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arXiv · 2301.09777

The entry sum of the inverse Cauchy matrix

Darij Grinberg
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Abstract

Let $x_{1},x_{2},\ldots,x_{n}$ be $n$ numbers, and $y_{1},y_{2},\ldots,y_{n}$ be $n$ further numbers chosen such that all $n^{2}$ pairwise sums $x_{i}+y_{j}$ are nonzero. Consider the $n\times n$-matrix \[ C:=\left( \dfrac{1}{x_{i}+y_{j}}\right) _{1\leq i\leq n,\ 1\leq j\leq n} = \begin{pmatrix} \dfrac{1}{x_{1}+y_{1}} & \dfrac{1}{x_{1}+y_{2}} & \cdots & \dfrac{1}{x_{1}+y_{n}}\\ \dfrac{1}{x_{2}+y_{1}} & \dfrac{1}{x_{2}+y_{2}} & \cdots & \dfrac{1}{x_{2}+y_{n}}\\ \vdots & \vdots & \ddots & \vdots\\ \dfrac{1}{x_{n}+y_{1}} & \dfrac{1}{x_{n}+y_{2}} & \cdots & \dfrac{1}{x_{n}+y_{n}} \end{pmatrix}. \] This matrix $C$ is known as the "Cauchy matrix", and has been studied for 180 years. A classical result says that if $C$ is invertible, then the sum of all entries of its inverse $C^{-1}$ is $\sum_{k=1}^{n}x_{k}+\sum_{k=1}^{n}y_{k}$. We give a simple and short proof of this result, and briefly discuss a "tropicalized" variant in which the entries $\dfrac{1}{x_i+y_j}$ are replaced by $ \min\left\{ x_{i},y_{j}\right\}$.

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Darij Grinberg. 2023-01-24. The entry sum of the inverse Cauchy matrix. https://doi.org/10.1007/s00283-023-10268-4

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