v2010.10.26 - Convex Optimization

2.13. DUAL CONE & GENERALIZED INEQUALITY 207Identify a sum of mutually orthogonal projections x z −f(z) ; in Moreau’sterms, z=P K x and −f(z)=P −K∗x . Then f(z)∈ K ∗ (E.9.2.2 no.4) and zis a solution to the complementarity problem iff it is a fixed point ofz = P K x = P K (z − f(z)) (464)Given that a solution exists, existence of a fixed point would be guaranteedby theory of contraction. [227, p.300] But because only nonexpansivity(Theorem E.9.3.0.1) is achievable by a projector, uniqueness cannot beassured. [203, p.155] Elegant proofs of equivalence between complementarityproblem (463) and fixed point problem (464) are provided by Németh[Wıκımization, Complementarity problem].2.13.10.1.4 Example. Linear complementarity problem. [87] [272] [311]Given matrix B ∈ R n×n and vector q ∈ R n , a prototypical complementarityproblem on the nonnegative orthant K = R n + is linear in w =f(z) :find z ≽ 0subject to w ≽ 0w T z = 0w = q + Bz(465)This problem is not convex when both vectors w and z are variable. 2.82Notwithstanding, this linear complementarity problem can be solved byidentifying w ← ∇f(z)= q + Bz then substituting that gradient into (463)find z ∈ Ksubject to ∇f(z) ∈ K ∗〈z , ∇f(z)〉 = 0(466)2.82 But if one of them is fixed, then the problem becomes convex with a very simplegeometric interpretation: Define the affine subsetA {y ∈ R n | By = w − q}For w T z to vanish, there must be a complementary relationship between the nonzeroentries of vectors w and z ; id est, w i z i =0 ∀i. Given w ≽0, then z belongs to theconvex set of solutions:z ∈ −K ⊥ R n + (w ∈ Rn +) ∩ A = R n + ∩ w ⊥ ∩ Awhere K ⊥ R n +(w) is the normal cone to R n + at w (453). If this intersection is nonempty, thenthe problem is solvable.