I tried to solve the following eigen equation of (1-1) on the way to one problem on the quantum mechanics , but the eigenvector seems to have an imaginary part (see 1-10) despite using the hermitian matrix. Is my answer correct?
【My question】
Solve the following eigen equation. $$\left(\begin{matrix}\varepsilon_1&\bar{\gamma}\\\gamma&\varepsilon_2\\\end{matrix}\right)\left(\begin{matrix}x\\y\\\end{matrix}\right)=\lambda\left(\begin{matrix}x\\y\\\end{matrix}\right)\tag{1-1}$$ Here, ${\varepsilon_1}$ and ${\varepsilon_2}$ are real number, $\gamma$ is a complex number, and $\bar{\gamma}$ is a conjugate complex number of the $\gamma$, these are regarded as the constant value.
【My answer】
(1)Eigenvalues
$$\left|\begin{matrix}\varepsilon_1-\lambda&\bar{\gamma}\\\gamma&\varepsilon_2-\lambda\\\end{matrix}\right|=0\tag{1-2}$$
therefore, $$\left(\lambda\ -\varepsilon_1\right)\left(\lambda-\varepsilon_2\right)-\left|\gamma\right|^2\ =\ 0 \tag{1-3}$$ $${\lambda\ }^2-\left(\varepsilon_1+\varepsilon_2\right)\lambda+\varepsilon_1\varepsilon_2-\left|\gamma\right|^2\ \ =0\tag{1-4}$$
Using the Quadratic formula we get, $${E_+}=\frac{\left({\varepsilon_1}+{\varepsilon_2}\right)\ +\sqrt{\left(\varepsilon_1+\varepsilon_2\right)^2\ +\ 4\left(\left|\gamma\right|^2-\ \varepsilon_1\varepsilon_2\right)}}{2}\ =\frac{\left(\varepsilon_1+\varepsilon_2\right)\ }{2}\ +\ \frac{D\ }{2}\tag{1-5}$$ and $${E_-}=\frac{\left({\varepsilon_1}+{\varepsilon_2}\right)\ -\sqrt{\left(\varepsilon_1+\varepsilon_2\right)^2\ +\ 4\left(\left|\gamma\right|^2-\ \varepsilon_1\varepsilon_2\right)}}{2}\ = \frac{\left(\varepsilon_1+\varepsilon_2\right)\ }{2}\ - \ \frac{D\ }{2} .\tag{1-6}$$
Here, $$D:= \sqrt{\left(\varepsilon_1+\varepsilon_2\right)^2\ +\ 4\left(\left|\gamma\right|^2-\ \varepsilon_1\varepsilon_2\right)} .\tag{1-7}$$
(2)Eigenvectors
Consider the 1 st row of the (1-1)
$$\ \ x\varepsilon_1\ +\bar{\gamma}y=\lambda x\ \tag{1-8}$$
therefore,
$$y=\frac{x\gamma}{\left|\gamma\right|^2}\left(\lambda-{\varepsilon_1}\right)
=-\ \frac{x\gamma}{2\left|\gamma\right|^2}\left(\left(\varepsilon_1\ -\ {\varepsilon_2}\right)\pm\ D\right)\tag{1-9}$$
therefore, the eigen vectors are
$$\left(\begin{matrix}1\\-\ \frac{\gamma}{2\left|\gamma\right|^2}\left(\left(\varepsilon_1\ -\ \varepsilon_2\right)\pm\ D\right)\\\end{matrix}\right)\tag{1-10}$$