v2009.01.01 - Convex Optimization

410 CHAPTER 5. EUCLIDEAN DISTANCE MATRIX
In terms of V N , the unit simplex (265) in R N−1 has an equivalent
representation:
S = {s ∈ R N−1 | √ 2V N s ≽ −e 1 } (987)
where e 1 is as in (814). Incidental to the EDM assertion, shifting the
unit-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 list
generating P ;
p : R N−1
⎛⎧
⎪⎨
p
⎜
⎝
⎪⎩
−β
e 1 − β
e 2 − β
.
e N−1 − β
⎫⎞
⎪⎬
⎟
⎠
⎪⎭
→
=
⎧
⎪⎨
⎪⎩
R n
x 1 − α
x 2 − α
x 3 − α
.
x N − α
⎫
⎪⎬
⎪⎭
(988)
5.9.1.0.5 Proof. EDM assertion.
(⇒) We demonstrate that if D is an EDM, then each distance-square ‖p(y)‖ 2
described by (985) corresponds to a point p in some embedded polyhedron
P − α . Assume D is indeed an EDM; id est, D can be made from some list
X of N unknown points in Euclidean space R n ; D = D(X) for X ∈ R n×N
as in (798). Since D is translation invariant (5.5.1), we may shift the affine
hull A of those unknown points to the origin as in (928). Then take any
point p in their convex hull (78);
P − α = {p = (X − Xb1 T )a | a T 1 = 1, a ≽ 0} (989)
where α = Xb ∈ A ⇔ b T 1 = 1. Solutions to a T 1 = 1 are: 5.43
a ∈
{e 1 + √ }
2V N s | s ∈ R N−1
(990)
where e 1 is as in (814). Similarly, b = e 1 + √ 2V N β .
P − α = {p = X(I − (e 1 + √ 2V N β)1 T )(e 1 + √ 2V N s) | √ 2V N s ≽ −e 1 }
= {p = X √ 2V N (s − β) | √ 2V N s ≽ −e 1 } (991)
5.43 Since R(V N )= N(1 T ) and N(1 T )⊥ R(1) , then over all s∈ R N−1 , V N s is a
hyperplane through the origin orthogonal to 1. Thus the solutions {a} constitute a
hyperplane orthogonal to the vector 1, and offset from the origin in R N by any particular
solution; in this case, a = e 1 .