How to show that $\begin{pmatrix} 0 & -x \\\ 1/x & 0\end{pmatrix}$ is conjugate to a rotation?

Question

Let $x >0$, and set $A=\begin{pmatrix} 0 & -x \\\ 1/x & 0\end{pmatrix}$.

Question: How to show that $A \in \operatorname{SL}_2(\mathbb R)$ is conjugate to an element of $\operatorname{SO}(2)$? That is, I am trying to show that there exist $C \in \operatorname{SL}_2(\mathbb R)$, and $Q \in \operatorname{SO}(2)$ such that $A=CQC^{-1}$.

Note that $A=\begin{pmatrix} x & 0 \\\ 0 & 1/x\end{pmatrix}R_{\pi/2}$, where Let $R_{\pi/2}=\begin{pmatrix} 0 & -1 \\\ 1 & 0\end{pmatrix}$ is a rotation by $\pi/2$. However, I don't see how that representation helps us.

Answer 1

$\pmatrix{1/\sqrt{x}\\ &\sqrt{x}}\pmatrix{0&-x\\ 1/x&0}\pmatrix{\sqrt{x}\\ &1/\sqrt{x}}=\pmatrix{0&-1\\ 1&0}=\pmatrix{\cos\frac\pi2&-\sin\frac\pi2\\ \sin\frac\pi2&\cos\frac\pi2}$.

Since all elements of $SO_2(\mathbb R)$ have unimodular eigenvalues, not every $A\in SL_2(\mathbb R)$ is similar to a special orthogonal matrix. In fact, $A\in SL_2(\mathbb R)$ is similar to some $Q\in SO_2(\mathbb R)$ if and only if $A$ is diagonalisable over $\mathbb C$ and its eigenvalues are unimodular. In this case, $Q$ is just the real Jordan form of $A$.