Consider the Stirling numbers of the first kind $s(i, j)$, and form a matrix $S_1(n),$ where the $(i, j)$th entry is $s(i, j)$. (IMPORTANT NOTE the indices start at $0,$ so this matrix is $(n+1)\times (n+1).$) This matrix is the inverse of the corresponding matrix for the Stirling numbers of the second kind, and the question applies to both of them.

Now, for the question: I have computed the logs of the singular values of these things (you have to do this to high precision, since the entries grow very fast). Here is the plot of the singular values for $n=100:$

Stirling Matrix log singular values

You will note two empirical facts: firstly, $1$ is a singular value, and secondly the singular values bigger than $1$ decay exponentially, as do the ones smaller than $1$, but with a different rate. Has this been observed? Can one prove it (or at least explain it)?

ADDITION In my comment on the answer, I speak of the shape of the curve. To underscore what I mean, here are two more interesting matrices: the binomial coefficient matrix (which should be, in principle, similar to the Stirling matrix, due to the recurrence relation), and the Hilbert matrix. Here are the singular value plots of both (in order), again logarithmically. You will note that all three curves are regular. While the Stirling curve is piecewise linear, the Binomial curve is convex (on the top segment) and the Hilbert curve is concave. This must mean something :)

binomial matrix log singular values

Hilbert matrix log singular values

  • $\begingroup$ Someone downvoted this? $\endgroup$ – Igor Rivin Dec 21 '17 at 15:04

I will make the first attempt, though my resulting bound can certainly be improved.

I will do the $N\times N$ Stirling matrix of the second kind. We have $S_2(n,k) = \left\{n\atop k\right\}$ for $0\leq n,k\leq N-1$, which satisfies the following recurrence relation:

$$\left\{n+1\atop k\right\} = k\left\{n\atop k\right\} + \left\{n\atop k-1\right\},\qquad n\geq 0, \quad k>0.$$

Therefore, the matrix $S_2$ satisfies the Sylvester matrix equation:

$$\underbrace{\begin{bmatrix}0 & 1 \\ & & \ddots \\ & & & 1 \\ -1 \end{bmatrix}}_{=A} S_2 - S_2\underbrace{\begin{bmatrix} 0 & 1 \\ & 1 & \ddots \\ & & \ddots & 1 \\ & & & N-1 \end{bmatrix}}_{=B} = \begin{bmatrix}0 & \ldots & 0 & 0 \\\vdots & \ddots & \vdots & \vdots \\0 & \ldots & 0 & 0 \\ \times & \ldots & \times & \times \end{bmatrix},$$ where the $\times$'s denote nonzero entries. We say that $S_2$ has an $(A,B)$-displacement rank of 1.

From now on assume that $N$ is an even integer. The same idea, with different details, works when $N$ is an odd integer.

Since the rhs of the above equation is of rank 1, the eigenvalues of $A$ lie at (shifted) roots-of-unity, and the eigenvalues of $B$ are in $\{0,\ldots,N-1\}$ we find that, see Corollary 2.2 of paper $$\sigma_{2k+1}(S_2) \leq \|V\|_2\|V^{-1}\|_2 Z_{2k}(\{0,\ldots,N-1\},F_+\cup F_-)\|S_2\|_2,$$ where $V$ is the eigenvector matrix for $B$, i.e., $B = V\Lambda_B V^{-1}$ and $\|\cdot\|_2$ is the spectral norm. Here, $Z_{k}(E,F)$ denotes a Zolotarev number (see Section 2 of paper) and

$$F_+ = \left\{e^{it}: t\in [\tfrac{\pi}{N},\pi-\tfrac{\pi}{N}\right\}, \quad F_- = \left\{e^{it}: t\in [-\pi + \tfrac{\pi}{N},-\tfrac{\pi}{N}\right\}.$$

I do not have bounds on $Z_{2k}(\{0,\ldots,N-1\},F_+\cup F_-)$ at hand and would have to work them out. Instead, to use a previous result, I will use the fact that

$$Z_{2k}(\{0,\ldots,N-1\},F_+\cup F_-)\leq Z_{2k}(\mathbb{R},F_+\cup F_-)$$

since $\{0,\ldots,N-1\}\subset \mathbb{R}$.

Therefore, when $N$ is an even integer, we obtain (see Lemma 5.1 of paper):

$$\sigma_{2k+1}(S_2) \leq 4\|V\|_2\|V^{-1}\|_2\left[\exp\left(\frac{\pi^2}{4\log(4N/\pi)}\right)\right]^{-k}\|S_2\|_2.$$

This explains the rapid decay of the singular values of $S_2$, provided $\|V\|_2\|V^{-1}\|_2$ does not grow rapidly with $N$. With a little more work, I believe that one can show that $\|V\|_2\|V^{-1}\|_2<8$ for any $N$ (this appears to be true numerically for $1<N<1000$). If so, an explicit bound on the singular values is (for $N$ even):
$$\sigma_{2k+1}(S_2) \leq 32\left[\exp\left(\frac{\pi^2}{4\log(4N/\pi)}\right)\right]^{-k}\|S_2\|_2, \qquad 0\leq 2k\leq N-1.$$

DISCLAIMER: This bound explains the geometric decay of the singular values of $S_2$ but the geometric rate is not tight. One would have to directly bound $Z_{2k}(\{0,\ldots,N-1\},F_+\cup F_-)$ to get a tighter bound. I do not know how to explain the two different rates of decay of the singular values yet.

  • $\begingroup$ Very nice argument! Ideally one would like to explain the shape of the curve, but that may be a whole'nother level of complexity... $\endgroup$ – Igor Rivin Nov 17 '17 at 0:51
  • $\begingroup$ Yes, I agree. This was just a first attempt. I will keep thinking about it. $\endgroup$ – alext87 Nov 17 '17 at 15:31
  • $\begingroup$ You can check out the edits for more examples of shapes... $\endgroup$ – Igor Rivin Nov 17 '17 at 16:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.