Finding a basis of a complex vector space over $\Bbb R$ given a basis over $\Bbb C$ Suppose $X$ is a vector space over $\mathbb C$ and  has as basis $\{e_1,e_2,\ldots,e_n\}$. Now regard $X$ as a vector space over $\mathbb R$.
What will be the basis?
My thoughts:
I considered $\mathbb C$ over $\mathbb C$ and $\mathbb C$ over $\mathbb R$.In the first case we have $(1,0)$ as basis and in the latter case we have $\{(1,0),(0,1)\}$ as basis i.e. $\{(1,0),(0,1)(1,0)=(0,1)\}$ as basis.
So may be the answer is $\{e_1,e_2,\ldots,e_n,ie_1,\ldots,ie_n\}$. How to justify the result if its true?
 A: Suppose $x\in X$.  Then $x = c_1 e_1 + \cdots+c_n e_n$ for some complex numbers $c_1,\ldots,c_n$.
For $k=1,\ldots,n$ write $c_k = a_k + i b_k$ where $a_k$ and $b_k$ are real.
Then
\begin{align}
x & = c_1 e_1 + \cdots+c_n e_n \\[8pt]
& = (a_1+ib_1)e_1 + \cdots + (a_n+ib_n)e_n \\[8pt]
& = a_1 e_1 + \cdots + a_n e_n + b_1(ie_1) + \cdots + b_n (ie_n).
\end{align}
So $x$ is a linear combination of $e_x,\ldots,e_n,ie_1,\ldots,ie_n$ with coefficients that are real.
Linear independence can be proved by considering almost the same sequence of equalities:
\begin{align}
0 & = a_1 e_1 + \cdots + a_n e_n + b_1(ie_1) + \cdots + b_n (ie_n) \tag 1 \\[8pt]
& = (a_1+ib_1)e_1 + \cdots + (a_n+ib_n)e_n \\[8pt]
& = c_1 e_1 + \cdots+c_n e_n
\end{align}
and that can be true only if $c_1=\cdots=c_n=0$, by linear independence of $e_1,\ldots,e_n$ over $\mathbb C$.  Hence $(1)$ can be true only if $a_1=\cdots=a_n=b_1=\cdots=b_n=0$.
A: In the case of $\mathbb{C}$ over $\mathbb{C}$, the basis would be $\{1\}$ because every element of $\mathbb{C}$ can be written as a $\mathbb{C}$-multiple of $1$. 
$$\mathbb{C}=\{z\times 1 : z \in \mathbb{C}\}$$ 
In the case of $\mathbb{C}$ over $\mathbb{R}$, the basis would be $\{1,\mathrm{i}\}$ because every element of $\mathbb{C}$ can be written as an $\mathbb{R}$-multiple of $1$ and $\mathrm{i}$.
$$\mathbb{C}=\{x\times 1 + y \times \mathrm{i} : x,y \in \mathbb{R}\}$$
If $\{{\bf v}_1,\ldots,{\bf v}_n\}$ is a basis for $V$ over $\mathbb{C}$ then $$V = \{a_1{\bf v}_1+\cdots+a_n{\bf v}_n: a_k \in \mathbb{C} \}$$
We can write each of the $a_k$ as $b_k+\mathrm{i}c_k$, where $b_k,c_k \in \mathbb{R}$. Hence
\begin{eqnarray*}
a_1{\bf v}_1+\cdots+a_n{\bf v}_n &=& 
(b_1+\mathrm{i}c_1){\bf v}_1+\cdots+(b_n+\mathrm{i}\mathrm{c}_n){\bf v}_n \\
&=& b_1{\bf v}_1+\cdots+b_n{\bf v}_n+c_1(\mathrm{i}{\bf v}_1)+\cdots+c_n(\mathrm{i}{\bf v}_n)
\end{eqnarray*}
We can take $\{{\bf v}_1,\ldots,{\bf v}_n,\mathrm{i}{\bf v}_1,\ldots,\mathrm{i}{\bf v}_n\}$ as a basis for $V$.
The final step is to show that 
$$V = \mathbb{R}\langle {\bf v}_1,\ldots,{\bf v}_n\rangle \oplus \mathbb{R}\langle \mathrm{i}{\bf v}_1,\ldots,\mathrm{i}{\bf v}_n\rangle$$
This is obvious since $\mathbb{i} \notin \mathbb{R}$.
