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 $A \in GL(4,\mathbb R)$ ?
 A: 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 $\det J_1=1$.
A: As a complement to @yths' nice answer: first observe that for any non-zero vector $v \in {\bf R}^4$, the vectors $v$ and $Jv$ are linearly independent (otherwise $Jv=c v$ for a real number $c$ with $c^2=-1$). So you can start building a basis of ${\bf R}^4$ by taking the first basis vector to be any non-zero vector $v$ and the second basis vector to be $J v$. Since the group generated by $J$ is finite (of order $4$), the span of $v$ and $Jv$ has a $J$-stable complement $U$, and choosing any non-zero vector $u \in U$ gives a basis $Jv,v,Ju,u$ of ${\bf R}^4$ with respect to which the matrix of $J$ is the matrix you call $J_0$. In other words, every $J$ with $J^2=-1$ is conjugate to $J_0$ (and in particular has determinant $1$).
