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Let $A_n$ be the following matrix for $p\leq 1$: \begin{equation} A_n = \begin{bmatrix} 1 & 2^{-p} & 3^{-p} & \dots & n^{-p} \\ 2^{-p} & 1 & 2^{-p} & \dots & (n-1)^{-p} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n^{-p} & (n-1)^{-p} & (n-2)^{-p} & \dots & 1. \end{bmatrix} \end{equation}

How do I show whether this matrix is positive definite or not? Numerically, it seems to be positive definite. I've tried to show this by showing the Cholesky decomposition exists, but the expressions I get are confusing.

amWhy
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oswinso
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  • @WillJagy So I've got $A_{ij} = \int_0^1 x^{(i - j + 1)^p-1} dx$, but I'm not sure how to break this up even further given $p>0$ is arbitrary...... – oswinso Mar 16 '22 at 00:29

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Let $B_n(x)$ be the $n\times n$ matrix whose $(i,j)$-th entry is $x^{|i-j|}$. It is known that $B_n(x)$ is positive definite for $|x|<1$. Now the fact that $A_n$ is positive definite follows easily from $$A_n=\frac1{\Gamma(p)}\int_0^1(-\log x)^{p-1}B_n(x)\,dx.$$

metamorphy
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