New inequalities for the Hadamard product of an M-matrix and its inverse

Abstract

For the Hadamard product $A\circ A^{-1}$ of an M-matrix A and its inverse $A^{-1}$, some new inequalities for the minimum eigenvalue of $A\circ A^{-1}$ are derived. Numerical example is given to show that the inequalities are better than some known results.

1 Introduction

The set of all $n\times n$ real matrices is denoted by $\mathbb{R}^{n\times n}$, and $\mathbb{C}^{n\times n}$ denotes the set of all $n\times n$ complex matrices.

A matrix $A=(a_{ij})\in\mathbb{R}^{n\times n}$ is called an M-matrix [1] if there exists a nonnegative matrix B and a nonnegative real number λ such that

$$\begin{aligned} A=\lambda I-B, \quad\lambda\geq\rho(B), \end{aligned}$$

where I is an identity matrix, $\rho(B)$ is a spectral radius of the matrix B. If $\lambda=\rho(B)$, then A is a singular M-matrix; if $\lambda>\rho(B)$, then A is called a nonsingular M-matrix. Denote by $M_{n}$ the set of all $n\times n$ nonsingular M-matrices. Let us denote

$$\begin{aligned} \tau(A)=\min\bigl\{ \operatorname{Re}(\lambda):\lambda\in\sigma(A)\bigr\} , \end{aligned}$$

and $\sigma(A)$ denotes the spectrum of A. It is known that [2] $\tau(A)=\frac{1}{\rho(A^{-1})}$ is a positive real eigenvalue of $A\in M_{n}$.

The Hadamard product of two matrices $A=(a_{ij})$ and $B=(b_{ij})$ is the matrix $A\circ B=(a_{ij}b_{ij})$. If A and B are M-matrices, then it is proved in [3] that $A\circ B^{-1}$ is also an M-matrix.

A matrix A is irreducible if there does not exist any permutation matrix P such that

$$\begin{aligned} PAP^{T}= \begin{bmatrix} A_{11}&A_{12}\\ 0 &A_{22} \end{bmatrix}, \end{aligned}$$

where $A_{11}$ and $A_{22}$ are square matrices.

For convenience, for any positive integer n, N denotes the set $\{1,2,\ldots,n\}$. Let $A=(a_{ij})\in\mathbb{R}^{n\times n}$ be a strictly diagonally dominant by row, for any $i\in N$, denote

$$\begin{aligned}& R_{i}=\sum _{k\neq i} |a_{ik}|,\qquad C_{i}=\sum _{k\neq i} |a_{ki}|,\qquad d_{i}=\frac{R_{i}}{|a_{ii}|},\qquad c_{i}=\frac{C_{i}}{|a_{ii}|},\quad i\in N; \\& s_{ji}=\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|d_{k}}{|a_{jj}|},\quad j\neq i,j\in N; \qquad s_{i}=\max _{j\neq i} \{s_{ij}\},\quad i\in N; \\& m_{ji}=\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|s_{ki}}{|a_{jj}|},\quad j\neq i,j\in N;\qquad m_{i}=\max _{j\neq i} \{m_{ij}\},\quad i\in N. \end{aligned}$$

Recently, some lower bounds for the minimum eigenvalue of the Hadamard product of an M-matrix and its inverse have been proposed. Let $A\in M_{n}$, it was proved in [4] that

$$\begin{aligned} 0< \tau\bigl(A\circ A^{-1}\bigr)\leq1. \end{aligned}$$

Subsequently, Fiedler and Markham [3] gave a lower bound on $\tau(A\circ A^{-1})$,

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr)\geq\frac{1}{n}, \end{aligned}$$

and conjectured that

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr)\geq\frac{2}{n}. \end{aligned}$$

Chen [5], Song [6] and Yong [7] have independently proved this conjecture.

In [8], Li et al. gave the following result:

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr)\geq\min _{i} \biggl\{ \frac {a_{ii}-s_{i}R_{i}}{1+\sum _{j\neq i}s_{ji}} \biggr\} . \end{aligned}$$

Furthermore, if $a_{11}=a_{22}=\cdots=a_{nn}$, they have obtained

$$\begin{aligned} \min_{i} \biggl\{ \frac{a_{ii}-s_{i}R_{i}}{1+\sum _{j\neq i}s_{ji}} \biggr\} \geq\frac{2}{n}. \end{aligned}$$

In this paper, we present some new lower bounds for $\tau(A\circ A^{-1})$. These bounds improve the results in [8–11].

2 Preliminaries and notations

In this section, we give some lemmas that involve inequalities for the entries of $A^{-1}$. They will be useful in the following proofs.

Lemma 2.1

[7]

If $A=(a_{ij})\in\mathbb{R}^{n\times n}$ is a strictly row diagonally dominant matrix, that is,

$$\begin{aligned} |a_{ii}|>\sum _{j\neq i}|a_{ij}|,\quad i\in N, \end{aligned}$$

then $A^{-1}=(b_{ij})$ exists, and

$$\begin{aligned} |b_{ji}|\leq\frac{\sum _{k\neq j}|a_{jk}|}{|a_{jj}|}|b_{ii}|,\quad j\neq i. \end{aligned}$$

Lemma 2.2

Let $A=(a_{ij})\in\mathbb{R}^{n\times n}$ be a strictly diagonally dominant M-matrix by row. Then, for $A^{-1}=(b_{ij})$, we have

$$\begin{aligned} b_{ji}\leq\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|s_{ki}}{a_{jj}}b_{ii}\leq m_{j}b_{ii},\quad j\neq i, i\in N. \end{aligned}$$

Proof

For $i\in N$, let

$$\begin{aligned} d_{k}(\varepsilon)=\frac{\sum _{l\neq k}|a_{kl}|+\varepsilon}{a_{kk}}, \end{aligned}$$

and

$$\begin{aligned} s_{ji}(\varepsilon)=\frac{|a_{ji}|+(\sum _{k\neq j,i}|a_{jk}|+\varepsilon)d_{k}(\varepsilon)}{|a_{jj}|},\quad j\neq i. \end{aligned}$$

Since A is strictly diagonally dominant, then $0< d_{k}<1$ and $0< s_{ji}<1$. Therefore, there exists $\varepsilon>0$ such that $0< d_{k}(\varepsilon)<1$ and $0< s_{ji}(\varepsilon)<1$. For any $i\in N$, let

$$\begin{aligned} S_{i}(\varepsilon)=\operatorname{diag} \bigl(s_{1i}(\varepsilon),\ldots ,s_{i-1,i}(\varepsilon), 1,s_{i+1,i}(\varepsilon), \ldots,s_{ni}(\varepsilon) \bigr). \end{aligned}$$

Obviously, the matrix $AS_{i}(\varepsilon)$ is also a strictly diagonally dominant M-matrix by row. Therefore, by Lemma 2.1, we derive the following inequality:

$$\begin{aligned} \frac{b_{ji}}{s_{ji}(\varepsilon)}\leq \frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|s_{ki}(\varepsilon)}{s_{ji}(\varepsilon)a_{jj}}b_{ii},\quad j\neq i, j\in N, \end{aligned}$$

i.e.,

$$\begin{aligned} b_{ji}\leq\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|s_{ki}(\varepsilon)}{a_{jj}}b_{ii},\quad j\neq i, j\in N. \end{aligned}$$

Let $\varepsilon\longrightarrow0$ to obtain

$$\begin{aligned} |b_{ji}|\leq\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|s_{ki}}{a_{jj}}b_{ii}\leq m_{j}b_{ii},\quad j\neq i, i\in N. \end{aligned}$$

This proof is completed. □

Lemma 2.3

Let $A=(a_{ij})\in\mathbb{R}^{n\times n}$ be a strictly row diagonally dominant M-matrix. Then, for $A^{-1}=(b_{ij})$, we have

$$\begin{aligned} \frac{1}{a_{ii}-\sum _{j\neq i}|a_{ij}|m_{ji}} \geq b_{ii}\geq\frac{1}{a_{ii}},\quad i\in N. \end{aligned}$$

Proof

Let $B=A^{-1}$. Since A is an M-matrix, then $B\geq0$. By $AB=I$, we have

$$\begin{aligned} 1=\sum _{j=1}^{n}a_{ij}b_{ji}=a_{ii}b_{ii}- \sum _{j\neq i}|a_{ij}|b_{ji},\quad i\in N. \end{aligned}$$

Hence

$$\begin{aligned} a_{ii}b_{ii}\geq1,\quad i\in N, \end{aligned}$$

that is,

$$\begin{aligned} b_{ii}\geq\frac{1}{a_{ii}},\quad i\in N. \end{aligned}$$

By Lemma 2.2, we have

$$\begin{aligned} 1 =&a_{ii}b_{ii}-\sum _{j\neq i}|a_{ij}|b_{ji}\\ \geq& a_{ii}b_{ii}-\sum _{j\neq i}|a_{ij}| \frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|s_{ki}}{a_{jj}}b_{ii}\\ =& \biggl(a_{ii}-\sum _{j\neq i}|a_{ij}|m_{ji} \biggr)b_{ii}, \end{aligned}$$

i.e.,

$$\begin{aligned} \frac{1}{a_{ii}-\sum _{j\neq i}|a_{ij}|m_{ji}}\geq b_{ii},\quad i\in N. \end{aligned}$$

Thus the proof is completed. □

Lemma 2.4

[12]

If $A^{-1}$ is a doubly stochastic matrix, then $Ae=e$, $A^{T}e=e$, where $e=(1,1,\ldots,1)^{T}$.

Lemma 2.5

[13]

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

$$\begin{aligned} \mathop{\bigcup_{i,j=1}}_{i\neq j}^{n} \biggl\{ z\in C:|z-a_{ii}||z-a_{jj}| \leq \biggl({x_{i}}\sum _{k\neq i}\frac{1}{x_{k}}|a_{ki}| \biggr) \biggl({x_{j}}\sum _{k\neq j}\frac{1}{x_{k}}|a_{kj}| \biggr) \biggr\} . \end{aligned}$$

Lemma 2.6

[3]

If P is an irreducible M-matrix, and $Pz\geq kz$ for a nonnegative nonzero vector z, then $\tau(P)\geq k$.

3 Main results

In this section, we give two new lower bounds for $\tau(A\circ A^{-1})$ which improve some previous results.

Theorem 3.1

Let $A=(a_{ij})\in\mathbb{R}^{n\times n}$ be an M-matrix, and suppose that $A^{-1}=(b_{ij})$ is doubly stochastic. Then

$$\begin{aligned} b_{ii}\geq\frac{1}{1+\sum _{j\neq i}m_{ji}},\quad i\in N. \end{aligned}$$

Proof

Since $A^{-1}$ is doubly stochastic and A is an M-matrix, by Lemma 2.4, we have

$$\begin{aligned} a_{ii}=\sum _{k\neq i}|a_{ik}|+1=\sum _{k\neq i}|a_{ki}|+1,\quad i\in N, \end{aligned}$$

and

$$\begin{aligned} b_{ii}+\sum _{j\neq i}b_{ji}=1,\quad i\in N. \end{aligned}$$

The matrix A is strictly diagonally dominant by row. Then, by Lemma 2.2, for $i\in N$, we have

$$\begin{aligned} 1 =&b_{ii}+\sum _{j\neq i}b_{ji}\leq b_{ii}+\sum _{j\neq i}\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|s_{ki}}{a_{jj}}b_{ii} \\ =& \biggl(1+\sum _{j\neq i}\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|s_{ki}}{a_{jj}} \biggr)b_{ii} \\ =& \biggl(1+\sum _{j\neq i}m_{ji} \biggr)b_{ii}, \end{aligned}$$

i.e.,

$$\begin{aligned} b_{ii}\geq\frac{1}{1+\sum _{j\neq i}m_{ji}},\quad i\in N. \end{aligned}$$

This proof is completed. □

Theorem 3.2

Let $A=(a_{ij})\in\mathbb{R}^{n\times n}$ be an M-matrix, and let $A^{-1}=(b_{ij})$ be doubly stochastic. Then

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr) \geq&\min _{i\neq j}\frac{1}{2} \biggl\{ a_{ii}b_{ii}+a_{jj}b_{jj}- \biggl[ (a_{ii}b_{ii}-a_{jj}b_{jj} )^{2} \\ &{}+4 \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr) \biggl(m_{j}\sum _{k\neq j}|a_{kj}|b_{jj} \biggr) \biggr]^{\frac{1}{2}} \biggr\} . \end{aligned}$$

(3.1)

Proof

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

We next assume that $n\geq2$.

Firstly, we assume that $A^{-1}$ is irreducible. By Lemma 2.4, we have

$$\begin{aligned} a_{ii}=\sum _{j\neq i}|a_{ij}|+1=\sum _{j\neq i}|a_{ji}|+1,\quad i\in N, \end{aligned}$$

and

$$\begin{aligned} a_{ii}>1,\quad i\in N. \end{aligned}$$

Let

$$\begin{aligned} m_{j}=\max _{i\neq j} \{m_{ji} \}=\max _{i\neq j} \biggl\{ \frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|s_{ki}}{a_{jj}} \biggr\} ,\quad j\in N. \end{aligned}$$

Since A is an irreducible matrix, then $0< m_{j}\leq1$. Let $\tau(A\circ A^{-1})=\lambda$, so that $0<\lambda<a_{ii}b_{ii}$, $i\in N$. Thus, by Lemma 2.5, there is a pair $(i,j)$ of positive integers with $i\neq j$ such that

$$\begin{aligned} |\lambda-a_{ii}b_{ii}||\lambda-a_{jj}b_{jj}| \leq& \biggl(m_{i}\sum _{k\neq i}\frac{1}{m_{k}}|a_{ki}b_{ki}| \biggr) \biggl(m_{j}\sum _{k\neq j}\frac{1}{m_{k}}|a_{kj}b_{kj}| \biggr) \\ \leq& \biggl(m_{i}\sum _{k\neq i}\frac{1}{m_{k}}|a_{ki}|m_{k}b_{ii} \biggr) \biggl(m_{j} \sum _{k\neq i}\frac{1}{m_{k}}|a_{kj}|m_{k}b_{jj} \biggr) \\ =& \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr) \biggl(m_{j}\sum _{k\neq j}|a_{kj}|b_{jj} \biggr). \end{aligned}$$

(3.2)

From inequality (3.2), we have

$$\begin{aligned} (\lambda-a_{ii}b_{ii}) (\lambda-a_{jj}b_{jj}) \leq \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr) \biggl(m_{j}\sum _{k\neq j}|a_{kj}|b_{jj} \biggr). \end{aligned}$$

(3.3)

Thus, (3.3) is equivalent to

$$\begin{aligned} \lambda \geq&\frac{1}{2} \biggl\{ a_{ii}b_{ii}+a_{jj}b_{jj}- \biggl[ (a_{ii}b_{ii}-a_{jj}b_{jj} )^{2}\\ &{}+4 \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr) \biggl(m_{j}\sum _{k\neq j}|a_{kj}|b_{jj} \biggr) \biggr]^{\frac{1}{2}} \biggr\} , \end{aligned}$$

that is,

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr) \geq&\frac{1}{2} \biggl\{ a_{ii}b_{ii}+a_{jj}b_{jj}- \biggl[ (a_{ii}b_{ii}-a_{jj}b_{jj} )^{2}\\ &{}+4 \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr) \biggl(m_{j}\sum _{k\neq j}|a_{kj}|b_{jj} \biggr) \biggr]^{\frac{1}{2}} \biggr\} \\ \geq&\min _{i\neq j}\frac{1}{2} \biggl\{ a_{ii}b_{ii}+a_{jj}b_{jj}- \biggl[ (a_{ii}b_{ii}-a_{jj}b_{jj} )^{2}\\ &{}+4 \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr) \biggl(m_{j}\sum _{k\neq j}|a_{kj}|b_{jj} \biggr) \biggr]^{\frac{1}{2}} \biggr\} . \end{aligned}$$

If A is reducible, without loss of generality, we may assume that A has the following block upper triangular form:

$$\begin{aligned} A= \begin{bmatrix} A_{11}&A_{12}&\cdots& A_{1s}\\ &A_{22}&\cdots& A_{2s}\\ & &\cdots& \cdots\\ & & & A_{ss} \end{bmatrix} \end{aligned}$$

with irreducible diagonal blocks $A_{ii}$, $i=1,2,\ldots,s$. Obviously, $\tau(A\circ A^{-1})=\min _{i}\tau(A_{ii}\circ A_{ii}^{-1})$. Thus, the problem of the reducible matrix A is reduced to those of irreducible diagonal blocks $A_{ii}$. The result of Theorem 3.2 also holds. □

Theorem 3.3

Let $A=(a_{ij})\in M_{n}$ and $A^{-1}=b_{ij}$ be a doubly stochastic matrix. Then

$$\begin{aligned} &\min _{i\neq j}\frac{1}{2} \biggl\{ a_{ii}b_{ii}+a_{jj}b_{jj}- \biggl[ (a_{ii}b_{ii}-a_{jj}b_{jj} )^{2}\\ &\qquad{}+4 \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr) \biggl(m_{j}\sum _{k\neq j}|a_{kj}|b_{jj} \biggr) \biggr]^{\frac{1}{2}} \biggr\} \\ &\quad\geq\min _{i} \biggl\{ \frac{a_{ii}-s_{i}R_{i}}{1+\sum _{j\neq i}s_{ji}} \biggr\} . \end{aligned}$$

Proof

Since $A^{-1}$ is a doubly stochastic matrix, by Lemma 2.4, we have

$$\begin{aligned} a_{ii}=\sum _{k\neq i}|a_{ik}|+1=\sum _{k\neq i}|a_{ki}|+1,\quad i\in N. \end{aligned}$$

For any $j\neq i$, we have

$$\begin{aligned} d_{j}-s_{ji} =&\frac{R_{j}}{a_{jj}}-\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|d_{k}}{a_{jj}} \\ =&\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|}{a_{jj}}-\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|d_{k}}{a_{jj}} \\ =&\frac{(1-d_{k})\sum _{k\neq j,i}|a_{jk}|}{a_{jj}}\geq0, \end{aligned}$$

or equivalently

$$\begin{aligned} d_{j}\geq s_{ji},\quad j\neq i, j\in N. \end{aligned}$$

(3.4)

So, we can obtain

$$\begin{aligned} m_{ji}=\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|s_{ki}}{a_{jj}}\leq\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|d_{k}}{a_{jj}}=s_{ji},\quad j \neq i, j\in N, \end{aligned}$$

(3.5)

and

$$\begin{aligned} m_{i}\leq s_{i},\quad i\in N. \end{aligned}$$

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

$$\begin{aligned} a_{ii}b_{ii}-m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \leq a_{jj}b_{jj}-m_{j}\sum _{k\neq j}|a_{kj}|b_{jj}. \end{aligned}$$

(3.6)

Thus, (3.6) is equivalent to

$$\begin{aligned} m_{j}\sum _{k\neq j}|a_{kj}|b_{jj} \leq a_{jj}b_{jj}-a_{ii}b_{ii}+m_{i} \sum _{k\neq i}|a_{ki}|b_{ii}. \end{aligned}$$

(3.7)

From (3.1) and (3.7), we have

$$\begin{aligned} & \frac{1}{2} \biggl\{ a_{ii}b_{ii}+a_{jj}b_{jj}- \biggl[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} +4 \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr) \biggl(m_{j}\sum _{k\neq j} |a_{kj}|b_{jj} \biggr) \biggr]^{\frac{1}{2}} \biggr\} \\ &\quad \geq\frac{1}{2} \biggl\{ a_{ii}b_{ii}+a_{jj}b_{jj}- \biggl[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} \\ &\qquad{} +4 \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr) \biggl(a_{jj}b_{jj}-a_{ii}b_{ii}+m_{i} \sum _{k\neq i} |a_{ki}|b_{ii} \biggr) \biggr]^{\frac{1}{2}} \biggr\} \\ &\quad = \frac{1}{2} \biggl\{ a_{ii}b_{ii}+a_{jj}b_{jj}- \biggl[(a_{ii}b_{ii}-a_{jj}b_{jj})^{2} \\ &\qquad{} +4 \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr)^{2}+4 \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr) (a_{jj}b_{jj} -a_{ii}b_{ii} ) \biggr]^{\frac{1}{2}} \biggr\} \\ &\quad=\frac{1}{2} \biggl\{ a_{ii}b_{ii}+a_{jj}b_{jj}- \biggl[ \biggl(a_{jj}b_{jj}- a_{ii}b_{ii}+2m_{i} \sum _{k\neq i}|a_{ki}|b_{ii} \biggr)^{2} \biggr]^{\frac{1}{2}} \biggr\} \\ &\quad=\frac{1}{2} \biggl\{ a_{ii}b_{ii}+a_{jj}b_{jj}- \biggl(a_{jj}b_{jj}- a_{ii}b_{ii}+2m_{i} \sum _{k\neq i}|a_{ki}|b_{ii} \biggr) \biggr\} \\ &\quad= a_{ii}b_{ii}-m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \\ &\quad= b_{ii} \biggl(a_{ii}-m_{i}\sum _{k\neq i}|a_{ki}| \biggr)\\ &\quad\geq\frac{a_{ii}-m_{i}R_{i}}{1+\sum _{j\neq i}m_{ji}}\\ &\quad\geq\frac{a_{ii}-s_{i}R_{i}}{1+\sum _{j\neq i}s_{ji}}. \end{aligned}$$

Thus we have

$$\begin{aligned} &\min _{i\neq j}\frac{1}{2} \biggl\{ a_{ii}b_{ii}+a_{jj}b_{jj}- \biggl[ (a_{ii}b_{ii}-a_{jj}b_{jj} )^{2}+4 \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr) \biggl(m_{j}\sum _{k\neq j}|a_{kj}|b_{jj} \biggr) \biggr]^{\frac{1}{2}} \biggr\} \\ &\quad\geq\min _{i} \biggl\{ \frac{a_{ii}-s_{i}R_{i}}{1+\sum _{j\neq i}s_{ji}} \biggr\} . \end{aligned}$$

This proof is completed. □

Remark 3.1

According to inequality (3.4), it is easy to know that

$$\begin{aligned} b_{ji}\leq\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|s_{ki}}{a_{jj}}b_{ii}\leq\frac{|a_{ji}|+\sum _{k\neq j,i}|a_{jk}|d_{k}}{a_{jj}}b_{ii},\quad j\in N. \end{aligned}$$

That is to say, the result of Lemma 2.2 is sharper than that of Theorem 2.1 in [8]. Moreover, the result of Theorem 3.2 is sharper than that of Theorem 3.1 in [8], respectively.

Theorem 3.4

Let $A=(a_{ij})\in\mathbb{R}^{n\times n}$ be an irreducible strictly row diagonally dominant M-matrix. Then

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr)\geq\min _{i} \biggl\{ 1-\frac{1}{a_{ii}} \sum _{j\neq i}|a_{ji}|m_{ji} \biggr\} . \end{aligned}$$

Proof

Since A is irreducible, then $A^{-1}>0$, and $A\circ A^{-1}$ is again irreducible. Note that

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr)=\tau\bigl(\bigl(A\circ A^{-1} \bigr)^{T}\bigr)=\tau\bigl(A^{T}\circ \bigl(A^{T} \bigr)^{-1}\bigr). \end{aligned}$$

Let

$$\begin{aligned} \bigl(A^{T}\circ\bigl(A^{T}\bigr)^{-1} \bigr)e=(t_{1},t_{2},\ldots,t_{n})^{T}, \end{aligned}$$

where $e=(1,1,\ldots,1)^{T}$. Without loss of generality, we may assume that $t_{1}=\min _{i} \{t_{i} \}$, by Lemma 2.2, we have

$$\begin{aligned} t_{1} =&\sum _{j=1}^{n}|a_{j1}b_{j1}|=a_{11}b_{11}- \sum _{j\neq 1}|a_{j1}|b_{j1} \\ \geq&a_{11}b_{11}-\sum _{j\neq 1}|a_{j1}| \frac{|a_{j1}|+\sum _{k\neq j,1}|a_{jk}|s_{k1}}{a_{jj}}b_{11} \\ =&a_{11}b_{11}-\sum _{j\neq1}|a_{j1}|m_{j1}b_{11} \\ =&\biggl(a_{11}-\sum _{j\neq1}|a_{j1}|m_{j1} \biggr)b_{11} \\ \geq&\frac{a_{11}-\sum _{j\neq1}|a_{j1}|m_{j1}}{a_{11}} \\ =&1-\frac{1}{a_{11}}\sum _{j\neq1}|a_{j1}|m_{j1}. \end{aligned}$$

Therefore, by Lemma 2.6, we have

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr)\geq\min _{i} \biggl\{ 1-\frac{1}{a_{ii}} \sum _{j\neq i}|a_{ji}|m_{ji} \biggr\} . \end{aligned}$$

This proof is completed. □

Remark 3.2

According to inequality (3.5), we can get

$$\begin{aligned} 1-\frac{1}{a_{ii}}\sum _{j\neq i}|a_{ji}|m_{ji} \geq1-\frac{1}{a_{ii}}\sum _{j\neq i}|a_{ji}|s_{ji}. \end{aligned}$$

That is to say, the bound of Theorem 3.4 is sharper than the bound of Theorem 3.5 in [8].

Remark 3.3

If A is an M-matrix, we know that there exists a diagonal matrix D with positive diagonal entries such that $D^{-1}AD$ is a strictly row diagonally dominant M-matrix. So the result of Theorem 3.4 also holds for a general M-matrix.

4 Example

Consider the following M-matrix:

$$A= \begin{bmatrix} 4&-1&-1&-1\\ -2&5&-1&-1\\ 0&-2&4&-1\\ -1&-1&-1&4 \end{bmatrix}. $$

Since $Ae=e$ and $A^{T}e=e$, $A^{-1}$ is doubly stochastic. By calculations we have

$$A^{-1}= \begin{bmatrix} 0.4000&0.2000&0.2000&0.2000\\ 0.2333&0.3667&0.2000&0.2000\\ 0.1667&0.2333&0.4000&0.2000\\ 0.2000&0.2000&0.2000&0.4000 \end{bmatrix}. $$

(1) Estimate the upper bounds for entries of $A^{-1}=(b_{ij})$ . If we apply Theorem 2.1(a) of [8], we have

$$A^{-1}\leq \begin{bmatrix} 1&0.6250&0.6375&0.6375\\ 0.7000&1&0.6500&0.6500\\ 0.5875&0.6875&1&0.6500\\ 0.6375&0.6250&0.6375&1 \end{bmatrix} \circ \begin{bmatrix} b_{11}&b_{22}&b_{33}&b_{44}\\ b_{11}&b_{22}&b_{33}&b_{44}\\ b_{11}&b_{22}&b_{33}&b_{44}\\ b_{11}&b_{22}&b_{33}&b_{44} \end{bmatrix}. $$

If we apply Lemma 2.2, we have

$$A^{-1}\leq \begin{bmatrix} 1&0.5781&0.5718&0.5750\\ 0.6450&1&0.5825&0.5850\\ 0.5093&0.6562&1&0.5750\\ 0.5718&0.5781&0.5718&1 \end{bmatrix} \circ \begin{bmatrix} b_{11}&b_{22}&b_{33}&b_{44}\\ b_{11}&b_{22}&b_{33}&b_{44}\\ b_{11}&b_{22}&b_{33}&b_{44}\\ b_{11}&b_{22}&b_{33}&b_{44} \end{bmatrix}. $$

Combining the result of Lemma 2.2 with the result of Theorem 2.1(a) of [8], we see that the result of Lemma 2.2 is the best.

By Theorem 2.3 and Lemma 3.2 of [8], we can get the following bounds for the diagonal entries of $A^{-1}$:

$$\begin{aligned}[b] &0.3419\leq b_{11}\leq0.5882;\qquad 0.3404\leq b_{22}\leq0.5128,\\ &0.3419\leq b_{33}\leq0.6061; \qquad 0.3404\leq b_{44}\leq0.5882. \end{aligned} $$

By Lemma 2.3 and Theorem 3.1, we obtain

$$\begin{aligned}& 0.3668\leq b_{11}\leq0.4397;\qquad 0.3556\leq b_{22}\leq0.3832, \\& 0.3668\leq b_{33}\leq0.4419;\qquad 0.3656\leq b_{44}\leq0.4415. \end{aligned}$$

(2) Lower bounds for $\tau(A\circ A^{-1})$.

By the conjecture of Fiedler and Markham, we have

$$\begin{aligned} \tau \bigl(A\circ A^{-1} \bigr)\geq\frac{2}{n}= \frac{1}{2}=0.5 . \end{aligned}$$

By Theorem 3.1 of [8], we have

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr)\geq\min _{i} \biggl\{ \frac{a_{ii}-s_{i}R_{i}}{1+\sum _{j\neq i}s_{ji}} \biggr\} =0.6624. \end{aligned}$$

By Corollary 2.5 of [9], we have

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr)\geq1-\rho^{2}(J_{A})=0.4145. \end{aligned}$$

By Theorem 3.1 of [10], we have

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr)\geq\min _{i} \biggl\{ \frac{a_{ii}-u_{i}R_{i}}{1+\sum _{j\neq i}u_{ji}} \biggr\} =0.8250. \end{aligned}$$

By Corollary 2 of [11], we have

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr)\geq\min _{i} \biggl\{ \frac{a_{ii}-w_{i}\sum _{j\neq i}\mid a_{ji}\mid}{1+\sum _{j\neq i}w_{ji}} \biggr\} =0.8321. \end{aligned}$$

If we apply Theorem 3.2, we have

$$\begin{aligned} \tau\bigl(A\circ A^{-1}\bigr) \geq&\min _{i\neq j}\frac{1}{2} \biggl\{ a_{ii}b_{ii}+a_{jj}b_{jj}- \biggl[ (a_{ii}b_{ii}-a_{jj}b_{jj} )^{2}\\ &{}+4 \biggl(m_{i}\sum _{k\neq i}|a_{ki}|b_{ii} \biggr) \biggl(m_{j}\sum _{k\neq j}|a_{kj}|b_{jj} \biggr) \biggr]^{\frac{1}{2}} \biggr\} =0.8456. \end{aligned}$$

The numerical example shows that the bound of Theorem 3.2 is better than these corresponding bounds in [8–11].