Proof of the conjecture that the kernel is of dimension 2

Solution 1:

Let $$\begin{split}A&=Q_2Q_1-{Q_1}^{-1}Q_2 \\ &=\begin{pmatrix}0&\Omega ^{-1}\sin(\Omega t)(I+P^{-1}DP)\\-(P^{-1}DP+I)\Omega\sin(\Omega t)&P^{-1}DP\cos(\Omega t)-\cos(\Omega t)P^{-1}DP\end{pmatrix}\end{split}$$ Then $$\begin{split}\phi(x)&=\pm\det(A-xI)\\ &=\det\bigg[-(P^{-1}DP+I)\sin(\Omega t)^2(I+P^{-1}DP) \\ &+x(P^{-1}DP\cos(\Omega t)-\cos(\Omega t)P^{-1}DP)-x^2I\bigg] \\ &=\det(-U+xV-x^2I) \end{split}$$

$\phi(0)=\det(-U)=0$ (because $I+P^{-1}DP$ is not invertible). Now $$\begin{split}\phi'(0)&=-\operatorname{tr}(\operatorname{adjoint}(U)V) \\ &=-\operatorname{tr}\bigg((\operatorname{adjoint}((P^{-1}DP+I)\sin(\Omega t)^2(I+P^{-1}DP)) \\ &\times (P^{-1}DP\cos(\Omega t)-\cos(\Omega t)P^{-1}DP)\bigg) \end{split}$$

After a change of basis , we may assume $P^{-1}DP=\operatorname{diag}(1,\cdots,1,-1)$, $$S=\operatorname{adjoint}(\sin(\Omega t)^{2})=[s_{i,j}],C=\cos(\Omega t)=[c_{i,j}]$$

Then $\operatorname{adjoint}(U)=2^{2n-2}s_{n,n}E_{n,n}$ and consequently (if $V=[v_{i,j}]$), $$\operatorname{tr}(\operatorname{adjoint}(U)V)=2^{2n-2}s_{n,n}v_{n,n}.$$ Yet $v_{n,n}=-c_{n,n}+c_{n,n}=0$. Thus $\phi'(0)=0$ and $x=0$ is (at least) a double eigenvalue.

We consider the generic case (with respect to $\Omega,P,t$ and not only $t$). To see that, generically, $\dim(\ker(A))$ is exactly $2$, it suffices to give an instance, in dimension $n$, where $A$ is diagonalizable and $0$ is an eigenvalue with exactly the multiplicity $2$. It is not difficult (take $P=I_n,...$) - For a rigorous reasoning, you must use the Zariski's topology-.