# New upper bounds for $$\|A^{-1}\|_{\infty}$$ of strictly diagonally dominant M-matrices

## Abstract

A new upper bound for the infinity norm of inverse matrix of a strictly diagonally dominant M-matrix is given, and the lower bound for the minimum eigenvalue of the matrix is obtained. Furthermore, an upper bound for the infinity norm of inverse matrix of a strictly α-diagonally dominant M-matrix is presented. Finally, we give numerical examples to illustrate our results.

## Introduction

Let $$R^{n\times n}$$ denote the set of all $$n\times n$$ real matrices, $$N=\{1, 2, \ldots, n\}$$ and $$A=(a_{ij})\in R^{n\times n}$$ ($$n\geq2$$). A matrix A is called a nonsingular M-matrix if there exist a nonnegative matrix B and some real number s such that

$$A=sI-B,\quad s>\rho(B),$$

where I is the identity matrix, $$\rho(B)$$ is the spectral radius of B. $$\tau(A)$$ denotes the minimum of all real eigenvalues of the nonsingular M-matrix A.

Very often in numerical analysis, one needs a bound for the condition number of a square $$n\times n$$ matrix A, $$\operatorname{Cond}(A)=\|A\|_{\infty}\cdot\|A^{-1}\|_{\infty}$$. Bounding $$\|A\|_{\infty}$$ is not usually difficult, but a bound of $$\|A^{-1}\|_{\infty}$$ is not usually available unless $$A^{-1}$$ is known explicitly.

However, if $$A=(a_{ij})\in R^{n\times n}$$ is a strictly diagonally dominant matrix, Varah  bound $$\|A^{-1}\|_{\infty}$$ quite easily by the following result:

$$\bigl\Vert A^{-1}\bigr\Vert _{\infty}\leq \frac{1}{\min_{i\in N} \{|a_{ii}|-\sum_{j\neq i}|a_{ij}| \}}.$$
(1)

### Remark 1



If the diagonal dominance of A is weak, i.e., $$\min_{i\in N} \{|a_{ii}|-\sum_{j\neq i}|a_{ij}| \}$$ is small, then using (1) in estimating $$\|A^{-1}\|_{\infty}$$, the bound may yield a large value.

In 2007, Cheng and Huang  presented the following results.

If $$A=(a_{ij})$$ is a strictly diagonally dominant M-matrix, then

$$\bigl\Vert A^{-1}\bigr\Vert _{\infty}\leq \frac{1}{a_{11}(1-u_{1}l_{1})}+\sum_{i=2}^{n} \Biggl[ \frac{1}{a_{ii}(1-u_{i}l_{i})} \prod_{j=1}^{i-1} \biggl(1+ \frac{u_{j}}{1-u_{j}l_{j}} \biggr) \Biggr].$$
(2)

If $$A=(a_{ij})$$ is a strictly diagonally dominant M-matrix, then the bound in (2) is sharper than that in Theorem 3.3 in , i.e.,

$$\frac{1}{a_{11}(1-u_{1}l_{1})}+\sum_{i=2}^{n} \Biggl[ \frac{1}{a_{ii}(1-u_{i}l_{i})} \prod_{j=1}^{i-1} \biggl(1+ \frac{u_{j}}{1-u_{j}l_{j}} \biggr) \Biggr]< \sum_{i=1}^{n} \Biggl[a_{ii} \prod_{j=1}^{i}(1-u_{j}) \Biggr]^{-1} .$$

In 2009, Wang  obtained the better result: Let $$A=(a_{ij})$$ be a strictly diagonally dominant M-matrix. Then

$$\bigl\Vert A^{-1}\bigr\Vert _{\infty}< \frac{1}{a_{11}(1-u_{1}l_{1})}+\sum_{i=2}^{n} \Biggl[ \frac{1}{a_{ii}(1-u_{i}l_{i})} \prod_{j=1}^{i-1} \frac{1}{1-u_{j}l_{j}} \Biggr].$$
(3)

In this paper, we present new upper bounds for $$\|A^{-1}\|_{\infty}$$ of a strictly (α-)diagonally dominant M-matrix A, which improved the above results. As an application, a lower bound of $$\tau(A)$$ is obtained.

For convenience, for $$i,j,k \in N$$, $$j\neq i$$, denote

\begin{aligned}& R_{i}(A)=\sum_{j\neq i} |a_{ij}|, \qquad C_{i}(A)=\sum_{j\neq i} |a_{ji}|,\qquad d_{i}=\frac{R_{i}(A)}{|a_{ii}|}, \\& J(A)=\{i\in N| d_{i}< 1\},\qquad u_{i}=\frac{\sum_{j=i+1}^{n} |a_{ij}|}{|a_{ii}|},\qquad l_{k}=\max _{k\leq i\leq n} \biggl\{ \frac{\sum_{k\leq j\leq n} |a_{ij}|}{|a_{ii}|} \biggr\} , \\& l_{n}=u_{n}=0, \qquad r_{ji}=\frac {|a_{ji}|}{|a_{jj}|-\sum_{k\neq{j,i}}|a_{jk}|},\qquad r_{i}=\max_{j\neq{i}}\{{r_{ji}}\}, \\& \sigma_{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}|\sigma_{ji}-\sum_{k\neq j, i} |a_{jk}|\sigma_{ki}} \biggr\} , \\& u_{ji}=\frac{|a_{ji}|+\sum_{k\neq{j,i}}|a_{jk}|\sigma_{ki}h_{i}}{|a_{jj}|},\qquad \omega_{ji}= \frac{|a_{ji}|+\sum_{k\neq{j,i}}|a_{jk}|u_{ki}}{|a_{jj}|}. \end{aligned}

We will denote by $$A^{(n_{1},n_{2})}$$ the principal submatrix of A formed from all rows and all columns with indices between $$n_{1}$$ and $$n_{2}$$ inclusively; e.g., $$A^{(2,n)}$$ is the submatrix of A obtained by deleting the first row and the first column of A.

### Definition 1



$$A=(a_{ij})\in R^{n\times n}$$ is a weakly chained diagonally dominant if for all $$i\in N$$, $$d_{i}\leq1$$ and $$J(A)\neq\phi$$, and for all $$i\in N$$, $$i\notin J(A)$$, there exist indices $$i_{1}, i_{2},\ldots,i_{k}$$ in N with $$a_{i_{r},i_{r+1}}\neq0$$, $$0\leq r\leq k-1$$, where $$i_{0}=i$$ and $$i_{k}\in J(A)$$.

### Definition 2



$$A=(a_{ij})\in R^{n\times n}$$ is called a strictly α-diagonally dominant matrix if there exists $$\alpha\in[0,1]$$ such that

$$|a_{ii}|> \alpha R_{i}(A)+(1- \alpha)C_{i}(A),\quad \forall i\in N.$$

## Upper bounds for $$\|A^{-1}\|_{\infty}$$ of a strictly diagonally dominant M-matrix

In this section, we give several bounds of $$\|A^{-1}\|_{\infty}$$ and $$\tau(A)$$ for a strictly diagonally dominant M-matrix A.

### Lemma 1



Let $$A=(a_{ij})$$ be a weakly chained diagonally dominant M-matrix, $$B=A^{(2,n)}$$, $$A^{-1}=(\alpha_{ij})$$, and $$B^{-1}=(\beta_{ij})$$. Then, for $$i,j=2,\ldots,n$$,

\begin{aligned}& \alpha_{11}=\frac{1}{\triangle},\qquad \alpha_{i1}= \frac{1}{\triangle}\sum_{k=2}^{n} \beta_{ik}(-a_{k1}),\qquad \alpha_{1j}= \frac{1}{\triangle}\sum_{k=2}^{n} \beta_{kj}(-a_{1k}), \\& \alpha_{ij}=\beta_{ij}+\alpha_{1j}\sum _{k=2}^{n}\beta_{ik}(-a_{k1}), \qquad \triangle=a_{11}-\sum_{k=2}^{n}a_{1k} \Biggl(\sum_{i=2}^{n}\beta _{ki}a_{i1} \Biggr)>0. \end{aligned}

Furthermore, if $$J(A)=N$$, then

$$\triangle\geq a_{11}(1-d_{1}l_{1})\geq a_{11}(1-d_{1}).$$

### Lemma 2



If $$A=(a_{ij})$$ is a strictly diagonally dominant M-matrix, then

$$\triangle\geq a_{11}(1-d_{1}l_{1})> a_{11}(1-d_{1})>0.$$

### Lemma 3

Let $$A=(a_{ij})$$ be a strictly diagonally dominant M-matrix. Then, for $$A^{-1}=(\alpha_{ij})$$,

$$\alpha_{ji}\leq\omega_{ji}\alpha_{ii}, \quad i, j\in N, j\neq i.$$

### Proof

This proof is similar to the one of Lemma 2 in . □

### Lemma 4

Let $$A=(a_{ij})$$ be a strictly diagonally dominant M-matrix. Then, for $$A^{-1}=(\alpha_{ij})$$,

$$\frac{1}{a_{ii}}\leq\alpha_{ii}\leq \frac{1}{a_{ii}-\sum_{j\neq i}|a_{ij}|\omega_{ji}},\quad i\in N.$$

### Proof

This proof is similar to the one of Lemma 2.3 in . □

### Lemma 5



Let $$A=(a_{ij})$$ be a weakly chained diagonally dominant M-matrix, $$A^{-1}=(\alpha_{ij})$$, and $$\tau=\tau(A)$$. Then

$$\tau\leq\min_{i\in N} \{a_{ii} \},\qquad \tau\leq\max _{i\in N} \biggl\{ \sum_{j\in N}a_{ij} \biggr\} ,\qquad \tau\geq\min_{i\in N} \biggl\{ \sum _{j\in N}a_{ij} \biggr\} ,\quad \frac{1}{M}\leq \tau\leq\frac{1}{m},$$

where

$$M=\max_{i\in N} \biggl\{ \sum_{j\in N} \alpha_{ij} \biggr\} =\bigl\Vert A^{-1}\bigr\Vert _{\infty},\qquad m=\min_{i\in N} \biggl\{ \sum _{j\in N}\alpha_{ij} \biggr\} .$$

### Theorem 1

Let $$A=(a_{ij})$$ be a strictly diagonally dominant M-matrix, $$B=A^{(2,n)}$$, $$A^{-1}=(\alpha_{ij})$$, and $$B^{-1}=(\beta_{ij})$$. Then

$$\bigl\Vert A^{-1}\bigr\Vert _{\infty}\leq \frac{1}{a_{11}-\sum_{j=2}^{n}|a_{1j}|\omega_{j1}}+\frac{1}{1-d_{1}l_{1}}\bigl\Vert B^{-1}\bigr\Vert _{\infty}.$$

### Proof

Let

$$\eta_{i}=\sum_{j=1}^{n} \alpha_{ij},\qquad M_{A}=\bigl\Vert A^{-1}\bigr\Vert _{\infty}, \qquad M_{B}=\bigl\Vert B^{-1}\bigr\Vert _{\infty}.$$

Then

$$M_{A}=\max_{i\in N}\{\eta_{i}\}, \qquad M_{B}=\max_{2\leq i\leq n} \Biggl\{ \sum _{j=2}^{n}\beta_{ij} \Biggr\} .$$

By Lemma 1, Lemma 2, and Lemma 4,

\begin{aligned} \eta_{1} =&\alpha_{11}+\sum _{j=2}^{n}\alpha_{1j} =\frac{1}{\triangle}+ \frac{1}{\triangle}\sum_{k=2}^{n}(-a_{1k}) \sum_{j=2}^{n}\beta_{kj} \leq \frac{1}{\triangle}+\frac{1}{\triangle}a_{11}d_{1}M_{B} \\ \leq&\frac{1}{\triangle}+\frac{d_{1}M_{B}}{1-d_{1}l_{1}} \leq\frac{1}{a_{11}-\sum_{j=2}^{n}|a_{1j}|\omega_{j1}}+ \frac{M_{B}}{1-d_{1}l_{1}}. \end{aligned}
(4)

Let $$2\leq i\leq n$$. Then, by Lemma 1 and Lemma 3,

\begin{aligned}& \sum_{k=2}^{n}\beta_{ik}(-a_{k1})= \triangle\cdot\alpha_{i1}\leq \triangle \omega_{i1} \alpha_{11}=\omega_{i1}< 1, \\& \alpha_{ij}=\beta_{ij}+\alpha_{1j}\sum _{k=2}^{n}\beta_{ik}(-a_{k1})\leq \beta_{ij}+\alpha_{1j}\omega_{i1}< \beta_{ij}+\alpha_{1j}. \end{aligned}

Therefore, for $$2\leq i\leq n$$, we have

\begin{aligned} \eta_{i} =&\alpha_{i1}+\sum _{j=2}^{n}\alpha_{ij} \leq \alpha_{11}\omega_{i1}+\sum_{j=2}^{n} (\beta_{ij}+\alpha_{1j}\omega _{i1} ) = \eta_{1}\omega_{i1}+M_{B} \leq \eta_{1}l_{1}+M_{B} \\ \leq& \biggl(\frac{1}{\triangle}+\frac{d_{1}M_{B}}{1-d_{1}l_{1}} \biggr)l_{1}+M_{B} \leq\frac{1}{\triangle}+\frac{M_{B}}{1-d_{1}l_{1}} \leq\frac{1}{a_{11}-\sum_{j=2}^{n}|a_{1j}|\omega_{j1}}+ \frac{M_{B}}{1-d_{1}l_{1}}. \end{aligned}
(5)

Furthermore, from (4) and (5), we obtain

$$M_{A}\leq \frac{1}{a_{11}-\sum_{j=2}^{n}|a_{1j}|\omega_{j1}}+\frac{1}{1-d_{1}l_{1}}\bigl\Vert B^{-1}\bigr\Vert _{\infty}.$$
(6)

The result follows. □

### Theorem 2

Let $$A=(a_{ij})$$ be a strictly diagonally dominant M-matrix. Then

$$\bigl\Vert A^{-1}\bigr\Vert _{\infty}\leq \frac{1}{a_{11}-\sum_{k=2}^{n}|a_{1k}|\omega_{k1}}+\sum_{i=2}^{n} \Biggl[ \frac{1}{a_{ii}-\sum_{k=i+1}^{n}|a_{ik}|\omega_{ki}} \prod_{j=1}^{i-1} \frac{1}{1-u_{j}l_{j}} \Biggr].$$
(7)

### Proof

The result follows by applying the principle of mathematical induction with respect to k on $$A^{(k,n)}$$ in (6). □

By Lemma 5 and Theorem 1, we can obtain a new bound of $$\tau(A)$$.

### Corollary 1

If $$A=(a_{ij})$$ is a strictly diagonally dominant M-matrix, then

$$\tau(A)\geq \Biggl\{ \frac{1}{a_{11}-\sum_{k=2}^{n}|a_{1k}|\omega_{k1}}+\sum_{i=2}^{n} \Biggl[ \frac{1}{a_{ii}-\sum_{k=i+1}^{n}|a_{ik}|\omega_{ki}} \prod_{j=1}^{i-1} \frac{1}{1-u_{j}l_{j}} \Biggr] \Biggr\} ^{-1}.$$

### Theorem 3

Let $$A=(a_{ij})$$ be a strictly diagonally dominant M-matrix. Then the bound in (7) is better than that in (3), i.e.,

\begin{aligned}& \frac{1}{a_{11}-\sum_{k=2}^{n}|a_{1k}|\omega_{k1}}+\sum_{i=2}^{n} \Biggl[ \frac{1}{a_{ii}-\sum_{k=i+1}^{n}|a_{ik}|\omega_{ki}} \prod_{j=1}^{i-1} \frac{1}{1-u_{j}l_{j}} \Biggr] \\& \quad \leq\frac{1}{a_{11}(1-u_{1}l_{1})}+\sum_{i=2}^{n} \Biggl[\frac{1}{a_{ii}(1-u_{i}l_{i})} \prod_{j=1}^{i-1} \frac{1}{1-u_{j}l_{j}} \Biggr]. \end{aligned}

### Proof

Since A is a strictly diagonally dominant matrix, so $$0\leq u_{j}$$, $$l_{j}<1$$ for all j. By the definition of $$u_{i}$$, $$l_{i}$$, $$\omega_{ki}$$, we have $$\omega_{ki}\leq l_{i}$$ and $$a_{ii}u_{i}=\sum_{k=i+1}^{n}|a_{ik}|$$ for all i. Obviously, the result follows. □

## Upper bounds for $$\|A^{-1}\|_{\infty}$$ of a strictly α-diagonally dominant M-matrix

In this section, we present an upper bound of $$\|A^{-1}\|_{\infty}$$ for a strictly α-diagonally dominant M-matrix A.

### Lemma 6



Let $$A, B\in R^{n\times n}$$. If A and $$A-B$$ are nonsingular, then

$$(A-B)^{-1}=A^{-1}+A^{-1}B \bigl(I-A^{-1}B\bigr)^{-1}A^{-1}.$$

### Lemma 7

Let $$A=(a_{ij})\in R^{n\times n}$$ be a strictly diagonally dominant M-matrix, and $$B=(b_{ij}) \in R^{n\times n}$$. If $$\varphi_{0} \cdot \|B\|_{\infty}<1$$, then $$\|A^{-1}B\|_{\infty}<1$$, where

$$\varphi_{0}=\frac{1}{a_{11}-\sum_{k=2}^{n}|a_{1k}|\omega_{k1}}+\sum_{i=2}^{n} \Biggl[ \frac{1}{a_{ii}-\sum_{k=i+1}^{n}|a_{ik}|\omega_{ki}} \prod_{j=1}^{i-1} \frac{1}{1-u_{j}l_{j}} \Biggr].$$

### Proof

By Theorem 2, we get

$$\bigl\Vert A^{-1}B\bigr\Vert _{\infty}\leq\bigl\Vert A^{-1}\bigr\Vert _{\infty}\|B\|_{\infty}\leq \varphi_{0} \|B\|_{\infty}< 1.$$

The result follows. □

### Lemma 8



If $$\|A^{-1}\|_{\infty}<1$$, then $$I-A$$ is nonsingular and

$$\bigl\Vert (I-A)^{-1}\bigr\Vert _{\infty}\leq \frac{1}{1-\|A\|_{\infty}}.$$

### Theorem 4

Let $$A=(a_{ij})\in R^{n\times n}$$ be a strictly α-diagonally dominant matrix, $$\alpha\in(0, 1]$$ and A be an M-matrix. If $$\{ i\in N| R_{i}(A)>C_{i}(A) \}\neq\emptyset$$, and

$$\varphi_{1}< \frac{1}{ \max_{1\leq i\leq n}\alpha(R_{i}(A)-C_{i}(A)) },$$

then

$$\bigl\Vert A^{-1}\bigr\Vert _{\infty}< \frac{\varphi_{1}}{1-\varphi_{1} \max_{1\leq i\leq n}\alpha(R_{i}(A)-C_{i}(A))},$$
(8)

where

\begin{aligned}& \varphi_{1}=\frac{1}{\nu_{1}-\sum_{k=2}^{n}|a_{1k}|\omega_{k1}}+\sum_{i=2}^{n} \Biggl[ \frac{1}{\nu_{i}-\sum_{k=i+1}^{n}|a_{ik}|\omega_{ki}} \prod_{j=1}^{i-1} \frac{1}{1-u_{j}l_{j}} \Biggr], \\& \nu_{i}=\max_{1\leq i\leq n} \bigl\{ a_{ii}, a_{ii}+ \alpha\bigl(R_{i}(A)-C_{i}(A)\bigr) \bigr\} . \end{aligned}

### Proof

Let $$A=B-C$$, where $$B=(b_{ij})$$, $$C=(c_{ij})$$, and

\begin{aligned}& b_{ij}=\left \{ \begin{array}{l@{\quad}l} a_{ii}+\alpha(R_{i}(A)-C_{i}(A)), & i=j, R_{i}(A)>C_{i}(A), \\ a_{ij}, &\mbox{otherwise}, \end{array} \right . \\& c_{ij}=\left \{ \begin{array}{l@{\quad}l} \alpha(R_{i}(A)-C_{i}(A)), & i=j, R_{i}(A)>C_{i}(A), \\ 0, &\mbox{otherwise}. \end{array} \right . \end{aligned}

For any $$i\in\{ i\in N| R_{i}(A)>C_{i}(A)\}$$, we get

$$b_{ii}=a_{ii}+\alpha\bigl(R_{i}(A)-C_{i}(A) \bigr)>R_{i}(A)=R_{i}(B).$$

For any $$i\in\{ i\in N| R_{i}(A)\leq C_{i}(A)\}$$, we have

$$b_{ii}=a_{ii}>\alpha R_{i}(A)+(1- \alpha)C_{i}(A) \geq R_{i}(A)=R_{i}(B).$$

Thus, B is a strictly diagonal dominant M-matrix. By Lemma 7, we get $$\|B^{-1}C\|_{\infty}<1$$. By Lemma 6, Lemma 8, and Theorem 2, we have

\begin{aligned} \bigl\Vert B^{-1}\bigr\Vert _{\infty} \leq& \frac{1}{b_{11}-\sum_{k=2}^{n}|a_{1k}|\omega_{k1}}+\sum_{i=2}^{n} \Biggl[ \frac{1}{b_{ii}-\sum_{k=i+1}^{n}|a_{ik}|\omega_{ki}} \prod_{j=1}^{i-1} \frac{1}{1-u_{j}l_{j}} \Biggr] \\ =& \frac{1}{\nu_{1}-\sum_{k=2}^{n}|a_{1k}|\omega_{k1}}+\sum_{i=2}^{n} \Biggl[ \frac{1}{\nu_{i}-\sum_{k=i+1}^{n}|a_{ik}|\omega_{ki}} \prod_{j=1}^{i-1} \frac{1}{1-u_{j}l_{j}} \Biggr]. \end{aligned}

Therefore

$$\bigl\Vert B^{-1}C\bigr\Vert _{\infty}\leq \varphi_{1} \max_{1\leq i\leq n}\alpha\bigl(R_{i}(A)-C_{i}(A) \bigr) .$$

Furthermore, we have

\begin{aligned} \bigl\Vert A^{-1}\bigr\Vert _{\infty} =&\bigl\Vert (B-C)^{-1}\bigr\Vert _{\infty}=\bigl\Vert B^{-1}+B^{-1}C\bigl(I-B^{-1}C\bigr)^{-1}B^{-1} \bigr\Vert _{\infty}\\ \leq&\bigl\Vert B^{-1}\bigr\Vert _{\infty}+\bigl\Vert B^{-1}C\bigr\Vert _{\infty}\cdot\bigl\Vert \bigl(I-B^{-1}C\bigr)^{-1}\bigr\Vert _{\infty}\cdot\bigl\Vert B^{-1}\bigr\Vert _{\infty}\\ \leq&\bigl\Vert B^{-1}\bigr\Vert _{\infty}+ \frac{\|B^{-1}C\|_{\infty}}{1-\|B^{-1}C\|_{\infty}} \bigl\Vert B^{-1}\bigr\Vert _{\infty}\\ =& \frac{\|B^{-1}\|_{\infty}}{1-\|B^{-1}C\|_{\infty}} \\ \leq&\frac{\varphi_{1}}{1-\varphi_{1} \max_{1\leq i\leq n}\alpha(R_{i}(A)-C_{i}(A))}. \end{aligned}

The result follows. □

## Numerical examples

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

### Example 1

Let

$$A=\left ( \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} 37&-1&-3&-1&-2&-4&-2&-3&-1&-5\\ -4&30&-1&-2&-3&-4&0&-1&-1&-3\\ -1&-3&30&-4&0&-2&-3&-2&-4&-5\\ -3&-5&-3&40&-1&-2&-3&-4&-2&-4\\ -5&-2&0&-5&25.01&-5&0&-1&-5&-2\\ -2&0&-2&-1&-4&30&-5&-2&-5&-3\\ 0&-3&-1&-1&-2&-4&40&-2&-3&-4\\ -1&-3&-2&-3&-2&-1&-2&40&-4&-1\\ -2&-4&-3&-1&-3&-3&-4&0&27&-2\\ -2&-1&0&-2&-4&-3&-1&0&-3&25 \end{array} \right ).$$

It is easy to see that A is a strictly diagonally dominant M-matrix. By calculations with Matlab 7.1, we have

\begin{aligned}& \bigl\Vert A^{-1}\bigr\Vert _{\infty}\leq100 \quad ( \mbox{by (1)}),\qquad \bigl\Vert A^{-1}\bigr\Vert _{\infty}\leq 11.2862\quad (\mbox{by (2)}), \\& \bigl\Vert A^{-1}\bigr\Vert _{\infty}\leq5.2305\quad ( \mbox{by (3)}),\qquad \bigl\Vert A^{-1}\bigr\Vert _{\infty}\leq 1.0003\quad (\mbox{by (7)}), \end{aligned}

respectively. It is obvious that the bound in (7) is the best result.

### Example 2

Let

$$A=\left ( \begin{array}{@{}c@{\quad}c@{\quad}c@{}} 2&-1&-1\\ -1&2&-1\\ -0.5&-0.5&2 \end{array} \right ).$$

It is easy to see that A is a strictly α-diagonally dominant M-matrix by taking $$\alpha=0.5$$, and A is not a strictly diagonally dominant matrix. Thus the bound of $$\|A^{-1}\|_{\infty}$$ cannot be estimated by (1), (2), and (3), but it can be estimated by (8). By (8), we get

$$\bigl\Vert A^{-1}\bigr\Vert _{\infty}\leq8.0322.$$

## References

1. Varah, JM: A lower bound for the smallest singular value of a matrix. Linear Algebra Appl. 11, 3-5 (1975)

2. Cheng, GH, Huang, TZ: An upper bound for A −1 of strictly diagonally dominant M-matrices. Linear Algebra Appl. 426, 667-673 (2007)

3. Shivakumar, PN, Williams, JJ, Ye, Q, Marinov, CA: On two-sided bounds related to weakly diagonally dominant M-matrices with application to digital circuit dynamics. SIAM J. Matrix Anal. Appl. 17, 298-312 (1996)

4. Wang, P: An upper bound for A −1 of strictly diagonally dominant M-matrices. Linear Algebra Appl. 431, 511-517 (2009)

5. Zhang, YL, Mo, HM, Liu, JZ: α-Diagonal dominance and criteria for generalized strictly diagonally dominant matrices. Numer. Math. 31, 119-128 (2009)

6. 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)

7. Li, YT, Chen, FB, Wang, DF: New lower bounds on eigenvalue of the Hadamard product of an M-matrix and its inverse. Linear Algebra Appl. 430, 1423-1431 (2009)

8. Xu, S: Theory and Methods about Matrix Computation. Tsinghua University Press, Beijing (1986)

## Acknowledgements

This work was supported by the National Natural Science Foundation of China (11361074, 71161020) and IRTSTYN, Applied Basic Research Programs of Science and Technology Department of Yunnan Province (2013FD002).

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Correspondence to Feng Wang.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

The authors contributed equally to this work. All authors read and approved the final manuscript.

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