v2010.10.26 - Convex Optimization

6.8. DUAL EDM CONE 535therefore the equality [100]EDM N = K 1 ∩ K 2 = S N h ∩ ( )S N⊥c − S N +(1268)whose veracity is intuitively evident, in hindsight, [89, p.109] from the mostfundamental EDM definition (891). Formula (1268) is not a matrix criterionfor membership to the EDM cone, it is not an EDM definition, and it isnot an equivalence between EDM operators or an isomorphism. Rather, it isa recipe for constructing the EDM cone whole from large Euclidean bodies:the positive semidefinite cone, orthogonal complement of the geometric centersubspace, and symmetric hollow subspace. A realization of this constructionin low dimension is illustrated in Figure 148 and Figure 149.The dual EDM cone follows directly from (1268) by standard propertiesof cones (2.13.1.1):EDM N∗ = K ∗ 1 + K ∗ 2 = S N⊥h − S N c ∩ S N + (1269)which bears strong resemblance to (1248).6.8.1.2 nonnegative orthant contains EDM NThat EDM N is a proper subset of the nonnegative orthant is not obviousfrom (1268). We wish to verifyEDM N = S N h ∩ ( )S N⊥c − S N + ⊂ RN×N+ (1270)While there are many ways to prove this, it is sufficient to show that allentries of the extreme directions of EDM N must be nonnegative; id est, forany particular nonzero vector z = [z i , i=1... N]∈ N(1 T ) (6.4.3.2),δ(zz T )1 T + 1δ(zz T ) T − 2zz T ≥ 0 (1271)where the inequality denotes entrywise comparison. The inequality holdsbecause the i,j th entry of an extreme direction is squared: (z i − z j ) 2 .We observe that the dyad 2zz T ∈ S N + belongs to the positive semidefinitecone, the doubletδ(zz T )1 T + 1δ(zz T ) T ∈ S N⊥c (1272)to the orthogonal complement (2000) of the geometric center subspace,while their difference is a member of the symmetric hollow subspace S N h .