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

Journal of Inequalities and Applications20152015:35

https://doi.org/10.1186/s13660-015-0555-1

  • Received: 17 September 2014
  • Accepted: 13 December 2014
  • Published:

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.

Keywords

  • M-matrix
  • Hadamard product
  • inequality
  • eigenvalue

MSC

  • 15A06
  • 15A18
  • 15A48

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 [811].

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 [811].

Declarations

Acknowledgements

The author is grateful to the referees for their useful and constructive suggestions. This research is supported by the Scientific Research Fund of Yunnan Provincial Education Department (2013C165).

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
Department of Engineering, Oxbridge College, Kunming University of Science and Technology, Kunming, Yunnan, 650106, P.R. China

References

  1. Berman, A, Plemmons, RJ: Nonnegative Matrices in the Mathematical Sciences. SIAM, Philadelphia (1979) MATHGoogle Scholar
  2. Horn, RA, Johnson, CR: Topics in Matrix Analysis. Cambridge University Press, Cambridge (1991) View ArticleMATHGoogle Scholar
  3. Fiedler, M, Markham, TL: An inequality for the Hadamard product of an M-matrix and inverse M-matrix. Linear Algebra Appl. 101, 1-8 (1988) View ArticleMATHMathSciNetGoogle Scholar
  4. Fiedler, M, Johnson, CR, Markham, TL, Neumann, M: A trace inequality for M-matrix and the symmetrizability of a real matrix by a positive diagonal matrix. Linear Algebra Appl. 71, 81-94 (1985) View ArticleMATHMathSciNetGoogle Scholar
  5. Chen, SC: A lower bound for the minimum eigenvalue of the Hadamard product of matrices. Linear Algebra Appl. 378, 159-166 (2004) View ArticleMATHMathSciNetGoogle Scholar
  6. Song, YZ: On an inequality for the Hadamard product of an M-matrix and its inverse. Linear Algebra Appl. 305, 99-105 (2000) View ArticleMATHMathSciNetGoogle Scholar
  7. Yong, XR: Proof of a conjecture of Fiedler and Markham. Linear Algebra Appl. 320, 167-171 (2000) View ArticleMATHMathSciNetGoogle Scholar
  8. Li, HB, Huang, TZ, Shen, SQ, Li, H: Lower bounds for the minimum eigenvalue of Hadamard product of an M-matrix and its inverse. Linear Algebra Appl. 420, 235-247 (2007) View ArticleMATHMathSciNetGoogle Scholar
  9. Zhou, DM, Chen, GL, Wu, GX, Zhang, XY: Some inequalities for the Hadamard product of an M-matrix and an inverse M-matrix. J. Inequal. Appl. 2013, 16 (2013) View ArticleMathSciNetGoogle Scholar
  10. Cheng, GH, Tan, Q, Wang, ZD: Some inequalities for the minimum eigenvalue of the Hadamard product of an M-matrix and its inverse. J. Inequal. Appl. 2013, 65 (2013) View ArticleMathSciNetGoogle Scholar
  11. Li, YT, Wang, F, Li, CQ, Zhao, JX: Some new bounds for the minimum eigenvalue of the Hadamard product of an M-matrix and an inverse M-matrix. J. Inequal. Appl. 2013, 480 (2013) View ArticleGoogle Scholar
  12. Yong, XR, Wang, Z: On a conjecture of Fiedler and Markham. Linear Algebra Appl. 288, 259-267 (1999) View ArticleMATHMathSciNetGoogle Scholar
  13. Horn, RA, Johnson, CR: Matrix Analysis. Cambridge University Press, Cambridge (1985) View ArticleMATHGoogle Scholar

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