6
$\begingroup$

(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: \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 (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: \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 & & 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 \end{equation}

and the four blocks are diagonal, for example: \begin{equation} \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} \end{equation}


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: \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}

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

(Note: I've already asked this question here but did not get any answer despite a bounty.)


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 $A$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Gabriel C. Drummond-Cole Nov 2 '14 at 10:17
  • $\begingroup$ @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. $\endgroup$ – anderstood Nov 2 '14 at 21:18
  • $\begingroup$ @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. $\endgroup$ – anderstood Nov 2 '14 at 21:24
  • $\begingroup$ 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. $\endgroup$ – Gabriel C. Drummond-Cole Nov 3 '14 at 6:24
  • $\begingroup$ @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 $Q_1$. $\endgroup$ – anderstood Nov 3 '14 at 14:38
2
$\begingroup$

Thanks to @Terry Tao's answer, I can now give a solution to the question. I am open to comments or precisions.

$Q_1(t)$ 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 $Q_2Q_1(t_1)-Q_1(-t_2)Q_2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.