Theorem: Sum of Elements in Inverse of Cauchy Matrix

Let $C_n$ be the Cauchy matrix of order $n$ given by:
:<nowiki>$C_n = \begin{bmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2 } & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2 } & \cdots & \dfrac 1 {x_2 + y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_m + y_1} & \dfrac 1 {x_m + y_2 } & \cdots & \dfrac 1 {x_m + y_n} \\
\end{bmatrix}$</nowiki>
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:
{{begin-eqn}}                        
{{eqn | l = C
      | r = <nowiki>\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}</nowiki>
      | c = Distinct values $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ required.
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \vec \bsone
      | r = \begin {pmatrix} 1 \\ \vdots \\ 1 \end{pmatrix} 
      | c = vector of all ones
}}
{{end-eqn}}
The sum $S$ of elements in $C^{-1}$ is the matrix product:
{{begin-eqn}}
{{eqn | n = 1
      | l = S
      | r = \vec \bsone^T C^{-1} \vec \bsone
}}
{{end-eqn}}
To identify vectors $\vec A$ and $\vec B$, the tool is:
{{begin-eqn}}
{{eqn | l = C 
      | r = -P V_x^{-1} V_y Q^{-1}
      | c = Vandermonde Matrix Identity for Cauchy Matrix
}}
{{end-eqn}}
Definitions of symbols:
{{begin-eqn}}
{{eqn | l = V_x
      | r = <nowiki>\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}</nowiki>
      | c = {{Defof|Vandermonde Matrix|subdef = Formulation 1}}
}}
{{eqn | l = \map p x
      | r = <nowiki>\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</nowiki> 
      | c = {{Defof|Polynomial over Complex Numbers}}
}}
{{eqn | l = P
      | r = <nowiki>\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}</nowiki>
      | c = {{Defof|Diagonal Matrix}}
}}
{{end-eqn}}
'''Compute the sum''' $S$:
{{begin-eqn}}
{{eqn | l = C^{-1}
      | r = - Q V_y^{-1} V_x P^{-1}
      | c = Inverse of Matrix Product
}}
{{eqn | l = S 
      | r = \vec \bsone^T C^{-1} \vec \bsone
      | c = from $(1)$
}}
{{eqn | r = \paren {-\vec {\bsone}^T Q V_y^{-1} } \paren {V_x P^{-1} \vec \bsone}
}}
{{end-eqn}}
Define vectors $\vec A$ and $\vec B$:
{{begin-eqn}}
{{eqn | n = 2
      | l = \vec A^T
      | r = -\vec \bsone^T Q V_y^{-1}
}}
{{eqn | n = 3
      | l = \vec B
      | r  = V_x P^{-1} \vec \bsone
      | c = Sum $S$ equals scalar product $\vec A^T \vec B$
}}
{{end-eqn}}
'''Compute''' $\vec A$:
{{begin-eqn}}
{{eqn | l = \vec A
      | r = -\paren {\vec \bsone^T Q V_y^{-1} }^T
      | c = from $(2)$
}}
{{eqn | r = -\paren {V_y^T}^{-1} Q^T \vec \bsone
      | c = Definition:Inverse Matrix and Transpose of Matrix Product
}}
{{end-eqn}}
Then $\vec A$ '''uniquely''' solves the problem 
{{begin-eqn}}
{{eqn | l = \begin{pmatrix} 1 & \cdots & y_1^{n - 1} \\ \vdots & \cdots & \vdots \\ 1 & \cdots & y_n^{n - 1} \end {pmatrix} \vec A 
      | r = -\begin{pmatrix} \map p {y_1} \\ \vdots \\ \map p {y_n} \\ \end {pmatrix}
      | c = coefficient matrix $V_y^T$
}}
{{end-eqn}}
The two polynomials
{{begin-eqn}}
{{eqn | l = \map p x
      | r = \prod_{j \mathop = 1}^n \paren {x - x_j}
}}
{{eqn | r = x^n + \sum_{i \mathop = 1}^n a_i x^{i - 1} 
}}
{{eqn | l = \map q y
      | r = \prod_{j \mathop = 1}^n \paren {y - y_j}
}}
{{eqn | r = y^n + \sum_{i \mathop = 1}^n b_i y^{i - 1}
}}
{{end-eqn}}
have coefficients expressed by symmetric functions:
{{begin-eqn}}
{{eqn | l = a_i
      | r = \paren {-1}^{n + 1 - i} e_{n + 1 - i} \paren {\set {x_1, \ldots, x_n} }
      | c = Viète's Formulas
}}
{{eqn | l = b_i
      | r = \paren {-1}^{n + 1 - i} e_{n + 1 - i} \paren {\set {y_1, \ldots, y_n} }
      | c = Viète's Formulas
}}
{{end-eqn}}
Solve in equation $\map q {y_j} = 0$ for highest power $y_j^n$ and replace in the equation for $\map p {y_j}$:
{{begin-eqn}}
{{eqn | l = \map p {y_j}
      | r = \sum_{i \mathop = 1}^n \paren {a_i - b_i } y_j^{i - 1}
      | c = for  $j = 1, \ldots, n$
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \vec A
      | r = \begin {pmatrix} b_1 - a_1 \\ \vdots \\ b_n - a_n \\ \end{pmatrix}
}}
{{eqn | r = <nowiki>\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}</nowiki>
      | c = The last entry contains the sum of the elements in $C^{-1}$: $S = x_1 + \cdots + x_n - \paren {y_1 + \cdots + y_n}$
}}
{{end-eqn}}
'''Compute''' $\vec B$:
{{begin-eqn}}
{{eqn | l = \vec B
      | r = V_x P^{-1} \vec { \bsone }
      | c = from $(3)$ 
}}
{{eqn | r = V_x \begin {pmatrix} \dfrac 1 {\map {p_1} {x_1} } \\ \vdots \\ \dfrac 1 {\map {p_n} {x_n} } \end{pmatrix} 
}}
{{eqn | r = \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}
}}
{{eqn | r = \begin {pmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ \end{pmatrix}
      | c = Summation of Powers over Product of Differences 
}}
{{end-eqn}}
'''Evaluate''' $S$:
{{begin-eqn}}
{{eqn | l = S
      | r = \vec A^T \vec B
}}
{{eqn | r = \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}
      | c = Entries $*$ contribute nothing because of matching zero in $\vec B$.
}}                                                                                                     
{{eqn | r = x_1 + \cdots + x_n - \paren {y_1 + \cdots + y_n}
}}
{{end-eqn}}
{{qed}}