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I cannot think of a finite vector space over the reals, because we must sum these two elements and get a new element also in the vector space. And over the reals, I can't think of a sum that, at some point, will return to a previous vector on the subspace.

Am I right? It also means that there's no finite vector space over the reals, right?

Guerlando OCs
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  • There is a vector space of dimension zero. But if the space has a non-zero vector $v$, you should be able to show that the vectors $rv: r\in \mathbb R$ are distinct for distinct $r$. – Mark Bennet Dec 07 '14 at 14:39

2 Answers2

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There is exactly one finite vector space over the reals: $\{0\}$. If you have any other vector space, it has to have cardinality (at least) that of the real numbers, since with a little bit of care you can produce an injective map from $\mathbb{R}$ into your vector space $V$ (think of using the scalars times any non-zero vector).

Simon Rose
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Yes, there is one (and up to isomorphy only one) finite vector space over the reals, namely $\{0\}$.

If there exists an element $x\in V$ with $x \neq 0$, then the vector space axioms force

$$ (a-b) x \neq 0 $$

for distinct $a,b \in \Bbb{R}$, because

$$ (a-b)^{-1} \cdot (a-b)x = x \neq 0. $$

But this yields $ax \neq bx$, so that the set

$$ \{a x \mid x \in \Bbb{R}\}\subset V $$

is infinite (even uncountable).

PhoemueX
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