Page 61 - Textos de Matemática Vol. 44

PARALLEL SUMS 51
(v) Let x ∈ V . By definition,
(A : B)(A : B)†x = PR(A:B)x.
Since A : B is hermitian and R(A : B) = R(A) ∩ R(B), from the proof of (iv) we have
(A : B)(A† + B†)PR(A:B)x = PR(A:B)x.
Then
(A : B)(A : B)†x = (A : B)(A† + B†)PR(A:B)x,
and, multiplying on the left by (A : B)† it follows (A : B)†x = PR(A:B)(A† + B†)PR(A:B)x.
Finally, as A†† = A, we obtain
(A : B) = (PR(A:B)(A† + B†)PR(A:B))†.
The following result is the previously referred solution for the finite case. Theorem 3.2. If E and F are projections, then 2E : F = E∧F is the projection
onto R(E) ∩ R(F ).
Proof. From Lemma 3.1 we know that R(E : F) = R(E) ∩ R(F) and also thatE:F =E∗ :F∗ =(E:F)∗.Hence,wejusthavetoprovethat2(E:F) is idempotent.
Let x ∈ V . Then,
2(E :F)2x=2(E :F)(F :E)x
= E(E + F)†F2(E + F)†E + F(E + F)†E2(E + F)†Fx = E(E + F)†F((E + F)†(E + F))x
= E(E + F)†FPR(E+F)x
= E(E + F)†Fx
= (E : F)x.
Finally, we observe that the extension of Theorem 3.2 to infinite-dimensional Hilbert spaces answers completely the question of Halmos. The proof is given as follows.