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Let $\boldsymbol{R}(u,v);~u,v\in\mathbb{R}$ be a Hermitian matrix (of Hermitian functions) with entries \begin{equation} r_{ij}(u,v) = 1 + Ae^{-2\pi i \phi_{ij}(ul_0 + vm_0)}; A\in\mathbb{R},l_0\in\mathbb{R},m_0\in\mathbb{R},\phi_{ij}\in\mathbb{Z}. \end{equation} Furthermore let, $0<A<1$, $l_0 \neq 0, m_0 \neq 0$, $\phi_{i,i}=0$ (diagonal), $\phi_{ij} \neq 0$ (non-diagonal), $\phi_{ij} > 0;~\forall j>i$, $\phi_{ij} = -\phi_{ji}$ and gcd($\{\phi_{ij}\}_{j>i}$) $=1$. If $\boldsymbol{G}(u,v)=\lambda(u,v)\mathbf{x}(u,v)\mathbf{x}^{H}(u,v)$, where $\lambda(u,v)$ is the largest eigenvalue of $\boldsymbol{R}(u,v)$, $\mathbf{x}(u,v)$ is its associated eigenvector (normalized) (of $\lambda(u,v)$) and $()^H$ is the Hermitian transpose then prove that the entries $g_{ij}(u,v)$ of $\boldsymbol{G}(u,v)$ are Hermitian functions that are periodic in the $u$ and $v$ direction with periods respectively equal to $\frac{1}{|l_0|}$ and $\frac{1}{|m_0|}$.

I arrived at this problem by studying ghost sources (radio interferometry) and alternating least squares calibration. I have a few final comments (based on experimental observation). The $\boldsymbol{R}(u,v)$ matrix seems to be rank 2 (how to prove this though?). Can an analytic expression for $\lambda(u,v)$ be derived (for any dimension of $\boldsymbol{R}(u,v)$) and can it aid in the proof? Also it seems that $\lambda(u,v)$ is always positive (how to prove this)? Once we have $\lambda(u,v)$? How do we derive the periodicity of $\mathbf{x}(u,v)$? Can anyone suggest an approach to derive this proof?

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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 $c\in\mathbb{Q}$? –  trienko Aug 23 '13 at 8:04
    
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? –  trienko Aug 23 '13 at 8:13
    
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? –  trienko Aug 23 '13 at 8:29
    
Please see my comments below regarding the rank of $\boldsymbol{R}$. –  trienko Aug 23 '13 at 10:18
    
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 $\lambda = 0$. –  trienko Aug 26 '13 at 13:41
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1 Answer

It's certainly not true that $R(u,v)$ has rank $2$. Consider the $3 \times 3$ case. One of the terms when $\det(R(u,v))$ is expanded out is $A^3 \exp(-2\pi i (\phi_{12} + \phi_{23} + \phi_{31}) (u l_0 + v m_0)$.
If $\phi_{12}, \phi_{23}, \phi_{31}$ are chosen properly, this can't be cancelled out by any of the other terms, and (for almost all $(u,l_0,v,m_0)$) the determinant is nonzero.

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Thank you very much for answering (I apologize for the wrong question), please (it would help me tremendously) give a numerical counter example. The weird thing is the matrices I am working with are all rank two. Meaning that there must be another property I am not seeing. –  trienko Aug 23 '13 at 7:47
    
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? –  trienko Aug 23 '13 at 10:16
    
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 $\boldsymbol{A}$?. –  trienko Aug 23 '13 at 12:03
    
Of course $\boldsymbol{\Phi}$ is the $\phi$ only part of $\boldsymbol{R}$. –  trienko Aug 23 '13 at 12:04
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