# skew-diagonalization of a matrix

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 $\begin{pmatrix}0 & t \\ -t^{-1}& 0 \end{pmatrix}$?

• Note that the word "conjugate" means different things in different contexts. It often means $A\mapsto P^{-1}AP$, but sometimes it means $A\mapsto P^TAP$. Which kind of conjugation are you referring to? Also, what do you mean by "skew-diagonalization"? Do you require $P$ to be skew symmetric? – user1551 May 15 '14 at 7:23
• @user1551 I think the name of $A\mapsto P^TAP$ is congruence, here "conjagate" means $A\mapsto P^{-1}AP$, and "skew-diagonalization" means it is conjugate to $\begin{pmatrix}0 & t\\ t^{-1}& 0\end{pmatrix}$ instead of $\begin{pmatrix}t & 0\\ 0& s\end{pmatrix}$. – user50402 May 15 '14 at 7:31

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.)