In Sheldon Axler's wonderful book Linear Algebra Done Right, one exercise goes like this: Is $\mathbb{R}^2$ a subspace of the complex vector space $\mathbb{C}^2$? My first naive guess was yes, since $\mathbb{R}^n$ make up a vector space and since $\mathbb{R}$ is a subset of $\mathbb{C}$. 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 $S$*and* a field $\mathbb{F}$. Usually we refer to the typical real vector space by just $\mathbb{R}^n$ and implicitly assume that $\mathbb{F} = \mathbb{R}$. To be more precise, let $(S, \mathbb{F})$ denote the vector space where elements $v \in S$ can be scaled by a scalar $a \in \mathbb{F}$. With this notation we can denote the typical real vector space by $(\mathbb{R}^n, \mathbb{R})$.

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 $\mathbb{C}^2$ we usually mean $(\mathbb{C}^2, \mathbb{C})$. So with the assumption that $\mathbb{F} = \mathbb{C}$, a more precise version of the question could be:

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

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

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.