Just wondering if a finite dimensional vector space could have two subspaces such that each of these subspaces has the same dimension but form vector spaces over different fields ?


Generally, vector spaces "come" with a pre-specified field $\mathbf{K}$ (via the scalar multiplication function, $\cdot\colon \mathbf{K}\times\mathbf{V}\to\mathbf{V}$), and when discussing "subspaces" we only refer to subspaces that are themselves vector spaces over that field.

About the only way to sort of make sense of what you write is the following observation: if $F$ and $K$ are fields, $F\subseteq K$ (for example, $F=\mathbb{R}$ and $K=\mathbb{C}$; or $F=\mathbb{Q}$ and $K=\mathbb{R}$), and $\mathbf{V}$ is a vector space over $K$, then by restricting the scalar multiplication to only allow scalars from $F$ you also get a vector space; formally, it is a different vector space from the original (because the scalar multiplication function is different), though it is of course closely related. Any $\mathbf{W}$ that was a subspace of $\mathbf{V}$ when considered as a $\mathbf{K}$-vector space will also be a subspace of $\mathbf{V}$ when considering both as $\mathbf{F}$-vector spaces, but in general you may have subsets that are subspaces as $\mathbf{F}$-vector spaces but not as $\mathbb{C}$-vector spaces. For example, $\mathbf{C}$ is a vector space over $\mathbf{C}$; the only subspaces are $\mathbb{C}$ and $\{\mathbf{0}\}$. However, you can also consider $\mathbf{C}$ as an $\mathbf{R}$-vector space (allowing only real "scalars"). It's a different vector space (as a complex vector space, $\mathbb{C}$ has dimension $1$; as a real vector space, it has dimension $2$); both subspaces-as-complex-vector-space are still subspaces-as-real-vector space (if it was closed under multiplication by any complex number, it is also closed under multiplication by any real number; and vector addition is unchanged). But now we have lots of subspaces-as-real-vector-space that are not subspaces-as-complex-vector-space. For example, $\mathbf{W}=\{a+ai\mid a\in\mathbb{R}\}$ is easily seen to be a subspace if you only allow reals as scalars, but is not a subspace when working over the complex numbers.

When we do that, however, we are really talking about two different vector spaces: $\mathbf{V}$-as-a-$\mathbf{K}$-vector-space, and $\mathbf{V}$-as-an-$\mathbf{F}$-vector-space. We refer to the second as "the vector space obtained by restriction of scalars" (we "close the door" on scalars that are in $\mathbf{K}$ but not in $\mathbf{F}$, and don't consider them at all; we restrict what may be considered as a scalar). Though the two are intimitaly connected, we don't mix the structures: we need to talk separately about dimensions-over-$\mathbf{F}$ and dimensions-over-$\mathbf{K}$, we need to talk separately about $\mathbf{F}$-linear-maps and $\mathbf{K}$-linear-maps, etc. There is a one-way street, as anything that is linear-relative-to-$\mathbf{K}$ is also linear-relative-to-$\mathbf{F}$, but the road doesn't go the other way.

(Of course, the same set may be a vector space over many different fields in many different ways that have no relation to one another; the same set may "work" as a 1-dimensional vector space over $\mathbb{Q}$, as a 10-dimensional vector space over $\mathbb{Q}$, or as an infinite-dimensional vector space over a field with $49$ elements, but these structures will generally have absolutely nothing to do with one another. It's really best to think of the vector space as consisting of four things: a field $\mathbf{K}$ (often understood from context, but not always), a set $\mathbf{V}$, a function $+\colon\mathbf{V}\times\mathbf{V}\to\mathbf{V}$, and a function $\cdot\colon \mathbf{K}\times\mathbf{V}\to\mathbf{V}$, that satisfy certain properties (the axioms). )

In that sense, your question is not even well-posed: when you discuss "vector spaces over different fields", are you assuming the vector addition is the same? Are you assuming any connection between the fields and/or the scalar multiplications? And when you talk about "subspace", is it subspace relative to one field (which) or both? Dimension relative to what field? I suspect that trying to make the question sensible will clarify what your confusion might be, and help you realize that what you were perhaps fuzzily envisioning cannot occur at all.

  • $\begingroup$ I think your post has clarified my misunderstanding. I had overlooked the fact that not the same operations would be defined on such subspaces if we restricted the field of the vector space. $\endgroup$ – Comic Book Guy Feb 8 '12 at 22:06

The definition of a subspace includes that the subspace inherits the operations defined on the vector space, and since these operations involve the field of scalars, the subspace is necessarily defined over the same field of scalars as the vector space of which it is a subspace. You can consider the sets underlying the subspaces as vector spaces over different fields by endowing them with different operations, but these vector spaces wouldn't be regarded as subspaces of the original vector space.


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