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

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

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

For example, it is known that the matrix is very ill-conditioned, i.e., [1]

$$\dfrac{\sigma_1}{\sigma_n} = \mathcal O \left( \frac{1+\sqrt{2})^{4n}}{\sqrt{n}} \right)$$

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 am 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.

  • $\begingroup$ 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. $\endgroup$
    – Suvrit
    Jul 18, 2013 at 17:45
  • $\begingroup$ Thanks, corrected. Yes, I agree that the singular values of H are the same as the eigenvalues of H (after ordering). $\endgroup$
    – alext87
    Jul 18, 2013 at 17:54

2 Answers 2


I came back to this a few months ago and I can now answer my own question. I hope it is appropriate to answer my own question given the length of time.

Bernhard Beckermann and I just submitted a paper [1] that shows that if $AX-XB = F$ with $A$ and $B$ normal matrices, then the singular values of $X$ can be bounded as $$ \sigma_{1+\nu k}(X) \leq Z_k(\sigma(A),\sigma(B))\|X\|_2, \qquad \nu = {\rm rank}(F), $$ where $\sigma(A)$ and $\sigma(B)$ are the spectra of $A$ and $B$, respectively, and $Z_k(E,F)$ is the so-called Zolotarev number. More precisely, $$ Z_k(E,F) = \inf_{r\in\mathcal{R}_{k,k}} \frac{\sup_{z\in E}|r(z)|}{\inf_{z\in F}|r(z)|}, $$ where $\mathcal{R}_{k,k}$ is the space of degree $(k,k)$ rational functions. This is called the third Zolotarev problem.

In the case of the Hilbert matrix, let $A = {\rm diag}(1/2,3/2,\ldots,n-1/2)$ and $B = -A$ then $$ AH - HB = \begin{bmatrix}1\\\vdots\\1 \end{bmatrix}\begin{bmatrix}1&\cdots&1 \end{bmatrix}. $$ Hence, $\nu=1$ and $$ \sigma_{1+k}(H) \leq Z_k([-n+1/2,-1/2],[1/2,n-1/2])\|H\|_2. $$ Fortunately, bounds on the Zolotarev numbers are well-studied (in [1] we give asymptotically tight bounds) and we conclude that $$ \sigma_{1+k}(H) \leq 4\left[{\rm exp}\left(\frac{\pi^2}{2\log(8n-4)}\right)\right]^{-2k}\|H\|_2. $$ If we seek the numerical rank of $H$ for some $0<\epsilon<1$, i.e., $k = {\rm rank}_\epsilon(H)$ if $k$ is the smallest integer such that $\sigma_{k+1}(H)\leq \epsilon\|H\|_2$, then from the bound on the singular values of $H$ we conclude that $$ {\rm rank}_\epsilon(H) \leq \Bigg\lceil \frac{\log(8n-4)\log(4/\epsilon)}{\pi^2} \Bigg\rceil. $$

This shows that Hilbert matrices are not only exponentially ill-conditioned with $n$, but its singular values decay geometrically to zero too. This methodology extends to any matrix with displacement structure such as Pick, Cauchy, Loewner, real Vandermonde, and positive definite Hankel matrices. For more details, see [1].

  • $\begingroup$ Interesting! Though these are upper bounds, not lower bounds as you asked in the original question, no? $\endgroup$
    – Suvrit
    Oct 1, 2016 at 20:07
  • $\begingroup$ It was stated a little backwards by my younger self: it's still about the numerical rank given by ${\rm rank}_\epsilon(H)$. The answer above gives an upper bound on ${\rm rank}_\epsilon(H)$ and hence $k$ (in my original question). Suppose that $k$ is an integer such that $\sigma_{k}(H)>\epsilon$ for some $0<\epsilon <1$, then from above and $\|H\|_2\leq \pi$ we know that $k \leq \lceil \log(8n-4)\log(4\pi/\epsilon)/\pi^2\rceil+1$. $\endgroup$
    – alext87
    Oct 1, 2016 at 20:40
  • $\begingroup$ I do not know how to get lower bounds on the singular values. This may involve developing lower bounds for so-called discrete Zolotarev numbers (if one goes for the above methodology). I am not sure if this is possible at the moment. For more details on discrete Zolotarev numbers, see: link.springer.com/article/10.1007/s00365-010-9087-6 $\endgroup$
    – alext87
    Oct 1, 2016 at 20:42
  • $\begingroup$ Hello, @alext87! Seen your lecture on low rank matrices in Gilbert Strang's course, and now I see you here in the comments. The Earth is round... or, at least, the internet is. $\endgroup$
    – Sl0wp0k3
    Sep 9, 2021 at 23:02
  • $\begingroup$ I am impressed that you noticed the connection. Yes, I did a lecture on singular value decay in Gilbert Strang's course based on this material. $\endgroup$
    – alext87
    Sep 23, 2021 at 12:31

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).

  • $\begingroup$ 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 mathoverflow.net/questions/136306/… $\endgroup$
    – Suvrit
    Jul 19, 2013 at 0:20
  • $\begingroup$ 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. $\endgroup$
    – alext87
    Jul 25, 2013 at 8:22

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