Exercise 8.C.7

Answers

Proof. By Exercise 4 in section 5B, we have

V = null P range P.

It is easy to check that if v range P then Pv = v. Thus

null P G(0,T) and  range P G(1,T).

(1) and 8.26 then imply that these inclusions are actually equalities, which gives the desired result. □

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2017-10-06 00:00
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