Vandermonde Matrix Identity for Cauchy Matrix

Theorem
Assume values $\set { x_1,\ldots,x_n,y_1,\ldots,y_n }$ are distinct in matrix

Then:

Definitions of Vandermonde matrices $V_x$, $V_y$ and diagonal matrices $P$, $Q$:


 * $\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

Definitions of polynomials $p$, $p_1$, $\ldots$, $p_n$:


 * $\displaystyle p(x) = \prod_{i \mathop = 1}^n \paren {x - x_i}$


 * $\displaystyle p_k(x) = \dfrac{ \map p x}{x-x_k} = \prod_{i \mathop = 1,i \mathop \ne k}^n \, \paren {x - x_i}$, $1 \mathop \le k \mathop \le n$

Proof
Matrices $P$ and $Q$ are invertible because all diagonal elements are nonzero.

For $1\le i \le n$ express polynomial $p_i$ as:


 * $\displaystyle \map {p_i} {x} = \sum_{k \mathop = 1}^n a_{ik} x^{k-1}$

Then:

Use second equation $\map {p_i} {y_j} = \dfrac{ \map {p} {y_j} }{y_j - x_i}$:

Also see

 * Vandermonde Determinant


 * Inverse of Vandermonde Matrix


 * Hilbert Matrix is Cauchy Matrix


 * Inverse of Cauchy Matrix


 * Inverse of Hilbert Matrix


 * Value of Cauchy Determinant


 * Sum of Elements in Inverse of Cauchy Matrix


 * Sum of Elements in Inverse of Hilbert Matrix