v2010.10.26 - Convex Optimization

A.3. PROPER STATEMENTS 607For A,B ∈ S n [393,6.2] (Theorem A.3.1.0.4)A ≽ B ⇒ trA ≥ trB (1489)A ≽ B ⇒ δ(A) ≽ δ(B) (1490)There is no converse, and restriction to the positive semidefinite conedoes not improve the situation. All-strict versions hold.A ≽ B ≽ 0 ⇒ rankA ≥ rankB (1491)A ≽ B ≽ 0 ⇒ detA ≥ detB ≥ 0 (1492)A ≻ B ≽ 0 ⇒ detA > detB ≥ 0 (1493)For A,B ∈ int S n + [34,4.2] [202,7.7.4]A ≽ B ⇔ A −1 ≼ B −1 , A ≻ 0 ⇔ A −1 ≻ 0 (1494)For A,B ∈ S n [393,6.2]A ≽ B ≽ 0 ⇒ √ A ≽ √ BA ≽ 0 ⇐ A 1/2 ≽ 0(1495)For A,B ∈ S n and AB = BA [393,6.2 prob.3]A ≽ B ≽ 0 ⇒ A k ≽ B k , k=1, 2,... (1496)A.3.1.0.1 Theorem. Positive semidefinite ordering of eigenvalues.For A,B∈ R M×M , place the eigenvalues of each symmetrized matrix intothe respective vectors λ ( 1(A 2 +AT ) ) , λ ( 1(B 2 +BT ) ) ∈ R M . Then [331,6]x T Ax ≥ 0 ∀x ⇔ λ ( A +A T) ≽ 0 (1497)x T Ax > 0 ∀x ≠ 0 ⇔ λ ( A +A T) ≻ 0 (1498)because x T (A −A T )x=0. (1432) 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 the1largest eigenvalue of symmetrized A while λ ( 1(B 2 +BT ) ) holds the largest1eigenvalue of symmetrized B , and so on. Then [202,7.7 prob.1 prob.9] forκ ∈ Rx T Ax ≥ x T Bx ∀x ⇒ λ ( A +A T) ≽ λ ( B +B T)x T Ax ≥ x T Ixκ ∀x ⇔ λ ( 12 (A +AT ) ) ≽ κ1(1499)