## 1 Introduction

Let $$\mathbb{C}^{n\times n}$$ ($$\mathbb{R}^{n\times n}$$) denote the set of all $$n\times n$$ complex (real) matrices, $$A=(a_{ij})\in \mathbb{C}^{n\times n}$$, $$N= \{1, 2, \ldots, n\}$$. We write $$A\geq 0$$ if all $$a_{ij}\geq0$$ ($$i,j\in N$$). A is called nonnegative if $$A\geq0$$. Let $$Z_{n}$$ denote the class of all $$n\times n$$ real matrices all of whose off-diagonal entries are nonpositive. A matrix A is called an M-matrix [1] if $$A\in Z_{n}$$ and the inverse of A, denoted by $$A^{-1}$$, is nonnegative. $$M_{n}$$ will be used to denote the set of all $$n\times n$$ M-matrices.

Let A be an M-matrix. Then there exists a positive eigenvalue of A, $$\tau(A) =\rho(A^{-1})^{-1}$$, where $$\rho(A^{-1})$$ is the spectral radius of the nonnegative matrix $$A^{-1}$$, $$\tau(A) = \min\{|\lambda| : \lambda\in\sigma(A)\}$$, $$\sigma(A)$$ denotes the spectrum of A. $$\tau(A)$$ is called the minimum eigenvalue of A [2, 3].

The Hadamard product of two matrices $$A=(a_{ij})\in \mathbb{R}^{n\times n}$$ and $$B=(b_{ij})\in\mathbb{R}^{n\times n}$$ is the matrix $$A\circ B=(a_{ij}b_{ij})\in\mathbb{R}^{n\times n}$$.

An $$n\times n$$ matrix A is said to be reducible if there exists a permutation matrix P such that

$$P^{T}AP=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} A_{11}&0 \\ A_{21}&A_{22} \end{array}\displaystyle \right ),$$

where $$A_{11}$$, $$A_{22}$$ are square matrices of order at least one. We call A irreducible if it is not reducible. Note that any nonzero $$1\times1$$ matrix is irreducible.

Estimating the bounds for the minimum eigenvalue $$\tau(A)$$ of an M-matrix A is an interesting subject in matrix theory, and it has important applications in many practical problems [46]. Hence, it is necessary to estimate the bounds for $$\tau(A)$$.

In [5], Shivakumar et al. obtained the following bound for $$\tau(A)$$: Let $$A = (a_{ij})\in R^{n\times n}$$ be a weakly chained diagonally dominant M-matrix. Then

$$r(A) \leq\tau(A) \leq R(A), \qquad \tau(A) \leq\min _{ i\in N} a_{ii}, \qquad \frac{1}{M} \leq\tau(A) \leq\frac{1}{m}.$$
(1)

Subsequently, Tian and Huang [7] provided a lower bound for $$\tau(A)$$ by using the spectral radius of the Jacobi iterative matrix $$J_{A}$$ of A: Let $$A = (a_{ij})\in R^{n\times n}$$ be an M-matrix and $$A^{-1} = (\alpha_{ij})$$. Then

$$\tau(A)\geq\frac{1}{1+(n-1)\rho(J_{A})}\frac{1}{\max_{i\in N}\{\alpha_{ii}\}}.$$
(2)

Recently, Li et al. [8] improved (2) and gave the following result: Let $$B = (b_{ij})\in\mathbb{R}^{n\times n}$$ be an M-matrix and $$B^{-1}= (\beta_{ij})$$. Then

$$\tau(B) \geq\frac{2}{\max_{i\neq j}\{ \beta_{ii}+\beta_{jj}+[(\beta_{ii}-\beta_{jj})^{2}+4(n-1)^{2}\beta_{ii}\beta _{jj}\rho^{2}(J_{B}) ]^{\frac{1}{2}}\}}.$$
(3)

In this paper, we continue to research the problems mentioned above. For an M-matrix B, we establish some new inequalities on the bounds for $$\tau(B)$$. Finally, some examples are given to illustrate our results.

For convenience, we employ the following notations throughout. Let $$A=(a_{ij})$$ be an $$n\times n$$ matrix. For $$i, j, k \in N$$, $$i\neq j$$, denote

\begin{aligned}& r_{ji}=\frac{|a_{ji}|}{|a_{jj}|-\sum_{k \neq j, i} |a_{jk}| }, \qquad r_{i}=\max _{j\neq i } \{{r_{ji}}\}, \\& m_{ji}= \frac{|a_{ji}|+\sum_{k \neq j, i} |a_{jk}|r_{i}}{|a_{jj}|}, \qquad h_{i}=\max_{j\neq i } \biggl\{ \frac{|a_{ji}|}{|a_{jj}|m_{ji}-\sum_{k \neq j, i} |a_{jk}| m_{ki}} \biggr\} , \\& u_{ji}=\frac{|a_{ji}|+\sum_{k \neq j, i} |a_{jk}|m_{ki} h_{i} }{|a_{jj}|}, \qquad u_{i}=\max _{j\neq i } \{u_{ij}\}. \end{aligned}

## 2 Main results

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

Let $$A\geq0$$ and $$D=\operatorname{diag}(a_{ii})$$. Denote $$C = A-D$$, $$\mathcal{J}_{A}=D_{1}^{-1}C$$, $$D_{1}=\operatorname{diag}(d_{ii})$$, where

$$d_{ii} = \left \{ \textstyle\begin{array}{l@{\quad}l} 1, & \mbox{if } a_{ii}=0, \\ a_{ii}, & \mbox{if }a_{ii}\neq0. \end{array}\displaystyle \right .$$

By the definition of $$\mathcal{J}_{A}$$, we obtain

$$\rho(\mathcal{J}_{A^{T}})=\rho\bigl(D_{1}^{-1}C^{T} \bigr)=\rho\bigl(CD_{1}^{-1}\bigr)=\rho\bigl( D_{1}^{-1}\bigl(CD_{1}^{-1} \bigr)D_{1} \bigr)=\rho\bigl(D_{1}^{-1}C\bigr)= \rho(\mathcal{J}_{A}).$$

### Lemma 1

[9]

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

$$\bigcup_{i} \biggl\{ z\in C :|z-a_{ii}| \leq x_{i}\sum_{j\neq i}\frac{1}{x_{j}}|a_{ji}|, i\in N \biggr\} .$$

### Lemma 2

[3]

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

$$\bigcup_{j\neq i} \biggl\{ z\in\mathbb{C}: |z-a _{ii}||z-a_{jj}| \leq \biggl( x_{i} \sum _{k \neq i}\frac{1}{x_{k}} |a_{ki}| \biggr) \biggl( x_{j} \sum_{l \neq j} \frac{1}{x_{l}} |a_{lj}| \biggr) \biggr\} .$$

### Lemma 3

[3]

Let $$A, B\in\mathbb{R}^{n\times n}$$, and let $$X, Y\in \mathbb{R}^{n\times n}$$ be diagonal matrices. Then

$$X(A\circ B)Y=(XAY)\circ B=(XA)\circ(BY)=(AY)\circ(XB)=A\circ (XBY).$$

### Lemma 4

[3]

Let $$A=(a_{ij})\in M_{n}$$. Then there exists a positive diagonal matrix X such that $$X^{-1}AX$$ is a strictly diagonally dominant M-matrix

### Lemma 5

[10]

Let $$A=(a_{ij})\in R^{n\times n}$$ be a strictly diagonally dominant matrix and let $$A^{-1}=(\alpha_{ij})$$. Then for all $$i\in N$$,

$$\alpha_{ij}\leq u_{ji} \alpha_{jj}, \quad j\in N, j\neq i.$$

### Theorem 1

Let $$A=(a_{ij}) \geq0$$, $$B=(b_{ij})\in M_{n}$$, and let $$B^{-1}=(\beta_{ij})$$. Then

$$\rho\bigl(A\circ B^{-1}\bigr)\leq\max _{1\leq i \leq n } \bigl\{ \bigl(a_{ii}+u_{i} \rho( \mathcal{J}_{A}) d_{ii} \bigr)\beta_{ii} \bigr\} .$$
(4)

### Proof

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

We next assume that $$n\geq2$$.

(i) First, we assume that A and B are irreducible matrices. Since B is an M-matrix, by Lemma 4, there exists a positive diagonal matrix X, such that $$X^{-1}BX$$ is a strictly row diagonally dominant M-matrix, and

$$\rho\bigl(A\circ B^{-1}\bigr)=\rho\bigl(X^{-1}\bigl(A\circ B^{-1}\bigr)X \bigr)=\rho\bigl(A\circ \bigl(X^{-1}BX \bigr)^{-1} \bigr).$$

Hence, for convenience and without loss of generality, we assume that B is a strictly diagonally dominant matrix.

On the other hand, since A is irreducible and so is $$\mathcal{J}_{A^{T}}$$. Then there exists a positive vector $$x=(x_{i})$$ such that $$\mathcal{J}_{A^{T}}x=\rho(\mathcal{J}_{A^{T}})x=\rho(\mathcal{J}_{A})x$$, thus, we obtain $$\sum_{j\neq i} a_{ji}x_{j}=\rho(\mathcal{J}_{A}) d_{ii}x_{i}$$.

Let $$\widetilde{A}=(\tilde{a}_{ij})=XAX^{-1}$$ in which X is the positive matrix $$X=\operatorname{diag}(x_{1}, x_{2}, \ldots, x_{n})$$. Then, we have

$$\widetilde{A}=(\tilde{a}_{ij})=XAX^{-1} = \left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} a_{11}&\frac{ a_{12}x_{1} }{ x_{2} }& \cdots& \frac{ a_{1n}x_{1} }{ x_{n} }\\ \frac{ a_{21}x_{2} }{ x_{1} }& a_{22} & \cdots& \frac{ a_{2n}x_{2} }{ x_{n} }\\ \vdots& \vdots& \ddots&\vdots\\ \frac{ a_{n1}x_{n} }{ x_{1} }& \frac{ a_{n2}x_{n} }{ x_{2} } & \cdots& a_{nn} \end{array}\displaystyle \right ).$$

From Lemma 3, we have

$$\widetilde{A}\circ B^{-1}=\bigl(XAX^{-1}\bigr)\circ B^{-1}=X\bigl(A\circ B^{-1}\bigr)X^{-1}.$$

Thus, we obtain $$\rho(\widetilde{A}\circ B^{-1})=\rho(A\circ B^{-1})$$. Let $$\lambda=\rho(\widetilde{A}\circ B^{-1})$$, so that $$\lambda\geq a_{ii}\beta_{ii}$$, $$\forall i \in N$$. By Lemma 1, there exists $$i_{0}\in N$$, such that

\begin{aligned} |\lambda-a_{i_{0}i_{0}}\beta_{i_{0}i_{0}}| \leq& u_{i_{0}} \sum _{t \neq i_{0}} \frac{1}{u_{t}} \tilde{a}_{ti_{0}} \beta_{ti_{0}} \leq u_{i_{0}} \sum_{t \neq i_{0}} \frac{1}{u_{t}} \tilde{a}_{ti_{0}} u_{ti_{0}} \beta_{i_{0} i_{0}} \\ \leq& u_{i_{0}} \sum_{t \neq i_{0}} \tilde{a}_{ti_{0}} \beta_{i_{0} i_{0}} = u_{i_{0}} \beta_{i_{0} i_{0}} \sum_{t \neq i_{0}} \frac{ a_{ti_{0}} x_{t} }{ x_{i_{0}}} = u_{i_{0}} \rho(\mathcal{J}_{A}) d_{i_{0}i_{0}} \beta_{i_{0} i_{0}}. \end{aligned}

Therefore,

$$\lambda\leq a_{i_{0}i_{0}}\beta_{i_{0}i_{0}}+ u_{i_{0}} \rho( \mathcal{J}_{A}) d_{i_{0}i_{0}}\beta_{i_{0} i_{0}} = \bigl(a_{i_{0}i_{0}}+ u_{i_{0}} \rho(\mathcal{J}_{A}) d_{i_{0}i_{0}}\bigr)\beta_{i_{0} i_{0}},$$

i.e.,

\begin{aligned} \rho\bigl(A\circ B^{-1}\bigr) \leq&\bigl(a_{i_{0}i_{0}}+ u_{i_{0}} \rho(\mathcal{J}_{A}) d_{i_{0}i_{0}}\bigr) \beta_{i_{0} i_{0}} \\ \leq&\max_{1\leq i \leq n } \bigl\{ \bigl(a_{ii}+u_{i} \rho(\mathcal{J}_{A}) d_{ii} \bigr)\beta_{ii} \bigr\} . \end{aligned}

(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 [1]). If we denote by $$T=(t_{ij})$$ the $$n\times n$$ permutation matrix with $$t_{12}=t_{23}=\cdots=t_{n-1,n}=t_{n1}=-1$$, the remaining $$t_{ij}$$ zero, then both $$A-\varepsilon T$$ and $$B+\varepsilon T$$ are irreducible matrices for any chosen positive real number ε, sufficiently small such that all the leading principal minors of $$B+\varepsilon T$$ are positive. Now, we substitute $$A-\varepsilon T$$ and $$B+\varepsilon T$$ for A and B, respectively, in the previous case, and then letting $$\varepsilon \rightarrow0$$, the result follows by continuity. □

### Theorem 2

Let $$B=(b_{ij})\in M_{n}$$ and $$B^{-1}=(\beta_{ij})$$. Then

$$\tau(B)\geq\frac{1}{ \max_{1\leq i \leq n } \{ (1+u_{i} (n-1 ))\beta_{ii} \} }.$$
(5)

### Proof

Let all entries of A in (4) be 1. Then $$a_{ii}=1$$ ($$\forall i\in N$$), $$\rho(\mathcal{J}_{A})=n-1$$. Therefore, by (4), we have

$$\tau(B)=\frac{1}{ \rho( B^{-1}) }\geq\frac{1}{ \max_{1\leq i \leq n } \{ (1+u_{i} (n-1 ))\beta_{ii} \} }.$$

The proof is completed. □

### Theorem 3

Let $$A=(a_{ij}) \geq0$$, $$B=(b_{ij})\in M_{n}$$, and let $$B^{-1}=(\beta_{ij})$$. Then

$$\rho\bigl(A\circ B^{-1}\bigr)\leq\frac{1}{2} \max _{i\neq j } \{ a_{ii}\beta_{ii}+ a_{jj} \beta_{jj}+\Delta_{ij} \},$$
(6)

where $$\Delta_{ij}=[(a_{ii}\beta_{ii}-a_{jj}\beta_{jj})^{2} +4u_{i}u_{j}\rho^{2}(\mathcal{J}_{A}) d_{ii}d_{jj}\beta_{ii}\beta_{jj}]^{\frac{1}{2}}$$.

### Proof

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

We next assume that $$n\geq2$$. For convenience and without loss of generality, we assume that B is a strictly row diagonally dominant matrix.

(i) First, we assume that A and B are irreducible matrices. Since A is irreducible and so is $$\mathcal{J}_{A^{T}}$$. Then there exists a positive vector $$y=(y_{i})$$ such that $$\mathcal{J}_{A^{T}}y=\rho(\mathcal{J}_{A^{T}})y=\rho(\mathcal{J}_{A})y$$, thus, we obtain

\begin{aligned}& \sum_{k\neq i} a_{ki}y_{k}=\rho( \mathcal{J}_{A}) d_{ii}y_{i}, \\& \sum _{k\neq j} a_{kj}y_{k}=\rho( \mathcal{J}_{A}) d_{jj}y_{j}. \end{aligned}

Let $$\widehat{A}=(\hat{a}_{ij})=YAY^{-1}$$ in which Y is the positive matrix $$Y=\operatorname{diag}(y_{1}, y_{2}, \ldots, y_{n})$$. Then, we have

$$\widehat{A}=(\hat{a}_{ij})=YAY^{-1} = \left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} a_{11}&\frac{ a_{12}y_{1} }{ y_{2} }& \cdots& \frac{ a_{1n}y_{1} }{ y_{n} }\\ \frac{ a_{21}y_{2} }{ y_{1} }& a_{22} & \cdots& \frac{ a_{2n}y_{2} }{ y_{n} }\\ \vdots& \vdots& \ddots&\vdots\\ \frac{ a_{n1}y_{n} }{ y_{1} }& \frac{ a_{n2}y_{n} }{ y_{2} } & \cdots& a_{nn} \end{array}\displaystyle \right ).$$

From Lemma 3, we get

$$\widehat{A}\circ B^{-1}=\bigl(YAY^{-1}\bigr)\circ B^{-1}=Y\bigl(A\circ B^{-1}\bigr)Y^{-1}.$$

Thus, we obtain $$\rho(\widehat{A}\circ B^{-1})=\rho(A\circ B^{-1})$$. Let $$\lambda=\rho(\widehat{A}\circ B^{-1})$$, so that $$\lambda\geq a_{ii}\beta_{ii}$$ ($$\forall i \in N$$). By Lemma 2, there exist $$i_{0}, j_{0}\in N$$, $$i_{0}\neq j_{0}$$, such that

$$|\lambda-a_{i_{0}i_{0}}\beta_{i_{0}i_{0}}||\lambda-a_{j_{0}j_{0}} \beta_{j_{0}j_{0}}| \leq \biggl(u_{i_{0}}\sum _{k \neq i_{0}} \frac{1}{u_{k}} \hat{a}_{ki_{0}} \beta_{ki_{0}} \biggr) \biggl(u_{j_{0}}\sum _{k \neq j_{0}} \frac{1}{u_{k}} \hat{a}_{kj_{0}} \beta_{kj_{0}} \biggr).$$

Note that

\begin{aligned}& u_{i_{0}}\sum_{k \neq i_{0}} \frac{1}{u_{k}} \hat{a}_{ki_{0}}\beta_{ki_{0}} \leq u_{i_{0}}\sum _{k \neq i_{0}}\frac{1}{u_{k}}\hat{a}_{ki_{0}} u_{ki_{0}} \beta_{i_{0}i_{0}} \leq u_{i_{0}}\beta_{i_{0}i_{0}}\sum _{k \neq i_{0}}\hat{a}_{ki_{0}} = u_{i_{0}} \beta_{i_{0}i_{0}}\rho(\mathcal{J}_{A}) d_{i_{0}i_{0}}, \\& u_{j_{0}}\sum_{k \neq j_{0}} \frac{1}{u_{k}} \hat{a}_{kj_{0}}\beta_{kj_{0}} \leq u_{j_{0}}\sum _{k \neq j_{0}}\frac{1}{u_{k}}\hat{a}_{kj_{0}} u_{kj_{0}} \beta_{j_{0}j_{0}} \leq u_{j_{0}}\beta_{j_{0}j_{0}}\sum _{k \neq j_{0}}\hat{a}_{kj_{0}} = u_{j_{0}} \beta_{j_{0}j_{0}}\rho(\mathcal{J}_{A}) d_{j_{0}j_{0}}. \end{aligned}

Hence, we obtain

$$\lambda\leq\frac{1}{2} ( a_{i_{0}i_{0}}\beta_{i_{0}i_{0}}+ a_{j_{0}j_{0}}\beta_{j_{0}j_{0}}+\Delta_{i_{0}j_{0}} ),$$

i.e.,

$$\rho\bigl(A\circ B^{-1}\bigr)\leq\frac{1}{2} ( a_{i_{0}i_{0}} \beta_{i_{0}i_{0}}+ a_{j_{0}j_{0}}\beta_{j_{0}j_{0}}+\Delta_{i_{0}j_{0}} ) \leq\frac{1}{2}\max_{i\neq j } \{ a_{ii} \beta_{ii}+ a_{jj}\beta_{jj}+\Delta_{ij} \} ,$$

where $$\Delta_{ij}=[(a_{ii}\beta_{ii}-a_{jj}\beta_{jj})^{2} +4u_{i}u_{j}\rho^{2}(\mathcal{J}_{A})d_{ii}d_{jj}\beta_{ii}\beta_{jj}]^{\frac{1}{2}}$$.

(ii) Now, assume that one of A and B is reducible. We substitute $$A-\varepsilon T$$ and $$B+\varepsilon T$$ for A and B, respectively, in the previous case (as in the proof of Theorem 1), and then letting $$\varepsilon\rightarrow0$$, the result follows by continuity. □

### Theorem 4

Let $$B=(b_{ij})\in M_{n}$$ and $$B^{-1}=(\beta_{ij})$$. Then

$$\tau( B)\geq\frac{2}{ \max_{i\neq j } \{ \beta_{ii}+ \beta_{jj}+\Delta_{ij} \} },$$
(7)

where $$\Delta_{ij}=[(\beta_{ii}-\beta_{jj})^{2} +4(n-1)^{2} u_{i}u_{j}\beta_{ii}\beta_{jj}]^{\frac{1}{2}}$$.

### Proof

Let all entries of A in (6) be 1. Then

$$a_{ii}=1\quad (\forall i\in N), \qquad \rho(\mathcal{J}_{A})=n-1, \qquad \Delta_{ij}=\bigl[(\beta_{ii}-\beta_{jj})^{2} +4(n-1)^{2} u_{i}u_{j}\beta_{ii} \beta_{jj}\bigr]^{\frac{1}{2}}.$$

Therefore, by (6), we have

$$\tau(B)=\frac{1}{ \rho( B^{-1}) }\geq \frac{2}{ \max_{i\neq j } \{ \beta_{ii}+ \beta_{jj}+\Delta_{ij} \} }.$$

The proof is completed. □

### Remark 1

We next give a simple comparison between (4) and (6), (5) and (7), respectively. For convenience and without loss of generality, we assume that, for $$i, j\in N$$, $$i\neq j$$,

$$a_{jj}\beta_{jj}+u_{j} d_{jj} \beta_{jj}\rho(\mathcal{J}_{A})\leq a_{ii} \beta_{ii}+u_{i} d_{ii} \beta_{ii}\rho( \mathcal{J}_{A}),$$

i.e.,

$$u_{j} d_{jj} \beta_{jj}\rho(\mathcal{J}_{A}) \leq a_{ii}\beta_{ii}-a_{jj}\beta_{jj}+u_{i} d_{ii} \beta_{ii}\rho(\mathcal{J}_{A}).$$

Hence,

\begin{aligned} \Delta_{ij} =&\bigl[(a_{ii}\beta_{ii}-a_{jj} \beta_{jj})^{2} +4u_{i}u_{j} \rho^{2}(\mathcal{J}_{A})d_{ii}d_{jj} \beta_{ii}\beta_{jj}\bigr]^{\frac {1}{2}} \\ \leq&\bigl[(a_{ii}\beta_{ii}-a_{jj} \beta_{jj})^{2} +4u_{i}\rho(\mathcal{J}_{A})d_{ii} \beta_{ii}\bigl( a_{ii}\beta_{ii}-a_{jj} \beta _{jj}+u_{i} d_{ii} \beta_{ii}\rho( \mathcal{J}_{A}) \bigr)\bigr]^{\frac{1}{2}} \\ =&a_{ii}\beta_{ii}-a_{jj}\beta_{jj}+2u_{i} d_{ii} \beta_{ii}\rho(\mathcal{J}_{A}). \end{aligned}

Further, we obtain

$$a_{ii}\beta_{ii}+a_{jj}\beta_{jj}+ \Delta_{ij}\leq 2a_{ii}\beta_{ii}+2u_{i} d_{ii} \beta_{ii}\rho(\mathcal{J}_{A}),$$

i.e.,

$$\rho\bigl(A\circ B^{-1}\bigr)\leq\frac{1}{2} \max _{i\neq j } \{ a_{ii}\beta_{ii}+ a_{jj} \beta_{jj}+\Delta_{ij} \} \leq \max_{1\leq i \leq n } \bigl\{ \bigl(a_{ii}+u_{i} \rho(\mathcal{J}_{A})d_{ii} \bigr)\beta_{ii} \bigr\} .$$

So, the bound in (6) is better than the bound in (4). Similarly, we can prove that the bound in (7) is better than the bound in (5).

## 3 Numerical examples

In this section, we present numerical examples to illustrate the advantages of our derived results.

### Example 1

Let

$$B=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 1.1 &-0.6 & -0.1 \\ -0.3 &1 & -0.6 \\ -0.2 &-0.4 &0.7 \end{array}\displaystyle \right ).$$

It is easy to see that B is an M-matrix. By calculations with Matlab 7.1, we have

\begin{aligned}& \tau(B)\geq0.10000000 \quad (\mbox{by (1)}), \qquad \tau(B)\geq 0.11396723 \quad (\mbox{by (2)}), \\& \tau(B)\geq 0.11582163 \quad (\mbox{by (3)}), \qquad \tau(B)\geq 0.11834016 \quad (\mbox{by (5)}), \\& \tau(B)\geq 0.13163534 \quad (\mbox{by (7)}), \end{aligned}

respectively. In fact, $$\tau(B)= 0.16213193$$. It is obvious that the bound in (7) is the best result.

### Example 2

Let

$$B=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 1 &-0.2 & -0.1 &-0.2 &-0.1 \\ -0.4 &1 & -0.2 &-0.1 &-0.1 \\ -0.3 &-0.2 &1 &-0.1 &-0.1 \\ -0.2 &-0.3 &-0.3 & 1 &-0.1 \\ -0.1 &-0.3 &-0.2 &-0.2 &1 \end{array}\displaystyle \right ).$$

It is easy to see that B is an M-matrix. By calculations with Matlab 7.1, we have

\begin{aligned}& \tau(B)\geq0.10000000 \quad (\mbox{by (1)}),\qquad \tau(B)\geq 0.16082517 \quad (\mbox{by (2)}), \\& \tau(B) \geq 0.16831778 \quad (\mbox{by (3)}), \qquad \tau(B)\geq 0.18147932 \quad (\mbox{by (5)}), \\& \tau(B)\geq 0.19169108 \quad (\mbox{by (7)}), \end{aligned}

respectively. In fact, $$\tau(B)= 0.25807710$$. It is obvious that the bound in (7) is the best result.