I'm trying to eigendecompose the following matrix $A$, i.e. to find $Q$ and $\Lambda$ such that $$ A = \begin{bmatrix} -\alpha & \alpha & -\gamma^{-1} & 0\\ \beta & -\beta & 0 & -\gamma^{-1}\\ -1 & 0 & \alpha & -\beta\\ 0 & -1 & -\alpha & \beta \end{bmatrix}=Q\Lambda Q^{-1} $$ where $\alpha>\beta>0$ and $\gamma>0$.

Notice that

- $A$ is a Hamiltonian matrix, i.e. $JA$ is symmetric where $J=\begin{pmatrix}0 & I_2 \\ -I_2 & 0\end{pmatrix}$, $I_2$ is the $2\times2$ identity matrix.
- The characteristic polynomial of a real Hamiltonian matrix is even. Thus, if $λ$ is an eigenvalue of $A$, then $−λ$, $\bar λ$ and $−\bar λ$ are also eigenvalues. It follows that $\text{trace} A=0$.
- $A$ can be written in block notation: $A=\begin{pmatrix}B & -\gamma^{-1}I_2 \\ -I_2 & -B^T \end{pmatrix}$, with $B=\begin{pmatrix}-\alpha & \alpha \\ \beta & -\beta\end{pmatrix}$.
- the elements on the antidiagonal are all $0$.
- sum of row $1$, sum of row $2$, sum of column $3$ and sum of column $4$ are all equal to $-\gamma^{-1}$.
- sum of row $3$ and sum of column $2$ are equal to $\alpha-\beta-1$.
- sum of row $4$ and sum of column $1$ are equal to $\beta-\alpha-1$.

can we use these facts to find $Q$ and $\Lambda$?

Moreover, the characteristic polynomial is $$ p_A(\lambda) = \lambda^4-[(\alpha+\beta)^2+2\gamma^{-1}]\lambda^2+\gamma^{-1}(2\alpha^2+2\beta^2+\gamma^{-1}) $$ hence $$ 2\lambda^2 = (\alpha+\beta)^2+2\gamma^{-1} \pm \sqrt{(\alpha+\beta)^4-4\gamma^{-1}(\alpha-\beta)^2} $$ and the eigenvalues are either real or complex depending on the values of the parameters.