v2010.10.26 - Convex Optimization

A.7. ZEROS 629For X,A∈ S M +[34,2.6.1 exer.2.8] [359,3.1]tr(XA) = 0 ⇔ XA = AX = 0 (1590)Proof. (⇐) Suppose XA = AX = 0. Then tr(XA)=0 is obvious.(⇒) Suppose tr(XA)=0. tr(XA)= tr( √ AX √ A) whose argument ispositive semidefinite by Corollary A.3.1.0.5. Trace of any square matrix isequivalent to the sum of its eigenvalues. Eigenvalues of a positive semidefinitematrix can total 0 if and only if each and every nonnegative eigenvalue is 0.The only positive semidefinite matrix, having all 0 eigenvalues, resides at theorigin; (confer (1614)) id est,√AX√A =(√X√A) T√X√A = 0 (1591)implying √ X √ A = 0 which in turn implies √ X( √ X √ A) √ A = XA = 0.Arguing similarly yields AX = 0.Diagonalizable matrices A and X are simultaneously diagonalizable if andonly if they are commutative under multiplication; [202,1.3.12] id est, iffthey share a complete set of eigenvectors.A.7.4.0.1 Example. An equivalence in nonisomorphic spaces.Identity (1590) leads to an unusual equivalence relating convex geometry totraditional linear algebra: The convex sets, given A ≽ 0{X | 〈X , A〉 = 0} ∩ {X ≽ 0} ≡ {X | N(X) ⊇ R(A)} ∩ {X ≽ 0} (1592)(one expressed in terms of a hyperplane, the other in terms of nullspace andrange) are equivalent only when symmetric matrix A is positive semidefinite.We might apply this equivalence to the geometric center subspace, forexample,S M c = {Y ∈ S M | Y 1 = 0}= {Y ∈ S M | N(Y ) ⊇ 1} = {Y ∈ S M | R(Y ) ⊆ N(1 T )}(1998)from which we derive (confer (994))S M c ∩ S M + ≡ {X ≽ 0 | 〈X , 11 T 〉 = 0} (1593)