Page 58 - Textos de Matemática Vol. 44

48 M. C. GOUVEIA
if A(A + B)−B is independent of the choice of the generalized inverse.
We notice that A(A + B)−B is independent of the choice of the generalized inverse if R(A) ⊂ R(A+B), R(A∗) ⊂ R(A∗ +B∗) or, equivalently, if R(B) ⊂ R(A + B), R(B∗) ⊂ R(A∗ + B∗).
Theorem 2.3. Let A, B be PSD matrices of the same order m × n. Then, for suitable choices of the generalized inverse,
(i) A ∓ B = B ∓ A.
(ii) A− + B− is a (1)-inverse of (A ∓ B).
(iii) R(A ∓ B) = R(A) ∩ R(B).
Proof. (i)
A ∓ B = A(A + B)−B
= A(A + B)−(A + B) − A(A + B)−A = A − A(A + B)−A
= (A + B)(A + B)−A − A(A + B)−A = B(A + B)−A = B ∓ A.
(ii)
(A ∓ B)(A− + B−)(A ∓ B) = A(A + B)−BB−B(A + B)−A+
B(A + B)−AA−A(A + B)−B =[A(A+B)− +B(A+B)−]B(A+B)−A = (A + B)(A + B)−A(A + B)−B
= A(A + B)−B = A ∓ B
(iii) By the definition it is clear that R(A ∓ B) ⊂ R(A) ∩ R(B). Conversely, let x ∈ R(A) ∩ R(B). Then
(A ∓ B)(A− + B−)x = A(A + B)−BB−x + B(A + B)−AA−x = A(A + B)−x + B(A + B)−x
= (A + B)(A + B)−x = x,
since x ∈ R(A) ∩ R(B) ⊂ R(A + B). Therefore, x ∈ R(A ∓ B) which leads
to R(A) ∩ R(B) ⊂ R(A ∓ B).
The formula for A ∓ B looses sense when we consider infinite Hilbert spaces. So, in this case, the parallel sum must be redefined.