Let's assume that:
$V$ is a real vector space,
$w:V\times V \to \Bbb R$ is a skew-symmetric form which non-degenerated (the last assumption means: $x\in V\ \&\ w(x,y)=0\ \forall y\in V\implies x=0$),
$J:V\to V$ is a complex structure on $V$, that is $J$ is linear and $J^2=-Id$,
$w$ and $J$ are compatible: $w(Jx,Jy)=w(x,y)$ for $x,y \in V$.
Is it then a function $\langle x,y\rangle=w(x,Jy)$ for $x,y \in V$ the inner product on $V$?
I want only to know if $\langle\cdot,\cdot\rangle$ is positive definite, because the other axioms of inner product are easy.
I'm afraid not. Counterexample: with respect to the standard basis of $V=\mathbb R^2$, let the matrix representations of $J$ and $w$ be $J=\pmatrix{0&-1\\ 1&0}$ and $W=J^T$. Then $$w(Jx,Jy)=y^TJ^TWJx=y^TWx=w(x,y),$$ but $$w(x,Jy)=y^TJ^TWx=y^T(J^T)^2x=-y^Tx$$ is the negative of the usual inner product.