }_{j\ne i}{u}_{ji}}\right\}.\hfill \end{array}$

So, the bound in (3) is bigger than the bound in (1) and the bound in (4) is bigger than the bound in (2).

Theorem 2 If $A=\left({a}_{ij}\right),B=\left({b}_{ij}\right)\in {M}_{n}$ and ${A}^{-1}=\left({\alpha }_{ij}\right)$, then
$\begin{array}{rl}\tau \left(B\circ {A}^{-1}\right)\ge & \underset{i\ne j}{min}\frac{1}{2}\left\{{\alpha }_{ii}{b}_{ii}+{\alpha }_{jj}{b}_{jj}-\left[{\left({\alpha }_{ii}{b}_{ii}-{\alpha }_{jj}{b}_{jj}\right)}^{2}\\ {+4\left({w}_{i}\sum _{k\ne i}|{b}_{ki}|{\alpha }_{ii}\right)\left({w}_{j}\sum _{k\ne j}|{b}_{kj}|{\alpha }_{jj}\right)\right]}^{\frac{1}{2}}\right\},\end{array}$

where ${w}_{i}$ ($i\in N$) is defined as in Theorem  1.

Proof It is evident that the result holds with equality for $n=1$.

We next assume that $n\ge 2$. For convenience and without loss of generality, we assume that A is a strictly row diagonally dominant matrix.

(i)First, we assume that A and B are irreducible matrices. Let ${R}_{j}^{\sigma }={\sum }_{k\ne j}|{a}_{jk}|{m}_{ki}{h}_{i}$, $j,i\in N$, $j\ne i$. Then, for any $j,i\in N$, $j\ne i$, we have
${R}_{j}^{\sigma }=\sum _{k\ne j}|{a}_{jk}|{m}_{ki}{h}_{i}\le |{a}_{ji}|+\sum _{k\ne j,i}|{a}_{jk}|{m}_{ki}{h}_{i}\le {R}_{j}<{a}_{jj}.$
Therefore, there exists a real number ${z}_{ji}$ ($0\le {z}_{ji}\le 1$) such that
$|{a}_{ji}|+\sum _{k\ne j,i}|{a}_{jk}|{m}_{ki}{h}_{i}={z}_{ji}{R}_{j}+\left(1-{z}_{ji}\right){R}_{j}^{\sigma },\phantom{\rule{1em}{0ex}}j,i\in N,j\ne i.$
Hence,
${w}_{ji}=\frac{{z}_{ji}{R}_{j}+\left(1-{z}_{ji}\right){R}_{j}^{\sigma }}{{a}_{jj}},\phantom{\rule{1em}{0ex}}j\in N.$
Let ${z}_{j}={max}_{i\ne j}{z}_{ji}$. Obviously, $0<{z}_{j}\le 1$ (if ${z}_{j}=0$, then A is reducible, which is a contradiction). Let
${w}_{j}=\underset{i\ne j}{max}\left\{{w}_{ji}\right\}=\frac{{z}_{j}{R}_{j}+\left(1-{z}_{j}\right){R}_{j}^{\sigma }}{{a}_{jj}},\phantom{\rule{1em}{0ex}}j\in N.$
Since A is irreducible, then ${R}_{j}>0$, ${R}_{j}^{\sigma }\ge 0$, and $0<{w}_{j}<1$. Let $\tau \left(B\circ {A}^{-1}\right)=\lambda$. By Lemma 6, there exist ${i}_{0},{j}_{0}\in N$, ${i}_{0}\ne {j}_{0}$ such that
$|\lambda -{\alpha }_{{i}_{0}{i}_{0}}{b}_{{i}_{0}{i}_{0}}||\lambda -{\alpha }_{{j}_{0}{j}_{0}}{b}_{{j}_{0}{j}_{0}}|\le \left({w}_{{i}_{0}}\sum _{k\ne {i}_{0}}\frac{1}{{w}_{k}}|{\alpha }_{k{i}_{0}}{b}_{k{i}_{0}}|\right)\left({w}_{{j}_{0}}\sum _{k\ne {j}_{0}}\frac{1}{{w}_{k}}|{\alpha }_{k{j}_{0}}{b}_{k{j}_{0}}|\right).$
And by Lemma 2, we have
$\begin{array}{r}\left({w}_{{i}_{0}}\sum _{k\ne {i}_{0}}\frac{1}{{w}_{k}}|{\alpha }_{k{i}_{0}}{b}_{k{i}_{0}}|\right)\left({w}_{{j}_{0}}\sum _{k\ne {j}_{0}}\frac{1}{{w}_{k}}|{\alpha }_{k{j}_{0}}{b}_{k{j}_{0}}|\right)\\ \phantom{\rule{1em}{0ex}}\le \left({w}_{{i}_{0}}\sum _{k\ne {i}_{0}}|{b}_{k{i}_{0}}|{\alpha }_{{i}_{0}{i}_{0}}\right)\left({w}_{{j}_{0}}\sum _{k\ne {j}_{0}}|{b}_{k{j}_{0}}|{\alpha }_{{j}_{0}{j}_{0}}\right).\end{array}$
Therefore,
$|\lambda -{\alpha }_{{i}_{0}{i}_{0}}{b}_{{i}_{0}{i}_{0}}||\lambda -{\alpha }_{{j}_{0}{j}_{0}}{b}_{{j}_{0}{j}_{0}}|\le \left({w}_{{i}_{0}}\sum _{k\ne {i}_{0}}|{b}_{k{i}_{0}}|{\alpha }_{{i}_{0}{i}_{0}}\right)\left({w}_{{j}_{0}}\sum _{k\ne {j}_{0}}|{b}_{k{j}_{0}}|{\alpha }_{{j}_{0}{j}_{0}}\right).$
Furthermore, we obtain
$\begin{array}{rl}\lambda \ge & \frac{1}{2}\left\{{\alpha }_{{i}_{0}{i}_{0}}{b}_{{i}_{0}{i}_{0}}+{\alpha }_{{j}_{0}{j}_{0}}{b}_{{j}_{0}{j}_{0}}-\left[{\left({\alpha }_{{i}_{0}{i}_{0}}{b}_{{i}_{0}{i}_{0}}-{\alpha }_{{j}_{0}{j}_{0}}{b}_{{j}_{0}{j}_{0}}\right)}^{2}\\ {+4\left({w}_{{i}_{0}}\sum _{k\ne {i}_{0}}|{b}_{k{i}_{0}}|{\alpha }_{{i}_{0}{i}_{0}}\right)\left({w}_{{j}_{0}}\sum _{k\ne {j}_{0}}|{b}_{k{j}_{0}}|{\alpha }_{{j}_{0}{j}_{0}}\right)\right]}^{\frac{1}{2}}\right\},\end{array}$
that is,
$\begin{array}{c}\tau \left(B\circ {A}^{-1}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\ge \frac{1}{2}\left\{{\alpha }_{{i}_{0}{i}_{0}}{b}_{{i}_{0}{i}_{0}}+{\alpha }_{{j}_{0}{j}_{0}}{b}_{{j}_{0}{j}_{0}}-\left[{\left({\alpha }_{{i}_{0}{i}_{0}}{b}_{{i}_{0}{i}_{0}}-{\alpha }_{{j}_{0}{j}_{0}}{b}_{{j}_{0}{j}_{0}}\right)}^{2}\hfill \\ {\phantom{\rule{2em}{0ex}}+4\left({w}_{{i}_{0}}\sum _{k\ne {i}_{0}}|{b}_{k{i}_{0}}|{\alpha }_{{i}_{0}{i}_{0}}\right)\left({w}_{{j}_{0}}\sum _{k\ne {j}_{0}}|{b}_{k{j}_{0}}|{\alpha }_{{j}_{0}{j}_{0}}\right)\right]}^{\frac{1}{2}}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\ge \underset{i\ne j}{min}\frac{1}{2}\left\{{\alpha }_{ii}{b}_{ii}+{\alpha }_{jj}{b}_{jj}-{\left[{\left({\alpha }_{ii}{b}_{ii}-{\alpha }_{jj}{b}_{jj}\right)}^{2}+4\left({w}_{i}\sum _{k\ne i}|{b}_{ki}|{\alpha }_{ii}\right)\left({w}_{j}\sum _{k\ne j}|{b}_{kj}|{\alpha }_{jj}\right)\right]}^{\frac{1}{2}}\right\}.\hfill \end{array}$

(ii) Now, assume that one of A and B is reducible. We substitute $A-ϵT$ and $B-ϵT$ for A and B, respectively, in the previous case, and then letting $ϵ\to 0$, the result follows by continuity. □

Corollary 3 If $A=\left({a}_{ij}\right)\in {M}_{n}$ and ${A}^{-1}=\left({\alpha }_{ij}\right)$, then
$\begin{array}{rl}\tau \left(A\circ {A}^{-1}\right)\ge & \underset{i\ne j}{min}\frac{1}{2}\left\{{\alpha }_{ii}{a}_{ii}+{\alpha }_{jj}{a}_{jj}-\left[{\left({\alpha }_{ii}{a}_{ii}-{\alpha }_{jj}{a}_{jj}\right)}^{2}\\ {+4\left({w}_{i}\sum _{k\ne i}|{a}_{ki}|{\alpha }_{ii}\right)\left({w}_{j}\sum _{k\ne j}|{a}_{kj}|{\alpha }_{jj}\right)\right]}^{\frac{1}{2}}\right\}.\end{array}$
Example 1 Let
$\begin{array}{c}A=\left(\begin{array}{cccccccccc}39& -16& -2& -3& -2& -5& -2& -3& -5& 0\\ -26& 44& -2& -4& -2& -1& 0& -2& -3& -3\\ -1& -9& 29& -3& -4& 0& -5& -4& -1& -1\\ -2& -3& -10& 36& -12& 0& -5& -1& -2& 0\\ 0& -3& -1& -9& 44& -16& -3& -4& -4& -3\\ -3& -4& -3& -4& -12& 48& -18& -1& 0& -2\\ -2& -1& -4& -3& -4& -16& 45& -9& -4& -1\\ -1& -2& -2& -2& -3& -1& -5& 38& -20& -1\\ -2& -1& 0& -3& -4& -5& -2& -10& 47& -19\\ -1& -4& -4& -4& 0& -3& -4& -3& -7& 31\end{array}\right),\hfill \\ B=\left(\begin{array}{cccccccccc}90& -3& -2& -7& -4& -7& -6& -3& -9& -3\\ -4& 100& -5& -4& -8& -7& -1& -9& -8& -8\\ -5& -9& 62& -4& -7& -9& -9& -1& -4& -8\\ -8& -8& -10& 99& 0& -6& -8& -9& -3& -6\\ -3& -8& -10& -6& 62& -3& -6& -7& -5& -1\\ -2& -3& -5& -10& -6& 55& -5& -1& -3& -10\\ -8& -5& -8& -8& -3& -3& 52& -6& -1& -4\\ -4& -5& -8& -4& -1& -1& -6& 57& -7& -7\\ -2& -1& -6& -10& -2& -6& -5& -9& 86& -5\\ -5& -7& -3& -9& -5& -7& -9& -5& -9& 72\end{array}\right).\hfill \end{array}$

It is easily proved that A and B are nonsingular M-matrices and A is a doubly stochastic matrix.

(i) If we apply Theorem 4.8 of , we have
$\tau \left(B\circ {A}^{-1}\right)\ge \underset{1\le i\le n}{min}\left\{\frac{{b}_{ii}-{m}_{i}{\sum }_{j\ne i}|{b}_{ji}|}{{a}_{ii}}\right\}=0.0027.$
If we apply Theorem 2.4 of , we have
$\tau \left(B\circ {A}^{-1}\right)\ge \left(1-\rho \left({J}_{A}\right)\rho \left({J}_{B}\right)\right)\underset{1\le i\le n}{min}\frac{{a}_{ii}}{{b}_{ii}}=0.3485.$
But, if we apply Theorem 1, we have
$\tau \left(B\circ {A}^{-1}\right)\ge \underset{1\le i\le n}{min}\left\{\frac{{b}_{ii}-{w}_{i}{\sum }_{j\ne i}|{b}_{ji}|}{{a}_{ii}}\right\}=0.0435.$
If we apply Corollary 1, we have
$\tau \left(B\circ {A}^{-1}\right)\ge \underset{1\le i\le n}{min}\left\{\frac{{b}_{ii}-{w}_{i}{\sum }_{j\ne i}|{b}_{ji}|}{1+{\sum }_{j\ne i}{w}_{ji}}\right\}=0.2172.$
If we apply Theorem 2, we have
$\begin{array}{rl}\tau \left(B\circ {A}^{-1}\right)\ge & \underset{i\ne j}{min}\frac{1}{2}\left\{{\alpha }_{ii}{b}_{ii}+{\alpha }_{jj}{b}_{jj}-\left[{\left({\alpha }_{ii}{b}_{ii}-{\alpha }_{jj}{b}_{jj}\right)}^{2}\\ {+4\left({w}_{i}\sum _{k\ne i}|{b}_{ki}|{\alpha }_{ii}\right)\left({w}_{j}\sum _{k\ne j}|{b}_{kj}|{\alpha }_{jj}\right)\right]}^{\frac{1}{2}}\right\}\\ =& 0.7212.\end{array}$
1. (ii)
If we apply Theorem 3.2 of , we get
$\tau \left(A\circ {A}^{-1}\right)\ge \underset{1\le i\le n}{min}\left\{\frac{{a}_{ii}-{u}_{i}{\sum }_{j\ne i}|{a}_{ji}|}{1+{\sum }_{j\ne i}{u}_{ji}}\right\}=0.3269.$

But, if we apply Corollary 2, we get
$\tau \left(A\circ {A}^{-1}\right)\ge \underset{1\le i\le n}{min}\left\{\frac{{a}_{ii}-{w}_{i}{\sum }_{j\ne i}|{a}_{ji}|}{1+{\sum }_{j\ne i}{w}_{ji}}\right\}=0.3605.$
If we apply Corollary 3, we get
$\begin{array}{rl}\tau \left(A\circ {A}^{-1}\right)\ge & \underset{i\ne j}{min}\frac{1}{2}\left\{{\alpha }_{ii}{a}_{ii}+{\alpha }_{jj}{a}_{jj}-\left[{\left({\alpha }_{ii}{a}_{ii}-{\alpha }_{jj}{a}_{jj}\right)}^{2}\\ {+4\left({w}_{i}\sum _{k\ne i}|{b}_{ki}|{\alpha }_{ii}\right)\left({w}_{j}\sum _{k\ne j}|{b}_{kj}|{\alpha }_{jj}\right)\right]}^{\frac{1}{2}}\right\}\\ =& 0.4072.\end{array}$

## Declarations

### Acknowledgements

The authors are very indebted to the referees for their valuable comments and corrections, which improved the original manuscript of this paper. This work was supported by the National Natural Science Foundation of China (11361074), IRTSTYN and Foundation of Yunnan University (2012CG017).

## Authors’ Affiliations

(1)
School of Mathematics and Statistics, Yunnan University, Kunming, Yunnan, 650091, P.R. China

## References 