By Barnett S.

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Choose B o to be an orthogonal projector on R(B)⊥ . Then it follows that o R((AA B ) ) + R(B) = R(A) + R(B). 3). This shows an interesting application of how to prove a statement by the aid of an adjoint transformation. Baksalary & Kala (1978), as well as several other authors, use a decomposition which is presented in the next theorem. 18. Let A, B and C be arbitrary transformations such that the spaces are well deﬁned, and let P be an orthogonal projector on R(C). Then V = B1 + B2 + R(P A) + ⊥ R(P ) , where B1 = R(P ) ∩ (R(P A) + R(P B))⊥ , B2 =(R(P A) + R(P B)) ∩ R(P A)⊥ .

4. Let A, B and C be arbitrary subspaces of Λ. Then (i) A + B=A + C ⇔ B = C; (ii) A + B⊆A + C ⇔ B ⊆ C. 3 (ii), has been applied to A + B = A + C. Thus (i) is proved, and (ii) follows analogously. Note that we do not have to assume B and C to be comparable. It is orthogonality between A and C, and B and C, that makes them comparable. Another important property of orthogonal subspaces not shared by disjoint subspaces may also be worth observing. 5. Let B and {Ai } be arbitrary subspaces of Λ such that B ⊥ Ai for all i.

Let A, B and C be any linear transformations such that AB and AC are deﬁned. Then (i) R(AB) = R(AC) if R(B) = R(C); (ii) R(AB) ⊆ R(AC) if R(B) ⊆ R(C). The next two lemmas comprise standard results which are very useful. In the proofs we accustom the reader with the technique of using inner products and adjoint transformations. 5. Let A be an arbitrary linear transformation. Then ⊥ N (A ) = R(A) Proof: Suppose y ∈ R(A)⊥ . By deﬁnition of R(A) for any z we have a vector x = Az ∈ R(A). Hence, 0 = (x, y) = (Az, y) = (z, A y) ⇒ A y = 0.