 Solution

The characteristic polynomial of $A$ isWhen $|\mathrm{tr}A|>2$, its discriminant is positive, so $A$ has two distinct real eigenvalues $\lambda ,{\lambda }^{-1}$ and two independent real eigenvectors. Let $P$ have those eigenvectors as columns. Rescaling one column makes $\det P=1$ without changing the diagonalization. Hence $P\in S{L}_2(\mathbb{R})$ andThis is the normal form of a hyperbolic element of PSL2(R).