Natural isomorphism between $\mathbb{C}\otimes_{\mathbb{R}}V$ and $V\oplus V.$ The title pretty much says it all. 
If $V$ is a finite-dimensional complex vector space, I want to find a natural complex linear isomorphism $\mathbb{C}\otimes_{\mathbb{R}}V \cong V\oplus V.$
I understand how to show that they are isomorphic using dimension but I can't come up with an explicit natural isomorphism.
Please can someone give me a hint?
Many thanks!
 A: Morally, we are splitting $V \otimes_\mathbb{R} \mathbb{C}$ into eigenspaces for the action of multiplication by $i$; however, this is confusing because there is already an action of multiplication by $i$ on $V$, and we don't mean this action. We can change notation to make the confusing part go away.
Write $\mathbb{C} = \mathbb{R}[x] / (x^2 + 1)$. Then $V \otimes_\mathbb{R} \mathbb{C}$ = $V \otimes_\mathbb{R} \mathbb{R}[x]/(x^2+1)$. Write $W$ for this vector space.
Now, multiplication by $x$ gives an endomorphism of $W$, annihilated by the polynomial $x^2 + 1$ and not any proper divisor of this polynomial, so it must split $W$ into $\pm i$ eigenspaces, which by counting are each one-dimensional.
A: Hint :
Consider the map $$c \otimes v \mapsto (\Re(c) v, \Im(c) v).$$ This map is clearly injective and since you have same dimensional vector spaces, it implies that it is also surjective. 
A: I am adding this as a separate answer from my other one because it is a little different in flavor; fundamentally these answers are the same, though.
Lemma: There is a natural isomorphism $\mathbb{C} \otimes_\mathbb{R} \mathbb{C} = \mathbb{C} \oplus \mathbb{C}$.
Proof: Write the second copy of $\mathbb{C}$ as $\mathbb{R}[x]/(x^2+1)$. Then 
\begin{align*}
\mathbb{C} \otimes \mathbb{R}[x]/(x^2+1) &= \mathbb{C}[x]/(x^2+1)\\
&= \mathbb{C}[x]/((x+i)(x-i))\\
&\stackrel{CRT}{=} \mathbb{C}[x]/(x+i) \oplus \mathbb{C}[x]/(x-i)\\
&=\mathbb{C} \oplus \mathbb{C}.
\end{align*}
Main result: For any complex vector space $V$, there is a natural isomorphism $V \otimes_\mathbb{R} \mathbb{C} = V \oplus V$.
Proof: One has
\begin{align*}
V \otimes_\mathbb{R} \mathbb{C} &= (V \otimes_\mathbb{C} \mathbb{C}) \otimes_\mathbb{R} \mathbb{C} \\
&= V \otimes_\mathbb{C} (\mathbb{C} \otimes_\mathbb{R} \mathbb{C})\\
&= V \otimes_\mathbb{C} (\mathbb{C} \oplus \mathbb{C})\\
&= V \oplus V
\end{align*}
as desired.
