v2006.03.09 - Convex Optimization

4.9. BRIDGE: CONVEX POLYHEDRA TO EDMS 273having N or fewer vertices embedded in an r-dimensional subspace A − αof R n , where α ∈ A = aff P and where the domain of linear surjection p(y)is the unit simplex S ⊂ R N−1+ shifted such that its vertex at the origin istranslated to −β in R N−1 . When β = 0, then α = x 1 .⋄In terms of V N , the unit simplex (241) in R N−1 has an equivalentrepresentation:S = {s ∈ R N−1 | √ 2V N s ≽ −e 1 } (629)where e 1 is as in (476). Incidental to the EDM assertion, shifting theunit-simplex domain in R N−1 translates the polyhedron P in R n . Indeed,there is a map from vertices of the unit simplex to members of the listgenerating P ;p : R N−1⎛⎧⎪⎨p⎜⎝⎪⎩−βe 1 − βe 2 − β.e N−1 − β⎫⎞⎪⎬⎟⎠⎪⎭→=⎧⎪⎨⎪⎩R nx 1 − αx 2 − αx 3 − α.x N − α⎫⎪⎬⎪⎭(630)4.9.1.0.4 Proof. EDM assertion.(⇒) We demonstrate that if D is an EDM, then each distance-square ‖p(y)‖ 2described by (627) corresponds to a point p in some embedded polyhedronP − α . Assume D is indeed an EDM; id est, D can be made from some listX of N unknown points in Euclidean space R n ; D =D(X) for X ∈ R n×Nas in (460). Since D is translation invariant (4.5.1), we may shift the affinehull A of those unknown points to the origin as in (572). Then take anypoint p in their convex hull (71);P − α = {p = (X − Xb1 T )a | a T 1 = 1, a ≽ 0} (631)where α = Xb ∈ A ⇔ b T 1 = 1. Solutions to a T 1 = 1 are: 4.35a ∈{e 1 + √ }2V N s | s ∈ R N−1(632)4.35 Since R(V N )= N(1 T ) and N(1 T )⊥ R(1) , then over all s∈ R N−1 , V N s is ahyperplane through the origin orthogonal to 1. Thus the solutions {a} constitute ahyperplane orthogonal to the vector 1, and offset from the origin in R N by any particularsolution; in this case, a = e 1 .