Let we have a 2x2 matrix $M$:
$$M=\begin{bmatrix} a & b \\b^* & -a\end{bmatrix}$$
which is written in basis (let's say) $\psi_1$ and $\psi_2$. i.e.
$$\begin{bmatrix} \psi_1^\dagger & \psi_1^\dagger \end{bmatrix}
\begin{bmatrix} a & b \\b^* & -a\end{bmatrix} \begin{bmatrix} \psi_1 \\ \psi_2 \end{bmatrix}$$
Eigen values of this matrix can be find by $|MI-\lambda|=0$ that are $\lambda_{1,2}=\pm\sqrt{a^2+bb^*}$. Eigenvectors can be written as:
for $\lambda=+\sqrt{a^2+bb*}$
$$(MI-\lambda_{1})\phi_{1}=0$$
$$\begin{bmatrix}a-\lambda_1 & b \\ b* & -a-\lambda_1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=0$$
$$(a-\lambda_1)x_1+bx_2=0$$
$$b^*x_1-(a-\lambda_1)x_2=0$$
$\phi_1$ can be written as:
$$\phi_1=\begin{bmatrix}\frac{bx_2}{\lambda_1 -a} \\ x_2\end{bmatrix}$$
normalization: $(\frac{b^2}{(\lambda_1-a)^2}+1)x_2^2=1$ gives:
$$\phi_1=\frac{1}{\sqrt{(\frac{b}{\lambda_1-a})^2+1}}\begin{bmatrix}\frac{b}{\lambda_1 -a} \\ 1\end{bmatrix}$$
similarly,
$$\phi_2=\frac{1}{\sqrt{(\frac{b}{\lambda_2-a})^2+1}}\begin{bmatrix}\frac{b}{\lambda_2 -a} \\ 1\end{bmatrix}$$
My question: I want to find relation between new basis ($\phi_{1,2}$) in which this $M$ matrix is diagonalized and the old basis ($\psi_{1,2}$). i.e. what are $A,B,C$ and $D$ if $$\phi_1=A\psi_1+B\psi_2$$ $$\phi_2=C\psi_1+D\psi_2$$