scholarly journals Karamardian Matrices: An Analogue of Q-Matrices

2021 ◽  
Vol 37 (37) ◽  
pp. 127-155
Author(s):  
K.C. Sivakumar ◽  
Sushmitha Parameswaran ◽  
Megan Wendler

A real square matrix $A$ is called a $Q$-matrix if the linear complementarity problem LCP$(A,q)$ has a solution for all $q \in \mathbb{R}^n$. This means that for every vector $q$ there exists a vector $x$ such that $x \geq 0, y=Ax+q\geq 0$, and $x^Ty=0$. A well-known result of Karamardian states that if the problems LCP$(A,0)$ and LCP$(A,d)$ for some $d\in \mathbb{R}^n, d >0$ have only the zero solution, then $A$ is a $Q$-matrix. Upon relaxing the requirement on the vectors $d$ and $y$ so that the vector $y$ belongs to the translation of the nonnegative orthant by the null space of $A^T$, $d$ belongs to its interior, and imposing the additional condition on the solution vector $x$ to be in the intersection of the range space of $A$ with the nonnegative orthant, in the two problems as above, the authors introduce a new class of matrices called Karamardian matrices, wherein these two modified problems have only zero as a solution. In this article, a systematic treatment of these matrices is undertaken. Among other things, it is shown how Karamardian matrices have properties that are analogous to those of $Q$-matrices. A subclass of a recently introduced notion of $P_{\#}$-matrices is shown to possess the Karamardian property, and for this reason we undertake a thorough study of $P_{\#}$-matrices and make some fundamental contributions.

1978 ◽  
Vol 18 (2) ◽  
pp. 161-168 ◽  
Author(s):  
J. Parida ◽  
B. Sahoo

Given an n × n matrix A, an n-dimensional vector q, and a closed, convex cone S of Rn, the generalized linear complementarity problem considered here is the following: find a z ∈ Rn such thatwhere s* is the polar cone of S. The existence of a solution to this problem for arbitrary vector q has been established both analytically and constructively for several classes of matrices A. In this note, a new class of matrices, denoted by J, is introduced. A is a J-matrix ifThe new class can be seen to be broader than previously studied classes. We analytically show that for any A in this class, a solution to the above problem exists for arbitrary vector q. This is achieved by using a result on variational inequalities.


2020 ◽  
Vol 2020 ◽  
pp. 1-8
Author(s):  
Yamna Achik ◽  
Asmaa Idmbarek ◽  
Hajar Nafia ◽  
Imane Agmour ◽  
Youssef El foutayeni

The linear complementarity problem is receiving a lot of attention and has been studied extensively. Recently, El foutayeni et al. have contributed many works that aim to solve this mysterious problem. However, many results exist and give good approximations of the linear complementarity problem solutions. The major drawback of many existing methods resides in the fact that, for large systems, they require a large number of operations during each iteration; also, they consume large amounts of memory and computation time. This is the reason which drives us to create an algorithm with a finite number of steps to solve this kind of problem with a reduced number of iterations compared to existing methods. In addition, we consider a new class of matrices called the E-matrix.


1993 ◽  
Vol 36 (2) ◽  
pp. 197-209 ◽  
Author(s):  
Vladimir Rakočević

Let X be an infinite-dimensional complex Banach space and denote the set of bounded (compact) linear operators on X by B(X) (K(X)). Let N(A) and R(A) denote, respectively, the null space and the range space of an element A of B(X). Set R(A∞)=∩nR(An) and k(A)=dim N(A)/(N(A)∩R(A∞)). Let σg(A) = ℂ\{λ∈ℂ:R(A−λ) is closed and k(A−λ)=0} denote the generalized (regular) spectrum of A. In this paper we study the subset σgb(A) of σg(A) defined by σgb(A) = ℂ\{λ∈ℂ:R(A−λ) is closed and k(A−λ)<∞}. Among other things, we prove that if f is a function analytic in a neighborhood of σ(A), then σgb(f(A)) = f(σgb(A)).


1979 ◽  
Vol 22 (1) ◽  
pp. 9-10 ◽  
Author(s):  
A. B. Thaheem

Let X be any real or complex Banach space. If T is a bounded linear operator on X then denote by N(T) the null space of T and by R(T) the range space of T.Now if X is finite dimensional and N(T) = N(T2) then also R(T) = R(T2). Therefore X admits a direct sum decomposition.Indeed it is easy to see that N(T) = N(T2) implies that and, using dimension theory of finite dimensional spaces, that N(T) and R(T) span the whole space (see, for example, (2, pp. 271–73))


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