Restrictions on vector spaces over arbitrary fields I haven't taken any abstract algebra class yet, and for those linear algebra classes I did take, the discussions are mainly focusing on real and complex vector spaces. From textbooks and other resources, I noticed that, in principle,  it is allowed to have vector spaces over other fields. So I tried to make up some random examples, but soon I found that the situation could become cumbersome quite easily:
We could definitely have $\mathbb{C}^n$ as a vector space over $\mathbb{C}$.
So I think it should also be fine to have $\mathbb{C}^n$ as a vector space over $\mathbb{R}$, and I checked all the properties for it to be a vector space, it seems fine indeed.
However, if we have something like $\mathbb{R}^n$ over $\mathbb{C}$, things are getting weird (as the scalar multiplication with usual rule can now take the vector out from $\mathbb{R}^n$), so I tried to modify the definition of the scalar multiplication, but it seems quite difficult (and useless) to turn such structure into a vector space.
I've also tried some other examples, including $\mathbb{R}^n$ over {0,1} where {0,1} is the field with arithmetic modulo 2. Things are getting worse here as the arithmetics for the field {0,1} are not even the same as for $\mathbb{R}$.
So I wonder, apart from explicitly checking the arithmetic properties for a structure to be vector space every time, if there are some restrictions or requirements that are needed to be satisfied between a vector space and the field containing its scalars?
 A: The first thing to keep in mind is that for $V$ to be a vector space over a field $\Bbb F$, the scalar multiplication $\cdot$ is a function of the form $$\cdot:\Bbb F \times V \to V.$$
That is, given any $a \in \Bbb F$ and $v \in V$, you require $a\cdot v \in V$.

The above definition already restricts you from making $\Bbb R^n$ a vector space over $\Bbb C$ in any "natural" way.

One more thing to note is the axiom that tells you $1\cdot v = v$ for any $v \in V$ where $1 \in \Bbb F$ is the multiplicative identity.
So, what do we get if we apply this to the case of $\Bbb R^n$ as a vector space over $\Bbb F_2$?
Well, we see that given any $v \in \Bbb R^n$, we must have
$$v + v = 1\cdot v + 1\cdot v = (1 + 1)\cdot v = 0\cdot v = 0.$$
(We have used the distributive axiom and the fact that $0\cdot v = 0$.)
What the above then tells you is that if you do want to make $\Bbb R^n$ a vector space over $\Bbb F_2$, you definitely cannot use the usual addition.

However, there is something artificial that you can always do. To avoid being too abstract, I shall demonstrate with the help of an example.
Suppose we wish to make $\Bbb R^n$ a vector space over $\Bbb C$.
It is well-known that $\Bbb R$ and $\Bbb C$ have the same cardinality. In particular, there exists a bijection $\sigma:\Bbb C^n \to \Bbb R^n$. (Note that this is purely a bijection of sets. Nothing about the function being continuous, linear, et cetera.)
Using this bijection, we now define addition $+$ on $\Bbb R^n$ and scalar multiplication $\cdot:\Bbb C\times \Bbb R^n \to \Bbb R^n$.
How? We do this using the bijection as follows:
Let $a \in \Bbb C$ and $v, w \in \Bbb R^n$, we then define
$$v + w = \sigma(\sigma^{-1}(v) + \sigma^{-1}(w)),$$
$$a\cdot v = \sigma(a\cdot\sigma^{-1}(v)).$$
(Note that the $+$ and $\cdot$ appearing on the right-hand side are those defined for $\Bbb C^n$.)
Now, one can check that this does indeed make $\Bbb R^n$ a vector space over $\Bbb C$.
Thus, what we essentially did in the above is that we "borrowed" the vector space structure of $\Bbb C^n$ over $\Bbb C$ and gave it to $\Bbb R^n$.

In general, if you wish to make a set $X$ a vector space over a field $\Bbb F$, you can first find a vector space $V$ (over $\Bbb F$) which has the same cardinality as that of $X$ and then use a bijection to make $X$ a vector space. (Of course, it is not always necessary that one can find such a $V$.)
A: For your example, notice that $\mathbb R $ is a subfield of $\mathbb C$. If $\mathbb R^n$ is a vector space over $\mathbb C$ of dimension $r$, then it is of dimension $2r$ as a vector space over $\mathbb R$. As dimension is an invariant of a vector space, $n$ must be an even integer. Then It is easy to define a $\mathbb C$-vector space structure of $\mathbb R^n$. E.g. when $n=2$, define $(a+bi) \cdot (c,d)=(ac-bd, ad+bc)$.
Edit: I assume that the restriction of $\mathbb C$ to $\mathbb R$ brings the $\mathbb C-$vector space $\mathbb R^n$ to the canonical $\mathbb R-$vector space $\mathbb R^n$.
