Harmonic APD relation between Hilbert and identity matrices

About 1 year old · traced to 2025

Let $H_n$ and $I_n$ be respectively the $n$ -th order Hilbert and identity matrices. Let $\det(H_n)$ be the determinant of $H_n$ , and let $\operatorname{APD}_{n-1}(H_n)$ and $\operatorname{APD}_{n-1}(I_n)$ denote their first appearance values.

Harmonic APD relation. For $n \geq 1$ ,

\operatorname{APD}_{n-1}(H_n)=\det(H_n)\,n\,\operatorname{APD}_{n-1}(I_n).

The source says this was confirmed by exact rational calculation for $n=1$ through $n=7$ ; it is therefore recorded as solved.

References

Primary source

Kenichi Takemura, “Alternating Power Difference and Matrix Symmetry: Closed-Form Formulas for the First Appearance Degree m_1”, arXiv:2512.18169 (2025).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A 2025 paper checked the formula only through size seven, while a reader has posted an unverified argument claiming it holds for every size.

The problem asserts the stated relation for every order $n \geq 1$ . Kenichi Takemura’s 2025 paper records exact rational confirmation for $n=1$ through $n=7$ , but that finite check does not establish the universal statement.

December 2025 exact checks

Takemura reports confirmation by exact rational calculation for $n=1,\ldots,7$ ; the paper presents broader APD formulas and conjectures, but no independently verified all- $n$ proof is identified here.

Posted attempt

A reader gives a purported complete proof for all $n$ , deriving a general adjugate formula and evaluating the Hilbert-matrix kernel; the attempt has not been independently verified.

Current status (as of August 2026): The relation is verified only for $n=1$ through $n=7$ in the retrieved primary source, while an all- $n$ proof has been claimed but remains unverified.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

For every $n\times n$ matrix $A$ , the determinant expansion gives

\sum_{m\ge0}\operatorname{APD}_m(A)\frac{z^m}{m!} =\det(e^{zA_{ij}}).

Because

(e^{zA_{ij}})=J_n+zA+O(z^2), \qquad J_n=\mathbf1\mathbf1^{\mathsf T},

and $J_n$ has rank one, determinant multilinearity yields the universal identity

\operatorname{APD}_{n-1}(A) =(n-1)!\mathbf1^{\mathsf T}\operatorname{adj}(A)\mathbf1. \tag{1}

Let

(H_n)_{ij}=\frac1{i+j-1}.

This is the Gram matrix of $1,x,\ldots,x^{n-1}$ in $L^2([0,1])$ , since

(H_n)_{ij}=\int_0^1x^{i+j-2}\,dx.

The all-ones vector represents evaluation at $x=1$ . Therefore $\mathbf1^{\mathsf T}H_n^{-1}\mathbf1$ is the value $K_n(1,1)$ of the reproducing kernel for polynomials of degree less than $n$ . An orthonormal basis is

\phi_k(x)=\sqrt{2k+1}\,P_k(2x-1), \qquad 0\le k<n,

where $P_k$ denotes the $k$ -th Legendre polynomial. Since $P_k(1)=1$ ,

\mathbf1^{\mathsf T}H_n^{-1}\mathbf1 =\sum_{k=0}^{n-1}\phi_k(1)^2 =\sum_{k=0}^{n-1}(2k+1) =n^2.

Equation (1) now gives

\operatorname{APD}_{n-1}(H_n) =(n-1)!\det(H_n)n^2 =n\,n!\det(H_n).

For the identity matrix,

\sum_{m\ge0}\operatorname{APD}_m(I_n)\frac{z^m}{m!} =(e^z-1)^{n-1}(e^z+n-1).

Hence

\operatorname{APD}_{n-1}(I_n)=n!,

and therefore

\operatorname{APD}_{n-1}(H_n) =n\det(H_n)\operatorname{APD}_{n-1}(I_n).

Both identities hold for all $n\ge2$ . For the endpoint $n=1$ , they also hold under the natural extension $\operatorname{APD}_0(A)=\sum_{\sigma\in S_1}\operatorname{sgn}(\sigma)=1$ .