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"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.

anderstood
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    Your statement seems to not be true when all $\omega_i$ are zero, and when $t = 0$, in which case $Q_1(0) = I$, and the kernel has dimension $2n$. Perhaps I'm misreading something... – John Hughes Oct 08 '14 at 00:10
  • @John No you're right, I've just edited to specify that $\omega_i>0$. – anderstood Oct 08 '14 at 00:12
  • Even so, you still get $Q_1(0) = I$ for $t = 0$, no? – John Hughes Oct 08 '14 at 00:25
  • @John you're still right... I've been working all day on the physical corresponding problem so I tend to forget all specifications, sorry. $t>0$. Edited. – anderstood Oct 08 '14 at 00:27
  • And you probably want the $\omega_i$ to be noncommensurable, or else $t = 2\pi u / \omega_0$, where $\omega_0$ is the smallest of the $\omega$s, in absolute value, and $u$ is the GCD of all the pairwise ratios $\omega_i/\omega_0$, will also give $Q_1(t) = I$. Perhaps it'd be easier just to say "for almost all $t$" :) – John Hughes Oct 08 '14 at 00:44
  • @John Haha now please stop finding mistakes in my question! :D Edited again! – anderstood Oct 08 '14 at 00:59

1 Answers1

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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.

anderstood
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John Hughes
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  • Thank you very much for your answer. I'll try it tomorrow. I'll also add an example of $\mathbf{Q}_1$ for $n=2$ to make it clearer. Again, you're note is correct, I meant $\mathbf{Q}_1^{-1}$. – anderstood Oct 08 '14 at 01:12
  • OK, this works. I edited your answer to improve it and accepted it. TY again. – anderstood Oct 08 '14 at 04:35
  • My pleasure. Sorry for making you do the editing work. Glad it worked out. – John Hughes Oct 08 '14 at 11:50
  • I made the question more complicated (in fact, I had forgotten a matrix $\mathbf{P}$ in $\mathbf{Q}_2$... which may have been fortunate), see http://math.stackexchange.com/questions/964373/proof-of-the-conjecture-that-the-kernel-is-of-dimension-2 if ever you want to have a look. There, I tried to extend your solution but did not manage to conclude due to the loss of the block-diagonal structure. – anderstood Oct 08 '14 at 21:52