v2009.01.01 - Convex Optimization

5.9. BRIDGE: CONVEX POLYHEDRA TO EDMS 411
that describes the domain of p(s) as the unit simplex
Making the substitution s − β ← y
S = {s | √ 2V N s ≽ −e 1 } ⊂ R N−1
+ (987)
P − α = {p = X √ 2V N y | y ∈ S − β} (992)
Point p belongs to a convex polyhedron P−α embedded in an r-dimensional
subspace of R n because the convex hull of any list forms a polyhedron, and
because the translated affine hull A − α contains the translated polyhedron
P − α (931) and the origin (when α∈A), and because A has dimension r by
definition (933). Now, any distance-square from the origin to the polyhedron
P − α can be formulated
{p T p = ‖p‖ 2 = 2y T V T NX T XV N y | y ∈ S − β} (993)
Applying (940) to (993) we get (985).
(⇐) To validate the EDM assertion in the reverse direction, we prove: If
each distance-square ‖p(y)‖ 2 (985) on the shifted unit-simplex S −β ⊂ R N−1
corresponds to a point p(y) in some embedded polyhedron P − α , then D
is an EDM. The r-dimensional subspace A − α ⊆ R n is spanned by
p(S − β) = P − α (994)
because A − α = aff(P − α) ⊇ P − α (931). So, outside the domain S − β
of linear surjection p(y) , the simplex complement \S − β ⊂ R N−1 must
contain the domain of the distance-square ‖p(y)‖ 2 = p(y) T p(y) to remaining
points 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; 5.44
−V T NDV N ∆ = Θ T pΘ p (995)
where 5.45 Θ p ∈ R m×N−1 for some m ≥ r . Because p(S − β) is a convex
polyhedron, it is necessarily a set of linear combinations of points from some
5.44 The antisymmetric part ( −V T N DV N − (−V T N DV N) T) /2 is annihilated by ‖p(y)‖ 2 . By
the same reasoning, any positive (semi)definite matrix A is generally assumed symmetric
because only the symmetric part (A +A T )/2 survives the test y T Ay ≥ 0. [176,7.1]
5.45 A T = A ≽ 0 ⇔ A = R T R for some real matrix R . [287,6.3]