# Geometric Interpretation of Properties of a Vector Space

I am currently undertaking a self-study on differential geometry and vector spaces, the latter of which is new to me. I was wondering whether the fact that every Vector space has a basis can be motivated geometrically. Also on a slightly deviant note, I was trying to understand how exactly the underlying field of a vector space dictates the properties of the vector space. For ex: The dimension is affected as can be concluded by the result that the dimension of $\mathbb R$ over $\mathbb R$ is $1$ whereas over $\mathbb Q$ is $|\mathbb R|$. Are there any other such influences? Does the kind of topology or geometry the field possesses induce some properties on the Vector space??

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Gerry's answer is nice. I will merely address the motivation question in the finite-dimensional case over $\mathbb{R}$ and add a silly comment.

For finite dimensions the geometric motivation is not difficult. Just draw the your proposed basis vectors (lets say $n=2$) then pick any point in the plane. An appropriate linear combination of the basis vectors will reach the point iff those vectors are in fact linearly independent (in this case not scalar multiples of one each other).

This picture lifts to any other abstract two-dimensional vector space because the coordinate maps preserve linear dependencies.

The picture for $2<n < \infty$ is similar.

Incidentally, if we think of $w=v_1+v_2$ as transport by $v_1$ then $v_2$ then in a vector space $w=v_2+v_1$ hence the order of transport is not important. When adding vectors tip-to-tail we can add the vectors in any order and still reach the same overall resultant. In contrast, on a curved space if we travel the flow of $v_1$ and then $v_2$ and compare it with the alternate flow given by $v_2$ then $v_1$ we need not reach the same point. Apparently curved space is not a vector space if we intend to identify flows with vector addition. This is a silly comment, but perhaps it is relevant to the ideas your exploring. Look up the introductory geometric comments about the Lie Bracket for details about what I'm saying here.

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Yeah, thanks it is relevant. I have not been exposed to Lie Groups as such. But I have been exposed to bits of manifold theory and my question was partly driven with that in mind. So is there any way to tie this up with the notion of tangent spaces of an abstract manifold?? D – Vishesh Oct 8 '12 at 14:32

The proof that every vector space has a basis depends on the Axiom of Choice, so I doubt it can be motivated at all, geometrically or otherwise. For finite-dimensional vector spaces, you're in luck: every finite-dimensional vector space over a field $F$ is isomorphic, as a vector space, to $F^n$. So if your field has a geometry, then your vector space inherits one.

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