Exercise 6.6.6

Answers

It’s enough to show that $R {(T)}^{⊥} = N (T)$ . If $x \in R {(T)}^{⊥}$ , we have

⟨ T (x), T (x) ⟩ = ⟨ x, T^{∗} T (x) ⟩ = ⟨ x, T (T^{∗} (x)) ⟩ = 0

and so $T (x) = 0$ . If $x \in N (T)$ , we have

⟨ x, T (y) ⟩ = ⟨ T^{∗} (x), y ⟩ = 0

since $T^{∗} (x) = 0$ by Theorem 6.15(c). Hence now we know that $R {(T)}^{⊥} = N (T)$ and $T$ is an orthogonal projection.

jephianlin

2011-06-27 00:00