I need to diagonalise a complex $2\times2$ matrix, $M$. I need the explicit diagonalisation transformation, i.e. a matrix, $S$ s.t. $S^{-1}MS= \text{diag}(m_1,m_2)$. I am talking about a matrix of the form:
$$M = \begin{pmatrix} a^2 & ic \\ ic & b^2 \end{pmatrix} $$ where $a,b,c$ are positive real numbers. We want to restrict ourselves to the assumption $(a^2-b^2)^2>4c^2$.
As one usually does, I caclulated the eigenvalues $m_{1/2}$, $$m_{1/2} =\frac{1}{2} \left( a^2+b^2 \pm \sqrt{(a^2-b^2)^2-4c^2} \right)$$ Our assumption means that these are real. Two linearly independent eigenvectors (one for each eigenvalue) are
$$\mathbf{x}_{1/2}= \begin{pmatrix} ic \\ m_{1/2}-a^2 \end{pmatrix} $$ It is quickly shown that the two are actually eigenvectors. Now I continued from there by normalising the eigenvectors, using the norm in $\mathbb{C}^2$,i.e. $\|\mathbf{x}_{1/2}\|^2 = c^2+(m_{1/2}-a^2)^2$. Having normalised my eigenvectors I wrote them in the rows of a matrix. Denoting $A_{1/2} = \|\mathbf{x}_{1/2}\|$ this gives $$\mathbf{S} = \begin{pmatrix} \frac{ic}{A_1} & \frac{m_1-a^2}{A_1}\\ \frac{ic}{A_2} & \frac{m_2-a^2}{A_2} \end{pmatrix} $$ However, when I calculated $S^{-1}MS$, the result was not diagonal (although there may of course be mistakes). I have no experience with diagonalising complex matrices and I know that it is not always possible do diagonalise a complex matrix. I really need to know if I just made a mistake in my calculation or if something more serious went wrong in my derivation of $S$.
Edit: Just found this related question Finding Eigensystem of Hermitian Matrix. However that is just a special case, does it work in the general case with $a,b,c>0$ as well?