Let $A=$ $\begin{bmatrix} \cos(\alpha) & -\sin(\beta) \\ \sin(\alpha) & \cos(\beta) \\ \end{bmatrix}$
If $\alpha =\beta$ then it is an orthogonal rotation and there are no real eigenvalues.
I cannot easily derive a nice rule for $A$ to have real eigenvalues when $\alpha \ne\beta$.
I am quite confident that when $\alpha$ and $\beta$ have a different sign, the matrix must have real eigenvalues.
But I get stuck in the geometric formulas. I would be glad if someone could point me to a source or just post a nice derivation.
My colleague Johan D. confirmed the calculation does not lead to a straightforward formula. I copy his calculation below:
$\det{\left(A-\lambda I\right)}=\left|\begin{matrix}\cos{\alpha}-\lambda&-\sin{\beta}\\+\sin{\alpha}&\cos{\beta-\lambda}\\\end{matrix}\right|=0$
$\left(\cos{\alpha}-\lambda\right)\left(\cos{\beta-\lambda}\right)-\sin{\alpha}\left(-\sin{\beta}\right)=0$
$\left(\cos{\alpha}-\lambda\right)\left(\cos{\beta-\lambda}\right)+\sin{\alpha}\sin{\beta}=0$
$\lambda^2-\left(\cos{\alpha}+\cos{\beta}\right)\lambda+\ \cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}=0$
$D=b^2-4ac=\left(\cos{\alpha}+\cos{\beta}\right)^2-4\left(\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}\right)$
$A$ has real eigenvalues if and only if $D=\left(\cos{\alpha}+\cos{\beta}\right)^2-4\left(\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}\right)\geq0$
In general: $\left(\cos{\alpha}-\cos{\beta}\right)^2\geq4\sin{\alpha}\sin{\beta}$
If $sgn\left(\sin{\alpha}\right)\neq sgn\left(\sin{\beta}\right)$ then $\sin{\alpha}\sin{\beta}\le0$ , hence $\left(\cos{\alpha}-\cos{\beta}\right)^2\geq0\geq4\sin{\alpha}\sin{\beta}$
so if $sgn\left(\sin{\alpha}\right)\neq sgn\left(\sin{\beta}\right)$, $A$ always has real eigenvalues