v2009.01.01 - Convex Optimization

A.7. ZEROS 573
For X,A∈ S M +
[31,2.6.1, exer.2.8] [312,3.1]
tr(XA) = 0 ⇔ XA = AX = 0 (1476)
Proof. (⇐) Suppose XA = AX = 0. Then tr(XA)=0 is obvious.
(⇒) Suppose tr(XA)=0. tr(XA)= tr( √ AX √ A) whose argument is
positive semidefinite by Corollary A.3.1.0.5. Trace of any square matrix is
equivalent to the sum of its eigenvalues. Eigenvalues of a positive semidefinite
matrix can total 0 if and only if each and every nonnegative eigenvalue
is 0. The only feasible positive semidefinite matrix, having all 0 eigenvalues,
resides at the origin; (confer (1500)) id est,
√
AX
√
A =
(√
X
√
A
) T√
X
√
A = 0 (1477)
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 and
only if they are commutative under multiplication; [176,1.3.12] id est, iff
they share a complete set of eigenvectors.
A.7.4.0.1 Example. An equivalence in nonisomorphic spaces.
Identity (1476) leads to an unusual equivalence relating convex geometry to
traditional linear algebra: The convex sets, given A ≽ 0
{X | 〈X , A〉 = 0} ∩ {X ≽ 0} ≡ {X | N(X) ⊇ R(A)} ∩ {X ≽ 0} (1478)
(one expressed in terms of a hyperplane, the other in terms of nullspace and
range) are equivalent only when symmetric matrix A is positive semidefinite.
We might apply this equivalence to the geometric center subspace, for
example,
S M c = {Y ∈ S M | Y 1 = 0}
= {Y ∈ S M | N(Y ) ⊇ 1} = {Y ∈ S M | R(Y ) ⊆ N(1 T )}
(1874)
from which we derive (confer (899))
S M c ∩ S M + ≡ {X ≽ 0 | 〈X , 11 T 〉 = 0} (1479)