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Nov 16, 2010 at 8:53 comment added Suvrit @Terry: thanks for the nice argument for an upper bound (and @Tracy, good point of invoking Cauchy's interlacing theorem to argue about the upper bound on the eigenvalues). As for the Fourier connection, indeed this matrix is generated by the posdef function $f(x)=1/(1+|x|)$. In fact, perhaps even more can be proved, because actually the said matrix is infinitely divisible.
Nov 15, 2010 at 4:32 comment added Tracy Hall It's even easier than that: once you've found a pair of indices giving distance order $1/n^2$, they define a $2 \times 2$ matrix with an eigenvalue of order $1/n^2$, and then by interlacing of eigenvalues for Hermitian (or real symmetric) matrices the entire matrix has an eigenvalue no larger than that.
Nov 14, 2010 at 21:29 history undeleted Terry Tao
Nov 14, 2010 at 21:29 history edited Terry Tao CC BY-SA 2.5
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Nov 14, 2010 at 21:17 history deleted Terry Tao
Nov 14, 2010 at 21:12 history answered Terry Tao CC BY-SA 2.5