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.
Best Answer
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 $$A_n=\frac1{\Gamma(p)}\int_0^1(-\log x)^{p-1}B_n(x)\,dx.$$