On unique solvability and Picard’s iterative method for absolute value equations

Abstract

In this paper, we deal with unique solvability and numerical solution ofabsolute value equations (AVE),Ax−B|x|=b, (A,B∈Rn×n,b∈Rn).Under some weaker conditions, a simple proof is given for unique solvabilityof AVE. Furthermore, we demonstrate with an example that these results arereliable to detect unique solvability of AVE. These results are also extendedto unique solvability of standard and horizontal linear complementarity prob-lems. Finally, we suggest a Picard iterative method to compute an approx-imated solution of some uniquely solvable AVE problems where its globallylinear convergence is guaranteed via one of our weaker sufficient condition.

A Note on Unique Solvability of the Generalized Absolute Value Matrix Equation

“…Theorem 4.9. [2] The GAVE has exactly one solution for any d if A T A − ||B|| 2 2 I is a P-matrix. Remark 4.10.…”

“…Remark 4.10. Achache et al [2] shows that condition of Theorem 4.9 is slighter weaker then condition of Theorem 4.8 with help of an example. Now we discussed all diferent types of AVEs separately.…”

“…By equivalent relation between AVE and LCP, necessary and sufficient conditions for the unique solvability of AVE and GAVE also can be used for establishing the unique solution results for the linear complementarity problem. Li et al [9] and Achache et al [2] provided the unique solvability conditions for the LCP and HLCP, based on the earlier unique solvability results of the AVEs, as follows: Theorem 9.1. [2,9] The LCP(q,M) has exactly one solution for any q, if any one of the following conditions holds:…”

“…Li et al [9] and Achache et al [2] provided the unique solvability conditions for the LCP and HLCP, based on the earlier unique solvability results of the AVEs, as follows: Theorem 9.1. [2,9] The LCP(q,M) has exactly one solution for any q, if any one of the following conditions holds:…”

“…Theorem 1. [1] The GAVE(1) has exactly one solution if the invertible matrix A satisfies the norm condition ||A −1 B|| < 1.…”