Understanding of Vector Space ($\mathbb{R} ^{n}$ is not a vector space over $\mathbb{C}$) $\mathbb{C} ^{n}$ is a vector space over $\mathbb{C}$, and $\mathbb{Q} ^{n}$ is a vector space over $\mathbb{Q}$. We remark that $\mathbb{R} ^{n}$ is not a vector space over $\mathbb{C}$. 
My question is that Why $\mathbb{R} ^{n}$ is not a vector space over $\mathbb{C}$? Can you explain detailed?
 A: At least under the standard scalar multiplication, if you multiplied a nonzero vector in $\mathbb{R}^n$ by a non-real complex scalar, you would get a vector which is not in $\mathbb{R}^n$ anymore, which would violate the vector space axioms. So at least with respect to the standard scalar multiplication, $\mathbb{R}^n$ can't be made a vector space over $\mathbb{C}$. 
However, when $n$ is even, there is a natural way to make $\mathbb{R}^n$ a vector space over $\mathbb{C}$: pair off components and think of each pair as a complex number, so that you essentially have $\mathbb{C}^{n/2}$. In this situation we would say things like $i \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$. I doubt there is any way to make this work when $n$ is odd, though.
You get the same problem if you try to make $\mathbb{Q}^n$ a vector space over $\mathbb{R}$ (and in this case there is absolutely no way to reconcile it, just for cardinality reasons).
A: What is a vector space?
A set $V$ is called vector space over the field $K$ if:
-$\forall v,w\in V, \ v+w \in V$  (of course you need to define the operation, and it has to have certain properties, look on your book. It doesn't matter for your question)
-$\forall k \in K,\forall v \in V,kv\in V$ (and again, you need to define this operation that is called scalar moltiplication)
If you look at your examples these properties are respected in all the cases except for the last: multiplying a real vector for a complex scalar you can get out of the vector space. For example consider the vector $\begin{pmatrix} 1 \\1\end{pmatrix} \in \mathbb{R}^2$ and the scalar $i\in \mathbb{C}$: $i \begin{pmatrix} 1 \\1\end{pmatrix}=\begin{pmatrix} i \\i\end{pmatrix}$ doesn't belong to $\mathbb{R}^2$ anymore. The same goes for $\mathbb{Q}^n$ over $\mathbb{R}$.
