$\newcommand\pair[1]{\left\langle #1 \right\rangle}$ Consider Minkowski plane $\Bbb L^2 = (\Bbb R^2, \pair{\cdot,\cdot}_L )$, where$\renewcommand\vec[1]{{\boldsymbol #1}}$ $$\pair{\vec{u},\vec{v}}_L\doteq u_1v_1-u_2v_2.$$If a linear map $\Lambda \colon \Bbb L^2 \to \Bbb L^2$ is self-adjoint with respect to $\pair{\cdot,\cdot}_L$, is it possible to give neat conditions on $\Lambda$ for the existence of an orthonormal basis of eigenvectors?
If ${\rm can} \doteq (\vec{e}_1,\vec{e}_2)$ is the standard basis, one can readily check that $[\Lambda]_{\rm can}$ is of the form $$[\Lambda]_{\rm can} = \begin{pmatrix} a & b \\ -b & d\end{pmatrix},$$for $a,b,d \in \Bbb R$. I can't interprete geometrically the condition $$\Delta = (a+d)^2 - 4(ad+b^2) = (a+2b-d)(a-2b-d) > 0.$$
Remark: In Greub's Linear Algebra it is proven that a self-adjoint linear operator $\Lambda\colon \Bbb R^n_\nu \to \Bbb R^n_\nu$ admits an orthogonal basis of eigenvectors, as long as $n \geq 3$ and $\pair{\vec{v},\Lambda\vec{v}}_\nu \neq 0$ for every lightlike vector $\vec{v}$. The proof is crazy, though, using topological arguments. I'm looking for some condition similar to that on lightlike vectors, in this border case.