# Proving that the kernel of this matrix is of dimension 2

(Edit : see at the bottom of the question for an additional surprising possible hint.)

Using a computational software program, I found that the kernel of the following matrix is of dimension 2 when $n\geqslant 2$ but I haven't managed to prove it: $$\text{for almost all } t_1>0,\quad \text{dim}\,\text{ker}\left(\mathbf{Q}_2\mathbf{Q}_1(t_1)-\mathbf{Q}_1(t_2)^{-1}\mathbf{Q}_2\right)\overbrace{=}^?\;2$$ where $t_2$ is chosen (assuming it exists) such that $$\text{det}\left(\mathbf{Q}_2\mathbf{Q}_1(t_1)-\mathbf{Q}_1(t_2)^{-1}\mathbf{Q}_2\right)=0$$ where

• $n\geqslant 2$
• $\mathbf{Q}_2$ is the following matrix: $$\mathbf{Q}_2=\begin{bmatrix} \mathbf{I}_n & \mathbf{0}_n \\ \mathbf{0}_n & \mathbf{P}^{-1}\begin{bmatrix}1 & && \\ & \ddots && \\ & & 1& \\ &&& -1 \end{bmatrix}\mathbf{P} \end{bmatrix}\in\mathbb{R}^{2n\times2n}$$ where $\mathbf{P}\in\mathbb{R}^{n\times n}$ is any invertible matrix.

• $\mathbf{Q}_1(t)$ is defined by:

$$\forall t>0,\quad\mathbf{Q}_1(t)=\begin{bmatrix}\textbf{cos}(\boldsymbol \Omega t) & \boldsymbol \Omega^{-1}\,\textbf{sin}(\boldsymbol \Omega t) \\ -\boldsymbol \Omega\,\textbf{sin}(\boldsymbol \Omega t) & \textbf{cos}(\boldsymbol \Omega t)\end{bmatrix}\in\mathbb{R}^{2n\times2n}$$ and: $$\boldsymbol\Omega=\begin{bmatrix} \omega_1 & & 0\\ & \ddots & \\ 0 & & \omega_n \end{bmatrix}\in\mathbb{R}^{n\times n},\quad \forall i\in\lbrace 1,\dots, n\rbrace, \omega_i>0$$

and the four blocks are diagonal, for example: $$\mathbf{cos}(\boldsymbol\Omega t)=\begin{bmatrix} \cos(\omega_1t) & &0 \\ & \ddots & \\ 0 & & \cos(\omega_n t) \end{bmatrix}\in\mathbb{R}^{n\times n}$$

Few 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: $$\mathbf{Q}_1(t)^{-1}=\begin{bmatrix}\textbf{cos}(\boldsymbol \Omega t) & -\boldsymbol \Omega^{-1}\,\textbf{sin}(\boldsymbol \Omega t) \\ \boldsymbol \Omega\,\textbf{sin}(\boldsymbol \Omega t) & \textbf{cos}(\boldsymbol \Omega t)\end{bmatrix}$$

Also, $\forall s,t,\ \mathbf{Q}_1(s+t)=\mathbf{Q}_1(s)\mathbf{Q}_1(t)=\mathbf{Q}_1(t)\mathbf{Q_1}(s)$.

Edit Additional possible hint: I've found that for $n=2$ (maybe it's more general), the follow results stands, if $A=\mathbf{Q}_2\mathbf{Q}_1(t_1)-\mathbf{Q}_1(t_2)^{-1}\mathbf{Q}_2$ and $\phi(x)=\det(A-xI)$: $$\phi'(0)^2=-\phi(0)\det(B)$$ where $B$ is the $A$ matrix if $\mathbf{P}=\mathbf{I}$... ! I have no clue where this comes from, maybe it's obvious for someone (maybe related to Lie algebras?).

This would conclude the proof as $\phi(0)=\det(A(t_1,t_2))=0$ for the appropriate $t_1,t_2$, so that $0$ would be a double eigenvalue of $A$.

• Sorry, this is sort of complicated so maybe I'm misreading it, but when $n=1$ it looks like $Q_2Q_1$ and $Q_1^{-1}Q_2$ both always have the same determinant but the kernel is almost never two dimensional. Can you tell me what I'm missing? Nov 2, 2014 at 10:17
• @GabrielC.Drummond-Cole $\forall n$, $\det(Q1)=1$ and $\det(Q2)=-1$ so $\det(Q_2Q_1)=\det(Q_1^{-1}Q2)=-1$ is always true. Nov 2, 2014 at 21:18
• @GabrielC.Drummond-Cole [failed to edit the previous comment within 5min...] It's quite complicated that's true, thank you for reading! When $n=1$, $Q_1(t)=\begin{bmatrix}\cos(\omega t)& \sin(\omega t)/\omega \\ -\omega \sin(\omega t) & \cos(\omega t)\end{bmatrix}$ and $Q_2=\operatorname{diag}(1,-1)$ and the computation yields, for this particular case, $Q_2\,Q_1 - Q_1^{-1}\,Q_2=0_{2}$, so the kernel is always of dimension 2. Nov 2, 2014 at 21:24
• But in the setup you use $Q_2 Q_1(t_1)- Q_1(t_2)^{-1}Q_2$ where $t_1$ and $t_2$ can be different as long as the determinant of that difference is zero. But for $t_1\ne t_2$ it looks like the difference of matrices is not identically zero. Nov 3, 2014 at 6:24
• @GabrielC.Drummond-Cole When $n=1$, $\det(Q_2Q_1(t_1)-Q_1(t_2)^{-1}Q_2)=2(\cos((t_1-t_2)\omega)-1)$ so you're right, $t_1=t_2$ may not be the only solution. In fact, I had only studied $n\geqslant 2$, I'll edit the question accordingly, TY. I'll also add another interesting property for $Q_1$. Nov 3, 2014 at 14:38

$Q_1(t)$ can be conjugated with by $\text{diag}(\omega_1,\dots,\omega_n,1,\dots, 1)$ which transforms it into a rotation matrix without affecting $Q_2$.
It is bit more tricky, but $Q_2$ can be conjugated with a diagonal matrix $\text{diag}(d_1,\dots,d_n,d_1,\dots,d_n)$ which transforms it into an orientation-reversing orthogonal matrix without affecting $Q_1(t)$. More precisely, this seems to be true for an open set of $P$ dense in the matrix vector space.
The conjugations do not depend on $t$, so at the end of the day, $Q_2Q_1(t_1)Q_2Q_1(t_1)$ is similar to an orientation-preserving orthogonal matrix, whose eigenvalues lie on the unit circle and can be $1$, $-1$ or appear by pair of complex conjugates. The product of the eigenvalues is $1$, so $-1$ has to appear an even number of times. The matrix $Q_2Q_1(t_1)Q_2Q_1(t_1)$ is a $2n\times 2n$ matrix, so $1$ also has to appear with an even multiplicity.
So if 1 is eigenvalue of $Q_2Q_1(t_1)Q_2Q_1(t_1)$, the corresponding eigenspace is of dimension greater than 2. I do not have the mathematical background to prove it (help is welcome), but I guess one concludes by proving that the set of eigenvalues with exactly two eigenvalues equal to 1 is dense in the set of eigenvalues with at least two eigenvalues equal to 1. But it is still possible to find some values of $P$ and $\omega_i$ such that 1 has a multiplicity of 4, or even 6, etc.
The last step is to remark that the eigenspace for the eigenvalue 1 of $Q_2Q_1(t_1)Q_2Q_1(t_1)$ is the eigenspace for the eigenvalue 0 of $Q_2Q_1(t_1)-(Q_2Q_1(t_2))^{-1}$ or of $Q_2Q_1(t_1)-Q_1(-t_2)Q_2$.