Pursuing my research, I am now looking for a proof of an extension of the problem proposed here and answered. It's an extension in the sense that I'm now considering two different $t_1$ and $t_2$. The "conjecture" still stands though.
"Experimentally", I found that the kernel (null space) of the following matrix is of dimension 2. I'd like to prove it, but haven't managed yet: \begin{equation} \text{for almost all } t_1>0,\quad \text{dim}\,\text{ker}\left(\mathbf{Q}_2\mathbf{Q}_1(t_1)-\mathbf{Q}_1(t_2)^{-1}\mathbf{Q}_2\right)\overbrace{=}^?\;2 \end{equation} where $t_2$ is chosen such that $$\text{det}\left(\mathbf{Q}_2\mathbf{Q}_1(t_1)-\mathbf{Q}_1(t_2)^{-1}\mathbf{Q}_2\right)=0$$ We assume that such a $t_2$ exist. where:
$\mathbf{Q}_2$ is the following matrix: \begin{equation} \mathbf{Q}_2=\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.
$\mathbf{Q}_1(t)$ is defined by:
\begin{equation} \forall t>0,\quad\mathbf{Q}_1(t)=\begin{bmatrix}\textbf{cos}(\boldsymbol \Omega t) & \boldsymbol \Omega^{-1}\,\textbf{sin}(\boldsymbol \Omega t) \\ -\boldsymbol \Omega\,\textbf{sin}(\boldsymbol \Omega t) & \textbf{cos}(\boldsymbol \Omega t)\end{bmatrix}\in\mathbb{R}^{2n\times2n} \end{equation} and: \begin{equation} \boldsymbol\Omega=\begin{bmatrix} \omega_1 & & \\ & \ddots & \\ & & \omega_n \end{bmatrix}\in\mathbb{R}^{n\times n},\quad \forall i\in\lbrace 1,\dots, n\rbrace, \omega_i>0 \end{equation}
and the four blocks are diagonal, for example: \begin{equation} \mathbf{cos}(\boldsymbol\Omega t)=\begin{bmatrix} \cos(\omega_1t) & & \\ & \ddots & \\ & & \cos(\omega_n t) \end{bmatrix}\in\mathbb{R}^{n\times n} \end{equation}
Interesting properties of $\mathbf{Q}_1$ and $\mathbf{Q}_2$:
Obviously, $\mathbf{Q}_2$ is invertible and $\mathbf{Q}_2=\mathbf{Q}_2^{-1}$.
Also, $\det(\mathbf{Q}_1)=1$ ($\omega_i>0$ and for proper $t>0$) and: \begin{equation} \mathbf{Q}_1(t)^{-1}=\begin{bmatrix}\textbf{cos}(\boldsymbol \Omega t) & -\boldsymbol \Omega^{-1}\,\textbf{sin}(\boldsymbol \Omega t) \\ \boldsymbol \Omega\,\textbf{sin}(\boldsymbol \Omega t) & \textbf{cos}(\boldsymbol \Omega t)\end{bmatrix} \end{equation}
Any clues would be greatly appreciated.
Why I'm not able to extend loup blanc's solution to the present problem with two different $t_1$ and $t_2$.
Let's define $A:=\mathbf{Q}_2\mathbf{Q}_1(t_1)-\mathbf{Q}_1(t_2)^{-1}\mathbf{Q}_2$ and $\phi(x):=\det(A-xI)$. The problem is to prove that $0$ is a zero of $\phi$ with multiplicity 2. By definition of $t_2$, $\phi(0)=\det(A)=0$. So I now have to prove that $\phi'(0)=0$ (and then than $\dim(\ker(A))\leqslant 2$).
$\phi'(0)=-\operatorname{tr}(\operatorname{adjoint}(A))$ from Jacobi's formula.
The calculation of $A$ gives: $$A=\begin{bmatrix} c_1-c_2 & \Omega^{-1}s_1+\Omega^{-1}s_2 P^{-1}DP \\ -P^{-1}DP\Omega s_1-\Omega s_2 & P^{-1}DPc_1-c_2P^{-1}DP\end{bmatrix}$$ where $c_i=\textbf{cos}(\Omega t_i)$, $s_i=\textbf{sin}(\Omega t_i)$ and $D=\operatorname{diag}(1,\dots,1,-1)$.
This time, the determinant of $A$ is harder to calculate because the upper-left block is non-zero. So I'm stuck with proving that $\operatorname{tr}(\operatorname{adjoint}(A))=0$. I also tried to apply things such as $A\operatorname{adjoint}(A)=\det(A)I$ but there is no mystery: I have to use the properties of $A$, and that's what I'm not managing to do.
EDIT I still did not manage to prove that for $t_1,t_2$ such that $\det(Q_2Q_1(t_2)Q_2Q_1(t_1)-I_{2n})=0$, $\dim(\ker(Q_2Q_1(t_2)Q_2Q_1(t_1)-I_{2n})=2$. I though of proving it by induction but I am still stuck with the inductive step. Do you think this is a good idea? Would you have some relevant references to help me?
