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I've been given the following question. My problem is that I'm not really sure what I'm suppose to do. Can someone help me getting started maybe just give me a theorem I could use.

Consider the set of functions $$ V := \{ f: \mathbb{R} \rightarrow \mathbb{C} \, \lvert \, f(-\pi) = f(\pi) = 0 \}$$ (i) Show that $V$ is a vector space with respect to the usual operations of addition and scalar multiplication.

Thanks in advance

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  • $\begingroup$ A vector space is closed under addition and scalar multiplication. Have you tried demonstrating either of these properties? $\endgroup$ – Mnifldz Feb 11 '15 at 22:16
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Let $f$ and $g$ be elements of your vector space $V$. Observe that

$$ (f+g)(\pi) \;\; =\;\; f(\pi) + g(\pi) \;\; =\;\; 0 \;\; f(-\pi) + g(-\pi) \;\; =\;\; (f+g)(-\pi) $$

hence $f+g \in V$ since it is closed under vector addition. Similarly if $a \in \mathbb{C}$ we have that $af(\pi) = 0 = af(-\pi)$ hence $V$ is closed under scalar multiplication. $V$ is therefore a vector space.

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  • $\begingroup$ To prove something is a vector space over a field $K$ you need to verify the $8$ axioms of a vector space. Here you are assuming the set of functions $\mathbb{R}\to \mathbb{C}$ is a vector space, and what you proved is that $V$ is a linear subspace. $\endgroup$ – Qidi Feb 11 '15 at 23:06

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