... equalities ab = 0 and ad+bc = 0cannot hold simultaneously. If ab = 0 and P is invertible, then either a = 0 and b = 0 or b = 0 and a = 0. In the first case 0 = ad + bc = bc and therefore, c ... Let V = V1⊕V2 and Vi= 0. Then 0 < |I −P12P21| ≤ 1 and the equality takes place if and only if V1⊥ V2.Proof. The operators P1 and P2are nonnegative definite and, therefore, ... since P1x ∈ V1 and P2x ∈ V2. Hence, x ⊥ V1 and x ⊥ V2 and, therefore, x = 0. Hence, A is positive definite and |A| > 0.For a basis of V take the union of bases of V1 and V2. In these...