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$$A=\left[\frac{1}{i+j}\right]=\left(\begin{matrix}\frac{1}{1+1}&\frac{1}{1+2}&\cdots&\frac{1}{1+n}&\\\frac{1}{2+1}&\frac{1}{2+2}&\cdots&\frac{1}{2+n}\\\vdots&\vdots&\ddots&\vdots\\\frac{1}{n+1}&\frac{1}{n+2}&\cdots&\frac{1}{n+n}\end{matrix}\right)$$

My textbook exercise says yes, but I can't prove it.

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1 Answer 1

Yes, this is similar to a Hilbert matrix, and a clever trick does it: Note that the $(j,k)$ element is $a_{jk}=\int_0^1 x^{j+k-1}\,dx$, and so $$\sum_{j,k} a_{jk}\bar u_ju_k=\int_0^1\Bigl|\sum_j u_jx^j\Bigr|^2x\,dx\ge0.$$

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