Spectral decomposition of a $4\times4$ real nonsymmetric matrix with unknown elements 
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.
 A: I don't know how explicit or simple you want the expressions for $\Lambda$ and $Q$ to be, but here is a description. Using your notations, we write $A$ as 
$\begin{bmatrix}
B&-\gamma^{-1}{\rm{I}}_2\\
-{\rm{I}}_2&-B^{\rm{T}}
\end{bmatrix}$
where 
$B=\begin{bmatrix}
-\alpha&\alpha\\
\beta&-\beta
\end{bmatrix}$.
Let $\lambda$ be an eigenvalue and 
$\begin{bmatrix}
\mathbf{v}_{2\times 1}\\
\mathbf{w}_{2\times 1}
\end{bmatrix}$
an eigenvector for it. We have
$$
\begin{bmatrix}
B&-\gamma^{-1}{\rm{I}}_2\\
-{\rm{I}}_2&-B^{\rm{T}}
\end{bmatrix}
\begin{bmatrix}
\mathbf{v}\\
\mathbf{w}
\end{bmatrix}
=\lambda
\begin{bmatrix}
\mathbf{v}\\
\mathbf{w}
\end{bmatrix}\Rightarrow
\begin{cases}
B\mathbf{v}-\gamma^{-1} \mathbf{w}=\lambda\mathbf{v}\\
-\mathbf{v}-B^{\rm{T}}\mathbf{w}=\lambda\mathbf{w}
\end{cases}
$$
Solving the second equation for $\mathbf{v}$ and substituting in the first equation yields 
$$
\mathbf{v}=-\left(B^{\rm{T}}+\lambda{\rm{I}_2}\right)\mathbf{w},\,\quad 
\left(BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}\right)\mathbf{w}=\mathbf{0}.
$$
Thus the eigenvalues are the roots of the quartic 
$$
{\rm{det}}\left(BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}\right)=0.
$$
Notice that changing $\lambda$ to $-\lambda$ changes the matrix 
$BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}$ to its transpose. Hence (as you mentioned) if $\lambda$ is an eigenvalue, so is $-\lambda$. Writing the roots of the quartic as $\lambda_1,-\lambda_1$ and $\lambda_2,-\lambda_2$, for each $\lambda_i$ the $2\times 2$ matrix 
$BB^{\rm{T}}+\lambda_i(B-B^{\rm{T}})+(\gamma^{-1}-\lambda_i^2)\,{\rm{I}_2}$
is not full rank and thus there are non-zero column vectors $\mathbf{w}_{ir}$ and $\mathbf{w}_{ic}$ ($i\in\{1,2\}$) with 
$$
\left(BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}\right)\mathbf{w}_{ic}=\mathbf{0}\quad
\mathbf{w}_{ir}^{\rm{T}}\left(BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}\right)=\mathbf{0}
$$
Plugging into previous equations, the corresponding eigenvalues could be computed. The  result is 
$$
Q=\begin{bmatrix}
-\left(B^{\rm{T}}+\lambda_1{\rm{I}_2}\right)\mathbf{w}_{1c}& 
-\left(B^{\rm{T}}-\lambda_1{\rm{I}_2}\right)\mathbf{w}_{1r}&-\left(B^{\rm{T}}+\lambda_2{\rm{I}_2}\right)\mathbf{w}_{2c}&
-\left(B^{\rm{T}}-\lambda_2{\rm{I}_2}\right)\mathbf{w}_{2r}\\
\mathbf{w}_{1c}&\mathbf{w}_{1r}&\mathbf{w}_{2c}&\mathbf{w}_{2r}
\end{bmatrix}
$$
which satisfies 
$$
A=Q
\begin{bmatrix}
\lambda_1& & &\\
&-\lambda_1& &\\
& & \lambda_2 & \\
& & & -\lambda_2
\end{bmatrix}
Q^{-1}.
$$
