v2007.09.13 - Convex Optimization

B.1. RANK-ONE MATRIX (DYAD) 519R(v)0 0 R(Ψ) = R(u)N(Ψ)= N(v T )N(u T )R N = R(v) ⊕ N(uv T )N(u T ) ⊕ R(uv T ) = R MFigure 112:The four fundamental subspaces [249,3.6] of any dyadΨ = uv T ∈R M×N . Ψ(x) ∆ = uv T x is a linear mapping from R N to R M . Themap from R(v) to R(u) is bijective. [247,3.1]When dyad uv T ∈R N×N is square, uv T has at least N −1 0-eigenvaluesand corresponding eigenvectors spanning v ⊥ . The remaining eigenvector uspans the range of uv T with corresponding eigenvalueλ = v T u = tr(uv T ) ∈ R (1392)Determinant is a product of the eigenvalues; so, it is always true thatdet Ψ = det(uv T ) = 0 (1393)When λ = 1, the square dyad is a nonorthogonal projector projecting onits range (Ψ 2 =Ψ ,E.6.2.1); a projector dyad. It is quite possible thatu∈v ⊥ making the remaining eigenvalue instead 0 ; B.2 λ = 0 together withthe first N −1 0-eigenvalues; id est, it is possible uv T were nonzero whileall its eigenvalues are 0. The matrix[ ] 1 [ 1 1]=−1[ 1 1−1 −1](1394)for example, has two 0-eigenvalues. In other words, eigenvector u maysimultaneously be a member of the nullspace and range of the dyad.The explanation is, simply, because u and v share the same dimension,dimu = M = dimv = N :B.2 A dyad is not always diagonalizable (A.5) because its eigenvectors are not necessarilyindependent.