v2009.01.01 - Convex Optimization

A.3. PROPER STATEMENTS 553
For A,B ∈ S n [344,6.2, prob.1] (Theorem A.3.1.0.4)
A ≽ B ⇒ trA ≥ trB (1389)
A ≽ B ⇒ δ(A) ≽ δ(B) (1390)
There is no converse, and restriction to the positive semidefinite cone
does not improve the situation. The all-strict versions hold. From
[344,6.2]
A ≽ B ≽ 0 ⇒ rankA ≥ rankB (1391)
A ≽ B ≽ 0 ⇒ detA ≥ detB ≥ 0 (1392)
A ≻ B ≽ 0 ⇒ detA > detB ≥ 0 (1393)
For A,B ∈ int S n + [31,4.2] [176,7.7.4]
For A,B ∈ S n [344,6.2]
A ≽ B ⇔ A −1 ≼ B −1 (1394)
A ≽ B ≽ 0 ⇒ √ A ≽ √ B (1395)
For A,B ∈ S n and AB = BA [344,6.2, prob.3]
A ≽ B ≽ 0 ⇒ A k ≽ B k , k=1, 2,... (1396)
A.3.1.0.1 Theorem. Positive semidefinite ordering of eigenvalues.
For A,B∈ R M×M , place the eigenvalues of each symmetrized matrix into
the respective vectors λ ( 1
(A 2 +AT ) ) , λ ( 1
(B 2 +BT ) ) ∈ R M . Then, [287,6]
x T Ax ≥ 0 ∀x ⇔ λ ( A +A T) ≽ 0 (1397)
x T Ax > 0 ∀x ≠ 0 ⇔ λ ( A +A T) ≻ 0 (1398)
because x T (A −A T )x=0. (1335) Now arrange the entries of λ ( 1
(A 2 +AT ) )
and λ ( 1
(B 2 +BT ) ) in nonincreasing order so λ ( 1
(A 2 +AT ) ) holds the
1
largest eigenvalue of symmetrized A while λ ( 1
(B 2 +BT ) ) holds the largest
1
eigenvalue of symmetrized B , and so on. Then [176,7.7, prob.1, prob.9]
for κ ∈ R
x T Ax ≥ x T Bx ∀x ⇒ λ ( A +A T) ≽ λ ( B +B T)
x T Ax ≥ x T Ixκ ∀x ⇔ λ ( 1
2 (A +AT ) ) ≽ κ1
(1399)