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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.

Asaf Shachar
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1 Answers1

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

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 $A$.

user1551
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  • Thanks, nice answer! Does the claim that any diagonalizable matrix with unimodular eigenvalues is similar (over $\mathbb R$) to an orthogonal matrix follow from the theorem on the real Jordan normal form? Or am I mistaken? – Asaf Shachar Jul 29 '20 at 16:59
  • @AsafShachar You may use real Jordan form. You may also use the usual complex Jordan form. Suppose $A\in SL_2(\mathbb R)$ is $\mathbb C$-diagonalisable and its eigenvalues have unit moduli. If the two real eigenvalues are real, they must be both equal to $1$ or both equal to $-1$. Hence $A=\pm I$. If $A$ has a conjugate pair of non-real eigenvalues $e^{\pm i\theta}$, then $A$ is similar to the rotation matrix $Q$ for angle $\theta$ over $\mathbb C$. But two real matrices that are similar over $\mathbb C$ must be similar over $\mathbb R$. Hence $A$ is similar to $Q$ over $\mathbb R$. – user1551 Jul 29 '20 at 17:14
  • OK, thanks! last (perhaps stupid) question: How do you see that if the eigenvalues of $A$ are $e^{\pm i\theta}$ then $A$ is similar to a rotation by angle $\theta$? I guess that this is some explicit computation that can be done once and for all? Oh, on a second thought, since these are two different eigenvalues, both matrices are just automatically diagonalizable over $\mathbb C$, right? – Asaf Shachar Jul 29 '20 at 17:22
  • @AsafShachar This is because both $Q$ and $A$ are similar to $\operatorname{diag}(e^{i\theta},e^{-i\theta})$. – user1551 Jul 29 '20 at 17:24