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Exercise 6.C.7

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Proof. Define $U = range P$ . Suppose $u \in U$ . There exists $v \in V$ such that $Pv = u$ . Then

u = Pv = P^{2} v = P (Pv) = Pu .

We’ve shown that $Pu = u$ for any $u \in U$ . Let $v \in V$ . By Exercise 4 in section 5B, we can write $v = u + w$ for some $u \in U$ and $w \in null P$ . Note that $null P \subset U^{⊥}$ . Thus

Pv = P (u + w) = Pu = u = P_{U} (u + w) = P_{U} v .

Therefore $P = P_{U}$ . □

celiopassos

2017-10-06 00:00

Exercise 6.C.7

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