v2010.10.26 - Convex Optimization

701U T = U † for orthonormal (including the orthogonal) matrices U . So, fororthonormal matrices U,Q and arbitrary A(UAQ T ) † = QA † U T (1875)E.0.0.0.2Prove:Exercise. Kronecker inverse.(A ⊗ B) † = A † ⊗ B † (1876)E.0.1Logical deductionsWhen A is invertible, A † = A −1 ; so A † A = AA † = I . Otherwise, forA∈ R m×n [153,5.3.3.1] [241,7] [295]g. A † A = I , A † = (A T A) −1 A T , rankA = nh. AA † = I , A † = A T (AA T ) −1 , rankA = mi. A † Aω = ω , ω ∈ R(A T )j. AA † υ = υ , υ ∈ R(A)k. A † A = AA † , A normall. A k† = A †k , A normal, k an integerEquivalent to the corresponding Moore-Penrose condition:1. A T = A T AA † or A T = A † AA T2. A †T = A †T A † A or A †T = AA † A †TWhen A is symmetric, A † is symmetric and (A.6)A ≽ 0 ⇔ A † ≽ 0 (1877)E.0.1.0.1 Example. Solution to classical linear equation Ax = b .In2.5.1.1, the solution set to matrix equation Ax = b was representedas an intersection of hyperplanes. Regardless of rank of A or its shape(fat or skinny), interpretation as a hyperplane intersection describing apossibly empty affine set generally holds. If matrix A is rank deficient orfat, there is an infinity of solutions x when b∈R(A). A unique solutionoccurs when the hyperplanes intersect at a single point.