v2010.10.26 - Convex Optimization

5.9. BRIDGE: CONVEX POLYHEDRA TO EDMS 463that describes the domain of p(s) as the unit simplexMaking the substitution s − β ← yS = {s | √ 2V N s ≽ −e 1 } ⊂ R N−1+ (1082)P − α = {p = X √ 2V N y | y ∈ S − β} (1087)Point p belongs to a convex polyhedron P−α embedded in an r-dimensionalsubspace of R n because the convex hull of any list forms a polyhedron, andbecause the translated affine hull A − α contains the translated polyhedronP − α (1026) and the origin (when α ∈ A), and because A has dimensionr by definition (1028). Now, any distance-square from the origin to thepolyhedron P − α can be formulated{p T p = ‖p‖ 2 = 2y T V T NX T XV N y | y ∈ S − β} (1088)Applying (1035) to (1088) we get (1080).(⇐) To validate the EDM assertion in the reverse direction, we prove: If eachdistance-square ‖p(y)‖ 2 (1080) on the shifted unit-simplex S −β ⊂ R N−1corresponds to a point p(y) in some embedded polyhedron P − α , thenD is an EDM. The r-dimensional subspace A − α ⊆ R n is spanned byp(S − β) = P − α (1089)because A − α = aff(P − α) ⊇ P − α (1026). So, outside domain S − βof linear surjection p(y) , simplex complement \S − β ⊂ R N−1 must containdomain of the distance-square ‖p(y)‖ 2 = p(y) T p(y) to remaining points insubspace A − α ; id est, to the polyhedron’s relative exterior \P − α . For‖p(y)‖ 2 to be nonnegative on the entire subspace A − α , −VN TDV N mustbe positive semidefinite and is assumed symmetric; 5.48−V T NDV N Θ T pΘ p (1090)where 5.49 Θ p ∈ R m×N−1 for some m ≥ r . Because p(S − β) is a convexpolyhedron, it is necessarily a set of linear combinations of points from some5.48 The antisymmetric part ( −V T N DV N − (−V T N DV N) T) /2 is annihilated by ‖p(y)‖ 2 . Bythe same reasoning, any positive (semi)definite matrix A is generally assumed symmetricbecause only the symmetric part (A +A T )/2 survives the test y T Ay ≥ 0. [202,7.1]5.49 A T = A ≽ 0 ⇔ A = R T R for some real matrix R . [331,6.3]