Open Access

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

Journal of Inequalities and Applications20152015:172

https://doi.org/10.1186/s13660-015-0696-2

Received: 4 January 2015

Accepted: 17 May 2015

Published: 30 May 2015

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.

Keywords

diagonal dominance M-matrixinfinity normupper boundminimum eigenvalue

MSC

15A4215A45

1 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 [1] 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

[2]

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 [2] 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 [3], 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 [4] 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

[3]

\(A=(a_{ij})\in R^{n\times n}\) 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

[5]

\(A=(a_{ij})\in R^{n\times n}\) is called a strictly α-diagonally dominant matrix if there exists \(\alpha\in[0,1]\) such that
$$ |a_{ii}|> \alpha R_{i}(A)+(1- \alpha)C_{i}(A),\quad \forall i\in N. $$

2 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

[2]

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

[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 [6]. □

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 [7]. □

Lemma 5

[3]

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. □

3 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

[8]

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

[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. □

4 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
$$ \bigl\Vert A^{-1}\bigr\Vert _{\infty}\leq8.0322. $$

Declarations

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).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
College of Science, Guizhou Minzu University

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