Timeline for The influence of eigendecomposition on the periodicity of a (rank 2) Hermitian matrix (of functions)
Current License: CC BY-SA 3.0
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| Aug 26, 2013 at 13:41 | comment | added | trienko | Since the rank of $\boldsymbol{R}$ is two, $|\boldsymbol{R}-I\lambda|=0$ can be written as $(\lambda^2-\textrm{tr}(\boldsymbol{R})\lambda+\sum_{i<j} \begin{vmatrix} r_{ii} & r_{ij} \\ r_{ji} & r_{jj} \end{vmatrix}\lambda)\lambda^{n-2}=0$. Solving for $\lambda$ produces $\lambda(u,v) = \frac{n(1+A)}{2} \pm$ $\frac{1}{2}\sqrt{[n^2-4 {n \choose 2}][1+A]^2+4\sum_{i>j}1+A^2+2A\cos(2\pi\phi_{ij}(u l_0+v m_0))}$ or $\lambda = 0$. | |
| Aug 23, 2013 at 10:18 | comment | added | trienko | Please see my comments below regarding the rank of $\boldsymbol{R}$. | |
| Aug 23, 2013 at 8:29 | comment | added | trienko | For the Hermitian part. The element $r_{ij}(-u,-v) = r_{ij}^*=r_{ji}$ (relation (1)) since $\boldsymbol{R}$ is a Hermitian matrix. The element $r_{ij}$ is linked to $g_{ij}$ and $r_{ji}$ is linked to $g_{ji}$. By construction $g_{ij}^*=g_{ji}$ (relation (2)) ($\lambda$ is real since $\boldsymbol{R}$ is Hermitian). Does relation (1) and (2) in some way imply $g_{ij}(-u,-v) = g_{ij}^*=g_{ji}$ as required? | |
| Aug 23, 2013 at 8:13 | comment | added | trienko | If $\lambda$ is periodic? Does it imply that $\mathbf{x}$ is also periodic, by the same reasoning as it is the solution of the equation $(\boldsymbol{R}-\boldsymbol{I}\lambda)\cdot\mathbf{x}=0$. Again how do I show that this period is unique? | |
| Aug 23, 2013 at 8:04 | comment | added | trienko | This comment is about the periodicity part of the proof. Since the eigenvalue of $\boldsymbol{R}$ is the solution of $|\boldsymbol{R} - \boldsymbol{I}\lambda|=0$, and $\boldsymbol{R}$ will have a period of (equal to the same matrix periodically [with period $\frac{1}{|l_0|}$ and $\frac{1}{|m_0|}$ due to the fact that gcd($\{\phi_{ij}\}_{j>i}$)=1] does it imply that $\lambda(u,v)$ has the same period as $\boldsymbol{R}$. The only problem is to show that this is the only possible period? How do I show that it can not be $c\frac{1}{|l_0|}$ and $c\frac{1}{|m_0|}$ where $c\in\mathbb{Q}$? | |
| Aug 22, 2013 at 15:06 | answer | added | Robert Israel | timeline score: 1 | |
| Aug 22, 2013 at 14:54 | review | First posts | |||
| Aug 22, 2013 at 15:56 | |||||
| Aug 22, 2013 at 14:38 | history | asked | trienko | CC BY-SA 3.0 |