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 $\map \det C$ is undefined or zero.

Preliminaries:

Vandermonde Matrix Identity for Cauchy Matrix supplies matrix equation


 * $(1): \quad - C = P V_x^{-1} V_y Q^{-1}$


 * Definitions of symbols:


 * $V_x = \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}$


 * $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}$


 * $P = \begin {pmatrix}

\map {p_1} {x_1} & \cdots & 0 \\ \vdots  & \ddots  & \vdots \\ 0       & \cdots  & \map {p_n} {x_n} \\ \end {pmatrix}$


 * $Q = \begin {pmatrix}

\map p {y_1} & \cdots  & 0 \\ \vdots & \ddots  & \vdots \\ 0      & \cdots  & \map p {y_n)} \\ \end {pmatrix}$


 * $1 \mathop \le k \mathop \le n$ Polynomials
 * $\ds \map p x = \prod_{i \mathop = 1}^n \paren {x - x_i}$
 * $\ds \map {p_k} x = \prod_{i \mathop = 1, i \mathop \ne k}^n \paren {x - x_i}$

Determinant of $C$ Calculation:

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


 * Details: Determinant $\map \det P$ 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):