This is a subject I've been working on for a very long time now, but still did not manage to fully understand the interesting properties of this matrix $\mathbf{A}$.

First, let's define two matrices:

$\mathbf{N}$ is the following matrix: \begin{equation} \mathbf{N}=\begin{bmatrix} \mathbf{I}_n & \mathbf{0}_n \\ \mathbf{0}_n & \mathbf{P}^{-1}\begin{bmatrix}1 & && \\ & \ddots && \\ & & 1& \\ &&& -1 \end{bmatrix}\mathbf{P} \end{bmatrix}\in\mathbb{R}^{2n\times2n} \end{equation} where $\mathbf{P}\in\mathbb{R}^{n\times n}$ is any invertible matrix.

with $\omega_i>0$ and $t>0$, the block-diagonal matrix: \begin{equation} \mathbf{S}(t)=\begin{bmatrix} \begin{bmatrix} \cos(\omega_1t) & \\& \ddots & \\ & & \cos(\omega_n t) \end{bmatrix} & \begin{bmatrix} \dfrac{\sin(\omega_1t)}{\omega_1} & \\& \ddots & \\ & & \dfrac{\sin(\omega_nt)}{\omega_n} \end{bmatrix} \\ \begin{bmatrix} -\omega_1 \sin(\omega_1t) & \\& \ddots & \\ & & -\omega_n\sin(\omega_n t) \end{bmatrix} & \begin{bmatrix} \cos(\omega_1t) & \\& \ddots & \\ & & \cos(\omega_n t) \end{bmatrix}\end{bmatrix}\in\mathbb{R}^{2n\times2n} \end{equation}

The eigenvalues of $\mathbf N$ are of course 1 (multiplicity $2n-1$) and $-1$ (multiplicity $1$). The eigenvalues of $\mathbf{S}(t)$, which is an exponential matrix, are the $n$ couples of the complex conjugates $(\exp(i\omega_jt),\overline{\exp(i\omega_jt)})$.

Now, we can define $\forall t>0$, $\mathbf{A}(t)=\mathbf N\mathbf S(t)$. We know that the product of the eigenvalues of $\mathbf{A}(t)$ is the product of those of $\mathbf{N}$ and $\mathbf S(t)$, i.e. $-1$.

I observe an interesting property but can't prove where it stems from:

- $1$ and $-1$ are eigenvalues of $\mathbf{A}(t)$ ($\forall t$);
- $1$ and $-1$ are $\color{red}{\text{not}}$ eigenvalues of $\mathbf{A}(t_2)\mathbf{A}(t)$ ($\forall t,t_2$, except maybe for specific values of $\mathbf P$ and $\omega_k$);
- $1$ and $-1$ are eigenvalues of $\mathbf{A}(t_3)\mathbf{A}(t_2)\mathbf{A}(t)$ ($\forall t,t_2,t_3$);
- $1$ and $-1$ are $\color{red}{\text{not}}$ eigenvalues of $\mathbf{A}(t_4)\mathbf{A}(t_3)\mathbf{A}(t_2)\mathbf{A}(t)$ ($\forall t,t_2,t_3,t_4$, except maybe for specific values of $\mathbf P$ and $\omega_k$);
- $\dots$

I managed to prove $1$ and $-1$ are eigenvalues of $\mathbf{A}(t)$ by considering $\mathbf{S}(t)\pm\operatorname{diag}(1,\dots,1,-1,\dots,1)$, calculating its kernel, and building the appropriate vectors (without having to calculate them *explicitly*).

Also, I understand that the product of the eigenvalues of $\mathbf{A}(t_2)\mathbf{A}(t)$ is 1, while that of $\mathbf{A}(t_3)\mathbf{A}(t_2)\mathbf{A}(t)$ is -1, but that does not prove anything.

**Questions**

1) Any suggestion to prove the framed observation would be very welcome: why are apparently 1 and -1 eigenvalues of $\prod_{i=1}^m A(t_i)$ if and only if $m$ is odd?

2) Also, I have the impression that there exists a powerful mathematical framework to study these matrices, but I can't figure out which one, as not being a mathematician; Lie algebra because $\mathbf S(t)$ is an exponential? Galois groups because the eigenvalues are complex conjugate? Zariski topology because @loup blanc mentioned it (see end of answer)?

A simple `Mathematica`

file to reproduce the results is available here. Just play with the arguments of `calculateEigenvals`

to change the dimension $n$ or/and the exponent $m$ (to prove: 1,-1 eigenvalues iff $m$ is odd).

Note that I have already asked the question on math.SX but have not got any answer.