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I am a beginning math student in linear algebra. We are going through Vector spaces and subspaces. The question is: Which of the following sets are vector spaces? Give a proof or a counterexample. The set of even functions: $\{f:\Bbb R \to\Bbb R \mid f(-x)=f(x)\text{ for all }x\in\Bbb R\}\;,$ with the usual scalar multiplication and addition of functions.

Is it ok to just write down the 10 axioms that make something a vector space, and then change the variable names to make it more my own? How does writing the axioms down make the proof true?

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    You need to assert that an axiom is true in your case of this set of even functions. Then check that your assertion is true.2012-09-16

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If you’ve already proved that the set of all functions from $\Bbb R$ to $\Bbb R$ is a vector space with the usual scalar multiplication and addition of functions, then you need only prove that this set is a subspace of that space. In order to do that, you must show that it’s non-empty, closed under scalar multiplication, and closed under vector addition. Call the set $V$.

  1. I’m sure that you’ll have no trouble finding an even function, so $V\ne\varnothing$.

  2. If $f$ is an even function and $\alpha\in\Bbb R$, is $\alpha f$ an even function? In other words, is it true for every $x\in\Bbb R$ that $(\alpha f)(-x)=(\alpha f)(x)$?

  3. If $f$ and $g$ are even functions, is $f+g$ an even function? In other words, is it true for every $x\in\Bbb R$ that $(f+g)(-x)=(f+g)(x)$?

In order to answer the last two questions, you simply have to use the definitions of scalar product and vector addition in $V$ to expand $(\alpha f)(x)$ and $(f+g)(x)$ in terms of $f(x)$ and $g(x)$.

If you have not already proved that the set of all functions from $\Bbb R$ to $\Bbb R$ is a vector space with the usual scalar multiplication and addition of functions, you’ve a lot more work to do: you have to check that all of the vector space axioms are satisfied by these operations.

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What you need to do here is verify that the vector space axioms apply to the set you have. I'll do two for you.

Let $V = \{f:\Bbb R \to\Bbb R \mid f(-x)=f(x)\text{ for all }x\in\Bbb R\}$

Let $f \in V$, $g \in V$. We want to show that $h = f + g$ is also in $V$:

$ h(-x) = f(-x) + g(-x) = f(x) + g(x) = h(x) $

Therefore, $h \in V$.

Now, let $f \in V$, $c \in \mathbb{R}$, and $h = cf$. We have:

$ h(-x) = cf(-x) = cf(x) = h(x) $

Therefore, $h \in V$.

Also note:

  • The constant function $0$ is in $V$.
  • If $f \in V$, so is $-f$.