Sum of Elements in Inverse of Cauchy Matrix

Theorem
Let $C_n$ be the Cauchy matrix of order $n$ given by:


 * $C_n = \begin{bmatrix}

\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_m + y_1} & \dfrac 1 {x_m + y_2 } & \cdots & \dfrac 1 {x_m + y_n} \\ \end{bmatrix}$

Let $C_n^{-1}$ be its inverse, from Inverse of Cauchy Matrix:
 * $b_{i j} = \dfrac {\ds \prod_{k \mathop = 1}^n \paren {x_j + y_k} \paren {x_k + y_i} } {\ds \paren {x_j + y_i} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {y_i - x_k} } }$

The sum of all the elements of $C_n^{-1}$ is:
 * $\ds \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k$

Proof
It suffices to prove the Theorem for Cauchy matrix:

The sum $S$ of elements in $C^{-1}$ will be shown to be a scalar product of two vectors $\vec A$ and $\vec B$.

The statement of the theorem is obtained by replacing $y_k \to -y_k$ in $C$ and in the sum $S$.

Let:

The sum $S$ of elements in $C^{-1}$ is the matrix product:

To identify vectors $\vec A$ and $\vec B$, the tool is:

Definitions of symbols:

Compute the sum $S$:

Define vectors $\vec A$ and $\vec B$:

Compute $\vec A$:

Then $\vec A$ uniquely solves the problem

The two polynomials

have coefficients expressed by symmetric functions:

Solve in equation $\map q {y_j} = 0$ for highest power $y_j^n$ and replace in the equation for $\map p {y_j}$:

Vector $\begin {pmatrix} \map p {y_1} \\ \vdots \\ -\map p {y_n} \\ \end {pmatrix}$ is a linear combination of the column vectors of matrix $V_y^T$ with weights (coefficients) $b_i - a_i$.

Then:

Compute $\vec B$:

Evaluate $S$: