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The $n\times n$ Hilbert matrix $H$ is defined as

$H_{ij} = \frac{1}{i+j-1}, \qquad 1\leq i,j\leq n$

What is known about the singular values $\sigma_1\geq\ldots\geq \sigma_n$ of $H$?

For example, it is known that the matrix is very ill-conditioned, i.e., $\sigma_1/\sigma_n = \mathcal{O}((1+\sqrt{2})^{4n}/\sqrt{n})$ [1].

But are there estimates for $\sigma_k$ for $2\leq k\leq n-1$? Or is there a bound on $1\leq k\leq n$ such that $\sigma_k > \epsilon$ for some $\epsilon>0$? I'm interested in this problem because I would like to know the numerical rank of $H$.

[1] J. Todd, "The condition of the finite segments of the Hilbert matrix." Contributions to the solution of systems of linear equations and the determination of eigenvalues, 39 (1954), pp. 109-116.

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You probably meant $\sigma_k > \epsilon$, no? (also, note that since the Hilbert matrix is symmetric positive definite, its singular values are the same as its eigenvalues. – Suvrit Jul 18 '13 at 17:45
Thanks, corrected. Yes, I agree that the singular values of H are the same as the eigenvalues of H (after ordering). – alext87 Jul 18 '13 at 17:54

1 Answer 1

While not a full characterization, the following result on $\sigma_n$ shows that $\sigma_k > \epsilon_n$, since $\sigma_n \ge \epsilon_n$.

Following my answer on this MO question here, we see that

\begin{equation*} \sigma_n \ge \frac{\det(H)}{\left[m + s/\sqrt{n-1}\right]^{n-1}} =: \epsilon_n, \end{equation*} where \begin{eqnarray*} m &=& \tfrac{\text{trace} H}{n} = \tfrac{1}{n}\left(\tfrac{H_{n-1/2}}{2} + \log 2\right),\\ s &=& \sqrt{\tfrac{\text{trace}H^2}{n} - m^2}, \end{eqnarray*} where $H_n$ is the nth Harmonic number).

Since it is known that $\det(H) = \frac{c_n^4}{c_{2n}}$, where $c_n := \prod_{i=1}^{n-1}i!$, the above bound on $\sigma_n$ is fully determined (I'm leaving an analytic / "closed-form" evaluation of $\text{trace}H^2$ as an exercise).

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Notice that $tr(H^2) = \sum_{ij} h_{ij}^2$, and curiously a week or so ago, other aspects of the matrix $[h_{ij}^2]$ were discussed at… – Suvrit Jul 19 '13 at 0:20
This is a nice lower bound on the smallest singular value, and I enjoyed the post on $H^2$. Thank you very much! Your answer shows when $H$ is numerically rank deficient, but not the numerical rank of H. Still this is a good step in that direction. – alext87 Jul 25 '13 at 8:22

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