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Exercise 6.2.17

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First, since $⟨ T (x), y ⟩ = 0$ for all $y$ , we have $T (x) = 0$ . Applying this argument to all $x$ we get $T (x) = 0$ for all $x$ . For the second version, by Exercise 6.1.9 we know that $T (x) = 0$ for all $x$ in some basis for $V$ . And this means $T = T_{0}$ by Theorem 2.6.

jephianlin

2011-06-27 00:00

Exercise 6.2.17

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