Conjugation map on a complex vector space Let $V$ be a complex vector space and $J:V\to V$ a map with the following properties:
(i) $J(v+w)=J(v)+J(w)$
(ii) $J(cv)=\overline{c}J(v)$
(iii) $J(J(v))=v$
for all $v,w\in V$ and $c\in\mathbb{C}$. Put $W=\{v\in V\,|\, J(v)=v\}$; this is a real vector space with respect to the operations in $V$ (I have shown this). From here, I am asked to show:
$\bullet$ For each $v\in V$, there exist unique vectors $w,u\in W$ such that $v=w+iu$.
The uniqueness is very straight-forward once you have the decomposition. I've been trying the usual kinds of tricks, like writing $v=\bigl(v-J(v)\bigr)+J(v)$, but this clearly doesn't work. And I understand morally what's happening: we're decomposing $V$ into its formal real part and formal imaginary part. I just don't see how to get the decomposition.
Thanks

For completeness, in order to get uniqueness, I would make an argument as follows.
If $z = x + iy = x' + iy'$ where $x, x', y, y' \in W$, then $(x - x') = i(y' - y)$. Since $W$ is a vector space, $x - x' \in W$, so $i(y' - y) \in W$ as well. But $W$ is a vector space over the reals, so if $i(y' - y) \in W$, $y' - y = 0$, which implies $x = x', y = y'$. 
