@0celo7 On $(V,J)$, define $(a+b\mathrm{i})v = av + bJv\forall a,b\in\mathbb{R}\forall v\in V$. This turns $V$ into a true complex vector space. Now the claim you want to prove reduces to the statement that all $\mathbb{K}$-vector spaces are linearly isomorphic to $\mathbb{K}^n$ for some $n$.
I wrote down how to multply an element of $V$ with a complex number, this means $V$ is a $\mathbb{C}$-vector space. Now, every $\mathbb{C}$-vector space has a complex linear isomorphism to $\mathbb{C}^m$ for some $m$.
Uh...I'm not sure what you mean by that. $V\neq V\otimes_\mathbb{R}\mathbb{C}$ just because the real dimensions don't match (the complexification has double the real dimension).
@ACuriousMind I mean that $V\neq V\oplus\mathrm{i}V$
@ACuriousMind Alright, but how does the desired equation actually follow? Is it because $\phi$ is $\mathbb{C}$-linear by definition, so $\phi(Jv)=\phi(\mathrm{i}v)=\mathrm{i}\phi(v)$?
It's about as satisfying as "$C^0$ functions need not be $C^1$ because of $|x|$"
@ACuriousMind Can one prove that $\lim_{x\to0}\sin 1/x$ does not exist via $\epsilon-\delta$? In class we constructed two sequences and found a contradiction.
