MathOverflow Asked by Rajesh D on November 3, 2021
Let $g:mathbb{T}tomathbb{R}$ and is given as $$g(x) = sumlimits_{etainmathbb{Z}}frac{1}{1+gamma eta^2}cos{2pieta x}$$
Consider the matrix $$G_{gamma} = [g(x_i-x_j)]_{1le i,jle n}$$
where $x_1,x_2,…x_n in (0,1)$ and pairwise distinct.
Due to Bochner’s theorem, $g(x)$ is a positive semi definite function and hence $G_{gamma}$ is a psd matrix.
Let $lambda_{min}(G_{gamma})$ denote the smallest eigenvalue of $G_{gamma}$.
I want to show that $G_{gamma}$ is infact positive definite and as $gammatoinfty$ $$lambda_{min}(G_{gamma}) = Theta(frac{1}{gamma})$$
As a sanity check, I have verified this using numerical computations on some examples.
Motivation : I want to come up with a similar looking formula in a generic dimension, that is for $mathbb{T}^m$. If I am able to prove this for $m=1$ dimension, then I will understand the mechanics of it so that might help me to come up with a $g:mathbb{T}^m to mathbb{R}$ such that $lambda_{min}(G_{gamma}) = Theta(frac{1}{gamma})$
Given any $n$ distinct points ${x_i/x_iin(0,1)}$ which are pairwise distinct. For any $c_i,i = 1,2,3,...n$, and not all zeros.
Using the given expression for $g(x)$ we can deduce that
$$sum_{i=1}^nsum_{j=1}^nc_ic_jg(x_i-x_j) = sum_{etainmathbb{Z}} left(frac{1}{1+gammaeta^2} left|sum_{i=1}^n c_i e^{2pi i eta x_i}right|^2 right)> 0$$
as $sum_{i=1}^n c_i e^{2pi i eta x_i}$ does not vanish simultaneously for all $etainmathbb{Z}$ and $frac{1}{1+gammaeta^2}>0forall eta inmathbb{Z}$.
Hence the matrix $G_{gamma}$ is positive definite.
Estimate on $lambda_{min}(G_{gamma})$ as $gammatoinfty$
Let $c = [c_1,c_2,...c_n]$ be such that $|c|_2 = 1$. Then $$c^TG_{gamma}c = sum_{etainmathbb{Z}} left(frac{1}{1+gammaeta^2} left|sum_{i=1}^n c_i e^{2pi i eta x_i}right|^2 right)> 0 .$$ As $|sum_{i=1}^n c_i e^{2pi i eta x_i}|^2 le n$ and we already know $sum_{i=1}^n c_i e^{2pi i eta x_i}$ does not vanish simultaneously for all $etainmathbb{Z}$, there exists constants $K_1$ and $K_2$ such that $$frac{K_1}{gamma} le c^TG_{gamma}c le frac{K_2}{gamma} mbox{ }forall cinmathbb{R}^m setminus{0}^m mbox{ and } |c|_2 = 1$$
Let $e(gamma)$ be the smallest eigenvector and as $|e(gamma)|_2 = 1$ $$lambda_{min}(G_{gamma}) = lambda_{min}(G_{gamma})e(gamma)^Te(gamma) = e(gamma)^TG_{gamma}e(gamma) = Theta(frac{1}{gamma})$$ So
$$lambda_{min}(G_{gamma}) = Theta(frac{1}{gamma})$$
Answered by Rajesh D on November 3, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP