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I can't seem to wrap my brain around this one, so I figured someone here could point out the connection I'm not making. I've been asked to prove that every real vector space other than the trivial one (V = {0}) has infinitely many vectors. This is intuitively true, but I haven't a clue how to prove it.

At the moment, I'm supposed to base my proof on the eight axioms of a vector space, so any help that remains within that limited field of knowledge would be apprectiated. Thanks much!

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    interestingly, there are finite non-trivial vector spaces. specifically, any finite field would be a candidate for such a vector space.2012-06-04

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If $V$ is not the trivial vector space let $v \in V$, $v\neq 0$. Then show that the vectors $\lambda v$ ($\lambda \in \mathbb R$) are all distinct.

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    @Auke: It is absolutely non-trivial that every vector space has a basis. Actually, this is equivalent to the axiom of choice; I wouldn't count that as following from the axioms of a vector space.2012-06-06
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Suppose that you know that the vector space contains some vector $a$. Do you know anything about vector spaces that would let you find other vectors that might be different from $a$? Perhaps many other vectors?

Do you know any way to make new vectors out of old vectors?

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For every vector space, $V$, over $\mathbb{R}$ it follows that if $dim(V)\neq 0$ then if it is not finite the claim is clear and if it is finite then $V$ is isomorphic to $\mathbb{R}^{dim(V)}$ hence the number of elements in $V$ is the same as in $\mathbb{R}^{dim(V)}$ (i.e. there is a bijection) and in particular it is not finite.

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    @KyleR. I didn't notice that that was the proof he needed...I think that in general this would be a good way to view the situation when he get to the material, though a bijection is almost trivial...2012-06-04
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Hint $\, $ If subgroup $\rm\:v\:\!\mathbb Z\:$ has finite order $\rm\:n\:$ then $\rm\:n\:\!v = 0\:\Rightarrow\:v = 0\:$ by scaling by $\rm\:n^{-1}\!\in\mathbb Q\subset \mathbb R.$