v2007.11.26 - Convex Optimization

508 APPENDIX A. LINEAR ALGEBRADiagonalizable matrices A,B∈R n×n commute if and only if they aresimultaneously diagonalizable. [150,1.3.12] A product of diagonalmatrices is always commutative.For A,B ∈ R n×n and AB = BAx T Ax ≥ 0, x T Bx ≥ 0 ∀x ⇒ λ(A+A T ) i λ(B+B T ) i ≥ 0 ∀i x T ABx ≥ 0 ∀x(1306)the negative result arising because of the schism between the productof eigenvalues λ(A + A T ) i λ(B + B T ) i and the eigenvalues of thesymmetrized matrix product λ(AB + (AB) T ) i. For example, X 2 isgenerally not positive semidefinite unless matrix X is symmetric; then(1287) applies. Simply substituting symmetric matrices changes theoutcome:For A,B ∈ S n and AB = BAA ≽ 0, B ≽ 0 ⇒ λ(AB) i =λ(A) i λ(B) i ≥0 ∀i ⇔ AB ≽ 0 (1307)Positive semidefiniteness of A and B is sufficient but not a necessarycondition for positive semidefiniteness of the product AB .Proof. Because all symmetric matrices are diagonalizable, (A.5.2)[251,5.6] we have A=SΛS T and B=T∆T T , where Λ and ∆ arereal diagonal matrices while S and T are orthogonal matrices. Because(AB) T =AB , then T must equal S , [150,1.3] and the eigenvaluesof A are ordered in the same way as those of B ; id est, λ(A) i =δ(Λ) iand λ(B) i =δ(∆) i correspond to the same eigenvector.(⇒) Assume λ(A) i λ(B) i ≥0 for i=1... n . AB=SΛ∆S T issymmetric and has nonnegative eigenvalues contained in diagonalmatrix Λ∆ by assumption; hence positive semidefinite by (1273). Nowassume A,B ≽0. That, of course, implies λ(A) i λ(B) i ≥0 for all ibecause all the individual eigenvalues are nonnegative.(⇐) Suppose AB=SΛ∆S T ≽ 0. Then Λ∆≽0 by (1273),and so all products λ(A) i λ(B) i must be nonnegative; meaning,sgn(λ(A))= sgn(λ(B)). We may, therefore, conclude nothing aboutthe semidefiniteness of A and B .