Guessing eigenvalue solution I am reading The Theory of Magnetism I, by Mattis. In Chapter 2, he proposes the following eigenproblem: 
$$ \left ( \begin{matrix} V & U \\ U^\dagger& V  \end{matrix} \right ) \left ( \begin{matrix} c_I \\ c_{II}  \end{matrix} \right ) = (E-2E_0) \left ( \begin{matrix} 1 & l^2 \\ (l^2)^* & 1  \end{matrix} \right ) \left ( \begin{matrix} c_I \\ c_{II}  \end{matrix} \right ) $$
He then mentions it is easy to guess the solutions are: 
$$ c_I = \pm c_{II}$$
How can you guess that? 
I don't think is relevant, but the context of the eigenvalue problem comes from the following variational problem: 
We have a Hamiltonian of a hydrogen molecule: 
$$H = H^0_1 + H^0_2  + H^\lambda $$
where: 
$$ H^0_1 = \left ( \frac{p_1^2}{2m}-\frac{e^2}{r_{1a}} \right ), \quad H^0_2 = \left ( \frac{p_2^2}{2m}-\frac{e^2}{r_{2b}} \right ), \quad H^\lambda =\left(  \frac{e^2}{R_{ab}}+\frac{e^2}{r_{12}}-\frac{e^2}{r_{1b}}-\frac{e^2}{r_{2a}} \right )$$
where (a,b) represent the nucluei of each hydrogen atom and the numbers 1,2 refer to their respective electron. 
The eigenfunctions of $H_a$ and $H_b$ are: $\phi_a(\vec{r_1})$ and $\phi_b(\vec{r_2})$ and we can choose $\Psi_I = \phi_a(\vec{r_1}) \phi_b(\vec{r_2})$ and $\Psi_{II} = \phi_a(\vec{r_2}) \phi_b(\vec{r_1})$
The functions $U$, $V$ and $l$ are: 
$$ l \equiv \int d^3 r \phi_a^*(\vec{r})\phi_b(\vec{r}) $$
$$ V \equiv \int d^3r_1d^3r_2 |\Psi_{II}|^2 H^\lambda =  \int d^3r_1d^3r_2 |\Psi_{I}|^2 H^\lambda $$
$$ U \equiv \int d^3r_1d^3r_2 \Psi_I^* \Psi_{II} H^\lambda$$
Then a variational function: $\Psi = c_I \Psi_I + c_{II} \Psi_{II}$ is chosen. And in order to determine the ground state we solve the followinf variational problem: 
$$E_{var} = \frac{\int d^r_1 d^3r_2 \Psi^* H \Psi}{\int d^r_1 d^3r_2 \Psi^*  \Psi}, \quad \frac{\partial E_{var}}{\partial c_{I,II}} = 0 $$
 A: Mathematically you can write
$$\left ( \begin{matrix} 1 & l^2 \\ (l^2)^* & 1  \end{matrix} \right )^{-1} \left ( \begin{matrix} V & U \\ U^\dagger& V  \end{matrix} \right ) \left ( \begin{matrix} c_I \\ c_{II}  \end{matrix} \right ) = (E-2E_0) \left ( \begin{matrix} c_I \\ c_{II}  \end{matrix} \right ).$$
The next step is to determine the inverse of the matrix $\left( \begin{matrix} 1 & l^2 \\ (l^2)^* & 1  \end{matrix} \right )^{-1}$ which is straight forward. This gives
$$\frac{1}{1 - |l^2|^2} \left ( \begin{matrix} 1 & - l^2 \\ - (l^2)^* & 1  \end{matrix} \right ) \left ( \begin{matrix} V & U \\ U^\dagger& V  \end{matrix} \right ) \left ( \begin{matrix} c_I \\ c_{II}  \end{matrix} \right ) = (E-2E_0) \left ( \begin{matrix} c_I \\ c_{II}  \end{matrix} \right ).$$
Taking the factor $1 - |l^2|^2$ to the rhs and multiplying the two matrices on the lhs gives
$$\left ( \begin{matrix} V - l^2 U^\dagger & U -l^2 V \\ U^\dagger -(l^2)^*V& V  - (l^2)^* U  \end{matrix} \right ) \left ( \begin{matrix} c_I \\ c_{II}  \end{matrix} \right ) = \lambda \left ( \begin{matrix} c_I \\ c_{II}  \end{matrix} \right ),$$ with $\lambda = (1 - |l^2|^2) (E - 2E_0)$ and this is the eigenproblem to be solved.
If you call $\alpha = V - l^2 U^\dagger, \beta = U - l^2 v$ the above matrix equation takes the form$$\left ( \begin{matrix} \alpha & \beta \\ \beta^*& \alpha^*  \end{matrix} \right ) \left ( \begin{matrix} c_I \\ c_{II}  \end{matrix} \right ) = \lambda \left ( \begin{matrix} c_I \\ c_{II}  \end{matrix} \right ).$$
Can you take it from here? 
