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

Abstract

Let A and B be nonsingular M-matrices. Some new lower bounds on the minimum eigenvalue $q(A\circ B^{-1})$ for the Hadamard product of A and $B^{-1}$ are given. These bounds improve the corresponding results of Chen (Linear Algebra Appl. 378:159-166, 2004) and Huang (Linear Algebra Appl. 428:1551-1559, 2008) and generalize the corresponding result of Xiang (Linear Algebra Appl. 367:17-27, 2003).

MSC:15A06, 15A15, 15A48.

1 Introduction

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

Let $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$ and $B=(b_{ij})\in {\mathbb{R}}^{n\times n}$. We write $A\ge B$ (>B) if $a_{ij}\ge b_{ij}$ ($>b_{ij}$) for all $1\le i\le n$, $1\le j\le n$. If $A\ge 0$ (>0), we say that A is a nonnegative (positive) matrix. The spectral radius of A is denoted by $\rho (A)$. Let A be an irreducible nonnegative matrix. It is well known that there exists a positive vector u such that $Au=\rho (A)u$, u being called a right Perron eigenvector of A. This guarantees that $\rho (A)\in \sigma (A)$, where $\sigma (A)$ denotes the spectrum of A.

The set $Z_n\subset {\mathbb{R}}^{n\times n}$ is defined by

$$Z_n\equiv \{A=(a_{ij})\in {\mathbb{R}}^{n\times n}:a_{ij}\le 0\text{ if }i\ne j,i,j=1,\dots ,n\}.$$

The simple sign patten of the matrices in $Z_n$ has many striking consequences. Let $A=(a_{ij})\in Z_n$ and suppose $A=\alpha I-P$ with $\alpha \in \mathbb{R}$ and $P\ge 0$. Then $\alpha -\rho (P)$ is an eigenvalue of A, every eigenvalue of A lies in the disc $\{z\in \mathbb{C}:|z-\alpha |\le \rho (P)\}$, and hence every eigenvalue λ of A satisfies $\text{Re}\lambda \ge \alpha -\rho (P)$. In particular, a matrix $A\in Z_n$ is called an M-matrix if $\alpha \ge \rho (P)$. If $\alpha >\rho (P)$, we call A nonsingular M-matrix, and denote the class of nonsingular M-matrices by $M_n$.

Let $A=(a_{ij})\in Z_n$, we denote $min\{\text{Re}(\lambda ):\lambda \in \sigma (A)\}$ by $q(A)$. The following simple facts are needed for our purpose in proving (see Problems 16 and 19 in Section 2.5 of [1]):

• (i) $q(A)\in \sigma (A)$; $q(A)$ is called a minimum eigenvalue of A.

• (ii) If $A\in M_n$ and $\rho (A^{-1})$ is the Perron eigenvalue of the nonnegative matrix $A^{-1}$, then $q(A)=\frac{1}{\rho (A^{-1})}$ is a positive real eigenvalue of A.

Let A be an irreducible nonsingular M-matrix. It is well known that there exists a positive vector u such that $Au=q(A)u$, u being called a right Perron eigenvector of A.

If $A=(a_{ij})\in M_n$, we write $C_A=D_A-A$, where $D_A=diag(a_{ii})$. Note that $a_{ii}>0$ for all $i\in N$ if $A\in M_n$. Thus we define the Jacobi iterative matrix of A by $J_A=D_A^{-1}{C}_A$. It is easy to check that $J_A$ is nonnegative and $\rho (J_A)<1$ (see [2]).

The Hadamard product of $A=(a_{ij})\in {\mathbb{C}}^{n\times n}$ and $B=(b_{ij})\in {\mathbb{C}}^{n\times n}$ is defined by $A\circ B\equiv (a_{ij}{b}_{ij})\in {\mathbb{C}}^{n\times n}$.

It has been noted [3, 4] that the Hadamard product $B\circ A^{-1}$ of an M-matrix B and the inverse of an M-matrix A is again an M-matrix.

In 1991, Horn et al. [[1], p. 375] showed the classical result: if $A=(a_{ij})\in M_n$, $B=(b_{ij})\in M_n$, $B^{-1}=({\beta }_{ij})$, then

$$q(A\circ B^{-1})\ge q(A)\underset{1\le i\le n}{min}{\beta }_{ii}.$$

(1.1)

Subsequently, Chen [5] improved the bound in (1.1) and obtained the following result:

$$q(A\circ B^{-1})\ge q(A)q(B)\underset{1\le i\le n}{min}\left\{(\frac{a_{ii}}{q(A)}+\frac{b_{ii}}{q(B)}-1)\frac{{\beta }_{ii}}{b_{ii}}\right\}.$$

(1.2)

In 2008, Huang [2] obtained the following result:

$$q(A\circ B^{-1})\ge \frac{1-\rho (J_A)\rho (J_B)}{1+{\rho }^2(J_B)}\underset{1\le i\le n}{min}\frac{a_{ii}}{b_{ii}}.$$

(1.3)

This bound in (1.3) improved the bound in (1.1) in some cases. For example, if

$$B=\left(\begin{array}{cc}4& 0\\ 0& 3\end{array}\right),A=\left(\begin{array}{cc}3& -1\\ 0& 2\end{array}\right),$$

then $q(A\circ B^{-1})=\frac{1-\rho (J_A)\rho (J_B)}{1+{\rho }^2(J_B)}{min}_{1\le i\le n}\frac{a_{ii}}{b_{ii}}=\frac{2}{3}\ge q(A){min}_{1\le i\le n}{\beta }_{ii}=\frac{1}{2}$. But $\frac{1-\rho (J_A)\rho (J_B)}{1+{\rho }^2(J_B)}{min}_{1\le i\le n}\frac{a_{ii}}{b_{ii}}\le q(A){min}_{1\le i\le n}{\beta }_{ii}$ in Example 2.1 in this paper.

In practice, the bound of $q(A\circ B^{-1})$ can give a rough estimate before actually solving it and can serve as a check of whether the solution technique for it actually resulted in valid solution. Besides, a good bound of $q(A\circ B^{-1})$ can also help us reduce the computational burden. Therefore, it is necessary to study the bound. In this paper, we present some new lower bounds of the minimum eigenvalue $q(A\circ B^{-1})$ for the Hadamard product of M-matrices, which improve (1.1), (1.2) and (1.3) and generalize the corresponding result of Xiang [6].

2 Main results

In this section, we state and prove our main results. Firstly, we give some lemmas.

Lemma 2.1 (See [[7], Theorem 11])

Let $A=(a_{ij})\in {\mathbb{C}}^{n\times n}$, with $n\ge 2$. Then, if λ is an eigenvalue of A, there is a pair $(r,q)$ of positive integers with $r\ne q$ ($1\le r,q\le n$) such that

$$|\lambda -a_{rr}|\cdot |\lambda -a_{qq}|\le \sum _{k\ne r}|a_{rk}|\cdot \sum _{l\ne q}|a_{ql}|.$$

Lemma 2.2 (See [[8], Lemma 2.2])

• (a) If $A=(a_{ij})$ is an $n\times n$ strictly diagonally dominant matrix by row, that is, $|a_{ii}|>{\sum }_{j\ne i}|a_{ij}|$ for any $i\in N$, then $A^{-1}=({\beta }_{ij})$ exists, and

  $$|{\beta }_{ji}|\le \frac{{\sum }_{k\ne j}|a_{jk}|}{|a_{jj}|}|{\beta }_{ii}|,\mathit{\text{for all }}j\ne i.$$

(b) If $A=(a_{ij})$ is an $n\times n$ strictly diagonally dominant matrix by column, that is, $|a_{ii}|>{\sum }_{j\ne i}|a_{ji}|$ for any $i\in N$, then $A^{-1}=({\beta }_{ij})$ exists, and

$$|{\beta }_{ij}|\le \frac{{\sum }_{k\ne j}|a_{kj}|}{|a_{jj}|}|{\beta }_{ii}|,\mathit{\text{for all }}j\ne i.$$

Proof We give a simple proof of (a) which is different from that in [8]. Similarly, one can prove (b). Firstly, we prove $|{\beta }_{ji}|\le |{\beta }_{ii}|$ for all $j\ne i$. Suppose not. Let $|{\beta }_{ji}|\ge |{\beta }_{ii}|$ for some j and $j\ne i$. We can then assume $|{\beta }_{ji}|\ge |{\beta }_{ki}|$ for all $k\in N$. Since $A{A}^{-1}=I$, we have ${\sum }_{k=1}^n{a}_{jk}{\beta }_{ki}=0$. Thus

$$|a_{jj}{\beta }_{ji}|\le \sum _{k\ne j}|a_{jk}{\beta }_{ki}|\le \sum _{k\ne j}|a_{jk}||{\beta }_{ji}|<|a_{jj}||{\beta }_{ji}|,$$

which is a contradiction. Hence, $|{\beta }_{ji}|\le |{\beta }_{ii}|$ holds for all pairs i, j. Thus

$$|a_{jj}{\beta }_{ji}|\le \sum _{k\ne j}|a_{jk}{\beta }_{ki}|\le \sum _{k\ne j}|a_{jk}||{\beta }_{ii}|,\text{for all }j\ne i,$$

that is,

$$|{\beta }_{ji}|\le \frac{{\sum }_{k\ne j}|a_{jk}|}{|a_{jj}|}|{\beta }_{ii}|,\text{for all }j\ne i.$$

 □

Theorem 2.3 Let $A=(a_{ij})\in M_n$, $B=(b_{ij})\in M_n$ and $B^{-1}=({\beta }_{ij})$. Then

$$\begin{array}{rcl}q(A\circ B^{-1})& \ge & \underset{i\ne j}{min}\frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-[{(a_{ii}{\beta }_{ii}-a_{jj}{\beta }_{jj})}^2\\ +{4\frac{{\beta }_{ii}{\beta }_{jj}}{b_{ii}{b}_{jj}}[b_{ii}-q(B)][a_{ii}-q(A)][b_{jj}-q(B)][a_{jj}-q(A)]]}^{\frac{1}{2}}\}.\end{array}$$

(2.1)

Proof If both A and B are irreducible. Let $v=(v_i)$ and $y=(y_i)$ be the right Perron eigenvectors of $B^T$ and A, respectively, i.e., $B^{T}v=q(B^T)v=q(B)v$, $Ay=q(A)y$. Define $C=B^{T}V$, where $V=diag(v_1,v_2,\dots ,v_n)$. It is easy to check that C is diagonally dominant by row. It follows from Lemma 2.2, for all $i\ne j$, we have

$$\frac{{\beta }_{ij}}{v_j}\le \frac{{\sum }_{k\ne j}|v_k{b}_{kj}|}{v_j{b}_{jj}}\frac{{\beta }_{ii}}{v_i}=\frac{(b_{jj}-q(B))v_j}{b_{jj}{v}_j}\frac{{\beta }_{ii}}{v_i}.$$

Thus

$${\beta }_{ij}\le \frac{(b_{jj}-q(B))v_j{\beta }_{ii}}{b_{jj}{v}_i}.$$

Let $s_j=\frac{(b_{jj}-q(B))v_j}{b_{jj}{y}_j}$, $S=diag(s_1,s_2,\dots ,s_n)$. Then $S>0$ and

$$S(A\circ B^{-1})S^{-1}=\left(\begin{array}{cccc}{a}_{11}{\beta }_{11}& s_1{a}_{12}{\beta }_{12}/s_2& \cdots & s_1{a}_{1n}{\beta }_{1n}/s_n\\ s_2{a}_{21}{\beta }_{21}/s_1& a_{22}{\beta }_{22}& \cdots & s_2{a}_{2n}{\beta }_{2n}/s_n\\ ⋮& ⋮& \ddots & ⋮\\ s_n{a}_{n1}{\beta }_{n1}/s_1& s_n{a}_{n2}{\beta }_{n2}/s_2& \cdots & a_{nn}{\beta }_{nn}\end{array}\right).$$

Hence, $\sigma (A\circ B^{-1})=\sigma (S(A\circ B^{-1})S^{-1})$. Since $q(A\circ B^{-1})$ is an eigenvalue of $A\circ B^{-1}$, we have

$$q(A\circ B^{-1})\in \sigma \left(S(A\circ B^{-1})S^{-1}\right).$$

Thus, by Lemma 2.1, there exists a pair $(i,j)$ of positive integers with $i\ne j$ ($1\le i,j\le n$) such that

$$\begin{array}{c}|q(A\circ B^{-1})-a_{ii}{\beta }_{ii}||q(A\circ B^{-1})-a_{jj}{\beta }_{jj}|\hfill \\ \le \sum _{k\ne i}\frac{|a_{ik}{\beta }_{ik}|}{s_k}{s}_i\sum _{l\ne j}\frac{|a_{jl}{\beta }_{jl}|}{s_l}{s}_j\hfill \\ \le s_i\sum _{k\ne i}\frac{|a_{ik}|(b_{kk}-q(B))v_k{\beta }_{ii}}{b_{kk}{v}_i}\frac{b_{kk}{y}_k}{(b_{kk}-q(B))v_k}{s}_j\sum _{l\ne j}\frac{|a_{jl}|(b_{ll}-q(B))v_l{\beta }_{jj}}{b_{ll}{v}_j}\frac{b_{ll}{y}_l}{(b_{ll}-q(B))v_l}\hfill \\ =s_i\frac{{\beta }_{ii}}{v_i}\sum _{k\ne i}|a_{ik}|y_k{s}_j\frac{{\beta }_{jj}}{v_j}\sum _{l\ne j}|a_{jl}|y_l\hfill \\ =\frac{(b_{ii}-q(B))v_i}{b_{ii}{y}_i}\frac{{\beta }_{ii}}{v_i}(a_{ii}-q(A))y_i\frac{(b_{jj}-q(B))v_j}{b_{jj}{y}_j}\frac{{\beta }_{jj}}{v_j}(a_{jj}-q(A))y_j\hfill \\ =\frac{{\beta }_{ii}{\beta }_{jj}}{b_{ii}{b}_{jj}}[b_{ii}-q(B)][a_{ii}-q(A)][b_{jj}-q(B)][a_{jj}-q(A)].\hfill \end{array}$$

From the above inequality and $0\le q(A\circ B^{-1})\le a_{ii}{\beta }_{ii}$, $\mathrm{\forall }i\in N$, we have

(2.2)

Thus, from inequality (2.2), we have

$$\begin{array}{rcl}q(A\circ B^{-1})& \ge & \frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-[{(a_{ii}{\beta }_{ii}-a_{jj}{\beta }_{jj})}^2\\ +{4\frac{{\beta }_{ii}{\beta }_{jj}}{b_{ii}{b}_{jj}}[b_{ii}-q(B)][a_{ii}-q(A)][b_{jj}-q(B)][a_{jj}-q(A)]]}^{\frac{1}{2}}\}\\ \ge & \underset{i\ne j}{min}\frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-[{(a_{ii}{\beta }_{ii}-a_{jj}{\beta }_{jj})}^2\\ +{4\frac{{\beta }_{ii}{\beta }_{jj}}{b_{ii}{b}_{jj}}[b_{ii}-q(B)][a_{ii}-q(A)][b_{jj}-q(B)][a_{jj}-q(A)]]}^{\frac{1}{2}}\}.\end{array}$$

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 condition (E17) of Theorem 6.2.3 of [9]). If we denote by $D=(d_{ij})$ the $n\times n$ permutation matrix with $d_{12}=d_{23}=\cdots =d_{n-1n}=d_{n1}=1$, 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 $t\to 0$, the result follows by continuity. □

Using ideas of the proof of Theorem 2.3, we next give a new proof of inequality (2.2) in [5]. Similar to the proof of Theorem 2.3, by the theorem of Gerschgorin, there exist some positive integers $i\in N$ such that

$$\begin{array}{rcl}|q(A\circ B^{-1})-a_{ii}{\beta }_{ii}|& \le & \sum _{k\ne i}\frac{|a_{ik}{\beta }_{ik}|}{s_k}{s}_i\\ \le & s_i\sum _{k\ne i}\frac{|a_{ik}|(b_{kk}-q(B))v_k{\beta }_{ii}}{b_{kk}{v}_i}\frac{b_{kk}{y}_k}{(b_{kk}-q(B))v_k}\\ =& s_i\frac{{\beta }_{ii}}{v_i}\sum _{k\ne i}|a_{ik}|y_k\\ =& \frac{(b_{ii}-q(B))v_i}{b_{ii}{y}_i}\frac{{\beta }_{ii}}{v_i}(a_{ii}-q(A))y_i\\ =& \frac{{\beta }_{ii}}{b_{ii}}[b_{ii}-q(B)][a_{ii}-q(A)].\end{array}$$

From the above inequality and $0\le q(A\circ B^{-1})\le a_{ii}{\beta }_{ii}$, $\mathrm{\forall }i\in N$, we have

$$a_{ii}{\beta }_{ii}-q(A\circ B^{-1})\le \frac{{\beta }_{ii}}{b_{ii}}[b_{ii}-q(B)][a_{ii}-q(A)].$$

(2.3)

Thus, from inequality (2.3), we have

$$\begin{array}{rcl}q(A\circ B^{-1})& \ge & a_{ii}{\beta }_{ii}-\frac{{\beta }_{ii}}{b_{ii}}[b_{ii}-q(B)][a_{ii}-q(A)]\\ =& q(A)q(B)\left\{(\frac{a_{ii}}{q(A)}+\frac{b_{ii}}{q(B)}-1)\frac{{\beta }_{ii}}{b_{ii}}\right\}\\ \ge & q(A)q(B)\underset{1\le i\le n}{min}\left\{(\frac{a_{ii}}{q(A)}+\frac{b_{ii}}{q(B)}-1)\frac{{\beta }_{ii}}{b_{ii}}\right\}.\end{array}$$

Remark 2.1 We next give a simple comparison between the upper bound in (2.1) and the upper bound in (1.2) and (1.1). Without loss of generality, for $i\ne j$, assume that

$$a_{ii}{\beta }_{ii}-\frac{{\beta }_{ii}}{b_{ii}}[b_{ii}-q(B)][a_{ii}-q(A)]\le a_{jj}{\beta }_{jj}-\frac{{\beta }_{jj}}{b_{jj}}[b_{jj}-q(B)][a_{jj}-q(A)].$$

(2.4)

Thus, we can write (2.4) equivalently as

$$\frac{{\beta }_{jj}}{b_{jj}}[b_{jj}-q(B)][a_{jj}-q(A)]\le a_{jj}{\beta }_{jj}-a_{ii}{\beta }_{ii}+\frac{{\beta }_{ii}}{b_{ii}}[b_{ii}-q(B)][a_{ii}-q(A)].$$

(2.5)

From (2.1), we have

$$\begin{array}{c}{(a_{ii}{\beta }_{ii}-a_{jj}{\beta }_{jj})}^2+4\frac{{\beta }_{ii}{\beta }_{jj}}{b_{ii}{b}_{jj}}[b_{ii}-q(B)][a_{ii}-q(A)][b_{jj}-q(B)][a_{jj}-q(A)]\hfill \\ \le {(a_{jj}{\beta }_{jj}-a_{ii}{\beta }_{ii})}^2+4\frac{{\beta }_{ii}}{b_{ii}}[b_{ii}-q(B)][a_{ii}-q(A)](a_{jj}{\beta }_{jj}-a_{ii}{\beta }_{ii})\hfill \\ +4{\left[\frac{{\beta }_{ii}}{b_{ii}}[b_{ii}-q(B)][a_{ii}-q(A)]\right]}^2\hfill \\ ={(a_{jj}{\beta }_{jj}-a_{ii}{\beta }_{ii}+2\frac{{\beta }_{ii}}{b_{ii}}[b_{ii}-q(B)][a_{ii}-q(A)])}^2.\hfill \end{array}$$

Thus, from (2.1), (2.5) and the above inequality, we have

$$\begin{array}{rcl}q(A\circ B^{-1})& \ge & \underset{i\ne j}{min}\frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-[{(a_{ii}{\beta }_{ii}-a_{jj}{\beta }_{jj})}^2\\ +{4\frac{{\beta }_{ii}{\beta }_{jj}}{b_{ii}{b}_{jj}}[b_{ii}-q(B)][a_{ii}-q(A)][b_{jj}-q(B)][a_{jj}-q(A)]]}^{\frac{1}{2}}\}\\ \ge & \underset{i\ne j}{min}\frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-a_{jj}{\beta }_{jj}+a_{ii}{\beta }_{ii}-2\frac{{\beta }_{ii}}{b_{ii}}[b_{ii}-q(B)][a_{ii}-q(A)]\}\\ \ge & q(A)q(B)\underset{1\le i\le n}{min}\left\{(\frac{a_{ii}}{q(A)}+\frac{b_{ii}}{q(B)}-1)\frac{{\beta }_{ii}}{b_{ii}}\right\}.\end{array}$$

Hence, the bound in (2.1) is sharper than the bound in (1.2). According to Remark 2.4 in [5], we know

$$q(A)q(B)\underset{1\le i\le n}{min}\left\{(\frac{a_{ii}}{q(A)}+\frac{b_{ii}}{q(B)}-1)\frac{{\beta }_{ii}}{b_{ii}}\right\}\ge q(A)\underset{1\le i\le n}{min}{\beta }_{ii}.$$

So, the bound in (2.1) is sharper than the bound in (1.1).

Theorem 2.4 Let $A=(a_{ij})\in M_n$, $B=(b_{ij})\in M_n$. Then

$$q(A\circ B^{-1})\ge (1-\rho (J_A)\rho (J_B))\underset{1\le i\le n}{min}\frac{a_{ii}}{b_{ii}}.$$

(2.6)

Proof Suppose that A and B are irreducible, $D_B$ is the diagonal matrix of B and $C_B=D_B-B$, then $D_B$ is a diagonal matrix with positive diagonal entries, $C_B$ is an irreducible nonnegative matrix and $J=D_B^{-1}{C}_B^T$ is again an irreducible nonnegative matrix. Since the Jacobi iterative matrix of B is $J_B=D_B^{-1}{C}_B$, we have

$$\rho (J_B)=\rho \left(D_B^{-1}{C}_B\right)=\rho \left({\left(D_B^{-1}{C}_B\right)}^T\right)=\rho \left(C_B^T{D}_B^{-1}\right)=\rho \left(D_B^{-1}{C}_B^T\right)=\rho (J).$$

(2.7)

By the Perron-Frobenius theorem on irreducible nonnegative matrices, there is a positive eigenvector $x={(x_1,x_2,\dots ,x_n)}^T$ such that $D_B^{-1}{C}_B^{T}x=\rho (J)x$. That is,

$$\sum _{k\ne i}\frac{|b_{ki}|x_k}{b_{ii}}=\rho (J)x_i\mathrm{\forall }i\in N.$$

(2.8)

Thus, we can write (2.8) equivalently as

$$\sum _{k\ne i}\frac{|b_{ki}|x_k}{b_{ii}{x}_i}=\rho (J)\mathrm{\forall }i\in N.$$

Set $X=diag(x_1,x_2,\dots ,x_n)$ and $\tilde{B}=XB$. It is easy to check that $\tilde{B}$ is a strictly diagonally dominant matrix by column. Let $B^{-1}=({\beta }_{ij})$. By Lemma 2.2, for all $i\ne j$ ($1\le i,j\le n$), we have

$${\beta }_{ij}{x}_j^{-1}\le \frac{{\sum }_{k\ne j}|b_{kj}|x_k}{b_{jj}{x}_j}{\beta }_{ii}{x}_i^{-1}=\rho (J){\beta }_{ii}{x}_i^{-1}.$$

Thus

$${\beta }_{ij}\le \rho (J){\beta }_{ii}\frac{x_j}{x_i}\mathrm{\forall }i\in N.$$

(2.9)

Combining (2.9) with (2.7), we get

$${\beta }_{ij}\le \rho (J_B){\beta }_{ii}\frac{x_j}{x_i}.$$

(2.10)

Since $B^{-1}B=I$, we obtain

$${\beta }_{ii}{b}_{ii}=1+\sum _{k\ne i}{\beta }_{ik}|b_{ki}|\ge 1\mathrm{\forall }i\in N.$$

Thus

$${\beta }_{ii}\ge \frac{1}{b_{ii}}\mathrm{\forall }i\in N.$$

(2.11)

Let $J_{A}y=\rho (J_A)y$ for positive vectors $y=(y_i)$. Set $S=diag(\frac{x_1}{y_1},\frac{x_2}{y_2},\dots ,\frac{x_n}{y_n})$, then $S>0$. Hence, $\sigma (A\circ B^{-1})=\sigma (S(A\circ B^{-1})S^{-1})$. Since $q(A\circ B^{-1})$ is an eigenvalue of $A\circ B^{-1}$, we have

$$q(A\circ B^{-1})\in \sigma \left(S(A\circ B^{-1})S^{-1}\right).$$

By the theorem of Gerschgorin and (2.10), there exist some positive integers $i\in N$ such that

$$\begin{array}{rcl}|q(A\circ B^{-1})-a_{ii}{\beta }_{ii}|& \le & \sum _{k\ne i}|a_{ik}{\beta }_{ik}|\frac{x_i{y}_k}{y_i{x}_k}\\ \le & \frac{x_i}{y_i}\sum _{k\ne i}|a_{ik}|\rho (J_B){\beta }_{ii}\frac{x_k}{x_i}\frac{y_k}{x_k}\\ =& \rho (J_B)\frac{{\beta }_{ii}}{y_i}\sum _{k\ne i}|a_{ik}|y_k\\ =& a_{ii}{\beta }_{ii}\rho (J_A)\rho (J_B).\end{array}$$

From the above inequality and $0\le q(A\circ B^{-1})\le a_{ii}{\beta }_{ii}$, $\mathrm{\forall }i\in N$, we have

$$a_{ii}{\beta }_{ii}-q(A\circ B^{-1})\le a_{ii}{\beta }_{ii}\rho (J_A)\rho (J_B).$$

(2.12)

Thus, from inequality (2.11) and (2.12), we have

$$\begin{array}{rcl}q(A\circ B^{-1})& \ge & a_{ii}{\beta }_{ii}-a_{ii}{\beta }_{ii}\rho (J_A)\rho (J_B)\\ =& (1-\rho (J_A)\rho (J_B))a_{ii}{\beta }_{ii}\\ \ge & (1-\rho (J_A)\rho (J_B))\frac{a_{ii}}{b_{ii}}\\ \ge & (1-\rho (J_A)\rho (J_B))\underset{1\le i\le n}{min}\frac{a_{ii}}{b_{ii}}.\end{array}$$

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 condition (E17) of Theorem 6.2.3 of [9]). If we denote by $D=(d_{ij})$ the $n\times n$ permutation matrix with $d_{12}=d_{23}=\cdots =d_{n-1n}=d_{n1}=1$, 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 $t\to 0$, the result follows by continuity. □

Remark 2.2 If $B\in M_n$ is a diagonal matrix, the equality of (2.6) holds. Thus the bound (2.6) is sharp. Since $1+{\rho }^2(J_B)\ge 1$, then

$$(1-\rho (J_A)\rho (J_B))\underset{1\le i\le n}{min}\frac{a_{ii}}{b_{ii}}\ge \frac{1-\rho (J_A)\rho (J_B)}{1+{\rho }^2(J_B)}\underset{1\le i\le n}{min}\frac{a_{ii}}{b_{ii}}.$$

The bound in (2.6) is sharper than the bound in (1.3).

If $B=A$, according to Theorem 2.4, we can deduce the following corollary.

Corollary 2.5 Let $B\in M_n$, then

$$q(B\circ B^{-1})\ge 1-{\rho }^2(J_B).$$

Remark 2.3 Corollary 2.5 is Theorem 2.8 of Xiang [6]. So, Theorem 2.4 generalizes Theorem 2.8 in [6].

If we apply Lemma 2.1 to $J=D_B^{-1}{C}_B^T$ and $J_B=D_B^{-1}{C}_B$, then we have

$$\begin{array}{c}{\rho }^2(J)\le \underset{i\ne j}{max}\sum _{k\ne i}\frac{|b_{ki}|}{b_{ii}}\cdot \sum _{l\ne j}\frac{|b_{lj}|}{b_{jj}},\hfill \\ {\rho }^2(J_B)\le \underset{i\ne j}{max}\sum _{k\ne i}\frac{|b_{ik}|}{b_{ii}}\cdot \sum _{l\ne j}\frac{|b_{jl}|}{b_{jj}}.\hfill \end{array}$$

Since $\rho (J_B)=\rho (J)$, then

$${\rho }^2(J_B)\le min\{\underset{i\ne j}{max}\sum _{k\ne i}\frac{|b_{ik}|}{b_{ii}}\cdot \sum _{l\ne j}\frac{|b_{jl}|}{b_{jj}},\underset{i\ne j}{max}\sum _{k\ne i}\frac{|b_{ki}|}{b_{ii}}\cdot \sum _{l\ne j}\frac{|b_{lj}|}{b_{jj}}\}.$$

(2.13)

From (2.13) we have the following corollary.

Corollary 2.6 Let $B=(b_{ij})\in M_n$, then

$$q(B\circ B^{-1})\ge 1-min\{\underset{i\ne j}{max}\sum _{k\ne i}\frac{|b_{ik}|}{b_{ii}}\cdot \sum _{l\ne j}\frac{|b_{jl}|}{b_{jj}},\underset{i\ne j}{max}\sum _{k\ne i}\frac{|b_{ki}|}{b_{ii}}\cdot \sum _{l\ne j}\frac{|b_{lj}|}{b_{jj}}\}.$$

Example 2.1 Let A and B be the same as in Example 2.1 in [10]:

$$B=\left(\begin{array}{cccc}4& -1& -1& -1\\ -2& 5& -1& -1\\ 0& -2& 4& -1\\ -1& -1& -1& 4\end{array}\right),A=\left(\begin{array}{cccc}1& -1/2& 0& 0\\ -1/2& 1& -1/2& 0\\ 0& -1/2& 1& -1/2\\ 0& 0& -1/2& 1\end{array}\right).$$

It is easy to check that $A,B\in M_4$. If we apply Theorem 5.7.31 of [1], we have

$$q(A\circ B^{-1})\ge q(A)\underset{1\le i\le n}{min}{\beta }_{ii}=0.07003.$$

If we apply Theorem 9 of [2], we have

$$q(A\circ B^{-1})\ge \frac{1-\rho (J_A)\rho (J_B)}{1+{\rho }^2(J_B)}\underset{1\le i\le n}{min}\frac{a_{ii}}{b_{ii}}=0.05229.$$

If we apply Theorem 2.1 of [10], we have

$$\begin{array}{r}q(A\circ B^{-1})\ge \underset{1\le i\le n}{min}\left\{\frac{a_{ii}-s_i{\sum }_{j\ne i}|a_{ji}|}{b_{ii}}\right\}=0.08.\end{array}$$

But if we apply Theorem 2.4, we have

$$q(A\circ B^{-1})\ge (1-\rho (J_A)\rho (J_B))\underset{1\le i\le n}{min}\frac{a_{ii}}{b_{ii}}=0.08291.$$

In fact, $q(A\circ B^{-1})=0.21478$. Example 2.1 shows that the bound in (2.6) is better than these corresponding bounds in [1, 2, 10].