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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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