Let $(V,g)$ be an $n$-dimensional real vector space with some scalar product $g$, $A\in GL(n)$ be a $g$-skew-adjoint, invertible real matrix (so $A^* = -A$) and $P = -A^2 = A\cdot A^*$. This matrix is positive definite, so it has a unique positive definite root $Q\in GL(n)$ with $P = Q^2 = Q\cdot Q$.
Now, one has to prove that $(Q^{-1}A)^2 = -id$ (meaning that $Q^{-1}A$ is an almost complex structure). However, I have not made any progress with it. It should be very easy to prove if one knows that $A$ and $Q$ commute, but I have not been able to prove this fact.
If one tries to diagonalize $Q$ as $D = U^{-1}\cdot Q\cdot U$ for some $U\in GL(n)$ and a diagonal matrix $D$, one sees $$ D^2 = U^{-1}\cdot Q^2\cdot U = U^{-1}\cdot P\cdot U = -U^{-1} \cdot A^2\cdot U = -(U^{-1}\cdot A\cdot U)^2,$$ meaning $(U^{-1}\cdot A\cdot U)^2 = - D^2$, so it would be tempting to assume that $U^{-1} \cdot A\cdot U = \pm i\cdot D$, which would of course make all calculations a bit more easy. However, I do not think that one can perform the last step, since $-D^2$ is not positive definite and thus could have a lot of roots.
$P$ is positive definite. Therefore $Q=\sqrt{P}=f(P)$ for some polynomial $f$. (More specifically, if $\lambda_1,\lambda_2,\ldots,\lambda_m$ are the distinct eigenvalues of $P$ and $f$ is the Lagrange interpolation polynomial that maps each $\lambda_i$ to $\sqrt{\lambda_i}$, then $f(P)=Q$.) In turn, $Q=f(P)=f(-A^2)$ is a polynomial in $A$. Hence $Q$ commutes with $A$. Consequently, $(Q^{-1}A)^2=Q^{-2}A^2=P^{-1}(-P)=-I$.