Exercise 1.3.13

Let S be a nonempty set and F a field. Prove that for any s0 S, {f (S,F) : f (s0) = 0}, is a subspace of (S,F).

Answers

Proof. It’s closed under addition since (f + g)(s0) = 0 + 0 = 0. It’s closed under scalar multiplication since cf(s0) = c0 = 0. And zero function is in the set. □

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2011-06-27 00:00
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