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?
Best Answer
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 $\mathbb{i} \notin \mathbb{R}$.