Value of Cauchy Determinant/Proof 2

Proof
Let:

To be proved:

Assume hereafter that set $\set {x_1,\ldots,x_n,y_1,\ldots,y_n}$ consists of distinct values, because otherwise $\det \paren {C}$ is undefined or zero.

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

Determinant of $C$ Calculation:

Lemma: $\det \paren {P} = \paren {-1}^m \paren { \det \paren {V_x} }^2$ where $m=\frac 1 2 n \paren {n-1}$


 * Details: Determinant $\det \paren {P}$ expands to:


 * Pair factors $\paren {x_r - x_s}$ and $\paren {x_s - x_r}$ into factor $- \paren {x_s - x_r}^2$, then:

Apply the Lemma to equation (2):