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Two vector spaces are said to be isomorphic iff there's an invertible linear map between them. It can be shown that isomorphic vectors spaces would have to have the same finite dimension or both be infinite dimensional. But what if they are over different fields? For example, would the trivial vector space over $\mathbb C$ be considered isomorphic to the trivial vector space over $\mathbb R$? Or would it not, since, if we let $T$ be the only possible linear map between them, $T(i0)\neq iT(0)=i0$, since $i0$ is not defined in the codomain (since $i$ is not a scalar in $\mathbb R$)?

Also, are there any other examples of vector spaces over different field that would be "isomorphic" like this?

Nishant
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    In the definition of a linear transformation $T:V \to W$, we require that $V$ and $W$ be vector spaces over the same field $F$. Otherwise, the condition $T(cx) = cT(x)$ would not make sense. – littleO Feb 13 '14 at 06:32
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    Hmm, that seems reasonable. Thanks! – Nishant Feb 13 '14 at 06:36

1 Answers1

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Morphisms of vector spaces (i.e., linear maps) and in particular isomorphisms, are only defined in the context (category) of vector spaces over a specific field$~K$. The mention of$~K$ is often omitted once the field is fixed, but it is still implicitly always part of all statements. So the very statement that two vector spaces are isomorphic implies that they are considered as vector spaces over the same field (one or both might allow being viewed as vectors space over another field, but that view is irrelevant to the isomorphism statement.)

Indeed if two spaces can both be viewed as vector spaces over two different fields $F,K$, then the fact that they are isomorphic over $F$ does not imply that they are isomorphic over $K$. For instance $\Bbb R^1$ and $\Bbb R^2$ are not isomorphic as vector spaces over$~\Bbb R$, but they are isomorphic as vector spaces over$~\Bbb Q$ (to those who accept the axiom of choice).

Marc van Leeuwen
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    Could you help me to understand how $\Bbb R^1$ and $\Bbb R^2$ are isomorphic as vector spaces over $~\Bbb Q$? Thanks – mathscrazy Dec 30 '17 at 06:19
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    @mathscrazy it is because $\mathbb R^1$ and $\mathbb R^2$ are both vector spaces over $\mathbb Q$ of dimension $\left|\mathbb R\right|$, whence they are isomorphic because they have the same dimension. In general, the proof that two vector spaces are isomorphic iff they have equal dimension requires the Axiom of Choice, which is why Marc van Leeuwen said that $\mathbb R^1$ and $\mathbb R^2$ are isomorphic over $\mathbb Q$ if you assume choice. – feralin Jan 06 '18 at 19:50
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    @feralin Thanks a lot for your comment. – mathscrazy Feb 13 '18 at 23:25