v2009.01.01 - Convex Optimization

A.3. PROPER STATEMENTS 551
Diagonalizable matrices A,B∈R n×n commute if and only if they are
simultaneously diagonalizable. [176,1.3.12] A product of diagonal
matrices is always commutative.
For A,B ∈ R n×n and AB = BA
x 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
(1379)
the negative result arising because of the schism between the product
of eigenvalues λ(A + A T ) i λ(B + B T ) i and the eigenvalues of the
symmetrized matrix product λ(AB + (AB) T ) i
. For example, X 2 is
generally not positive semidefinite unless matrix X is symmetric; then
(1360) applies. Simply substituting symmetric matrices changes the
outcome:
For A,B ∈ S n and AB = BA
A ≽ 0, B ≽ 0 ⇒ λ(AB) i =λ(A) i λ(B) i ≥0 ∀i ⇔ AB ≽ 0 (1380)
Positive semidefiniteness of A and B is sufficient but not a necessary
condition for positive semidefiniteness of the product AB .
Proof. Because all symmetric matrices are diagonalizable, (A.5.2)
[287,5.6] we have A=SΛS T and B=T∆T T , where Λ and ∆ are
real diagonal matrices while S and T are orthogonal matrices. Because
(AB) T =AB , then T must equal S , [176,1.3] and the eigenvalues of
A are ordered in the same way as those of B ; id est, λ(A) i =δ(Λ) i and
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 is
symmetric and has nonnegative eigenvalues contained in diagonal
matrix Λ∆ by assumption; hence positive semidefinite by (1347). Now
assume A,B ≽0. That, of course, implies λ(A) i λ(B) i ≥0 for all i
because all the individual eigenvalues are nonnegative.
(⇐) Suppose AB=SΛ∆S T ≽ 0. Then Λ∆≽0 by (1347),
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 about
the semidefiniteness of A and B .