0
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(1 1/2 1/3 1/4 ... 1/(n-1) 1/n) (1/2 1/3 1/4 1/5 ... 1/n 1/(n+1)) (1/3 1/4 1/5 1/6 ... 1/(n+1) 1/(n+2)) (1/4 1/5 1/6 1/7 ... 1/(n+2) 1/(n+3)) (...................................) (1/(n-3) 1/(n-2) 1/(n-1) 1/n ... 1/(2n-5) 1/(2n-4)) (1/(n-2) 1/(n-1) 1/n 1/(n+1) ... 1/(2n-4) 1/(2n-3)) (1/(n-1) 1/n 1/(n+1) 1/(n+2) ... 1/(2n-3) 1/(2n-2)) (1/n 1/(n+1) 1/(n+2) 1/(n+3) ... 1/(2n-2) 1/(2n-1))

I need to prove that it is positive. But I can't complete it.

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  • $\begingroup$ sorry for antilatex $\endgroup$ – Andrey Komisarov Apr 22 at 21:43
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    $\begingroup$ Is that the Hilbert matrix? en.wikipedia.org/wiki/Hilbert_matrix $\endgroup$ – Lord Shark the Unknown Apr 22 at 21:44
  • $\begingroup$ @LordSharktheUnknown yes $\endgroup$ – Andrey Komisarov Apr 22 at 21:45
  • $\begingroup$ So, if is a natural form of matrix, I think it is enough. $\endgroup$ – Andrey Komisarov Apr 22 at 21:57
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    $\begingroup$ An inner product with this matrix is $\int_0^1p(t)q(t)\,dt$, where $p$ and $q$ are degree $n$ or less polynomials with real coefficients. $\endgroup$ – amd Apr 22 at 22:01

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