Is this matrix positive semi-definite?

Question

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

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

Harald Hanche-Olsen · Answer 1 · 2015-11-09T16:42:06.480

4

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 $$\sum_{j,k} a_{jk}\bar u_ju_k=\int_0^1\Bigl|\sum_j u_jx^j\Bigr|^2x^{-1}\,dx\ge0.$$

edited Nov 09 '15 at 16:42

answered Oct 17 '12 at 16:43

Harald Hanche-Olsen

31,960

Please check this proof http://math.stackexchange.com/questions/1515638/sum-limits-j-k-a-jk-bar-u-ju-k-mathop-limits-int-01-sum/1515997?noredirect=1#comment3094802_1515997 By this proof, your proof is'nt true. – H.... Nov 08 '15 at 21:19
1

@H.S Right, but that was just a simple typo. Fixed now. – Harald Hanche-Olsen Nov 09 '15 at 16:42
check this...http://math.stackexchange.com/questions/1515954/can-we-prove-that-all-eigenvalues-of-a-are-positive/1516004#1516004 – Black-horse Nov 09 '15 at 16:47
@Harald Hanche-Olsen - Thanks. – H.... Nov 09 '15 at 17:06
@soumitra Yes, that is another, perfectly valid, way of doing it. – Harald Hanche-Olsen Nov 09 '15 at 17:09

Is this matrix positive semi-definite?

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