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v2007.11.26 - Convex Optimization

v2007.11.26 - Convex Optimization

v2007.11.26 - Convex Optimization

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5.9. BRIDGE: CONVEX POLYHEDRA TO EDMS 369Making the substitution s − β ← yP − α = {p = X √ 2V N y | y ∈ S − β} (920)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 (860) and the origin (when α∈A), and because A has dimension r bydefinition (862). Now, any distance-square from the origin to the polyhedronP − α can be formulated{p T p = ‖p‖ 2 = 2y T V T NX T XV N y | y ∈ S − β} (921)Applying (869) to (921) we get (913).(⇐) To validate the EDM assertion in the reverse direction, we prove: Ifeach distance-square ‖p(y)‖ 2 (913) on the shifted unit-simplex S −β ⊂ R N−1corresponds to a point p(y) in some embedded polyhedron P − α , then Dis an EDM. The r-dimensional subspace A − α ⊆ R n is spanned byp(S − β) = P − α (922)because A − α = aff(P − α) ⊇ P − α (860). So, outside the domain S − βof linear surjection p(y) , the simplex complement \S − β ⊂ R N−1 mustcontain the domain of the distance-square ‖p(y)‖ 2 = p(y) T p(y) to remainingpoints in the subspace 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 ,−V T N DV N must be positive semidefinite and is assumed symmetric; 5.43−V T NDV N ∆ = Θ T pΘ p (923)where 5.44 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 somelength-N list because every convex polyhedron having N or fewer verticescan be generated that way (2.12.2). Equivalent to (913) are{p T p | p ∈ P − α} = {p T p = y T Θ T pΘ p y | y ∈ S − β} (924)5.43 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. [150,7.1]5.44 A T = A ≽ 0 ⇔ A = R T R for some real matrix R . [251,6.3]

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