Is R^2 a subspace of C^2?

In Sheldon Axler's wonderful book Linear Algebra Done Right, one exercise goes like this: Is R2\mathbb{R}^2 a subspace of the complex vector space C2\mathbb{C}^2? My first naive guess was yes, since Rn\mathbb{R}^n make up a vector space and since R\mathbb{R} is a subset of C\mathbb{C}. This guess turned out to be wrong due to a subtle but important detail. Here I'd like to offer a proper answer to this question.

The key to answering this question for me was to recall that a vector space is actually defined by a set SSand a field F\mathbb{F}. Usually we refer to the typical real vector space by just Rn\mathbb{R}^n and implicitly assume that F=R\mathbb{F} = \mathbb{R}. To be more precise, let (S,F)(S, \mathbb{F}) denote the vector space where elements vSv \in S can be scaled by a scalar aFa \in \mathbb{F}. With this notation we can denote the typical real vector space by (Rn,R)(\mathbb{R}^n, \mathbb{R}).

With this clarification in mind, the question mentioned above seems ill-defined: There is no mention of what fields to use. When referring to the complex vector space C2\mathbb{C}^2 we usually mean (C2,C)(\mathbb{C}^2, \mathbb{C}). So with the assumption that F=C\mathbb{F} = \mathbb{C}, a more precise version of the question could be:

Is (R2,C)(\mathbb{R}^2, \mathbb{C}) a subspace of (C2,C)(\mathbb{C}^2, \mathbb{C})?

With this formulation the question is much easier to reason about. In order for (R2,C)(\mathbb{R}^2, \mathbb{C}) to be a valid vector space it needs to be closed under scalar multiplication. Now that we have made clear that F=C\mathbb{F} = \mathbb{C} it is fairly easy to see that this is not the case. Suppose a=iCa = i \in \mathbb{C} and v=(1,1)R2v = (1,1) \in \mathbb{R}^2 then av=(i,i)R2av = (i, i) \notin \mathbb{R}^2. So (R2,C)(\mathbb{R}^2, \mathbb{C}) is not a vector space and therefore cannot be a subspace of (C2,C)(\mathbb{C}^2, \mathbb{C}).

The important insight here is that whenever you deal with a mix of real and complex vector spaces, you have to be careful about what "scalar" means, i.e. what field is used.