v2010.10.26 - Convex Optimization

2.9. POSITIVE SEMIDEFINITE (PSD) CONE 1212.9.2.1 rank ρ subset of the positive semidefinite coneFor the same reason (closure), this applies more generally; for 0≤ρ≤M{A ∈ SM+ | rankA=ρ } = { A ∈ S M + | rankA≤ρ } (216)For easy reference, we give such generally nonconvex sets a name: rank ρsubset of a positive semidefinite cone. For ρ < M this subset, nonconvex forM > 1, resides on the positive semidefinite cone boundary.2.9.2.1.1 Exercise. Closure and rank ρ subset.Prove equality in (216).For example,∂S M + = { A ∈ S M + | rankA=M− 1 } = { A ∈ S M + | rankA≤M− 1 } (217)In S 2 , each and every ray on the boundary of the positive semidefinite conein isomorphic R 3 corresponds to a symmetric rank-1 matrix (Figure 43), butthat does not hold in any higher dimension.2.9.2.2 Subspace tangent to open rank ρ subsetWhen the positive semidefinite cone subset in (216) is left unclosed as inM(ρ) { A ∈ S N + | rankA=ρ } (218)then we can specify a subspace tangent to the positive semidefinite coneat a particular member of manifold M(ρ). Specifically, the subspace R Mtangent to manifold M(ρ) at B ∈ M(ρ) [186,5, prop.1.1]R M (B) {XB + BX T | X ∈ R N×N } ⊆ S N (219)has dimension(dim svec R M (B) = ρ N − ρ − 1 )= ρ(N − ρ) +2ρ(ρ + 1)2(220)Tangent subspace R M contains no member of the positive semidefinite coneS N + whose rank exceeds ρ .Subspace R M (B) is a hyperplane supporting S N + when B ∈ M(N −1).Another good example of tangent subspace is given inE.7.2.0.2 by (2000);R M (11 T ) = S N⊥c , orthogonal complement to the geometric center subspace.(Figure 148, p.532)