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v2007.09.13 - Convex Optimization

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For A,B ∈ R n×n x T Ax ≥ x T Bx ∀x ⇒ trA ≥ trB (1285)A.3. PROPER STATEMENTS 493For 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 (1278)AB = BA ⇒ λ(AB) i ≥ 0, λ(A) i λ(B) i ≥ 0 ∀i ⇔ AB ≽ 0 (1279)For A,B ∈ S n [298,6.2]A ≽ 0 ⇒ trA ≥ 0 (1280)A ≽ 0, B ≽ 0 ⇒ trA trB ≥ tr(AB) ≥ 0 (1281)Because A ≽ 0, B ≽ 0 ⇒ λ(AB) = λ( √ AB √ A) ≽ 0 by (1264) andCorollary A.3.1.0.5, then we have tr(AB) ≥ 0.For A,B,C ∈ S n (Löwner)A ≽ 0 ⇔ tr(AB)≥ 0 ∀B ≽ 0 (322)A ≼ B , B ≼ C ⇒ A ≼ C (1282)A ≼ B ⇔ A + C ≼ B + C (1283)A ≼ B , A ≽ B ⇒ A = B (1284)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. For A,B ∈ S n [298,6.2, prob.1] (Theorem A.3.1.0.4)A ≽ B ⇒ trA ≥ trB (1286)A ≽ B ⇒ δ(A) ≽ δ(B) (1287)There is no converse, and restriction to the positive semidefinite conedoes not improve the situation. The all-strict versions hold. From[298,6.2]A ≽ B ≽ 0 ⇒ rankA ≥ rankB (1288)A ≽ B ≽ 0 ⇒ detA ≥ detB ≥ 0 (1289)A ≻ B ≽ 0 ⇒ detA > detB ≥ 0 (1290)For A,B ∈ int S n + [27,4.2] [149,7.7.4]A ≽ B ⇔ A −1 ≼ B −1 (1291)

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