Let $V$ be a real vector space equipped with a symplectic form $\omega : V\times V \to \mathbb{R}$. An almost complex structure $J$ is compatible with $\omega$ if $\omega(J(u), J(v)) = \omega(u, v)$ for all $u, v \in V$. In addition, $J$ is said to tame $\omega$ if $J$ is compatible with $\omega$ and $\omega(u, J(u)) > 0$ for all $u \in V\setminus\{0\}$.
Given a compatible almost complex structure $J$ which tames $\omega$, we have another compatible almost complex structure given by $-J$ and satisfies $\omega(u, -J(u)) < 0$ for all $u \in V\setminus\{0\}$.
In both of these cases, the resulting bilinear form $\omega(\cdot, J\cdot)$ is definite; is this always the case? Alternatively phrased,
Given an almost complex structure $J$ which is compatible with the symplectic form $\omega$, must one of $J$ or $-J$ tame $\omega$?
What if $J(e_1)=e_2$, $J(e_3)=e_4$, and $\omega=dx_1\wedge dx_3+dx_2\wedge dx_4$?