v2009.01.01 - Convex Optimization

552 APPENDIX A. LINEAR ALGEBRA
For A,B ∈ S n and A ≽ 0, B ≽ 0 (Example A.2.1.0.1)
AB = BA ⇒ λ(AB) i =λ(A) i λ(B) i ≥ 0 ∀i ⇒ AB ≽ 0 (1381)
AB = BA ⇒ λ(AB) i ≥ 0, λ(A) i λ(B) i ≥ 0 ∀i ⇔ AB ≽ 0 (1382)
For A,B ∈ S n [177,4.2.13]
A ≽ 0, B ≽ 0 ⇒ A ⊗ B ≽ 0 (1383)
A ≻ 0, B ≻ 0 ⇒ A ⊗ B ≻ 0 (1384)
and the Kronecker product is symmetric.
For A,B ∈ S n [344,6.2]
A ≽ 0 ⇒ trA ≥ 0 (1385)
A ≽ 0, B ≽ 0 ⇒ trA trB ≥ tr(AB) ≥ 0 (1386)
Because A ≽ 0, B ≽ 0 ⇒ λ(AB) = λ( √ AB √ A) ≽ 0 by (1367) and
Corollary A.3.1.0.5, then we have tr(AB) ≥ 0.
A ≽ 0 ⇔ tr(AB)≥ 0 ∀B ≽ 0 (335)
For A,B,C ∈ S n (Löwner)
A ≼ B , B ≼ C ⇒ A ≼ C
A ≼ B ⇔ A + C ≼ B + C
A ≼ B , A ≽ B ⇒ A = B
A ≼ A
(transitivity)
(additivity)
(antisymmetry)
(reflexivity)
(1387)
For A,B ∈ R n×n x T Ax ≥ x T Bx ∀x ⇒ trA ≥ trB (1388)
Proof. x T Ax≥x T Bx ∀x ⇔ λ((A −B) + (A −B) T )/2 ≽ 0 ⇒
tr(A+A T −(B+B T ))/2 = tr(A −B)≥0. There is no converse.