Prove that the kernel is of dimension 2 "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>0,\quad \text{dim}\,\text{ker}\left(\mathbf{Q}_2\mathbf{Q}_1(t)-\mathbf{Q}_1(t)^{-1}\mathbf{Q}_2\right)\overbrace{=}^?\;2 
\end{equation}
where:


*

*$\mathbf{Q}_2$ is the identity matrix everywhere except in $(2n,2n)$:
\begin{equation}
\mathbf{Q}_2=\begin{bmatrix} 1 & & \\  & \ddots &   \\ & & 1 & \\ & & & -1  \end{bmatrix}\in\mathbb{R}^{2n\times2n}
\end{equation}

*$\mathbf{Q}_1(t)$ 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}
where (... sorry...):
\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}
The initial equation can therefore also be written as:
\begin{equation}
\text{ker}\left(\mathbf{Q}_2\mathbf{Q}_1(t)-\mathbf{Q}_1(t)^{-1}\mathbf{Q}_2\right)=\text{ker}\left(\mathbf{Q}_2\mathbf{Q}_1(t)\mathbf{Q}_2\mathbf{Q}_1(t)-\mathbf{1}_{2n}\right)
\end{equation}
So another way of solving the problem is to prove that 1 is an eigenvalue of $\mathbf{Q}_2\mathbf{Q}_1(t)\mathbf{Q}_2\mathbf{Q}_1(t)$ with a multiplicity of 2. But I'm not sure this helps...
Any clues would be greatly appreciated.
 A: OK. Here's a half-baked kind of answer. 
First, re-order your indices (i.e., perform a change of basis in which the change is just a shuffle): Group indices 1 and n+1 together, then 2 and n+2, and so on. Once you've done that, your matrix $Q_1$ is just a bunch of $2 \times 2$ matrices down the diagonal. $Q_2$ doesn't change at all.  I suspect that a typical $2 \times 2$ block $M$ of $Q_1$ has the property that for almost all $t$, $M - M^{-1}$ is nonsingular, so there are no nullvectors to be found there. So only the last $2 \times 2$ block is interesting. You might check this out; maybe I've misunderstood the shape of the matrices a bit. 
Note: Your "hint" section has a displayed formula for $Q_1$ that looks wrong; I think this is the formula for $Q_1^{-1}$. 
EDIT by the OP
Let's illustrate this solution with $n=2$: once the indices reordered,$Q_2$ writes:
\begin{equation}
\begin{bmatrix}\mathbf{M}_1 & 0 \\ 0 &  \mathbf{M}_2\end{bmatrix} \end{equation}
where 
\begin{equation}
\mathbf{M}_i = \begin{bmatrix} \cos(\omega_i t) & -\dfrac{1}{\omega_i}\sin(\omega_i t) \\ \omega_1\sin(\omega_i t) & \cos(\omega_i t)\end{bmatrix}
\end{equation}
When introduced in $Q_2Q_1-Q_1^{-1}Q_2$, the only block which is singular is the last one:
\begin{equation} \begin{bmatrix} 1 &0\\ 0 & -1\end{bmatrix}\mathbf{M}_n - \mathbf{M}_{n}^{-1}\begin{bmatrix} 1 &0\\ 0 & -1\end{bmatrix}
\end{equation}
which is... equal to $\mathbf{0}_2$!
So the conjecture was good for $t$ s.t. $Q_2$ is not singular.
