example of complex structure with negative determinant is it possible to find a matrix $J_1 \in GL(4,\mathbb R)$ such that $\det J_1=-1 $ and $J_1^2=-\operatorname{id}$ ?
if it is, how can we prove that every matrix $M \in GL(4,\mathbb R)$ such that $M^2=-\operatorname{id}$ is conjugate to either $J_1$ or $J_0$, where
$$ J_0 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & -1 & 0 \\ \end{pmatrix},$$
namely $M = A^{-1}J_{0,1}A$ for some $A \in GL(4,\mathbb R)$ ?

It is impossible. 
Let $J_1\in GL_4\mathbb R$ satisfy $J_1^2=-I$. Put $m(t)=t^2+1$ and denote by $\Phi(t) = \det(tI-J_1)$. Since $J_1$ is a real matrix, $m(t)$ is the minimal polynomial of $J_1$. Now the roots of $\Phi(t)$ agree with ones of $m(t)$ (up to multiplicities). Since $\Phi(t)$ has only real coefficients and a multiple of $m(t)$, 
$$\Phi(t)=(t^2+1)m(t)=(t^2+1)^2 = t^4+2t^2+1.$$
Thus $\det J_1=1$.
