It suffices to prove the Theorem for Cauchy matrix: {{begin-eqn}}                         {{eqn | l = C       | r = \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}       | 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 = \paren { \begin{smallmatrix} 1 \\ \vdots \\ 1 \end{smallmatrix}  }       | 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 = -PV_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 =              \paren {\begin{smallmatrix}             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{smallmatrix} },             \quad              V_y =              \paren {\begin{smallmatrix}             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{smallmatrix} }       | c = Vandermonde matrices }} {{eqn | l = \map p {x}       | r =              \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        | c = Polynomials   }} {{eqn | l = P       | r = \paren {\begin{smallmatrix}                     \map {p_1} {x_1} &  \cdots & 0 \\                     \vdots           & \ddots  & \vdots \\                     0                & \cdots  & \map {p_n} {x_n} \\                     \end{smallmatrix} },                      \quad             Q =              \paren {\begin{smallmatrix}                     \map p {y_1}  & \cdots  & 0 \\                     \vdots        & \ddots  & \vdots \\                     0             & \cdots  & \map p {y_n} \\                     \end{smallmatrix} }       | c = Diagonal matrices }} {{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 = Equation (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 = Equation (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 $\paren { \begin{smallmatrix} -\map p {y_1} \\ \vdots \\ -\map p {y_n} \\ \end{smallmatrix} }$ 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 = \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}       | 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 = Equation (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}              \displaystyle\sum_{k \mathop = 1}^n \frac {1}{ \map {p_k} {x_k} } \\             \vdots \\             \displaystyle\sum_{k \mathop = 1}^n \frac { x_n^{n-1} }{ \map {p_n} {x_n} } \\             \end{pmatrix} }} {{eqn | r = \paren {                \begin{smallmatrix}                0 \\               0 \\               \vdots \\               1 \\               \end{smallmatrix}             }        | 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 = \paren {                \begin{smallmatrix}                * \\               * \\               \vdots \\               x_1 + \cdots + x_n - \paren { y_1 + \cdots + y_n }  \\               \end{smallmatrix}              }^T              \paren {                \begin{smallmatrix}                0 \\               0 \\               \vdots \\               1 \\               \end{smallmatrix}              }       | c = Entries $*$ don't contribute because of matching zero in $\vec B$. }}                                                                                                      {{eqn | r = x_1 + \cdots + x_n - \paren { y_1 + \cdots + y_n} }} {{end-eqn}} {{qed}}