# 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:

 $\ds C$ $=$ $\ds \begin{pmatrix} \dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2 } & \cdots & \dfrac 1 {x_1 - y_n} \\ \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2 } & \cdots & \dfrac 1 {x_2 - y_n} \\ \vdots & \vdots & \ddots & \vdots \\ \dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2 } & \cdots & \dfrac 1 {x_n - y_n} \\ \end {pmatrix}$ Distinct values $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ required.

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:

 $\ds \vec \bsone$ $=$ $\ds \begin {pmatrix} 1 \\ \vdots \\ 1 \end{pmatrix}$ vector of all ones

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

 $\text {(1)}: \quad$ $\ds S$ $=$ $\ds \vec \bsone^T C^{-1} \vec \bsone$

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

 $\ds C$ $=$ $\ds -P V_x^{-1} V_y Q^{-1}$ Vandermonde Matrix Identity for Cauchy Matrix

Definitions of symbols:

 $\ds V_x$ $=$ $\ds \begin {pmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n - 1} & x_2^{n - 1} & \cdots & x_n^{n - 1} \\ \end {pmatrix}, \quad V_y = \begin {pmatrix} 1 & 1 & \cdots & 1 \\ y_1 & y_2 & \cdots & y_n \\ \vdots & \vdots & \ddots & \vdots \\ y_1^{n - 1} & y_2^{n - 1} & \cdots & y_n^{n - 1} \\ \end {pmatrix}$ Definition of Vandermonde Matrix $\ds \map p x$ $=$ $\ds \prod_{i \mathop = 1}^n \paren {x - x_i}, \quad \map {p_k} x = \prod_{i \mathop = 1,i \mathop \ne k}^n \, \paren {x - x_i} ,\quad 1 \mathop \le k \mathop \le n$ Definition of Polynomial over Complex Numbers $\ds P$ $=$ $\ds \begin {pmatrix} \map {p_1} {x_1} & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \map {p_n} {x_n} \\ \end {pmatrix}, \quad Q = \begin {pmatrix} \map p {y_1} & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \map p {y_n} \\ \end {pmatrix}$ Definition of Diagonal Matrix

Compute the sum $S$:

 $\ds C^{-1}$ $=$ $\ds - Q V_y^{-1} V_x P^{-1}$ Inverse of Matrix Product $\ds S$ $=$ $\ds \vec \bsone^T C^{-1} \vec \bsone$ from $(1)$ $\ds$ $=$ $\ds \paren {-\vec {\bsone}^T Q V_y^{-1} } \paren {V_x P^{-1} \vec \bsone}$

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

 $\text {(2)}: \quad$ $\ds \vec A^T$ $=$ $\ds -\vec \bsone^T Q V_y^{-1}$ $\text {(3)}: \quad$ $\ds \vec B$ $=$ $\ds V_x P^{-1} \vec \bsone$ Sum $S$ equals scalar product $\vec A^T \vec B$

Compute $\vec A$:

 $\ds \vec A$ $=$ $\ds -\paren {\vec \bsone^T Q V_y^{-1} }^T$ from $(2)$ $\ds$ $=$ $\ds -\paren {V_y^T}^{-1} Q^T \vec \bsone$ Definition:Inverse Matrix and Transpose of Matrix Product

Then $\vec A$ uniquely solves the problem

 $\ds \begin{pmatrix} 1 & \cdots & y_1^{n - 1} \\ \vdots & \cdots & \vdots \\ 1 & \cdots & y_n^{n - 1} \end {pmatrix} \vec A$ $=$ $\ds -\begin{pmatrix} \map p {y_1} \\ \vdots \\ \map p {y_n} \\ \end {pmatrix}$ coefficient matrix $V_y^T$

The two polynomials

 $\ds \map p x$ $=$ $\ds \prod_{j \mathop = 1}^n \paren {x - x_j}$ $\ds$ $=$ $\ds x^n + \sum_{i \mathop = 1}^n a_i x^{i - 1}$ $\ds \map q y$ $=$ $\ds \prod_{j \mathop = 1}^n \paren {y - y_j}$ $\ds$ $=$ $\ds y^n + \sum_{i \mathop = 1}^n b_i y^{i - 1}$

have coefficients expressed by symmetric functions:

 $\ds a_i$ $=$ $\ds \paren {-1}^{n + 1 - i} e_{n + 1 - i} \paren {\set {x_1, \ldots, x_n} }$ Viète's Formulas $\ds b_i$ $=$ $\ds \paren {-1}^{n + 1 - i} e_{n + 1 - i} \paren {\set {y_1, \ldots, y_n} }$ Viète's Formulas

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

 $\ds \map p {y_j}$ $=$ $\ds \sum_{i \mathop = 1}^n \paren {a_i - b_i } y_j^{i - 1}$ for $j = 1, \ldots, n$

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:

 $\ds \vec A$ $=$ $\ds \begin {pmatrix} b_1 - a_1 \\ \vdots \\ b_n - a_n \\ \end{pmatrix}$ $\ds$ $=$ $\ds \begin{pmatrix} \paren {-1}^n \paren {e_n \paren {\set {y_1,\dots,y_n} } - e_n \paren {\set {x_1,\dots,x_n} } } \\ \vdots \\ \paren {-1}^1 \paren {e_1 \paren {\set {y_1, \dots, y_n} } - e_1 \paren {\set {x_1, \dots, x_n} } } \\ \end {pmatrix}$ The last entry contains the sum of the elements in $C^{-1}$: $S = x_1 + \cdots + x_n - \paren {y_1 + \cdots + y_n}$

Compute $\vec B$:

 $\ds \vec B$ $=$ $\ds V_x P^{-1} \vec { \bsone }$ from $(3)$ $\ds$ $=$ $\ds V_x \begin {pmatrix} \dfrac 1 {\map {p_1} {x_1} } \\ \vdots \\ \dfrac 1 {\map {p_n} {x_n} } \end{pmatrix}$ $\ds$ $=$ $\ds \begin {pmatrix} \ds \sum_{k \mathop = 1}^n \frac 1 {\map {p_k} {x_k} } \\ \vdots \\ \ds \sum_{k \mathop = 1}^n \frac {x_n^{n - 1} } { \map {p_n} {x_n} } \\ \end{pmatrix}$ $\ds$ $=$ $\ds \begin {pmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ \end{pmatrix}$ Summation of Powers over Product of Differences

Evaluate $S$:

 $\ds S$ $=$ $\ds \vec A^T \vec B$ $\ds$ $=$ $\ds \begin {pmatrix} * \\ * \\ \vdots \\ x_1 + \cdots + x_n - \paren { y_1 + \cdots + y_n } \\ \end {pmatrix} ^T \begin {pmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ \end {pmatrix}$ Entries $*$ contribute nothing because of matching zero in $\vec B$. $\ds$ $=$ $\ds x_1 + \cdots + x_n - \paren { y_1 + \cdots + y_n}$

$\blacksquare$