Exercise 7.A.4

Answers

Proof. Since ${(T^{∗})}^{∗} = T$ , the statements in (a) and (b) are equivalent, so we give a single proof.

This follows directly from 7.7 (b) and 6.46 (b). We have

\begin{aligned} T is injective & ⟺ null T = 0 \\ ⟺ {(range T^{∗})}^{⊥} = {0} \\ ⟺ range T^{∗} = V \\ ⟺ T^{∗} is surjective. \end{aligned}

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2017-10-06 00:00