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Suppose I want to show that two infinite-dimensional vector spaces are not isomorphic to each other. This is easy if my vector spaces are finite-dimensional as I just find a basis for each and show they are of different size, since finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.

If the vector spaces have different cardinality then of course we would not be able to find a bijection between them, so they are not isomorphic, but what if the cardinality is the same? I know I need to show that every bijection is not linear, but that sounds like a lot of work. What is the best way to go about it generally?

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1 Answer

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Two vector spaces (over the same field) are isomorphic iff they have the same dimension - even if that dimension is infinite.

Actually, in the high-dimensional case it's even simpler: if $V, W$ are infinite-dimensional vector spaces over a field $F$ with $\mathrm{dim}(V), \mathrm{dim}(W)\ge \vert F\vert$, then $V\cong W$ iff $\vert V\vert=\vert W\vert$. In particular, if $F=\mathbb{Q}$, two infinite-dimensional vector spaces over $F$ are isomorphic iff they have the same cardinality.

So, for example:

• As vector spaces over $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$ are isomorphic (this assumes the axiom of choice).

• The algebraic numbers $\overline{\mathbb{Q}}$ are not isomorphic to the complex numbers $\mathbb{C}$ as vector spaces over $\mathbb{Q}$, since the former is countable while the latter is uncountable.

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EDIT: All of this assumes the axiom of choice - without which, the idea of "dimension" doesn't really make sense. See the comments for a bit more about this.

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