'''Preliminaries''': Vandermonde Matrix Identity for Cauchy Matrix supplies matrix equation :$\displaystyle (1)\quad - C = PV_x^{-1} V_y Q^{-1}$ :Definitions of symbols: ::$\displaystyle V_x=\paren {\begin{smallmatrix} 1         & 1         & \cdots & 1 \\ x_1       & x_2       & \cdots & x_n \\ \vdots    & \vdots    & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \\ \end{smallmatrix} },\quad  V_y=\paren {\begin{smallmatrix} 1         & 1         & \cdots & 1 \\ y_1       & y_2       & \cdots & y_n \\ \vdots    & \vdots    & \ddots & \vdots \\ y_1^{n-1} & y_2^{n-1} & \cdots & y_n^{n-1} \\ \end{smallmatrix} }$ Vandermonde matrices ::$\displaystyle P= \paren {\begin{smallmatrix} p_1(x_1) &  \cdots & 0 \\ \vdots   & \ddots  & \vdots \\ 0        & \cdots  & p_n(x_n) \\ \end{smallmatrix} }, \quad Q= \paren {\begin{smallmatrix} p(y_1)  & \cdots  & 0 \\ \vdots  & \ddots  & \vdots \\ 0       & \cdots  & p(y_n) \\ \end{smallmatrix} }$  Diagonal matrices ::$\displaystyle  p(x) = \prod_{i \mathop = 1}^n \paren {x - x_i}, \quad \displaystyle  p_k(x) = \prod_{i \mathop = 1,i \mathop \ne k}^n \, \paren {x - x_i}, \quad 1 \mathop \le k \mathop \le n$ Polynomials '''Compute the inverse''' $C^{-1}$ for: {{begin-eqn}} {{eqn | l = C       | r = \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_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\             \end{bmatrix}       | c = Assume $\set {x_1,\ldots,x_n,y_1,\ldots,y_n}$ has distinct elements. }} {{end-eqn}} Replacement $y_k \to -y_k$ then gives the inverse $C_n^{-1}$ in the theorem. Inverse of Matrix Product applied to equation (1) gives:  :$ C^{-1} = -Q V_y^{-1} V_x P^{-1}$  Let ${\vec K}_1,\ldots,{\vec K}_n$ denote the columns of the $n\times n$ identity matrix. Then $n\times n$ matrix $B = C^{-1}$ has entries $b_{ij} = {\vec K}_i^T C^{-1} {\vec K}_j$. Hold fixed until the end of the proof the row and column index symbols $i$ and $j$. Define column vectors $\vec A$, $\vec B$ so that $b_{ij} = {\vec A}^T \vec B$: :$\displaystyle     \vec A = \paren { {\vec K}_i^T \, Q \, V_y^{-1} }^T,    \quad      \vec B = -V_x \, P^{-1} \, {\vec K}_j $   Define $u = x_j$ and simplify: {{begin-eqn}} {{eqn | l = \vec A        | r = {\map p {y_i} } \, \paren { V_y^{-1} }^T \, {\vec K}_i        | c = Transpose of Matrix Product }} {{eqn | r = \dfrac{ \map p {y_i} }{\det \paren {V_y} } \paren { \adj {V_y} }^T {\vec K}_i       | c = Matrix Product with Adjugate Matrix }} {{eqn | r = \dfrac{ \map p {y_i} }{\det \paren {V_y} }         \begin{bmatrix}            {\mathbf {Cofactor} } \paren {V_y,1,i } \\            \vdots \\            {\mathbf {Cofactor} } \paren {V_y,n,i } \\          \end{bmatrix}       | c = ${\mathbf {Cofactor} } \paren {M,r,s}$ denotes cofactor $M_{rs}$ }} {{eqn | l = \vec B       | r = -\dfrac{1}{\map {p_j} {x_j} } \, V_x \, {\vec K}_j }} {{eqn | r = -\dfrac{1}{\map {p_j} {x_j} } \,          \begin{bmatrix}            1 \\            u \\            \vdots \\            u^{n-1} \\          \end{bmatrix}       | c = Symbol $u = x_j$. }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = A^T\, B       | r = \dfrac             { -\map p {y_i}  }             {\map {p_j} {x_j} \det \paren {V_y} }               \sum_{k \mathop = 1}^n {\mathbf {Cofactor} } \paren {V_y,k,i } \, u^{k-1}       | c = A Cofactor expansion  along column $i$. }} {{eqn | r = \dfrac             { -\map p {y_i}  }             {\map {p_j} {x_j} \det \paren {V_y} }          \det \paren { \begin{matrix}          1         & \cdots & 1             & 1         & 1             & \cdots & 1 \\         y_1       & \cdots & y_{i-1}       & u         & y_{i+1}       & \cdots & y_n \\         \vdots    & \cdots & \vdots        &\vdots     & \vdots        & \cdots & \vdots \\         y_1^{n-1} & \cdots & y_{i-1}^{n-1} & u^{n-1}   & y_{i+1}^{n-1} & \cdots & y_n^{n-1} \\         \end{matrix}  }  }} {{eqn | r = \dfrac             { -\map p {y_i}  }             {\map {p_j} {x_j} }             \,\,              \dfrac             {               \left.                 \paren { \det \paren {V_y} }               \right \vert_{y_i \mathop = u}             }             { \det \paren {V_y} } }} {{end-eqn}} Simplify the fraction on the right: {{begin-eqn}} {{eqn | l = (2)\quad \displaystyle             \dfrac             { \displaystyle               \left.                  \paren { \det \paren {V_y} }                \right\vert_{y_i \mathop = u}              }             { \displaystyle \det \paren {V_y} }       | r = \displaystyle             \dfrac             {\displaystyle               \left.                  \paren { \prod_{1 \mathop \le m \mathop \lt k \mathop \le n} \paren { y_k - y_m  } }                \right\vert_{ y_i \mathop = u }             }             {\displaystyle               \prod_{1 \mathop \le m \mathop \lt k \mathop \le n} \paren { y_k - y_m  }             }       | c = Vandermonde Determinant }} {{end-eqn}} Define sets $D,A,B,C$ with disjoint decomposition $D = A \cup B \cup C$: {{begin-eqn}} {{eqn | l = D        | r = \set { \paren { m,k } : 1 \le m \lt k \le n } }} {{eqn | l = A        | r = \set { \paren { m,k } \in D : m \neq i \mbox{ and } k \neq i } }} {{eqn | l = B       | r = \set { \paren { m,k } \in D : m = i } }} {{eqn | l = C       | r = \set { \paren { m,k } \in D : k = i } }} {{end-eqn}} Use $\prod_{D} = \prod_{A} \prod_{B} \prod_{C}$ to convert the numerator and denominator in the right side of (2). Then: {{begin-eqn}} {{eqn | l = \displaystyle             \dfrac             { \displaystyle               \left.                  \paren { \det \paren {V_y} }                \right\vert_{y_i \mathop = u}              }             { \displaystyle \det \paren {V_y} }       | r = \displaystyle             \dfrac             {\displaystyle                \prod_{k \mathop = 1,\, k \mathop \neq i}^n \paren { u - y_k }              }             {\displaystyle                \prod_{k \mathop = 1,\, k \mathop \neq i}^n \paren { y_i - y_k }               }       | c = Common factors canceled in (2).  }} {{end-eqn}} Replacement of ${ \map p {y_i}  }$ and ${ \map {p_j} {x_j} }$ by products gives: {{begin-eqn}} {{eqn | l = (3)\quad \displaystyle             b_{ij}        | r =              \paren {-1}             \,             \dfrac             { \displaystyle \prod_{k \mathop = 1}^n \paren {y_i - x_k } }             { \displaystyle \prod_{k \mathop = 1,\, k \mathop \neq j}^n \paren { x_j - x_k } }              \,\,             \prod_{k \mathop = 1,\, k \mathop \neq i}^n              \paren {                \dfrac { x_j - y_k }{ y_i - y_k }             }       | c = Replaced $u = x_j$. }} {{end-eqn}}   After replacement $y_k \to -y_k$ and canceling a common factor, ''Knuth's original identity'' (1997) becomes: {{begin-eqn}} {{eqn | l = \displaystyle (4)\quad b_{ij}       | r = \displaystyle   \dfrac   { \displaystyle \prod_{k \mathop = 1}^n \paren { x_k - y_i } }   { \displaystyle \prod_{k \mathop = 1,\, k \mathop \neq j}^n \paren { x_j - x_k } }   \dfrac   { \displaystyle \prod_{k \mathop = 1,\, k \mathop \neq i}^n \paren { x_j - y_k } }   { \displaystyle \prod_{k \mathop = 1,\, k \mathop \neq i}^n \paren { - y_i + y_k } } }} {{end-eqn}} In (3), factor $(-1)^{n+1}$ from the numerator and $(-1)^{n-1}$ from the denominator. Then $(-1)^{n+1} = (-1)^{n-1}$ verifies that (3) matches (4).{{qed}}