Page 115 - Complementarity and Variational Inequalities in Electronics

106 Complementarity and Variational Inequalities in Electronics

Remark 28. i) Recall that a matrix M ∈ R n×n is copositive plus on K if

(∀x ∈ K) : Mx,x ≥ 0

and
T
(x ∈ K, Mx,x = 0) =⇒ x ∈ ker{M + M }.
In this case,

T
∗
B(M,K) ={x ∈ K : Mx ∈ K and x ∈ ker{M + M }}.
ii) Note that the approach developed in [47] and [48] for complementarity
systems originated the approach developed in [41] for variational inequalities.
We note that Theorem 8 recovers both solvability results and unifies both ap-
proaches (anyway in the framework of finite-dimensional problems).

Example 56. Let
⎛ ⎞
1 2 0
M = ⎝ −2 0 0 ⎠
⎟
⎜
4 0 1
and set

3
K = R .
+
The matrix M is positive copositive. Indeed,
2
Mx,x = (x 1 + x 3 ) + 2x 1 x 3 .
We thus have

(∀x ∈ K) : Mx,x ≥ 0.
We may check that

3
T
3
K ∩ ker(M + M ) ={x ∈ R : x 1 = x 3 = 0}={x ∈ R : Mx,x = 0}.
+ +
T
∗
The matrix is thus copositive plus. If x ∈ K ∩ ker(M + M ), then Mx ∈ K =
3
R , and thus
+
3
B(M,K) ={x ∈ R : x 1 = x 3 = 0}.
+
The condition
(∀v ∈ B(M,K), v
= 0) : q,v > 0