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.

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

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