The entry sum of the inverse Cauchy matrix

Abstract

Let $x_1,x_2,\dots ,x_n$ be $n$ numbers, and $y_1,y_2,\dots ,y_n$ 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 $\frac{1}{x_i+y_j}$ are replaced by $\min \left\{x_i,y_j\right\}$.

7 pages. Expository note