A modified iteration for total asymptotically nonexpansive mappings in Hadamard spaces

1.   Introduction

The linear complementarity problem is to find a vector $x\in R^{n}$ satisfying

where $A$ is an $n\times n$ real matrix, and $q\in R^{n}$. Usually, it is denoted by $LCP(A, q)$.

$LCP(A, q)$ arises in many applications such as finding a Nash equilibrium point of a bimatrix game, the network equilibrium problem, the contact problem etc. For details, see ^[1]. Moreover, it has been shown that the $LCP(A, q)$ has a unique solution for any vector $q\in R^{n}$ if and only if $A$ is a $P$-matrix, where an $n\times n$ real matrix $A$ is called a $P$-matrix if all its principal minors are positive. Therefore, the class of $P$-matrices plays an important role in $LCP(A, q)$; see ^[2,3,4].

The cases when the matrix $A$ for the $LCP(A, q)$ belongs to $P$-matrices or some subclasses of $P$-matrices have been widely studied, such as $B$-matrices ^[5,6], $SB$-matrices ^[7], $DB$-matrices ^[8] and so on ^{[9,10,11,12,13]}. The class of $B_1$-matrices is a subclass of $P$-matrices that contains $B$-matrices, which was proposed by Pe$\tilde{n}$a ^[14]. However, the error bound for the linear complementarity problem of $B_1$-matrices has not been reported yet.

At the end of this section, the structure of the article is given. Some properties of $B_1$-matrices are given in Section 2. In Section 3, the infinity norm upper bound for the inverse of $B_1$-matrices is obtained. In Section 4, the error bound for the linear complementarity problem corresponding to $B_1$-matrices is proposed.

2.   Some properties for $B_1$-matrices

In this section, some properties for $B_1$-matrices are proposed. To begin with, some notations, definitions and lemmas are listed as follows.

Let $n$ be an integer number, $N = \{1, 2, \ldots, n\}$ and $C^{n\times n}$ be the set of all complex matrices of order $n$.

$r_{i}(A) = \sum\limits_{j\neq i}|a_{ij}|,$

$N_{1}(A) = \{i\in N:|a_{ii}|\leq r_{i}(A)\}$,

$N_{2}(A) = \{i\in N:|a_{ii}| > r_{i}(A)\}$,

$p_{i}(A) = \sum\limits_{{j\in N_{1}(A)}\backslash\{i\}}{|a_{ij}|}+\sum\limits_{{j\in N_{2}(A)}\backslash\{i\}}{\frac{r_{j}(A)}{|a_{jj}|}}|a_{ij}|$,

$r_{i_{A}}^{+} = max\{0, a_{ij}|i\neq j\}$,

$B^+ = (b_{ij}^{+})_{1\leq i, j \leq n} = \left[\begin{matrix} a_{11}-r_{1_{A}}^{+} \qquad a_{12}-r_{1_{A}}^{+} & \cdots & a_{1n}-r_{1_{A}}^{+}\\ a_{21}-r_{2_{A}}^{+} \qquad a_{22}-r_{2_{A}}^{+} & \ddots & a_{2n}-r_{2_{A}}^{+}\\ \vdots & \ddots & \vdots \\ a_{n1}-r_{n_{A}}^{+} \qquad a_{n2}-r_{n_{A}}^{+} & \cdots & a_{nn}-r_{n_{A}}^{+} \end{matrix} \right]$,

$p_{i}(B^{+}) = \sum\limits_{{j\in N_{1}(A)}\backslash\{i\}}{|a_{ij}-r_{i_{A}}^{+}|}+\sum\limits_{{j\in N_{2}(A)}\backslash \{i\}}{\frac{r_{j}(B^{+})}{|a_{jj}-r_{j_{A}}^{+}|}}|a_{ij}-r_{i_{A}}^{+}|.$

Definition 2.1. ^[2] A matrix $A = (a_{ij})\in C^{n\times n}$ is a $P$-matrix if all its principal minors are positive.

Definition 2.2. ^[14] A matrix $A = (a_{ij})\in C^{n\times n}$ is an $SDD_{1}$ by rows if for each $i \in N_{1}(A)$,

Definition 2.3. ^[14] A matrix $A = (a_{ij})\in C^{n\times n}$ is a $B_{1}$-matrix if for all $i\in N$,

Lemma 2.1. ^[14] If a matrix $A = (a_{ij})\in C^{n\times n}$ is an $SDD_{1}$ by rows, then it is also a nonsingular $H$-matrix.

Lemma 2.2. ^[15] If a matrix $A = (a_{ij})\in C^{n\times n}$ is an $H$-matrix with positive diagonal entries, then it is also a $P$-matrix.

In the following, some properties of $B_1$-matrices are derived.

First of all, utilizing Definitions 2.2 and 2.3, Theorems 2.1–2.4 can be easily obtained.

Theorem 2.1. Let matrix $A = (a_{ij})\in C^{n\times n}$ be a $B_{1}$-matrix. Then, $B^{+}$ is an $SDD_{1}$ matrix.

Example 2.1. Let matrix

We write $A = B^{+}+C$, where

and

By calculation, we have that

and

From Definition 2.3, it is easy to obtain that $A$ is a $B_{1}$-matrix, and $B^{+}$ is an $SDD_{1}$ matrix since $|b_{22}^{+}| = 2 > 0.0300 = p_{2}(B^{+}) = \frac{r_{1}(B^{+})}{|a_{11}-r_{1_{A}}^{+}|}|a_{21}-r_{2_{A}}^{+}|+\frac{r_{3}(B^{+})}{|a_{33}-r_{3_{A}}^{+}|}|a_{23}-r_{2_{A}}^{+}|$ from Definition 2.2.

However, Theorem 2.1 is not true on the contrary, and the following Example 2.2 illustrates this fact.

Example 2.2. Let us consider the matrix

We write $A = B^{+}+C$, where

and

By calculation, we have that

From Definitions 2.2 and 2.3, we obtain that $B^{+}$ is an $SDD_{1}$ matrix since $|b_{22}^{+}| = 3 > 0 = p_{2}(B^{+}) = \frac{r_{1}(B^{+})}{|a_{11}-r_{1_{A}}^{+}|}|a_{21}-r_{2_{A}}^{+}|+\frac{r_{3}(B^{+})}{|a_{33}-r_{3_{A}}^{+}|}|a_{23}-r_{2_{A}}^{+}|$, but $A$ is not a $B_{1}$-matrix since $a_{11}-r_{1_{A}}^{+} = -4 < 0 = p_{1}(B^{+}) = |a_{12}-r_{1_{A}}^{+}|+\frac{r_{3}(B^{+})}{|a_{33}-r_{3_{A}}^{+}|}|a_{13}-r_{1_{A}}^{+}|$.

Motivated by Example 2.2, one can easily obtain Theorem 2.2.

Theorem 2.2. Let matrix $A = (a_{ij})\in C^{n\times n}$ be a $B_{1}$-matrix if and only if $B^{+}$ is an $SDD_{1}$ matrix with positive diagonal entries.

Note that $B^{+}$ is a $Z$-matrix with positive diagonal entries from the definition of $B^{+}$, and it is easy to obtain Theorem 2.3.

Theorem 2.3. Let matrix $A = (a_{ij})\in C^{n\times n}$ be a $B_{1}$-matrix if and only if $B^{+}$ is a $B_{1}$-matrix.

Example 2.3. Let us consider the matrix

We write $A = B^{+}+C$, where

and

By calculation, we have that

and

From Definition 2.3, we get that $A$ is a $B_{1}$-matrix, and $B^{+}$ is also a $B_{1}$-matrix since $b_{11}^{+}-r_{1_{B^{+}}}^{+} = 1 > 0.5100 = p_{1}(V^{+})$, $b_{22}^{+}-r_{2_{B^{+}}}^{+} = 3 > 2 = p_{2}(V^{+})$ and $b_{33}^{+}-r_{3_{B^{+}}}^{+} = 100 > 1 = p_{3}(V^{+})$, where $p_{i}(V^{+}) = p_{i}(B^{+})$ since $r_{i_{B^{+}}}^{+} = 0$. Consequently, Theorem 2.3 is shown to be valid by the example provided in Example 2.3.

Note that when $A$ is a $Z$-matrix, we have that $r_{i_{A}}^{+} = 0$ and $B^{+}$ = $A$, and therefore, it is easy to obtain Theorem 2.4.

Theorem 2.4. If $A = (a_{ij})\in C^{n\times n}$ is a $Z$-matrix with positive diagonal entries, then $A$ is a $B_1$-matrix if and only if $A$ is an $SDD_1$ matrix.

Example 2.4. Let us consider the $Z$-matrix with positive diagonal entries

We write $A = B^{+}+C$, where

and

That is, $A$ = $B^{+}$. By calculation, we have that

and

From Definition 2.3, we get that $A$ is a $B_{1}$-matrix, and $A$ is also an $SDD_{1}$ matrix since $|a_{11}| = 1 > 0 = p_{1}(A)$. Therefore, Example 2.4 illustrates that Theorem 2.4 is valid.

Next, some properties between $B_{1}$-matrix and nonnegative diagonal matrix, $P$-matrix are proposed.

Theorem 2.5. If $A = (a_{ij})\in C^{n\times n}$ is a $B_{1}$-matrix, and $D$ is a nonnegative diagonal matrix of the same order, then $A+D$ is a $B_{1}$-matrix.

Proof. Let $D = diag(d_{1}, d_{2}, \ldots, d_{n})$ with $d_{i}\geq 0$, and $C = A+D$ with $C = (c_{ij})\in C^{n\times n}$, where

Next, let us prove $c_{ii}-r_{i_{C}}^{+} > p_{i}(V^{+})$ for all $i\in N$, where $V^{+} = (v_{ij})\in C^{n\times n}$ with $v_{ij} = c_{ij}-r_{i_{C}}^{+}$, and $r_{i_{C}}^{+} = max\{0, c_{ij}|i\neq j\} = r_{i_{A}}^{+}$.

Since $A$ is a $B_1$-matrix, for all $i\in N$,

For $i\in N_{1}({V^{+}})\subseteq N_{1}({B^{+}})$,

By (2.3), we deduce that $c_{ii}-r_{i_{C}}^{+} > p_i(B^{+})\geq p_{i}(V^{+})$, and this proof is completed.

Example 2.5. Let the matrix

and

By calculation, we have that

and

From Definition 2.3, we get that $A$ is a $B_{1}$-matrix. Further, Theorem 2.5 demonstrates that $A+D$ satisfies the conditions required for a $B_1$-matrix.

Theorem 2.6. If $A = (a_{ij})\in C^{n\times n}$ is a $B_{1}$-matrix, then we write $A$ as $A = B+C$, where $B$ is a $Z$-matrix with positive diagonal entries, and $C$ is a nonnegative matrix of rank 1. In particular, if $A$ is a $B_{1}$-matrix and $Z$-matrix, then $C$ is a zero matrix.

Proof. Let us define $B = (b_{ij})\in C^{n\times n}$ with $b_{ij} = a_{ij}-r_{i_{A}}^{+}$. Taking into account that $b_{ij}\leq 0$ from definition of $r_{i_{A}}^{+}$, and $b_{ii} > 0$ since $A$ is a $B_{1}$-matrix, then $B$ is a $Z$-matrix with positive diagonal entries.

Let $C = (c_{ij})\in C^{n\times n}$ with $c_{ij} = r_{i_{A}}^{+}$, and obviously $C$ is a nonnegative matrix of rank 1. Hence, $A$ can be decomposed as $A = B+C$.

Theorem 2.7. If $A = (a_{ij})\in C^{n\times n}$ is a $B_{1}$-matrix, then $B^{+}$ is a $P$-matrix.

Proof. Since $A$ is a $B_1$-matrix, it is easy to obtain that $B^{+}$ is an $H$-matrix by Theorem 2.1 and Lemma 2.1. $a_{ii}-r_{i_{A}}^{+} > p_{i}(B^{+})\geq 0$, and it is equivalent to $b_{ii}^{+} > 0$ for all $i\in N$. We conclude that $B^{+}$ is a $P$-matrix from Lemma 2.2.

Example 2.6. Let the matrix

We write $A = B^{+}+C$, where

and

By calculation, we have that

and

From Definition 2.3, we get that $A$ is a $B_{1}$-matrix, and by Definition 2.1, one can easily verify that $B^{+}$ is a $P$-matrix. Therefore, Example 2.6 illustrates that Theorem 2.7 is valid.

Theorem 2.8. If $A = (a_{ij})\in C^{n\times n}$ is a $B_{1}$-matrix, and $C$ is a nonnegative matrix of the form

where $C = (c_{ij})\in C^{n\times n}$ with $c_{ij} = r_{i_{A}}^{+}$, then $A+C$ is a $B_1$-matrix.

Proof. Note that for each $i\in N$, $r_{i_{A+C}}^{+} = r_{i_{A}}^{+}+c_i$, and moreover, $(A+C)^{+} = B^{+}$. Since $A$ is a $B_1$-matrix, from Theorem 2.3, we have that $B^{+}$ is a $B_1$-matrix, and then $(A+C)^{+}$ is also a $B_1$-matrix. Therefore, $A+C$ is a $B_1$-matrix.

3.   Infinity norm upper bound for the inverse of $B_{1}$-matrix

In this section, an infinity norm upper bound for the inverse of $B_1$-matrices is obtained. Before that, some lemmas and theorems are listed.

Lemma 3.1. ^[10] If $P = (p_1, \ldots, p_n)^{T}e$, where $e = (1, \ldots, 1)$ and $p_1, \ldots, p_n\geq 0$, then

where $I$ is the $n\times n$ identity matrix.

Theorem 3.1. ^[16] Let matrix $A = (a_{ij})\in C^{n\times n}$ be an $SDD_1$ matrix, and then

where

$H_{i} = |a_{ii}|-\sum\limits_{j\in N_{1}(A)\backslash\{i\}}{|a_{ij}|}-\sum\limits_{j\in N_{2}(A)\backslash\{i\}}(\frac{p_{j}(A)}{|a_{jj}|}+\varepsilon)|a_{ij}|,$ $i\in N_{1}(A)$,

$Q_{i} = \varepsilon (|a_{ii}|-\sum\limits_{j\in N_{2}(A)\backslash\{i\}}|a_{ij}|)+\sum\limits_{j\in N_{2}(A)\backslash\{i\}}{\frac{r_{j}(A)-p_{j}(A)}{|a_{jj}|}}|a_{ij}|,$ $i\in N_{2}(A)$,

and $\varepsilon$ satisfies $0 < \varepsilon < \min_{i\in N}{\frac{|a_{ii}|-p_{i}(A)}{\sum\limits_{j\in N_{2}(A)\backslash\{i\}}{|a_{ij}|}}}$.

Theorem 3.2. Let $A = (a_{ij})\in C^{n\times n}$ be a $B_1$-matrix, and then

where $B^{+} = (b_{ij}^{+})_{1\leq i, j \leq n}$ with $b_{ij}^{+} = a_{ij}-r_{i_{A}}^{+},$

$T_{i} = {|a_{ii}-r_{i_{A}}^{+}|-\sum\limits_{j\in N_{1}(B^{+})\backslash\{i\}}{|a_{ij}-r_{i_{A}}^{+}|}-\sum\limits_{j\in N_{2}(B^{+})\backslash\{i\}}(\frac{p_{j}(B^{+})}{|a_{jj}-r_{j_{A}}^{+}|}+\varepsilon)|a_{ij}-r_{i_{A}}^{+}|}$, $i\in N_{1}(B^{+}),$

$M_{i} = {\varepsilon (|a_{ii}-r_{i_{A}}^{+}|-\sum\limits_{j\in N_{2}(B^{+})\backslash\{i\}}|a_{ij}-r_{i_{A}}^{+}|)+\sum\limits_{j\in N_{2}(B^{+})\backslash\{i\}}{\frac{r_{j}(B^{+})-p_{j}(B^{+})}{|a_{jj}-r_{j_{A}}^{+}|}}|a_{ij}-r_{i_{A}}^{+}|}$, $i\in N_{2}(B^{+})$,

and $\varepsilon$ satisfies $0 < \varepsilon < \min_{i\in N}{\frac{|a_{ii}-r_{i_{A}}^{+}|-p_{i}(B^{+})}{\sum\limits_{j\in N_{2}(A)\backslash\{i\}}{|a_{ij}-r_{i_{A}}^{+}|}}}$.

Proof. Since $A$ is a $B_1$-matrix, let $A = B^{+}+C$, with $B^{+} = (b_{ij}^{+})\in C^{n\times n}$ and $b_{ij}^{+} = a_{ij}-r_{i_{A}}^{+}$, $C = (c_{ij})\in C^{n\times n}$ with $c_{ij} = r_{i_{A}}^{+}$. From Theorem 2.1 and Lemma 2.1, $B^{+}$ is an $H$-matrix, and it is also a nonsingular $M$-matrix. Then, $B^{+}$ has nonnegative inverse. According to $A = B^{+}+C = B^{+}(I+(B^{+})^{-1}C)$, it holds that $||A^{-1}||_{\infty} \leq ||(I+(B^{+})^{-1}C)^{-1}||_{\infty} ||(B^{+})^{-1}||_{\infty}$. Observe that the matrix $C$ is nonnegative, and $(B^{+})^{-1}\geq 0$. Then, $(B^{+})^{-1}C$ can be written as $(p_1, p_2, \ldots, p_n)^{T}e,$ where $p_i\geq0$ and $e = (1, 1, \ldots, 1)$, for $i = 1, 2, \ldots, n$, and by Lemma 3.1,

However, $B^{+}$ is an $SDD_1$ matrix, by Theorem 3.1,

where

$T_{i} = {|a_{ii}-r_{i_{A}}^{+}|-\sum\limits_{j\in N_{1}(B^{+})\backslash\{i\}}{|a_{ij}-r_{i_{A}}^{+}|}-\sum\limits_{j\in N_{2}(B^{+})\backslash\{i\}}(\frac{p_{j}(B^{+})}{|a_{jj}-r_{j_{A}}^{+}|}+\varepsilon)|a_{ij}-r_{i_{A}}^{+}|}$, $i\in N_{1}(B^{+}),$

$M_{i} = {\varepsilon (|a_{ii}-r_{i_{A}}^{+}|-\sum\limits_{j\in N_{2}(B^{+})\backslash\{i\}}|a_{ij}-r_{i_{A}}^{+}|)+\sum\limits_{j\in N_{2}(B^{+})\backslash\{i\}}{\frac{r_{j}(B^{+})-p_{j}(B^{+})}{|a_{jj}-r_{j_{A}}^{+}|}}|a_{ij}-r_{i_{A}}^{+}|}$, $i\in N_{2}(B^{+})$,

and $\varepsilon$ satisfies $0 < \varepsilon < \min_{i\in N}{\frac{|a_{ii}-r_{i_{A}}^{+}|-p_{i}(B^{+})}{\sum\limits_{j\in N_{2}(A)\backslash\{i\}}{|a_{ij}-r_{i_{A}}^{+}|}}}$.

By (3.4) and (3.5), we get (3.3).

4.   Error bound for the linear complementarity problem of $B_{1}$-matrices

In this section, before an error bound for the linear complementarity problem corresponding to $B_1$-matrices is proposed, some lemmas are listed.

Lemma 4.1. ^[11] Let $\gamma > 0$ and $\eta \geq 0$, and then for any $x\in [0, 1]$,

Lemma 4.2. ^[12] Let $A = (a_{ij})\in C^{n\times n}$ be an $SDD_1$ matrix with positive diagonal entries, and then $A_{D} = I-D+DA$ is also an $SDD_1$ matrix, where $D = diag(d_{i})$ with $0\leq d_{i}\leq 1$.

Theorem 4.1. Let $A = (a_{ij})\in C^{n\times n}$($n\geq 2$) be an $SDD_1$ matrix with positive diagonal entries, and $A_{D} = I-D+DA$ is also an $SDD_1$ matrix, where $D = diag(d_{i})$ with $0\leq d_{i}\leq 1$. Then,

where

$L_{i} = {|d_{i}a_{ii}|-\sum\limits_{j\in N_{1}(A_{D})\backslash\{i\}}{|d_{i}a_{ij}|}-\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}(\frac{p_{j}(A)}{|a_{jj}|}+\varepsilon)|d_{i}a_{ij}|}, \qquad i\in N_{1}(A_{D})$,

$G_{i} = {\varepsilon (|d_{i}a_{ii}|-\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}|d_{i}a_{ij}|)+\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}{\frac{d_{j}r_{j}(A)}{|1-d_{j}+d_{j}a_{jj}|}}|d_{i}a_{ij}|-\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}{\frac{p_{j}(A)}{|a_{jj}|}}{|d_{i}a_{ij}|}}, \qquad i\in N_{2}(A_{D})$,

and $\varepsilon$ satisfies $0 < \varepsilon < \min_{i\in N}{\frac{|1-d_{i}+d_{i}a_{ii}|-p_{i}(A_{D})}{\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}{|d_{i}a_{ij}|}}}.$

Proof. Let $A_{D} = I-D+DA$ be an $SDD_1$ matrix, where

By Theorem 3.1,

where

$H_{i} = |1-d_{i}+d_{i}a_{ii}|-\sum\limits_{j\in N_{1}(A_{D})\backslash\{i\}}{|d_{i}a_{ij}|}-\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}(\frac{p_{j}(A_{D})}{|1-d_{j}+d_{j}a_{jj}|}+\varepsilon)|d_{i}a_{ij}|,$ $i\in N_{1}(A_{D})$,

$Q_{i} = \varepsilon (|1-d_{i}+d_{i}a_{ii}|-\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}|d_{i}a_{ij}|)+\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}{\frac{r_{j}(A_{D})-p_{j}(A_{D})}{|1-d_{j}+d_{j}a_{jj}|}}|d_{i}a_{ij}|,$ $i\in N_{2}(A_{D})$,

and $\varepsilon$ satisfies $0 < \varepsilon < \min_{i\in N}{\frac{|1-d_{i}+d_{i}a_{ii}|-p_{i}(A_{D})}{\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}{|d_{i}a_{ij}|}}}$.

For $i\in N_{1}(A_{D})$ and from $0\leq d_{i}\leq 1$, we obtain that

which means that $i\in N_{1}(A)$, that is, $N_{1}(A_{D})\subseteq N_{1}(A)$. Then,

By Lemma 4.1,

Then, by (4.2)–(4.5), we get (4.1).

Example 4.1. Let

and $D = diag(d_{i})$ with $d_{i} = 0.9000$. Then, we have

By calculation, $N_{1}(A_{D}) = \{1, 3\}$, $N_{2}(A_{D}) = \{2, 4\}$, $p_{2}(A) = 4$, $p_{4}(A) = 6.6150$ and $0 < \varepsilon < 0.0541$. We choose $\varepsilon = 0.0540$. Then, $L_{1} = 0.3789$, $L_{3} = 0.4689$, $G_{2} = 0.3949$ and $G_{4} = 0.3364$. Hence, $||A_{D}^{-1}||_{\infty} \leq 2.9727$, and the true value is $||A_{D}^{-1}||_{\infty} = 0.3188.$

Next, an error bound for the linear complementarity problem corresponding to $B_1$-matrices is proposed.

Theorem 4.2. Let $A = (a_{ij})\in C^{n\times n}$($n\geq 2$) be a $B_{1}$-matrix satisfying the hypotheses of Theorem 4.1. Then,

where

$B^{+} = (b_{ij}^{+})\in C^{n\times n}$ with $b_{ij}^{+} = a_{ij}-r_{i_{A}}^{+},$

$F_{i} = |d_{i}b^{+}_{ii}|-\sum\limits_{j\in N_{1}(B^{+}_{D})\backslash\{i\}}{|d_{i}b_{ij}^{+}|}-\sum\limits_{j\in N_{2}(B^{+}_{D})\backslash\{i\}}(\frac{p_{j}(B^{+})}{|b_{jj}^{+}|}+\varepsilon)|d_{i}b_{ij}^{+}|, \qquad i\in N_{1}(B^{+}_D)$,

$Z_{i} = \varepsilon (|d_{i}b^{+}_{ii}|-\sum\limits_{j\in N_{2}(B^{+}_{D})\backslash\{i\}}|d_{i}b^{+}_{ij}|)+\sum\limits_{j\in N_{2}(B^{+}_{D})\backslash\{i\}}{\frac{d_{j}r_{j}(B^{+})}{|1-d_{j}+d_{j}b_{jj}^{+}|}}|d_{i}b_{ij}^{+}|-\sum\limits_{j\in N_{2}(B^{+}_{D})\backslash\{i\}}{\frac{p_{j}(B^{+})}{|b_{jj}^{+}|}}{|d_{i}b_{ij}^{+}|}, \qquad i\in N_{2}(B^{+}_D)$,

and $\varepsilon$ satisfies $0 < \varepsilon < \min_{i\in N}{\frac{|1-d_{i}+d_{i}b^{+}_{ii}|-p_{i}(B_{D}^{+})}{\sum\limits_{j\in N_{2}(B^{+}_D)\backslash\{i\}}{|d_{i}b^{+}_{ij}|}}}.$

Proof. Let $A = B^{+}+C$, where $B^{+} = (b_{ij}^{+})\in C^{n\times n}$ with $b_{ij}^{+} = a_{ij}-r_{i_{A}}^{+}$, $C = (c_{ij})\in C^{n\times n}$ with $c_{ij} = r_{i_{A}}^{+}$. $B^{+}$ is an $SDD_{1}$ matrix with positive diagonal entries. Thus for each diagonal matrix $D = diag(d_{i})$ with $0\leq d_{i}\leq 1$,

where $B_{D}^{+} = I-D+DB^{+}$ and $C_{D} = DC$. Similar to the proof of Theorem 3.2,

Notice that $B^{+}$ is an $SDD_{1}$ matrix, and by Lemma 4.2, $B_{D}^{+} = I-D+DB^{+}$ is also an $SDD_{1}$ matrix. Hence, by (4.1), it holds that

where

$F_{i} = |d_{i}b^{+}_{ii}|-\sum\limits_{j\in N_{1}(B^{+}_{D})\backslash\{i\}}{|d_{i}b_{ij}^{+}|}-\sum\limits_{j\in N_{2}(B^{+}_{D})\backslash\{i\}}(\frac{p_{j}(B^{+})}{|b_{jj}^{+}|}+\varepsilon)|d_{i}b_{ij}^{+}|, \qquad i\in N_{1}(B^{+}_D)$,

$Z_{i} = \varepsilon (|d_{i}b^{+}_{ii}|-\sum\limits_{j\in N_{2}(B^{+}_{D})\backslash\{i\}}|d_{i}b^{+}_{ij}|)+\sum\limits_{j\in N_{2}(B^{+}_{D})\backslash\{i\}}{\frac{d_{j}r_{j}(B^{+})}{|1-d_{j}+d_{j}b_{jj}^{+}|}}|d_{i}b_{ij}^{+}|-\sum\limits_{j\in N_{2}(B^{+}_{D})\backslash\{i\}}{\frac{p_{j}(B^{+})}{|b_{jj}^{+}|}}{|d_{i}b_{ij}^{+}|}, \qquad i\in N_{2}(B^{+}_D)$,

and $\varepsilon$ satisfies $0 < \varepsilon < \min_{i\in N}{\frac{|1-d_{i}+d_{i}b^{+}_{ii}|-p_{i}(B_{D}^{+})}{\sum\limits_{j\in N_{2}(B^{+}_D)\backslash\{i\}}{|d_{i}b^{+}_{ij}|}}}.$

Example 4.2. Let the matrix

and

where we set $D = diag(d_{i})$ with $d_{i} = 0.7000$. Then,

By the definitions of $B$-matrix and $B_1$-matrix, it is easy to get that $A$ is not a $B$-matrix but is a $B_1$-matrix. Therefore, the existing bounds (such as the bound (13) in Theorem 4 ^[10]) cannot be used to compute the error bound for the linear complementarity problem for matrix $A$. However, the error bound for the linear complementarity problem for matrix $A$ can be computed by Theorem 4.2.

By simple calculation, $N_{1}(B^{+}_{D}) = \{3, 4\}$, $N_{2}(B^{+}_{D}) = \{1, 2\}$, $p_{1}(B^{+}) = 2.5000$, $p_{2}(B^{+}) = 1.5000$ and $0 < \varepsilon < 0.1084$. Let $\varepsilon = 0.1083$. Then, from our bound in Theorem 4.2, the error bound for the linear complementarity problem for matrix $A$ is given as $\max_{d\in [0, 1]^{n}}||(I-D+DA)^{-1}||_{\infty}\leq 5.7186$, and the true value is $||(I-D+DA)^{-1}||_{\infty} = 0.3359$.

Example 4.3. Consider the matrix

and we write $A = B^{+}+C$, where

It is easy to verify that $A$ is a $B$-matrix. Then, it is also a $B_1$-matrix ^[14]. By the bound (13) in Theorem 4 ^[10], we have

By simple calculation, we have that

and $p_{1}(B^{+}) = 0.1568$, $p_{2}(B^{+}) = 0.0552$, $p_{3}(B^{+}) = 0.0221$ and $0 < \varepsilon < 0.8154$. Let $\varepsilon = 0.8153$, and then from our bound in Theorem 4.2, we get that $\max_{d\in [0, 1]^{n}}||(I-D+DA)^{-1}||_{\infty}\leq 24.2275 < 50.$ Therefore, Example 4.3 shows that the error bound of a $B_{1}$-matrix is sharper than the error bound of a $B$-matrix under some cases.

5.   Conclusions

In this paper, some properties for $B_1$-matrices and the infinity norm upper bound for the inverse of $B_1$-matrices are presented. Based on these results, the error bound for the linear complementarity problem of $B_1$-matrices is obtained. Moreover, numerical examples are also presented to illustrate the corresponding results.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This work was partly supported by the National Natural Science Foundations of China (31600299), Natural Science Basic Research Program of Shaanxi, China (2020JM-622).

Conflict of interest

The authors declare there is no conflict of interest.