12
The formula appears in
Marvin Marcus "determinants of sums". College mathematics journal, March 1990.
https://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/determinants-of-sums
The author notes that it is hard to track it's origins, but mentions previous references in the text, in particular to a book of himself in 1975.
2
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 ...
2
One partial answer, for which you don't need the knowledge that the spectra are complete sets of roots of unity.
First step : verify that the group $G$ generated by $A$ and $B$ is finite (it should be contained a permutation group of $n$ elements). For this, explore the free group in two letters $a$ and $b$ (together with $a^{-1}$ and $b^{-1}$). Consider ...
2
Here's a possible method:
We might as well assume that $A$ is already a permutation matrix -- just conjugate $A$ and $B$ by the same unitary matrix $C$ and relabel.
Now if we are looking to conjugate $B$ to a permutation matrix AND retain $A$ as a permutation matrix, then we'll need to conjugate by something that centralizes $A$ -- so our conjugating ...
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