v2010.10.26 - Convex Optimization

198 CHAPTER 2. CONVEX GEOMETRY2.13.9.4.2 Example. Monotone nonnegative cone.[61, exer.2.33] [353,2] Simplicial cone (2.12.3.1.1) K M+ is the cone of allnonnegative vectors having their entries sorted in nonincreasing order:K M+ {x | x 1 ≥ x 2 ≥ · · · ≥ x n ≥ 0} ⊆ R n += {x | (e i − e i+1 ) T x ≥ 0, i = 1... n−1, e T nx ≥ 0}= {x | X † x ≽ 0}(429)x T y =a halfspace-description where e i is the i th standard basis vector, and where 2.79X †T [ e 1 −e 2 e 2 −e 3 · · · e n ] ∈ R n×n (430)For any vectors x and y , simple algebra demandsn∑x i y i = (x 1 − x 2 )y 1 + (x 2 − x 3 )(y 1 + y 2 ) + (x 3 − x 4 )(y 1 + y 2 + y 3 ) + · · ·i=1+ (x n−1 − x n )(y 1 + · · · + y n−1 ) + x n (y 1 + · · · + y n ) (431)Because x i − x i+1 ≥ 0 ∀i by assumption whenever x∈ K M+ , we can employdual generalized inequalities (323) with respect to the selfdual nonnegativeorthant R n + to find the halfspace-description of dual monotone nonnegativecone K ∗ M+ . We can say xT y ≥ 0 for all X † x ≽ 0 [sic] if and only ifid est,wherey 1 ≥ 0, y 1 + y 2 ≥ 0, ... , y 1 + y 2 + · · · + y n ≥ 0 (432)x T y ≥ 0 ∀X † x ≽ 0 ⇔ X T y ≽ 0 (433)X = [e 1 e 1 +e 2 e 1 +e 2 +e 3 · · · 1 ] ∈ R n×n (434)Because X † x≽0 connotes membership of x to pointed K M+ , then by(297) the dual cone we seek comprises all y for which (433) holds; thus itshalfspace-descriptionK ∗ M+ = {y ≽K ∗ M+0} = {y | ∑ ki=1 y i ≥ 0, k = 1... n} = {y | X T y ≽ 0} ⊂ R n(435)2.79 With X † in hand, we might concisely scribe the remaining vertex- andhalfspace-descriptions from the tables for K M+ and its dual. Instead we use dualgeneralized inequalities in their derivation.