How is a vector space over C also a vector space over R? Please note that this is a homework problem, so if possible don't give me the whole answer.
The problem is as follows:

Let $V$ be a vector space over $\mathbb{C}$ of dimension $n$. Because $\mathbb{C}$ contains $\mathbb{R}$, 
  $V$ is also naturally a $\mathbb{R}$-vector space. What is its dimension over $\mathbb{R}$? Prove your answer.

I don't understand the basic premise of the problem. How is $V$ a vector space over $\mathbb{R}$? If $V$ is a vector space over $\mathbb{C}$, then any $v \in V$ can be written as $v = (v_1, v_2 \cdots v_n)$ where $v_i \in \mathbb{C}$ for any $i$, i.e. $v \in \mathbb{C}^n$. But $v_i \in \mathbb{C}$ does not imply that $v_i \in \mathbb{R}$, since $\mathbb{R} \subseteq \mathbb{C}$, so $v$ is not necessarily in $\mathbb{R}^n$. This is what confuses me, because the problem states that $V$ being a vector space over $\mathbb{C}$ means it's also a vector space over $\mathbb{R}$. Could someone please explain this to me?
 A: $\mathbb{C}$ is $1$-dimensional as a $\mathbb{C}$-space and $2$-dimensional as a $\mathbb{R}$-space, right? In this sense, why should a $n$-dimensional $\mathbb{C}$-space be $n$-dimensional as a $\mathbb{R}$-space?
Let's try to answer your question.
Suppose one has a field $K$ and a subfield $F\subseteq K$. Naturally, $K$ can be seen as a $F$-vector space (it satisfies the axioms for being a $F$-vector space trivially).
Now, let $V$ be a $K$-vector space, which means that $V$ satisfies all the axioms of being a vector space over $K$. Since $F\subseteq K$, $V$ trivially also satisfies all axioms of being a vector space over $F$, can you see it?
In particular, if $K=\mathbb{C}$ and $F=\mathbb{R}$, it follows that a $\mathbb{C}$-vector space $V$ is also a $\mathbb{R}$-vector space.
Now you're convinced that $V$ is also a vector space over $\mathbb{R}$, I invite you to think about the dimensions.

 Notice that $\dim_{\mathbb R}\mathbb{C}=2$ with basis $\{1,i\}$. If $\dim_{\mathbb C} V=n$ and if $B_1=\{v_1,\cdots, v_n\}$ is a basis for $V$ over $\mathbb{C}$, then every $v\in V$ can be written as $v=\sum_{j=1}^n a_jv_j$ for some $a_j\in\mathbb{C}$. Besides, each $a_j$ can be written as $a_j=x_j+iy_j$, with $x_j, y_j\in\mathbb{R}$. So for every $v\in V$ there exists $x_1,  \cdots, x_n, y_1, \cdots, y_n\in\mathbb{R}$ such that
 $$v=\sum_{j=1}^n (x_j+iy_j)v_j = \sum_{j=1}^n x_jv_j + \sum_{j=1}^n y_j (iv_j),$$
 which means that $B_2=\{v_1,\cdots, v_n, iv_1, \cdots, iv_n\}$ spans $V$ (over $\mathbb{R}$). Now you just need show that $B_2$ is linearly independent (over $\mathbb{R}$) to conclude that it is a basis for $V$ over $\mathbb{R}$. Therefore $\dim_{\mathbb R} V= 2n$.

Ps.: This fact can be generalized as follows. If $F\subseteq K$ are fields and $V$ is a $K$-vector space, then
$$\dim_F V = \dim_K V \cdot \dim_F K,$$
where $\dim_Y X$ denotes the cardinality of a basis for $X$ as a $Y$-vector space.
A: Hint: Can you see $\mathbb{C}$ as a vector space over $\mathbb{R}$?
