Chaotic performance and circuitry implement of piecewise logistic-like mapping

Caiwen Chen; Tianxiu Lu; Ping Gao; Caiwen Chen; Tianxiu Lu; Ping Gao

doi:10.3934/era.2025006

Electronic Research Archive

2025, Volume 33, Issue 1: 102-120. doi: 10.3934/era.2025006

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Theory article

Chaotic performance and circuitry implement of piecewise logistic-like mapping

1.
College of Mathematics and Statistics, Sichuan University of Science and Engineering, Zigong 643000, China
2.
Sichuan Province University Key Laboratory of Bridge Non-destruction Detecting and Engineering Computing, Zigong 643000, China

Received: 10 November 2024 Revised: 18 December 2024 Accepted: 03 January 2025 Published: 14 January 2025

Discrete chaotic systems are now a meaningful research area due to their intricate dynamical characteristics. This paper introduces a novel piecewise logistic-like mapping chaotic system and rigorously establishes its Devaney chaoticity through mathematical proofs. Experimental findings demonstrate that, compared to the traditional logistic mapping, the logistic-like mapping exhibits more complex dynamic behaviors, such as bifurcation diagram, Lyapunov exponents, permutation entropy, sensitivity, and distribution of function sequences. Furthermore, in order to implement the suggested new chaotic system, a simulation circuit is designed, and a PSIM simulation model is established to validate feasibility of the simulation circuit.

Keywords:

Citation: Caiwen Chen, Tianxiu Lu, Ping Gao. Chaotic performance and circuitry implement of piecewise logistic-like mapping[J]. Electronic Research Archive, 2025, 33(1): 102-120. doi: 10.3934/era.2025006

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Abstract

1. Introduction

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

$\begin{eqnarray} x\geq 0, Ax+q\geq 0, x^{T}(Ax+q) = 0, \end{eqnarray}$

(1.1)

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 $i \in N_{1}(A)$ ,

$\begin{eqnarray} |a_{ii}| > p_i(A). \end{eqnarray}$

(2.1)

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

$\begin{eqnarray} a_{ii}-r_{i_{A}}^{+} > p_{i}(B^{+}). \end{eqnarray}$

(2.2)

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

$\begin{eqnarray*} A = \left[\begin{array}{ccc} 2 &-1 &0\\ 1 &3 &-2\\ 9 &10 &110 \end{array}\right]. \end{eqnarray*}$

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

$\begin{eqnarray*} B^{+} = \ \left[ \begin{array}{ccc} 2 &-1 &0 \\ 0 &2 &-3 \\ -1 &0 &100 \ \end{array} \right], \end{eqnarray*}$

and

$\begin{eqnarray*} C = \ \left[ \begin{array}{ccc} 0 &0 &0 \\ 1 &1 &1 \\ 10 &10 &10 \ \end{array} \right]. \end{eqnarray*}$

By calculation, we have that

$\begin{eqnarray*} N_{1}(B^{+}) = \{2\}, N_{2}(B^{+}) = \{1, 3\}, \end{eqnarray*}$

$\begin{eqnarray*} a_{11}-r_{1_{A}}^{+} = 2 > 1 = p_{1}(B^{+}) = |a_{12}-r_{1_{A}}^{+}|+\frac{r_{3}(B^{+})}{|a_{33}-r_{3_{A}}^{+}|}|a_{13}-r_{1_{A}}^{+}|, \end{eqnarray*}$

$\begin{eqnarray*} a_{22}-r_{2_{A}}^{+} = 2 > 0.0300 = p_{2}(B^{+}) = \frac{r_{1}(B^{+})}{|a_{11}-r_{1_{A}}^{+}|}|a_{21}-r_{2_{A}}^{+}|+\frac{r_{3}(B^{+})}{|a_{33}-r_{3_{A}}^{+}|}|a_{23}-r_{2_{A}}^{+}|, \end{eqnarray*}$

and

$\begin{eqnarray*} a_{33}-r_{3_{A}}^{+} = 100 > 0.5000 = p_{3}(B^{+}) = |a_{32}-r_{3_{A}}^{+}|+\frac{r_{1}(B^{+})}{|a_{11}-r_{1_{A}}^{+}|}|a_{31}-r_{3_{A}}^{+}|. \end{eqnarray*}$

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

$\begin{eqnarray*} A = \left[\begin{array}{ccc} -3 &1 &1\\ -4 &-2 &1\\ 3 &1 &-2 \end{array}\right]. \end{eqnarray*}$

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

$\begin{eqnarray*} B^{+} = \ \left[ \begin{array}{ccc} -4 &0 &0 \\ -5 &-3 &0 \\ 0 &-2 &-5 \ \end{array} \right], \end{eqnarray*}$

and

$\begin{eqnarray*} C = \ \left[ \begin{array}{ccc} 1 &1 &1 \\ 1 &1 &1 \\ 3 &3 &3 \ \end{array} \right]. \end{eqnarray*}$

By calculation, we have that

$\begin{eqnarray*} N_{1}(B^{+}) = \{2\}, N_{2}(B^{+}) = \{1, 3\}. \end{eqnarray*}$

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

$\begin{eqnarray*} A = \ \left[ \begin{array}{cccccc} 75 &74 &23 \\ 10 &15 &12 \\ 9 &10 &110 \ \end{array} \right]. \end{eqnarray*}$

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

$\begin{eqnarray*} B^{+} = \ \left[ \begin{array}{cccccc} 1 &0 &-51 \\ -2 &3 &0 \\ -1 &0 &100 \ \end{array} \right], \end{eqnarray*}$

and

$\begin{eqnarray*} C = \ \left[ \begin{array}{ccc} 74 &74 &74 \\ 12 &12 &12 \\ 10 &10 &10 \ \end{array} \right]. \end{eqnarray*}$

By calculation, we have that

$\begin{eqnarray*} N_{1}(B^{+}) = \{1\}, N_{2}(B^{+}) = \{2, 3\}, \end{eqnarray*}$

$\begin{eqnarray*} a_{11}-r_{1_{A}}^{+} = 1 > 0.5100 = p_{1}(B^{+}) = \frac{r_{2}(B^{+})}{|a_{22}-r_{2_{A}}^{+}|}|a_{12}-r_{1_{A}}^{+}|+\frac{r_{3}(B^{+})}{|a_{33}-r_{3_{A}}^{+}|}|a_{13}-r_{1_{A}}^{+}|, \end{eqnarray*}$

$\begin{eqnarray*} a_{22}-r_{2_{A}}^{+} = 3 > 2 = p_{2}(B^{+}) = |a_{21}-r_{2_{A}}^{+}|+\frac{r_{3}(B^{+})}{|a_{33}-r_{3_{A}}^{+}|}|a_{23}-r_{2_{A}}^{+}|, \end{eqnarray*}$

and

$\begin{eqnarray*} a_{33}-r_{3_{A}}^{+} = 100 > 1 = p_{3}(B^{+}) = |a_{31}-r_{3_{A}}^{+}|+\frac{r_{2}(B^{+})}{|a_{22}-r_{2_{A}}^{+}|}|a_{32}-r_{3_{A}}^{+}|. \end{eqnarray*}$

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

$\begin{eqnarray*} A = \ \left[ \begin{array}{cccccc} 1 &-3 &-2 \\ 0 &1 &0 \\ 0 &0 &1 \ \end{array} \right]. \end{eqnarray*}$

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

$\begin{eqnarray*} B^{+} = \ \left[ \begin{array}{cccccc} 1 &-3 &-2 \\ 0 &1 &0 \\ 0 &0 &1 \ \end{array} \right], \end{eqnarray*}$

and

$\begin{eqnarray*} C = \ \left[ \begin{array}{ccc} 0 &0 &0 \\ 0 &0 &0 \\ 0 &0 &0 \ \end{array} \right]. \end{eqnarray*}$

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

$\begin{eqnarray*} N_{1}(B^{+}) = \{1\}, N_{2}(B^{+}) = \{2, 3\}, \end{eqnarray*}$

$\begin{eqnarray*} a_{11}-r_{1_{A}}^{+} = 1 > 0 = p_{1}(B^{+}), \end{eqnarray*}$

$\begin{eqnarray*} a_{22}-r_{2_{A}}^{+} = 1 > 0 = p_{2}(B^{+}), \end{eqnarray*}$

and

$\begin{eqnarray*} a_{33}-r_{3_{A}}^{+} = 1 > 0 = p_{3}(B^{+}). \end{eqnarray*}$

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

$\begin{align*} \begin{split} c_{ij} = \left \{ \begin{array}{ll} a_{ii}+d_{i}, & i = j,\\ a_{ij}, & i\neq j. \\ \end{array} \right. \end{split} \end{align*}$

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 $i\in N$ ,

$\begin{eqnarray} c_{ii}-r_{i_{C}}^{+} = a_{ii}+d_{i}-r_{i_{A}}^{+}\geq a_{ii}-r_{i_{A}}^{+} > p_{i}(B^{+}). \end{eqnarray}$

(2.3)

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

$\begin{eqnarray*} p_i(V^{+})& = &\sum\limits_{j\in N_{1}(V^{+})\backslash\{i\}}{|v_{ij}}|+\sum\limits_{j\notin N_{1}(V^{+})\cup\{i\}}\frac{r_{j}(V^+)}{|v_{jj}|}{ |v_{ij}|}\\ & = &\sum\limits_{j\in N_{1}(V^{+})\backslash\{i\}}{|a_{ij}-r_{i_{A}}^{+}}|+\sum\limits_{j\notin N_{1}(V^{+})\cup\{i\}}\frac{r_{j}(B^+)}{|a_{jj}-r_{j_{A}}^{+}+d_{j}|}{ |a_{ij}-r_{i_{A}}^{+}|}\\ &\leq& \sum\limits_{j\in N_{1}(B^{+})\backslash\{i\}}{|a_{ij}-r_{i_{A}}^{+}}|+\sum\limits_{j\notin N_{1}(B^{+})\cup\{i\}}\frac{r_{j}(B^+)}{|a_{jj}-r_{j_{A}}^{+}|}{ |a_{ij}-r_{i_{A}}^{+}|}\\ & = &p_{i}(B^{+}). \end{eqnarray*}$

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

$\begin{eqnarray*} A = \ \left[ \begin{array}{cccccc} 75 &74 &23 \\ 10 &15 &12 \\ 9 &10 &110 \ \end{array} \right], \end{eqnarray*}$

and

$\begin{eqnarray*} D = \ \left[ \begin{array}{cccccc} 5 &0 &0 \\ 0 &15 &0 \\ 0 &0 &0 \ \end{array} \right]. \end{eqnarray*}$

By calculation, we have that

$\begin{eqnarray*} N_{1}(B^{+}) = \{1\}, N_{2}(B^{+}) = \{2, 3\}, \end{eqnarray*}$

$\begin{eqnarray*} a_{11}-r_{1_{A}}^{+} = 1 > 0.5100 = p_{1}(B^{+}), \end{eqnarray*}$

$\begin{eqnarray*} a_{22}-r_{2_{A}}^{+} = 3 > 2 = p_{2}(B^{+}), \end{eqnarray*}$

and

$\begin{eqnarray*} a_{33}-r_{3_{A}}^{+} = 100 > 1 = p_{3}(B^{+}). \end{eqnarray*}$

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

$\begin{eqnarray*} A = \ \left[ \begin{array}{cccccc} 1 &0.1 &0.1 \\ 3 &4 &2 \\ 0.1 &0.1 &1 \ \end{array} \right]. \end{eqnarray*}$

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

$\begin{eqnarray*} B^{+} = \ \left[ \begin{array}{cccccc} 0.9 &0 &0 \\ 0 &1 &-1 \\ 0 &0 &0.9 \ \end{array} \right], \end{eqnarray*}$

and

$\begin{eqnarray*} C = \ \left[ \begin{array}{ccc} 0.1 &0.1 &0.1 \\ 3 &3 &3 \\ 0.1 &0.1 &0.1 \ \end{array} \right]. \end{eqnarray*}$

By calculation, we have that

$\begin{eqnarray*} N_{1}(B^{+}) = \{2\}, N_{2}(B^{+}) = \{1, 3\}, \end{eqnarray*}$

$\begin{eqnarray*} a_{11}-r_{1_{A}}^{+} = 0.9000 > 0 = p_{1}(B^{+}) = |a_{12}-r_{1_{A}}^{+}|+\frac{r_{3}(B^{+})}{|a_{33}-r_{3_{A}}^{+}|}|a_{13}-r_{1_{A}}^{+}|, \end{eqnarray*}$

$\begin{eqnarray*} a_{22}-r_{2_{A}}^{+} = 1 > 0 = p_{2}(B^{+}) = \frac{r_{1}(B^{+})}{|a_{11}-r_{1_{A}}^{+}|}|a_{21}-r_{2_{A}}^{+}|+\frac{r_{3}(B^{+})}{|a_{33}-r_{3_{A}}^{+}|}|a_{23}-r_{2_{A}}^{+}|, \end{eqnarray*}$

and

$\begin{eqnarray*} a_{33}-r_{3_{A}}^{+} = 0.9000 > 0 = p_{3}(B^{+}) = |a_{32}-r_{3_{A}}^{+}|+\frac{r_{1}(B^{+})}{|a_{11}-r_{1_{A}}^{+}|}|a_{31}-r_{3_{A}}^{+}|. \end{eqnarray*}$

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

$\begin{eqnarray*} C = \ \left[ \begin{array}{cccccc} c_1 &c_1 &\cdots &c_1 \\ c_2 &c_2 &\cdots &c_2 \\ \vdots &\vdots &\cdots &\vdots \\ c_n &c_n &\cdots &c_n \end{array} \right], \end{eqnarray*}$

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

$\begin{eqnarray} ||(I+P)^{-1}||_{\infty} \leq n-1, \end{eqnarray}$

(3.1)

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

$\begin{eqnarray} ||A^{-1}||_{\infty} \leq \frac{\max \{1, \max_{i\in N_{2}(A)} \frac{p_{i}(A)}{|a_{ii}|}+\varepsilon\}} {\min \{\min_{i\in N_{1}(A)}{H_{i}}, \min_{i\in N_{2}(A)}{Q_{i}}\}}, \end{eqnarray}$

(3.2)

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

$\begin{eqnarray} ||A^{-1}||_{\infty} \leq (n-1) \frac{\max \{1, \max_{i\in N_{2}(B^{+})}(\frac{p_{i}(B^{+})}{|a_{ii}-r_{i_{A}}^{+}|}+\varepsilon)\}} {\min \{\min_{i\in N_{1}(B^{+})}{T_{i}}, \min_{i\in N_{2}(B^{+})}{M_{i}}\}}, \end{eqnarray}$

(3.3)

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,

$\begin{eqnarray} ||(I+(B^{+})^{-1}C)^{-1}||_{\infty} \leq n-1 . \end{eqnarray}$

(3.4)

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

$\begin{eqnarray} ||(B^{+})^{-1}||_{\infty}\leq \frac{\max \{1, \max_{i\in N_{2}(B^{+})}\frac{p_{i}(B^{+})}{{|a_{ii}-r_{i_{A}}^{+}|}}+\varepsilon\}} {\min \{\min_{i\in N_{1}(B^{+})}{T_{i}}, \min_{i\in N_{2}(B^{+})}{M_{i}}\}}, \end{eqnarray}$

(3.5)

where

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 $x\in [0, 1]$ ,

$\begin{eqnarray*} \frac{1}{1-x+x\gamma}\leq \frac{1}{\min\{\gamma, 1\}}, \qquad \frac{\eta x}{1-x+x\gamma} \leq \frac{\eta}{\gamma}. \end{eqnarray*}$

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,

$\begin{eqnarray} ||A_{D}^{-1}||_{\infty}\leq \frac{\max \{1, \max_{i\in N_{2}(A_{D})} \frac{p_{i}(A)}{|a_{ii}|}+\varepsilon\}} {\min \{\min_{i\in N_{1}(A_{D})}{L_{i}},\min_{i\in N_{2}(A_{D})}{G_{i}}\}}, \end{eqnarray}$

(4.1)

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

$\begin{align*} \begin{split} A_{D} = \left \{ \begin{array}{ll} 1-d_i+d_{i}a_{ii}, & i = j,\\ d_{i}a_{ij}, & i\neq j. \\ \end{array} \right. \end{split} \end{align*}$

By Theorem 3.1,

$\begin{eqnarray} ||A_{D}^{-1}||_{\infty} \leq \frac{\max \{1, \max_{i\in N_{2}(A_{D})} \frac{p_{j}(A_{D})}{|1-d_{j}+d_{j}a_{jj}|}+\varepsilon\}} {\min \{\min_{i\in N_{1}(A_{D})}{H_{i}}, \min_{i\in N_{2}(A_{D})}{Q_{i}}\}}, \end{eqnarray}$

(4.2)

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

$\begin{eqnarray*} |d_{i}a_{ii}|\leq |1-d_{i}+d_{i}a_{ii}|\leq r_{i}(A_{D}) = d_{i} r_{i}(A), \end{eqnarray*}$

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

$\begin{eqnarray*} p_{i}(A_{D})& = &\sum\limits_{j\in N_{1}(A_{D})\backslash\{i\}}|d_{i}a_{ij}|+\sum\limits_{j\notin N_{1}(A_{D}) \bigcup\{i\}}{\frac{d_{j}r_{j}(A)}{|1-d_{j}+d_{j}a_{jj}|}}|d_{i}a_{ij}|\\ & = &d_{i}(\sum\limits_{j\in N_{1}(A_{D})\backslash\{i\}}|a_{ij}|+\sum\limits_{j\notin N_{1}(A_{D}) \bigcup\{i\}}{\frac{d_{j}r_{j}(A)}{|1-d_{j}+d_{j}a_{jj}|}}|a_{ij}|)\\ &\leq& d_{i}(\sum\limits_{j\in N_{1}(A)\backslash\{i\}}|a_{ij}|+\sum\limits_{j\notin N_{1}(A) \bigcup\{i\}}{\frac{r_{j}(A)}{|a_{jj}|}}|a_{ij}|)\\ & = &d_{i}p_{i}(A).\\ \end{eqnarray*}$

By Lemma 4.1,

$\begin{eqnarray} \frac{p_{i}(A_{D})}{|1-d_{i}+d_{i}a_{ii}|}\leq \frac{d_{i}p_{i}(A)}{1-d_{i}+d_{i}a_{ii}}\leq \frac{p_{i}(A)}{|a_{ii}|}, \end{eqnarray}$

(4.3)

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

(4.4)

$\begin{eqnarray} \frac{1}{Q_{i}}& = &\frac{1}{\varepsilon (|1-d_{i}+d_{i}a_{ii}|-\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}|d_{i}a_{ij}|)+\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}{\frac{r_{j}(A_{D})-p_{j}(A_{D})}{|1-d_{j}+d_{j}a_{jj}|}}|d_{i}a_{ij}|}\\ &\leq& \frac{1}{\varepsilon [|d_{i}a_{ii}|-\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}|d_{i}a_{ij}|]+\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}{\frac{d_{j}r_{j}(A)}{|1-d_{j}+d_{j}a_{jj}|}}|d_{i}a_{ij}|-\sum\limits_{j\in N_{2}(A_{D})\backslash\{i\}}{\frac{p_{j}(A)}{|a_{jj}|}}{|d_{i}a_{ij}|}}\\ & = &\frac{1}{G_{i}} \leq \frac{1}{\min{G_{i}}}, \qquad i\in N_{2}(A_{D}). \end{eqnarray}$

(4.5)

Then, by (4.2)–(4.5), we get (4.1).

Example 4.1. Let

$\begin{eqnarray*} A = \ \left[ \begin{array}{cccccc} 16 &-8 &4.1 &8\\ 0 &8 &3.1 &1\\ -8 &8 &20 &8\\ 1 &1.2 &5 &8\ \end{array} \right], \end{eqnarray*}$

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

$\begin{eqnarray*} A_{D} = I-D+DA = \ \left[ \begin{array}{cccccc} 14.5 &-7.2 &3.69 &7.2\\ 0 &7.3 &2.79 &0.9\\ -7.2 &7.2 &18.1 &7.2\\ 0.9 &1.08 &4.5 &7.3\ \end{array} \right]. \end{eqnarray*}$

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,

$\begin{eqnarray*} \max\limits_{d\in [0,1]^{n}}||(I-D+DA)^{-1}||_{\infty}\leq \max _{d\in [0,1]^{n}}\frac{(n-1)\max \{1, \max_{i\in N_{2}(B^{+}_{D})} (\frac{p_{i}(B^{+})}{|b^{+}_{ii}|}+\varepsilon)\}} {\min \{\min_{i\in N_{1}(B^{+}_D)}{F_{i}}, \min_{i\in N_{2}(B^{+}_{D})}{Z_{i}}\}}, \end{eqnarray*}$

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 $0\leq d_{i}\leq 1$ ,

$\begin{eqnarray*} A_{D} = I-D+DA = (I-D+DB^{+})+DC = B_{D}^{+}+C_{D}, \end{eqnarray*}$

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

$\begin{eqnarray*} ||A_{D}^{-1}||_{\infty}\leq ||[I+(B_{D}^{+})^{-1}C_{D}]||_{\infty} ||(B_{D}^{+})^{-1}||_{\infty}\leq (n-1) ||(B_{D}^{+})^{-1}||_{\infty}. \end{eqnarray*}$

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

$\begin{eqnarray*} ||(B_{D}^{+})^{-1}||_{\infty}\leq \frac{\max \{1, \max_{i\in N_{2}(B^{+}_D)} (\frac{p_{i}(B^{+})}{|b^{+}_{ii}|}+\varepsilon)\}} {\min \{\min_{i\in N_{1}(B^{+}_D)}{F_{i}}, \min_{i\in N_{2}(B^{+}_D)}{Z_{i}}\}}, \end{eqnarray*}$

where

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

$\begin{eqnarray*} A = \ \left[ \begin{array}{cccccc} 8 &-2 &-1 &-1\\ 4 &13 &4 &5\\ -8 &-8 &15 &-8\\ -4 &-4 &-2 &6 \ \end{array} \right], \end{eqnarray*}$

and

$\begin{eqnarray*} B^{+} = \ \left[ \begin{array}{cccccc} 8 &-2 &-1 &-1 \\ -1 &8 &-1 &0\\ -8 &-8 &15 &-8\\ -4 &-4 &-2 &6\ \end{array} \right], \end{eqnarray*}$

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

$\begin{eqnarray*} B_{D}^{+} = I-D+DB^{+} = \ \left[ \begin{array}{cccccc} 5.9 &-1.4 &-0.7 &-0.7 \\ -0.7 &5.9 &-0.7 &0\\ -5.6 &-5.6 &10.8 &-5.6\\ -2.8 &-2.8 &-1.4 &4.5\ \end{array} \right]. \end{eqnarray*}$

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

$\begin{eqnarray*} A = \ \left[ \begin{array}{cccccc} 0.5 &-0.24 &-0.22\\ -0.05 &0.2 &0.01\\ 0.01 &-0.06 &0.2\ \end{array} \right], \end{eqnarray*}$

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

$\begin{eqnarray*} B^{+} = \ \left[ \begin{array}{cccccc} 0.5 &-0.24 &-0.22 \\ -0.06 &0.19 &0\\ 0 &-0.07 &0.19\ \end{array} \right]. \end{eqnarray*}$

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

$\begin{eqnarray*} \max\limits_{d\in [0,1]^{n}}||(I-D+DA)^{-1}||_{\infty}\leq 50. \end{eqnarray*}$

By simple calculation, we have that

$\begin{eqnarray*} B_{D}^{+} = I-D+DB^{+} = \ \left[ \begin{array}{cccccc} 0.5005 &-0.2398 &-0.2198 \\ -0.0599 &0.1908 &0\\ 0 &-0.0699 &0.1908 \\ \end{array} \right], \end{eqnarray*}$

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.

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