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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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    $\begingroup$ take a vector $e\neq 0$ and show that $ne\neq me$ provided $n\neq m$ for all $n,m\in \mathbb{N}$. $\endgroup$
    – AnonymousCoward
    Jun 4, 2012 at 18:20
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    $\begingroup$ interestingly, there are finite non-trivial vector spaces. specifically, any finite field would be a candidate for such a vector space. $\endgroup$
    – akkkk
    Jun 4, 2012 at 18:26

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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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  • $\begingroup$ Thanks for that—one of those moments of momentary idiocy if I've ever had one. $\endgroup$
    – Kyle R.
    Jun 4, 2012 at 18:34
  • $\begingroup$ Specifically, since $v\neq0$, $v$ surely has a nonzero coordinate, and assume its value is $v_i$. Say there are $\lambda,\lambda'$ with $\lambda\neq\lambda'$ and assume $\lambda v=\lambda' v$. Then it follows that the $i$th coordinate is equal to $\lambda v_i=\lambda' v_i$, such that $\lambda=\lambda'$ which is a contradiction. $\endgroup$
    – akkkk
    Jun 4, 2012 at 18:35
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    $\begingroup$ @Auke: When speaking of coordinates you assume that $V$ is $\mathbb R^n$ for some $n$. However, it is possible to show $\lambda v = \mu v \Rightarrow \lambda = \mu$ using only the axioms from the definition of a vector space. $\endgroup$
    – marlu
    Jun 4, 2012 at 18:40
  • $\begingroup$ @marlu: all vector spaces have one or more coordinate mappings. $\endgroup$
    – akkkk
    Jun 5, 2012 at 13:23
  • $\begingroup$ @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. $\endgroup$
    – marlu
    Jun 6, 2012 at 10:20
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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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    $\begingroup$ That went beyond the axioms of vector space as well as my general understanding. $\endgroup$
    – Kyle R.
    Jun 4, 2012 at 18:34
  • $\begingroup$ @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... $\endgroup$
    – Belgi
    Jun 4, 2012 at 18:36
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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 $\rm\:n^{-1}\!\in\mathbb Q\subset \mathbb R.$

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