After invoking a recursion relation for Hankel determinants in my answer to a (mostly unrelated) question, I started wondering what else I could use this recursion for, and stumbled upon some results that surprised me. The proofs are purely computational, and I'm hoping someone can provide a more conceptual understanding.

So: Let $f(m)$ be a function defined on positive integers. The corresponding Hankel matrices are:

$${\cal H}(m,n)=\pmatrix{ f(m)&f(m+1)&\ldots&f(m+n)\cr f(m+1)&f(m+2)&\ldots &f(m+n+1)\cr \vdots&\vdots&&\vdots\cr f( m+n)&f( m+n+1)&\ldots&f(m+2n)\cr}$$ and the Hankel determinants are $H(m,n)=det({\cal H}(m,n))$.

In particular, set: $$H(m,n)=\hbox{ the Hankel determinant associated to $f(m)=1/m$}$$ $$J(m,n)=\hbox{ the Hankel determinant associated to $f(m)=m!$}$$ $$K(m,n)=\hbox{ the Hankel determinant associated to $f(m)=1/m!$}$$

Also, let $c(n)=\prod_{i=1}^{n-1}i!$

Then I can show that: $$H(m,n)={c(m+n-1)^2c(n)^2\over c(m-1)c(m+2n-1)}$$ $$J(m,n)={c(n)c(m+n)\over c(m)}$$ $$K(m,n)=\pm{c(n)c(m+n-1)\over c(m+2n-1)}$$ where the $\pm$ sign on $K(m,n)$ is $+$ or $-$ depending on whether $n$ is a square mod $4$.

These calculations immediately yield two theorems, each of which seems to cry out for a deeper explanation:

Theorem 1. $H(m,n)=J(m-1,n)K(m,n)$

Theorem 2. The $H(m,n)$ and $K(m,n)$ are all reciprocals of integers.

Question: Aside from the fact that they fall out of a calculation, why should we expect these theorems to be true?

For example, for Theorem 1, is there some natural interpretation of the product of two Hankel matrices that puts this in context? Or for Theorem 2, are the reciprocals of the $H(m,n)$ and $K(m,n)$ counting something?

Remark 1: Theorem 2 is certainly well-known for the $H(1,n)$, which are the determinants of the Hilbert matrices. I'm not sure whether it's well-known for the rest of the $H(m,n)$ or for the $K(m,n)$.

Remark 2: Theorem 2 would follow from the stronger statement that the Hankel matrices ${\cal H}(m,n)$ and ${\cal K}(m,n)$ (that is, the matrices of which the $H(m,n)$ and $K(m,n)$ are the determinants) have inverses with all integer entries. This, again, is well-known for $H(1,n)$ at least. I have both the outline of an argument and extensive numerical evidence to suggest that it is true for all the ${\cal H}(m,n)$ and all the ${\cal K}(m,n)$ but I fear that even if the argument pans out, it won't provide much conceptual understanding.

  • $\begingroup$ did you have a look at arxiv.org/abs/math/9902004 and arxiv.org/abs/math/0503507? $\endgroup$ – Martin Rubey Sep 14 '16 at 7:51
  • $\begingroup$ @MartinRubey: I had a quick look at the first and found many wonderful things, but not the answer I was looking for. (Though it's possible that my quick look was too quick.) I didn't know about the second but will spend some time with it now. Thanks for mentioning it. $\endgroup$ – Steven Landsburg Sep 14 '16 at 15:17
  • $\begingroup$ I notice that theorem 1 is about entrywise products. Just by curiosity: does a similar result hold numerically if, say, you square all entries of H, J, K? If so, there should be something much deeper here! $\endgroup$ – Wolfgang Sep 15 '16 at 20:52

After comparing Steven's original example with his new example, I believe I have an interesting generalization of both. Let $c \in \mathbb{C}$. Define the following three functions: $$h(m)=\frac{1}{m-1+c},k(m) = \frac{1}{\Gamma(m+c)},$$ $$j(m)=\frac{h(m+1)}{k(m+1)}=\Gamma(m+c).$$

Let $\mathcal{H}(m,n),\mathcal{J}(m,n),\mathcal{K}(m,n)$ be the Hankel matrices corresponding to $h,j,k$, respectively. Let $H(m,n),J(m,n),K(m,n)$ be the respective determinants. Then the following holds: $$H(m,n) =\pm J(m-1,n) K(m,n).$$

  • When $c=1$, we recover the example from the original post.
  • When $c=\frac{1}{2}$, we recover Steven's second example, from his answer to the post (up to some normalization).

Now I turn to prove the generalization.

Lemma 1: Let $V$ be an inner-product space. Let $\{v_i\}_{i=1}^{n},\{\tilde{v}_i\}_{i=1}^{n}$ be two pairs of sequences of linearly independent vectors in $V$, spanning the same subspace. Similarly, let $\{u_i\}_{i=1}^{n},\{\tilde{u}_i\}_{i=1}^{n}$ be two pairs of sequences of linearly independent vectors in $V$, spanning the same subspace.

There exist unique scalars $x_{i,j}, y_{i,j}$ such that $\tilde{v}_i = \sum x_{i,k} v_k$ and $\tilde{u}_i = \sum y_{i,k} u_k$.

Define the following 4 matrices:

  • $A_{i,j} = \langle \tilde{v}_i, \tilde{u}_j \rangle$, $B_{i,j} = \langle v_i, u_j \rangle$
  • $X_{i,j} = x_{i,j}, Y_{i,j} = y_{i,j}$

Then $$A = X B Y^{T}.$$

Lemma 2: $\det(\binom{i+j+m}{i}_{0\le i,j \le n}) = 1$ for any $m \in \mathbb{C}$.

We work in the inner-product space $L^2([0,1])$. Note that $$\mathcal{H}(m,n)_{i,j} = \langle x^{i+m+c-2}, x^{j} \rangle$$ and that $$\mathcal{K}(m,n)_{i,j} = \langle \frac{x^{i+m+c-2}}{\Gamma(i+m-1+c)}, \frac{(1-x)^j}{\Gamma(j+1)} \rangle.$$

We are in a situation where we may apply Lemma 1 with $A=\mathcal{H}(m,n)$ and $ B=\mathcal{K}(m,n)$, which yields $$\mathcal{H}(m,n)= \text{Diag}((\Gamma(i+m-1+c))_{0\le i\le n}) \times \mathcal{K}(m,n) \times (\Gamma(i+1)\binom{j}{i}(-1)^i)_{0\le i,j \le n},$$ hence $$(*) \frac{H(m,n)}{K(m,n)} = \det \text{Diag}((\Gamma(i+m-1+c))_{0\le i\le n}) \det (\Gamma(i+1)\binom{j}{i}(-1)^i)_{0\le i,j \le n}.$$

On the other hand, $$\mathcal{J}(m-1,n) = (\Gamma(m-1+c+i+j))_{0 \le i,j \le n}) = \text{Diag}((\Gamma(i+m-1+c))_{0\le i\le n}) (\binom{m-2+i+j+c}{j})_{0 \le i,j \le n}) \text{Diag}(\Gamma(j+1)_{0\le j\le n}),$$ which, combined with Lemma 2, yields $$(**) J(m-1,n) = \det \text{Diag}((\Gamma(i+m-1+c))_{0\le i\le n}) \det\text{Diag}(\Gamma(j+1)_{0\le j\le n}).$$ We finish by noting that $$\det (\Gamma(i+1)\binom{j}{i}(-1)^i)_{0\le i,j \le n} = (-1)^{\binom{n+1}{2}} \prod_{i=0}^{n} \Gamma(i+1) = (-1)^{\binom{n+1}{2}} \det \text{Diag}(\Gamma(j+1)_{0\le j\le n}),$$ which implies that $(*)$ and $(**)$ coincide up to sign. $\blacksquare$

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    $\begingroup$ For $K(m,n)$, one can use the beta function to get an integral representation (this is again "non-symmetric"). $\endgroup$ – js21 Sep 14 '16 at 9:32
  • $\begingroup$ @js21 Thanks, after row and column operations the beta function indeed allows for a similar proof for the integrality of $K(m,n)^{-1}$. I will rewrite my answer later today. $\endgroup$ – Ofir Gorodetsky Sep 14 '16 at 12:56
  • $\begingroup$ @js21 I was able to incorporate your idea in order to find a simple proof for Theorem 1 (which does not use Legendre polynomials, as in my "proof" for Theorem 2). It turns out that my proof for Theorem 2 was flawed, and so I removed it. $\endgroup$ – Ofir Gorodetsky Sep 14 '16 at 19:04
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    $\begingroup$ Thanks for this. I'm still trying to decide (particularly with regard to Theorem 1) whether it's actually more conceptual than my original proof by recursion. I think I'd be a lot more satisfied if your method yielded a substantially more general theorem, of which Theorem 1 is a special case. $\endgroup$ – Steven Landsburg Sep 15 '16 at 20:10
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    $\begingroup$ @OfirGorodetsky: Here is another example: Let $j(m)=(2m)!/m!$. Let $k(m)=\prod_{i=1}^m1/(2i-1)$. Let $h(m)=j(m-1)k(m)$. Then with $J,K,H$ the corresponding families of Hankel determinants, we have $H(m,n)=J(m-1,n)K(m,n)$. It's not immediately obvious to me whether this is a special case of your generalization. I'd be glad to know either way. $\endgroup$ – Steven Landsburg Sep 17 '16 at 6:40

Not yet a fully satisfying answer, but Theorem 1 does generalize.

Given two sequences $j$ and $k$, define the sequence $h$ by $$h(m)=j(m-1)k(m)$$ Let $H$, $J$ and $K$ be the corresponding families of Hankel determinants. Say that $(j,k)$ is a good example if, for all $m$ and $n$, $$H(m,n)=\pm J(m-1,n)K(m,n)$$

Theorem 1 above is the statement that if we set $j(m)=m!$, and $k(m)=1/m!$, then $(j,k)$ is a good example. This continues to surprise me, because it requires a lot of seemingly miraculous cancellation.

I now have other examples of good pairs, but not yet a good way to describe them all. However, the following will give the flavor: Let $$j(m)=(2m)!/(m!)\qquad k(m)=\prod_{i=1}^m 1/(2i-1)$$ Then $(j,k)$ a good example.

My hope continues to be that the existence of this and related good examples will shed some light on Theorem 1.

  • $\begingroup$ Since $$j(m) = \Gamma(m+\frac{1}{2}) \cdot \frac{4^m}{\sqrt{\pi}}$$ and $$k(m) = \frac{1}{\Gamma(m+\frac{1}{2})} \cdot \frac{\sqrt{\pi}}{2^m},$$ I decided to look at the pairs $$(j(m),k(m)) = (\Gamma(m+c), \frac{1}{\Gamma(m+c)})$$ for some complex $c$, which turn out to be good pairs for any $c$. Constant factors and exponential factors preserve "good"-ness of pairs, so your example is recovered from this. $\endgroup$ – Ofir Gorodetsky Sep 17 '16 at 14:30
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    $\begingroup$ @OfirGorodetsky : Your new family of examples is great. I have several more examples, but I believe they all fit into your family (modulo reparameterization, multiplication by constants and exponential factors, etc). Thanks very much for this. I want to spend a little time today fully digesting it $\endgroup$ – Steven Landsburg Sep 17 '16 at 15:11

In as far as the computation goes, these determinants all fall out easily from the method shown in this paper.

When inductive proofs work, this technique is simpler to use than any other listed in this paper.

On the basis of the solution and comments above, it only remains to verify Theorem 2 to the effect that $\frac1{\det(H^{-1})}$ is integral.

Proof of Theorem 2. If the entries of the "Hilbert matrix" are given by $$H_{ij}=\frac{1}{i+j+m-1}$$ then its inverse has an explicit form given by: \begin{align*} (H^{-1})_{ij} &= (-1)^{i+j}(m+i+j-1) \times \\ &{{n+m+i-1}\choose{n-j}}{{n+m+j-1}\choose{n-i}}{{m+i+j-2}\choose{i-1}}{{m+i+j-2}\choose{j-1}}. \end{align*} It follows that the entries in the inverse matrix are all integers. Therefore, $$\det(H)=\frac1{\det(H^{-1})}$$ explains why these are reciprocals of integers. The proof follows. $\square$

  • $\begingroup$ Ah, yes, that's pretty close to the method I used --- the key recursion is the equation dubbed Lewis in that paper. Unfortunately, this doesn't answer my question. $\endgroup$ – Steven Landsburg Sep 14 '16 at 15:18
  • $\begingroup$ I have added a formula for the general case, this should help. $\endgroup$ – T. Amdeberhan Sep 15 '16 at 12:31

IMHO Theorem 1 is a lucky coincidence. These Hankel determinants do not so much depend on the values of their entries but rather on the linear functionals which they generate.

More precisely: Let $a(n)$ be a sequence of real numbers with $a(0)=1$ such that no Hankel determinant vanishes. Define a linear functional $F$ on the polynomials by $F(x^n)=a(n).$ Let $p(n,x)$ be the uniquely determined set of monic polynomials which are orthogonal with respect to $F.$

By Favard’s theorem they satisfy $p(n,x)=(x-s(n-1))p(n-1,x)-t(n-2)p(n-2,x).$

Then the Hankel determinant $\det(a(i+j))_{i,j=0}^n= \prod_{i=1}^n\prod_{j=0}^{i-1}t(j)$ only depends on the numbers $t(n).$ These well-known results can for example be found here.

For $a_{1}(n,m)=\frac{m}{n+m}$, $a_{2}(n,m)=\frac{(n+m)!}{m!}$ and $a_{3}(n,m)=\frac{m!}{(n+m)!}$ the corresponding $t’s$ are $t_{1}(n,m)=\frac{(n+1)^2(n+m)^2}{(2n+m)(2n+m+1)^2(2n+m+2)}$, $t_{2}(n,m-1)=(n+1)(n+m)$ and $t_{3}(n,m)=\frac{(n+1)(n+m)}{(2n+m)(2n+m+1)^2(2n+m+2)}.$ Therefore $t_{1}(n,m)=t_2(n,m-1) t_{3}(n,m),$ which is the reason for Theorem 1.


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