v2007.09.13 - Convex Optimization

2.13. DUAL CONE & GENERALIZED INEQUALITY 171Because x i − x i+1 ≥ 0 ∀i by assumption whenever x ∈ K M+ , we can employdual generalized inequalities (279) with respect to the self-dual nonnegativeorthant R n + to find the halfspace-description of the dual cone K ∗ M+ . We cansay x T 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 (373)x T y ≥ 0 ∀X † x ≽ 0 ⇔ X T y ≽ 0 (374)X = [e 1 e 1 +e 2 e 1 +e 2 +e 3 · · · 1 ] ∈ R n×n (375)Because X † x ≽ 0 connotes membership of x to pointed K M+ , then by(258) the dual cone we seek comprises all y for which (374) 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(376)The monotone nonnegative cone and its dual are simplicial, illustrated fortwo Euclidean spaces in Figure 50.From2.13.6.1, the extreme directions of proper K M+ are respectivelyorthogonal to the facets of K ∗ M+ . Because K∗ M+ is simplicial, theinward-normals to its facets constitute the linearly independent rows of X Tby (376). Hence the vertex-description for K M+ employs the columns of Xin agreement with Cone Table S because X † =X −1 . Likewise, the extremedirections of proper K ∗ M+ are respectively orthogonal to the facets of K M+whose inward-normals are contained in the rows of X † by (370). So thevertex-description for K ∗ M+ employs the columns of X†T . 2.13.9.4.2 Example. Monotone cone.(Figure 51, Figure 52) Of nonempty interior but not pointed, the monotonecone is polyhedral and defined by the halfspace-descriptionK M ∆ = {x ∈ R n | x 1 ≥ x 2 ≥ · · · ≥ x n } = {x ∈ R n | X ∗T x ≽ 0} (377)Its dual is therefore pointed but of empty interior, having vertex-descriptionK ∗ M = {X ∗ b ∆ = [e 1 −e 2 e 2 −e 3 · · · e n−1 −e n ]b | b ≽ 0 } ⊂ R n (378)