Constructing a simple isomorphism from $\mathbb{R^{2n}}$ to $\mathbb{C^n}$ I need to construct an isomorphism from $\mathbb{R^{2n}}$ onto $\mathbb{C^n}$ (both over $\mathbb{R}$). I know couple of things:


*

*This isomorphism will be a (bijective) linear map between two vector spaces that have equal dimensions.

*The linear map is uniquely determined by action on the bases.


So I specify bases for both spaces, say, let $\{r_1,\dots,r_{2n}\}$ be a basis for $\mathbb{R^{2n}}$ and $\{c_1,\dots,c_{2n}\}$ be a basis for $\mathbb{C^n}$, where
\begin{align*}
r_1&=(1,0,0,\dots,0),\\
r_2&=(0,1,0,\dots,0),\ldots,\\
c_1&=(1,0,0,\dots,0),\\
c_2&=(i,0,0,\dots,0),\\
c_3&=(0,1,0,\dots,0),\\
c_4&=(0,i,0,\dots,0),\ldots.
\end{align*}
Then let the transformation be defined by $r_i\mapsto c_i$.
I could prove that it's linear, but I was wondering if that suffices to solve the problem, or not?
Thanks!

Yes, that's all.
You don't even need to prove linearity, as any mapping defined on a basis uniquely extends to a linear mapping, and such is an isomorphism iff the image of the (or any) basis is again a basis in the target space.
However, if you want, you can write this up explicitly as
$$(a_1,b_1,a_2,b_2,\dots,a_n,b_n)\mapsto (a_1+ib_1,\,a_2+ib_2,\dots,\,a_n+ib_n)\,.$$
