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As I was going over the classification theorem for closed surfaces today, the text I'm reading gave another example of a classification theorem: finite dimensional vector spaces are classified by their dimension. As a point of fact, I think that this is slightly wrong -- if I understand what the author was trying to say I think that finite dimensional vector spaces are classified by their dimension and their base field. Then he's just talking about that theorem that says that any $\Bbb F$-vector space with dimension $n$ is isomorphic to $\Bbb F^n$.

His use of the word "finite" though has me wondering, does the same thing hold for infinite dimensional vector spaces? Can we "classify" infinite dimensional vector spaces by their dimensions (meaning $\aleph_0$, $\aleph_1$, etc) and base fields? Or is there something more complex that happens for infinite dimensional spaces?

user302008
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    When you get into infinite dimensional vector spaces, what become very interesting are topological properties, more than algebaical properties. Is it complete? can it be normed? Does it have a countable dense subset? Is it separated? – Tryss Jan 01 '16 at 03:54
  • I agree with you that the base field matters. If an isomorphism $f\colon V\to W$ is defined as a linear bijection, then $V$ and $W$ cannot be isomorphic if they are not vector spaces over the same field. This is because the scaling condition $f(\lambda v)=\lambda f(v)$ would not even make sense otherwise. The notion of dimension also depends on the base field (e.g. the dimension of $\mathbb{C}^n$ is $2n$ over $\mathbb{R}$ but $n$ over $\mathbb{C}$). Hence, writing something like $\dim V = \dim W$ already assumes that $V$ and $W$ have the same base field. – prt13463 Sep 30 '17 at 00:21

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Yes, vector spaces are always classified by their dimension, and the argument is as you would expect: if $\text{dim }V=\text{dim }W$, then we can choose a bijection between a basis of $V$ and a basis of $W$, and this bijection induces an isomorphism $V\cong W$. (Note, however, that the existence of a basis for every (infinite-dimensional) vector space requires the use of the Axiom of Choice/Zorn's Lemma.)

An interesting application of this fact is that the additive group of $\mathbb{C}$ is isomorphic to the additive group of $\mathbb{R}$ - indeed, both are $\mathfrak{c}$-dimensional $\mathbb{Q}$-vector spaces, where $\mathfrak{c}$ denotes the cardinality of the continuum.

As @Tryss points out in the comments, there are different notions of isomorphism for infinite-dimensional vector spaces decorated with additional mathematical structure, e.g., normed linear spaces (vector spaces equipped with a norm), Banach spaces (complete normed linear spaces), inner product spaces (vector spaces equipped with an inner product), Hilbert spaces (complete inner product spaces) etc. Hilbert spaces are completely classified up to isomorphism by their dimension (which is at most countable assuming separability) but Banach spaces are not. However, this is a separate mathematical theory, so I won't comment further unless you would like me to (and in the meantime, would direct you to the relevant Wikipedia articles).

Hope that helps!

Amitesh Datta
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    R is not Aleph-1 dimensional over Q unless CH holds. – DanielWainfleet Jan 01 '16 at 10:34
  • Hi @user254665, thanks for the comment! You're right, of course. I replaced $\aleph_1$ with $\mathfrak{c}$, the cardinality of the continuum. Happy new year! – Amitesh Datta Jan 01 '16 at 12:38
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    You're welcome. BTW, If we drop the Axiom of Choice it is relatively consistent (with ZF) that there is a vector space with 2 Hamel bases that are cardinally incomparable. – DanielWainfleet Jan 01 '16 at 12:52