v2009.01.01 - Convex Optimization

2.9. POSITIVE SEMIDEFINITE (PSD) CONE 111
2.9.2.1 rank ρ subset of the positive semidefinite cone
For the same reason (closure), this applies more generally; for 0≤ρ≤M
{
A ∈ S
M
+ | rankA=ρ } = { A ∈ S M + | rankA≤ρ } (198)
For easy reference, we give such generally nonconvex sets a name:
rank ρ subset of a positive semidefinite cone. For ρ < M this subset,
nonconvex for M > 1, resides on the positive semidefinite cone boundary.
2.9.2.1.1 Exercise. Closure and rank ρ subset.
Prove equality in (198).
For example,
∂S M + = { A ∈ S M + | rankA=M− 1 } = { A ∈ S M + | rankA≤M− 1 } (199)
In S 2 , each and every ray on the boundary of the positive semidefinite cone
in isomorphic R 3 corresponds to a symmetric rank-1 matrix (Figure 37), but
that does not hold in any higher dimension.
2.9.2.2 Subspace tangent to open rank-ρ subset
When the positive semidefinite cone subset in (198) is left unclosed as in
M(ρ) ∆ = { A ∈ S M + | rankA=ρ } (200)
then we can specify a subspace tangent to the positive semidefinite cone
at a particular member of manifold M(ρ). Specifically, the subspace R M
tangent to manifold M(ρ) at B ∈ M(ρ) [160,5, prop.1.1]
R M (B) ∆ = {XB + BX T | X ∈ R M×M } ⊆ S M (201)
has dimension
(
dim svec R M (B) = ρ M − ρ − 1 )
= ρ(M − ρ) +
2
ρ(ρ + 1)
2
(202)
Tangent subspace R M contains no member of the positive semidefinite cone
S M + whose rank exceeds ρ .
Subspace R M (B) is a supporting hyperplane when B ∈ M(M −1).
Another good example of tangent subspace is given inE.7.2.0.2 by (1876);
R M (11 T ) = S M⊥
c , orthogonal complement to the geometric center subspace.
(Figure 124, p.480)