Let $V\cong \mathbb{R}^4$ be a Euclidean space, and let $\omega \in \bigwedge^2 V^*$ such that $||\omega||^2=2$ and assume further that $\omega$ s self-dual or anti-self-dual, i.e. $\ast \omega = \pm \omega$ (where $\ast$ is the Hodge star). Define the operator $I: V \to V$ by the property $(Ix,y)=\omega(x,y)$, then it is straightforward that $I^*=-I$. If one writes out $I$ in an orthonormal basis then one can see that $I$ is orthogonal and conclude $I^2=-\mathrm{Id}$ (i.e. $I$ is an arthogonal complex structure on $V$).
Question: is there a coordinate-free way to deduce the last fact?