skew-diagonalization of a matrix

Question

I think about the skew-diagonalization of a matrix, for example, let $A=\begin{pmatrix}a & b \\ c& d \end{pmatrix}\in SL(2,\mathbb{R})$ , if $trace(A)=0$, is it conjugate to $\begin{pmatrix}0 & t \\ -t^{-1}& 0 \end{pmatrix}$?

Answer 1

Yes. Every traceless matrix in $SL(2,\mathbb R)$ is similar over $\mathbb{C}$ to $\operatorname{diag}(i,-i)$, and hence they are all similar to each other over $\mathbb{C}$. Therefore they are similar over $\mathbb R$ too. (See q57242: Similar matrices and field extensions.)