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Exercise 6.C.9

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Proof. Suppose $U$ is invariant under $T$ . Let $v \in V$ . Then, by part (b) in 6.55, $P_{U} v \in U$ , which implies that $T P_{U} v \in U$ (since $U$ is invariant under $T$ ), which implies that $P_{U} T P_{U} v = T P_{U} v$ . Thus $P_{U} T P_{U} = T P_{U}$ .

Conversely, suppose $P_{U} T P_{U} = T P_{U}$ . Let $u \in U$ . Then

T_{U} = T P_{U} u = P_{U} T P_{U} u \in U .

Therefore $U$ is invariant under $T$ . □

celiopassos

2017-10-06 00:00

Exercise 6.C.9

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