v2007.09.13 - Convex Optimization

6.8. DUAL EDM CONE 425whose veracity is intuitively evident, in hindsight, [61, p.109] from themost fundamental EDM definition (705). Formula (1070) is not a matrixcriterion for membership to the EDM cone, and it is not an equivalencebetween EDM operators. Rather, it is a recipe for constructing the EDMcone whole from large Euclidean bodies: the positive semidefinite cone,orthogonal complement of the geometric center subspace, and symmetrichollow subspace. A realization of this construction in low dimension isillustrated in Figure 104 and Figure 105.The dual EDM cone follows directly from (1070) by standard propertiesof cones (2.13.1.1):EDM N∗ = K ∗ 1 + K ∗ 2 = S N⊥h − S N c ∩ S N + (1071)which bears strong resemblance to (1050).6.8.1.2 nonnegative orthantThat EDM N is a proper subset of the nonnegative orthant is not obviousfrom (1070). We wish to verifyEDM N = S N h ∩ ( )S N⊥c − S N + ⊂ RN×N+ (1072)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 ) (6.5.3.1),δ(zz T )1 T + 1δ(zz T ) T − 2zz T ≥ 0 (1073)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 (1074)to the orthogonal complement (1763) of the geometric center subspace, whiletheir difference is a member of the symmetric hollow subspace S N h .Here is an algebraic method provided by Trosset to prove nonnegativity:Suppose we are given A∈ S N⊥c and B = [B ij ]∈ S N + and A −B ∈ S N h .Then we have, for some vector u , A = u1 T + 1u T = [A ij ] = [u i + u j ] andδ(B)= δ(A)= 2u . Positive semidefiniteness of B requires nonnegativityA −B ≥ 0 because(e i −e j ) T B(e i −e j ) = (B ii −B ij )−(B ji −B jj ) = 2(u i +u j )−2B ij ≥ 0 (1075)