trienko
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Is Rellich's function valued theorem valid for a rank defficient function valued matrix?
Can some of the eigenvalues be zero? So the rank is never full rank?
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Is Rellich's function valued theorem valid for a rank defficient function valued matrix?
Does Rellich's theorem guarantee global analyticity? (or is it only local)
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Is Rellich's function valued theorem valid for a rank defficient function valued matrix?
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The influence of eigendecomposition on the periodicity of a (rank 2) Hermitian matrix (of functions)
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 $\lambda = 0$.
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The influence of eigendecomposition on the periodicity of a (rank 2) Hermitian matrix (of functions)
Of course $\boldsymbol{\Phi}$ is the $\phi$ only part of $\boldsymbol{R}$.
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The influence of eigendecomposition on the periodicity of a (rank 2) Hermitian matrix (of functions)
Let $\mathcal{X}$ denote the set consisting of all square sub-matrices of $\boldsymbol{\Phi}$ (including $\boldsymbol{\Phi}$ itself) [obtained by deleting the same rows and columns <i.e. row 1 and column 1 and row 2 and column 2> of $\boldsymbol{\Phi}$]. Let $|\boldsymbol{A}|_{D} = \sum_{i=1}^n a_{i i+1}$ (where $n$ is the dimension of $\boldsymbol{A}$). Is a sufficient condition for $\boldsymbol{R}$ to be rank 2 that $\forall\boldsymbol{A}\in\mathcal{X}$ $|\boldsymbol{A}|_D = a_{1n}$, where $n$ is the dimension of $\boldsymbol{A}$?.
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The influence of eigendecomposition on the periodicity of a (rank 2) Hermitian matrix (of functions)
Please see my comments below regarding the rank of $\boldsymbol{R}$.
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The influence of eigendecomposition on the periodicity of a (rank 2) Hermitian matrix (of functions)
I have checked your equation and as you said there is another restriction needed to be rank 2 (for 3 dimensional case). The equation $\phi_{13} - \phi_{12} - \phi_{23} = 0$. What is the general formula to make $\boldsymbol{R}$ rank 2 for any dimension? Is the original matrix positive semi-definite (so there is a largest eigenvalue)? Is the original proposition still true even though the matrix as stated above is not rank 2?
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The influence of eigendecomposition on the periodicity of a (rank 2) Hermitian matrix (of functions)
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?
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The influence of eigendecomposition on the periodicity of a (rank 2) Hermitian matrix (of functions)
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?
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The influence of eigendecomposition on the periodicity of a (rank 2) Hermitian matrix (of functions)
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 $c\in\mathbb{Q}$?