The entry sum of the inverse Cauchy matrix (2301.09777v1)

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}$.