For $A = \left(\begin{array}{ll} a & b \\ c & d \end{array}\right)$ in $\mathcal{M}_2(\mathbb{R})$, let $\mathcal{H}_A$ be the quadric with equation $\psi_A(x,y,z) = 0$ where $\psi_A(x,y,z)$ is the real part of the determinant of $\left(\begin{array}{cc} a-x-\mathrm{i}z & b-y \\ c+y & d-x-\mathrm{i}z \end{array}\right)$, and let $Z_A$ be the intersection of $\mathcal{H}_A$ with the plane $x = (a+d)/2$. If the matrix $A$ has two non-real eigenvalues, how can one see the eigenvalues of $A$ on $\mathcal{H}_A$? (One may consider the intersection of $Z_A$ with the plane with equation $y = 0$.) Can one see a basis of eigenvectors using $\mathcal{H}_A$?
For $A = \left(\begin{array}{ll} a & b \\ c & d \end{array}\right)$ in $\mathcal{M}_2(\mathbb{R})$, let $\mathcal{H}_A$ be the quadric with equation $\psi_A(x,y,z) = 0$ where $\psi_A(x,y,z)$ is the real part of the determinant of $\left(\begin{array}{cc} a-x-\mathrm{i}z & b-y \\ c+y & d-x-\mathrm{i}z \end{array}\right)$, and let $Z_A$ be the intersection of $\mathcal{H}_A$ with the plane $x = (a+d)/2$.
If the matrix $A$ has two non-real eigenvalues, how can one see the eigenvalues of $A$ on $\mathcal{H}_A$? (One may consider the intersection of $Z_A$ with the plane with equation $y = 0$.) Can one see a basis of eigenvectors using $\mathcal{H}_A$?