Decomposition of trace zero real matrix with purely imaginary eigenvalues I would like to prove that for a matrix $A\in sl(2,\mathbb{R})$, i.e. a real matrix with $\text{tr}(A)=0$, if the eigenvalues of $A$ are $\pm i\alpha$, for $\alpha\in \mathbb{R}$, then there is a real matrix $M$ with $\det(M)>0$ and
$$
MAM^{-1}=\alpha J
$$
where $J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$.
It seems like it should be easy, but I am having some difficulty proving it by hand. I am wondering if there is a more abstract solution (I don't know any Lie theory). Thanks for any help!
 A: You cannot prove it because it is false. Consider $A=J^T$ and $\alpha=1$.
The equation $MJ^TM^{-1}=J$ implies that $MJ^T=JM$, or, using coordinates,
$$
\pmatrix{b&-a\\ d&-c}
=\pmatrix{a&b\\ c&d}\pmatrix{0&-1\\ 1&0}
=\pmatrix{0&1\\ -1&0}\pmatrix{a&b\\ c&d}
=\pmatrix{c&d\\ -a&-b}.
$$
Hence $b=c,\,a=-d$ and $\det(M)=ad-bc=-a^2-b^2\le0$.
The statement is true, however, if $\alpha\ne0$ and one can pick the sign of $\alpha$ at will. In other words, one can pick a real matrix $M$ with positive determinant such that $MAM^{-1}=\beta J$ for some $\beta\in\{-\alpha ,\alpha\}$.
Presumably $\alpha$ is nonzero, otherwise the statement is false when $A$ is a nilpotent Jordan block. When $\alpha$ is indeed nonzero, both $A$ and $\alpha J$ have the same spectrum consisting of two distinct complex eigenvalues. Therefore they are similar over $\mathbb C$ (because they are both similar to $\operatorname{diag}(i\alpha,-i\alpha)$). However, two matrices that are similar over an extension field must be similar over the ground field (because they must have the same rational canonical form). Therefore $A$ is similar to $\alpha J$ over $\mathbb R$. That is, there exists some real invertible matrix $P$ such that $PAP^{-1}=\alpha J$. Now, if $\det P>0$, we can pick $M=P$ and $\beta=\alpha$. If $\det P<0$, we can pick $M=\pmatrix{1&0\\ 0&-1}P$ and $\beta=-\alpha$.
The sign of $\beta$ is uniquely determined by $A$. That is, if $MAM^{-1}=\beta J\ne0$, you cannot find another real matrix $M_1$ with positive determinant such that $M_1AM_1^{-1}=-\beta J$, because, as shown in the above, $-J=J^T$ is not similar to $J$ via a change-of-basis matrix with positive determinant.
