v2010.10.26 - Convex Optimization

4.1. CONIC PROBLEM 281faces of S 3 + correspond to faces of S 3 + (confer Table 2.9.2.3.1):k dim F(S+) 3 dim F(S 3 +) dim F(S 3 + ∋ rank-k matrix)0 0 0 0boundary 1 1 1 12 2 3 3interior 3 3 6 6Integer k indexes k-dimensional faces F of S 3 + . Positive semidefinitecone S 3 + has four kinds of faces, including cone itself (k = 3,boundary + interior), whose dimensions in isometrically isomorphicR 6 are listed under dim F(S 3 +). Smallest face F(S 3 + ∋ rank-k matrix)that contains a rank-k positive semidefinite matrix has dimensionk(k + 1)/2 by (222).For A equal to intersection of m hyperplanes having independentnormals, and for X ∈ S 3 + ∩ A , we have rankX ≤ m ; the analogueto (272).Proof. With reference to Figure 81: Assume one (m = 1) hyperplaneA = ∂H intersects the polyhedral cone. Every intersecting planecontains at least one matrix having rank less than or equal to 1 ; id est,from all X ∈ ∂H ∩ S 3 + there exists an X such that rankX ≤ 1. Rank 1is therefore an upper bound in this case.Now visualize intersection of the polyhedral cone with two (m = 2)hyperplanes having linearly independent normals. The hyperplaneintersection A makes a line. Every intersecting line contains at least onematrix having rank less than or equal to 2, providing an upper bound.In other words, there exists a positive semidefinite matrix X belongingto any line intersecting the polyhedral cone such that rankX ≤ 2.In the case of three independent intersecting hyperplanes (m = 3), thehyperplane intersection A makes a point that can reside anywhere inthe polyhedral cone. The upper bound on a point in S+ 3 is also thegreatest upper bound: rankX ≤ 3.