Some new inequalities for the Hadamard product of M-matrices

Abstract

If A and B are $n\times n$ nonsingular M-matrices, a new lower bound for the minimum eigenvalue $\tau (B\circ A^{-1})$ for the Hadamard product of B and $A^{-1}$ is derived. As a consequence, a new lower bound for the minimum eigenvalue $\tau (A\circ A^{-1})$ for the Hadamard product of A and its inverse $A^{-1}$ is given. Theoretical results and an example demonstrate that the new bounds are better than some existing ones.

MSC:15A06, 15A18, 15A48.

1 Introduction

For convenience, for any positive integer n, let $N=\{1,2,\dots ,n\}$ throughout. 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 a nonnegative matrix if $a_{ij}\ge 0$. 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.

$Z_n$ denotes 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 λ such that $A=\lambda I-B$ and $\lambda \ge \rho (B)$, 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 let $\tau (A)=min\{Re(\lambda ):\lambda \in \sigma (A)\}$. 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 $A\ge B$, then $\tau (A)\ge \tau (B)$.

• (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.

For two matrices $A=(a_{ij})$ and $B=(b_{ij})$, the Hadamard product of A and B is the matrix $A\circ B=(a_{ij}{b}_{ij})$. If A and B are two nonsingular M-matrices, then $B\circ A^{-1}$ is also a nonsingular M-matrix [2].

Let $A,B\in M_n$ and $A^{-1}=({\beta }_{ij})$, in [[1], Theorem 5.7.31] the following classical result is given:

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

(1.1)

Huang [[3], Theorem 9] improved this result and obtained the following result:

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

(1.2)

where $\rho (J_A)$, $\rho (J_B)$ are the spectral radii of $J_A$ and $J_B$.

The lower bound (1.1) is simple, but not accurate enough. The lower bound (1.2) is difficult to evaluate.

Recently, Li [[4], Theorem 2.1] improved these two results and gave a new lower bound for $\tau (B\circ A^{-1})$, that is,

$$\tau (B\circ A^{-1})\ge \underset{i}{min}\left\{\frac{b_{ii}-s_i{\sum }_{k\ne i}|b_{ki}|}{a_{ii}}\right\},$$

(1.3)

where $r_{li}=\frac{|a_{li}|}{|a_{ll}|-{\sum }_{k\ne l,i}|a_{lk}|}$, $l\ne i$; $r_i={max}_{l\ne i}\{r_{li}\}$, $i\in N$; $s_{ji}=\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|r_k}{a_{jj}}$, $j\ne i$, $j\in N$; $s_i={max}_{j\ne i}\{s_{ij}\}$, $i\in N$.

For an M-matrix A, Fiedler et al. showed in [5] that $\tau (A\circ A^{-1})\le 1$. Subsequently, Fiedler and Markham [[2], Theorem 3] gave a lower bound on $\tau (A\circ A^{-1})$,

$$\tau (A\circ A^{-1})\ge \frac{1}{n},$$

(1.4)

and proposed the following conjecture:

$$\tau (A\circ A^{-1})\ge \frac{2}{n}.$$

(1.5)

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

Li [[8], Theorem 3.1] obtained the following result:

$$\tau (A\circ A^{-1})\ge \underset{i}{min}\left\{\frac{a_{ii}-t_i{R}_i}{1+{\sum }_{j\ne i}{t}_{ji}}\right\},$$

(1.6)

which only depends on the entries of $A=(a_{ij})$, where $R_i={\sum }_{k\ne i}|a_{ik}|$; $d_i=\frac{R_i}{|a_{ii}|}$, $i\in N$; $t_{ji}=\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|d_k}{|a_{jj}|}$, $j\ne i$, $j\in N$; $t_i={max}_{j\ne i}\{t_{ij}\}$, $i\in N$.

Li [[9], Theorem 3.2] improved the bound (1.6) and obtained the following result:

$$\tau (A\circ A^{-1})\ge \underset{i}{min}\left\{\frac{a_{ii}-m_i{R}_i}{1+{\sum }_{j\ne i}{m}_{ji}}\right\},$$

(1.7)

where $r_{li}=\frac{|a_{li}|}{|a_{ll}|-{\sum }_{k\ne l,i}|a_{lk}|}$, $l\ne i$; $r_i={max}_{l\ne i}\{r_{li}\}$, $i\in N$; $m_{ji}=\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|r_i}{|a_{jj}|}$, $j\ne i$, $j\in N$; $m_i={max}_{j\ne i}\{m_{ij}\}$, $i\in N$.

Recently, Li [[10], Theorem 3.2] improved the bound (1.7) and gave a new lower bound for $\tau (A\circ A^{-1})$, that is,

$$\tau (A\circ A^{-1})\ge \underset{i}{min}\left\{\frac{a_{ii}-T_i{R}_i}{1+{\sum }_{j\ne i}{T}_{ji}}\right\},$$

(1.8)

where $T_{ji}=min\{m_{ji},s_{ji}\}$, $j\ne i$; $T_i={max}_{j\ne i}\{T_{ij}\}$, $i\in N$.

In the present paper, we present a new lower bound on $\tau (B\circ A^{-1})$. As a consequence, we present a new lower bound on $\tau (A\circ A^{-1})$. These bounds improve several existing results.

The following is the list of notations that we use throughout: For $i,j,k,l\in N$,

$$\begin{array}{c}{R}_i=\sum _{k\ne i}|a_{ik}|,C_i=\sum _{k\ne i}|a_{ki}|,d_i=\frac{R_i}{|a_{ii}|},{\hat{c}}_i=\frac{C_i}{|a_{ii}|};\hfill \\ r_{li}=\frac{|a_{li}|}{|a_{ll}|-{\sum }_{k\ne l,i}|a_{lk}|},l\ne i;r_i=\underset{l\ne i}{max}\{r_{li}\},i\in N;\hfill \\ c_{il}=\frac{|a_{il}|}{|a_{ll}|-{\sum }_{k\ne l,i}|a_{kl}|},l\ne i;c_i=\underset{l\ne i}{max}\{c_{il}\},i\in N;\hfill \\ m_{ji}=\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|r_i}{|a_{jj}|},j\ne i;m_i=\underset{j\ne i}{max}\{m_{ij}\},i\in N;\hfill \\ s_{ji}=\frac{|a_{ji}|+{\sum }_{k\ne j,i}|a_{jk}|r_k}{|a_{jj}|},j\ne i;s_i=\underset{j\ne i}{max}\{s_{ij}\},i\in N;\hfill \\ T_{ji}=min\{m_{ji},s_{ji}\},j\ne i;T_i=\underset{j\ne i}{max}\{T_{ij}\},i\in N.\hfill \end{array}$$

2 Some lemmas and the main results

In order to prove our results, we first give some lemmas.

Lemma 2.1 [11]

If $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$ is an M-matrix, then there exists a diagonal matrix D with positive diagonal entries such that $D^{-1}AD$ is a strictly row diagonally dominant M-matrix.

Lemma 2.2 [1]

Let $A,B=(a_{ij})\in {\mathbb{C}}^{n\times n}$ and suppose that $D\in {\mathbb{C}}^{n\times n}$ and $E\in {\mathbb{C}}^{n\times n}$ are diagonal matrices. Then

$$D(A\circ B)E=(DAE)\circ B=(DA)\circ (BE)=(AE)\circ (DB)=A\circ (DBE).$$

Lemma 2.3 [10]

If $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$ is a strictly row diagonally dominant M-matrix, then $A^{-1}=({\beta }_{ij})$ satisfies

$${\beta }_{ji}\le T_{ji}{\beta }_{ii},i,j\in N,i\ne j.$$

Lemma 2.4 [12]

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

Lemma 2.5 [9]

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

$${\beta }_{ii}\ge \frac{1}{a_{ii}},i\in N.$$

Lemma 2.6 [10]

If $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$ is an M-matrix and $A^{-1}=({\beta }_{ij})$ is a doubly stochastic matrix, then

$${\beta }_{ii}\ge \frac{1}{1+{\sum }_{j\ne i}{T}_{ji}},i\in N.$$

Lemma 2.7 [13]

Let $A=(a_{ij})\in {\mathbb{C}}^{n\times n}$, and let $x_1,x_2,\dots ,x_n$ be positive real numbers. Then all the eigenvalues of A lie in the region

$$\underset{i\ne j}{\overset{n}{\bigcup _{i,j=1}}}\{z\in \mathbb{C}:|z-a_{ii}||z-a_{jj}|\le (x_i\sum _{k\ne i}\frac{1}{x_k}|a_{ki}|)(x_j\sum _{k\ne j}\frac{1}{x_k}|a_{kj}|)\}.$$

Theorem 2.1 Let $A,B=(b_{ij})\in {\mathbb{R}}^{n\times n}$ be two nonsingular M-matrices, and let $A^{-1}=({\beta }_{ij})$. Then

$$\begin{array}{rcl}\tau (B\circ A^{-1})& \ge & \underset{i\ne j}{min}\frac{1}{2}\{b_{ii}{\beta }_{ii}+b_{jj}{\beta }_{jj}-[{(b_{ii}{\beta }_{ii}-b_{jj}{\beta }_{jj})}^2\\ {+4(T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|b_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}.\end{array}$$

(2.1)

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

We next assume that $n\ge 2$.

If A is an M-matrix, then by Lemma 2.1 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 and satisfies

$$\tau (B\circ A^{-1})=\tau \left(D^{-1}(B\circ A^{-1})D\right)=\tau (B\circ {\left(D^{-1}AD\right)}^{-1}).$$

So, for convenience and without loss of generality, we assume that A is a strictly row diagonally dominant M-matrix. Therefore, $0<T_i<1$, $i\in N$.

If $B\circ A^{-1}$ is irreducible, then B and A are irreducible. Let $\tau (B\circ A^{-1})=\lambda$, so that $0<\lambda <b_{ii}{\beta }_{ii}$, $\mathrm{\forall }i\in N$. Thus, by Lemma 2.7, there is a pair $(i,j)$ of positive integers with $i\ne j$ such that

$$|\lambda -b_{ii}{\beta }_{ii}||\lambda -b_{jj}{\beta }_{jj}|\le (T_i\sum _{k\ne i}\frac{1}{T_k}|b_{ki}{\beta }_{ki}|)(T_j\sum _{k\ne j}\frac{1}{T_k}|b_{kj}{\beta }_{kj}|).$$

Observe that

$$\begin{array}{r}(T_i\sum _{k\ne i}\frac{1}{T_k}|b_{ki}{\beta }_{ki}|)(T_j\sum _{k\ne j}\frac{1}{T_k}|b_{kj}{\beta }_{kj}|)\\ \le (T_i\sum _{k\ne i}\frac{1}{T_k}|b_{ki}|T_{ki}{\beta }_{ii})(T_j\sum _{k\ne j}\frac{1}{T_k}|b_{kj}|T_{kj}{\beta }_{jj})\\ \le (T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|b_{kj}|{\beta }_{jj}).\end{array}$$

Thus, we have

$$|\lambda -b_{ii}{\beta }_{ii}||\lambda -b_{jj}{\beta }_{jj}|\le (T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|b_{kj}|{\beta }_{jj}).$$

Then we have

$$\lambda \ge \frac{1}{2}\{b_{ii}{\beta }_{ii}+b_{jj}{\beta }_{jj}-{[{(b_{ii}{\beta }_{ii}-b_{jj}{\beta }_{jj})}^2+4(T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|b_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}.$$

That is,

$$\begin{array}{rcl}\tau (B\circ A^{-1})& \ge & \frac{1}{2}\{b_{ii}{\beta }_{ii}+b_{jj}{\beta }_{jj}-[{(b_{ii}{\beta }_{ii}-b_{jj}{\beta }_{jj})}^2\\ {+4(T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|b_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}\\ \ge & \underset{i\ne j}{min}\frac{1}{2}\{b_{ii}{\beta }_{ii}+b_{jj}{\beta }_{jj}-[{(b_{ii}{\beta }_{ii}-b_{jj}{\beta }_{jj})}^2\\ {+4(T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|b_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}.\end{array}$$

Now, assume that $B\circ A^{-1}$ is reducible. It is 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 [14]). If we denote by $D=(d_{ij})$ the $n\times n$ permutation matrix with $d_{12}=d_{23}=\cdots =d_{n-1,n}=d_{n1}=1$, 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 $t⟶0$, the result follows by continuity. □

Theorem 2.2 Let $A,B=(b_{ij})\in {\mathbb{R}}^{n\times n}$ be two nonsingular M-matrices, and let $A^{-1}=({\beta }_{ij})$. Then

$$\begin{array}{r}\underset{i\ne j}{min}\frac{1}{2}\{b_{ii}{\beta }_{ii}+b_{jj}{\beta }_{jj}-{[{(b_{ii}{\beta }_{ii}-b_{jj}{\beta }_{jj})}^2+4(T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|b_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}\\ \ge \underset{1\le i\le n}{min}\left\{\frac{b_{ii}-s_i{\sum }_{k\ne i}|b_{ki}|}{a_{ii}}\right\}.\end{array}$$

Proof Since $T_{ji}=min\{m_{ji},s_{ji}\}$, $j\ne i$, $T_i={max}_{j\ne i}\{T_{ij}\}$, so $T_i\le s_i$, $i\in N$. Without loss of generality, for $i\ne j$, assume that

$$b_{ii}{\beta }_{ii}-T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii}\le b_{jj}{\beta }_{jj}-T_j\sum _{k\ne j}|b_{kj}|{\beta }_{jj}.$$

(2.2)

Thus, (2.2) is equivalent to

$$T_j\sum _{k\ne j}|b_{kj}|{\beta }_{jj}\le T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii}+b_{jj}{\beta }_{jj}-b_{ii}{\beta }_{ii}.$$

(2.3)

From (2.1) and (2.3), we have

$$\begin{array}{r}\frac{1}{2}\{b_{ii}{\beta }_{ii}+b_{jj}{\beta }_{jj}-{[{(b_{ii}{\beta }_{ii}-b_{jj}{\beta }_{jj})}^2+4(T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|b_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}\\ \ge \frac{1}{2}\{b_{ii}{\beta }_{ii}+b_{jj}{\beta }_{jj}-[{(b_{ii}{\beta }_{ii}-b_{jj}{\beta }_{jj})}^2\\ {+4(T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})(T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii}+b_{jj}{\beta }_{jj}-b_{ii}{\beta }_{ii})]}^{\frac{1}{2}}\}\\ =\frac{1}{2}\{b_{ii}{\beta }_{ii}+b_{jj}{\beta }_{jj}-[{(b_{ii}{\beta }_{ii}-b_{jj}{\beta }_{jj})}^2\\ {+4{(T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})}^2+4(T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})(b_{jj}{\beta }_{jj}-b_{ii}{\beta }_{ii})]}^{\frac{1}{2}}\}\\ =\frac{1}{2}\{b_{ii}{\beta }_{ii}+b_{jj}{\beta }_{jj}-{\left[{(b_{jj}{\beta }_{jj}-b_{ii}{\beta }_{ii}+2T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})}^2\right]}^{\frac{1}{2}}\}\\ =\frac{1}{2}\{b_{ii}{\beta }_{ii}+b_{jj}{\beta }_{jj}-(b_{jj}{\beta }_{jj}-b_{ii}{\beta }_{ii}+2T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})\}\\ =b_{ii}{\beta }_{ii}-T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii}\\ ={\beta }_{ii}(b_{ii}-T_i\sum _{k\ne i}|b_{ki}|)\\ \ge {\beta }_{ii}(b_{ii}-s_i\sum _{k\ne i}|b_{ki}|)\\ \ge \frac{b_{ii}-s_i{\sum }_{k\ne i}|b_{ki}|}{a_{ii}}.\end{array}$$

Thus, we have

$$\begin{array}{rcl}\tau (B\circ A^{-1})& \ge & \underset{i\ne j}{min}\frac{1}{2}\{b_{ii}{\beta }_{ii}+b_{jj}{\beta }_{jj}-[{(b_{ii}{\beta }_{ii}-b_{jj}{\beta }_{jj})}^2\\ {+4(T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|b_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}\\ \ge & \underset{1\le i\le n}{min}\left\{\frac{b_{ii}-s_i{\sum }_{k\ne i}|b_{ki}|}{a_{ii}}\right\}.\end{array}$$

This proof is completed. □

Remark 2.1 Theorem 2.2 shows that the result of Theorem 2.1 is better than the result of Theorem 2.1 in [4].

If $A=B$, according to Theorem 2.1, we can obtain the following corollary.

Corollary 2.1 Let $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$ be an M-matrix, and let $A^{-1}=({\beta }_{ij})$ be a doubly stochastic matrix. Then

$$\begin{array}{rcl}\tau (A\circ A^{-1})& \ge & \underset{i\ne j}{min}\frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-[{(a_{ii}{\beta }_{ii}-a_{jj}{\beta }_{jj})}^2\\ {+4(T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|a_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}.\end{array}$$

(2.4)

Theorem 2.3 Let $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$ be an M-matrix, and let $A^{-1}=({\beta }_{ij})$ be a doubly stochastic matrix. Then

$$\begin{array}{r}\underset{i\ne j}{min}\frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-{[{(a_{ii}{\beta }_{ii}-a_{jj}{\beta }_{jj})}^2+4(T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|a_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}\\ \ge \underset{i}{min}\left\{\frac{a_{ii}-T_i{R}_i}{1+{\sum }_{j\ne i}{T}_{ji}}\right\}.\end{array}$$

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

$$a_{ii}=\sum _{k\ne i}|a_{ik}|+1=\sum _{k\ne i}|a_{ki}|+1,i\in N.$$

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

$$a_{ii}{\beta }_{ii}-T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii}\le a_{jj}{\beta }_{jj}-T_j\sum _{k\ne j}|a_{kj}|{\beta }_{jj}.$$

(2.5)

Thus, (2.5) is equivalent to

$$T_j\sum _{k\ne j}|a_{kj}|{\beta }_{jj}\le a_{jj}{\beta }_{jj}-a_{ii}{\beta }_{ii}+T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii}.$$

(2.6)

From (2.4) and (2.6), we have

$$\begin{array}{r}\frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-{[{(a_{ii}{\beta }_{ii}-a_{jj}{\beta }_{jj})}^2+4(T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|a_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}\\ \ge \frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-[{(a_{ii}{\beta }_{ii}-a_{jj}{\beta }_{jj})}^2\\ {+4(T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii})(a_{jj}{\beta }_{jj}-a_{ii}{\beta }_{ii}+T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii})]}^{\frac{1}{2}}\}\\ =\frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-[{(a_{ii}{\beta }_{ii}-a_{jj}{\beta }_{jj})}^2\\ {+4{(T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii})}^2+4(T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii})(a_{jj}{\beta }_{jj}-a_{ii}{\beta }_{ii})]}^{\frac{1}{2}}\}\\ =\frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-{\left[{(a_{jj}{\beta }_{jj}-a_{ii}{\beta }_{ii}+2T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii})}^2\right]}^{\frac{1}{2}}\}\\ =\frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-(a_{jj}{\beta }_{jj}-a_{ii}{\beta }_{ii}+2T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii})\}\\ =a_{ii}{\beta }_{ii}-T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii}\\ ={\beta }_{ii}(a_{ii}-T_i\sum _{k\ne i}|a_{ki}|)\\ \ge \frac{a_{ii}-T_i{R}_i}{1+{\sum }_{j\ne i}{T}_{ji}}.\end{array}$$

Thus, we have

$$\begin{array}{rcl}\tau (A\circ A^{-1})& \ge & \underset{i\ne j}{min}\frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-[{(a_{ii}{\beta }_{ii}-a_{jj}{\beta }_{jj})}^2\\ {+4(T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|a_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}\\ \ge & \underset{i}{min}\left\{\frac{a_{ii}-T_i{R}_i}{1+{\sum }_{j\ne i}{T}_{ji}}\right\}.\end{array}$$

This proof is completed. □

Remark 2.2 Theorem 2.3 shows that the result of Corollary 2.1 is better than the result of Theorem 3.2 in [10].

3 Example

For convenience, we consider that the M-matrices A and B are the same as the matrices of [4].

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

• (1) We consider the lower bound for $\tau (B\circ A^{-1})$.

If we apply (1.1), we have

$$\tau (B\circ A^{-1})\ge \tau (B)\underset{1\le i\le n}{min}{\beta }_{ii}=0.07.$$

If we apply (1.2), we have

$$\tau (B\circ A^{-1})\ge \frac{1-\rho (J_A)\rho (J_B)}{1+{\rho }^2(J_A)}\underset{1\le i\le n}{min}\frac{b_{ii}}{a_{ii}}=0.048.$$

If we apply (1.3), we have

$$\tau (B\circ A^{-1})\ge \underset{i}{min}\left\{\frac{b_{ii}-s_i{\sum }_{k\ne i}|b_{ki}|}{a_{ii}}\right\}=0.08.$$

If we apply Theorem 2.1, we have

$$\begin{array}{rcl}\tau (B\circ A^{-1})& \ge & \underset{i\ne j}{min}\frac{1}{2}\{b_{ii}{\beta }_{ii}+b_{jj}{\beta }_{jj}-[{(b_{ii}{\beta }_{ii}-b_{jj}{\beta }_{jj})}^2\\ {+4(T_i\sum _{k\ne i}|b_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|b_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}=0.1753.\end{array}$$

In fact, $\tau (B\circ A^{-1})=0.2148$.

• (2) We consider the lower bound for $\tau (A\circ A^{-1})$.

If we apply (1.5), we have

$$\tau (A\circ A^{-1})\ge \frac{2}{n}=\frac{1}{2}=0.5.$$

If we apply (1.6), we have

$$\tau (A\circ A^{-1})\ge \underset{i}{min}\left\{\frac{a_{ii}-t_i{R}_i}{1+{\sum }_{j\ne i}{t}_{ji}}\right\}=0.6624.$$

If we apply (1.7), we have

$$\tau (A\circ A^{-1})\ge \underset{i}{min}\left\{\frac{a_{ii}-m_i{R}_i}{1+{\sum }_{j\ne i}{m}_{ji}}\right\}=0.7999.$$

If we apply (1.8), we have

$$\tau (A\circ A^{-1})\ge \underset{i}{min}\left\{\frac{a_{ii}-T_i{R}_i}{1+{\sum }_{j\ne i}{T}_{ji}}\right\}=0.85.$$

If we apply Corollary 2.1, we have

$$\begin{array}{rcl}\tau (A\circ A^{-1})& \ge & \underset{i\ne j}{min}\frac{1}{2}\{a_{ii}{\beta }_{ii}+a_{jj}{\beta }_{jj}-[{(a_{ii}{\beta }_{ii}-a_{jj}{\beta }_{jj})}^2\\ {+4(T_i\sum _{k\ne i}|a_{ki}|{\beta }_{ii})(T_j\sum _{k\ne j}|a_{kj}|{\beta }_{jj})]}^{\frac{1}{2}}\}=0.9755.\end{array}$$

In fact, $\tau (A\circ A^{-1})=0.9755$.

Remark 3.1 The numerical example shows that the bounds of Theorem 2.1 and Corollary 2.1 are sharper than those of Theorem 2.1 in [4] and Theorem 3.2 in [10].