Vandermonde Matrix Identity for Cauchy Matrix/Examples

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Examples of Use of Vandermonde Matrix Identity for Cauchy Matrix

$3 \times 3$ Matrix

Illustrate $3 \times 3$ case for Vandermonde Matrix Identity for Cauchy Matrix and Value of Cauchy Determinant:

\(\ds C\) \(=\) \(\ds \begin {pmatrix} \dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \dfrac 1 {x_1 - y_3} \\ \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \dfrac 1 {x_2 - y_3} \\ \dfrac 1 {x_3 - y_1} & \dfrac 1 {x_3 - y_2} & \dfrac 1 {x_3 - y_3} \\ \end{pmatrix}\) Values $\set { x_1, x_2, x_3, y_1, y_2, y_3 }$ assumed distinct.

Cauchy matrix of order $3$

Then:

\(\ds C\) \(=\) \(\ds -P V_x^{-1} V_y Q^{-1}\) Vandermonde Matrix Identity for Cauchy Matrix
\(\ds \map \det C\) \(=\) \(\ds (-1)^3 \dfrac { \paren { x_{3} - x_{1} } \paren { x_{3} - x_{2} } \paren { x_{2} - x_{1} } \quad \paren { y_{3} - y_{1} } \paren { y_{3} - y_{2} } \paren { y_{2} - y_{1} } }{ \paren { x_{1} - y_{1} } \paren { x_{1} - y_{2} } \paren { x_{1} - y_{3} } \quad \paren { x_{2} - y_{1} } \paren { x_{2} - y_{2} } \paren { x_{2} - y_{3} } \quad \paren { x_{3} - y_{1} } \paren { x_{3} - y_{2} } \paren { x_{3} - y_{3} } }\) Determinant Product Theorem


$n \times n$ Matrix

The methods of the $3 \times 3$ example apply unchanged for the general $n \times n$ Cauchy matrix:

Assume values $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ are distinct. Then:

$\map \det {\begin{smallmatrix} \dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\ \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\ \vdots & \vdots & \cdots & \vdots \\ \dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\ \end{smallmatrix} } = \paren {-1}^n \dfrac {\ds \prod_{1 \mathop \le j \mathop < i \mathop \le n} \paren {x_i - x_j} \quad \prod_{1 \mathop \le j \mathop < i \mathop \le n} \paren {y_i - y_j} } {\ds \prod_{i \mathop = 1}^n \prod_{j \mathop = 1}^n \paren {x_i - y_j} }$ Value of Cauchy Determinant

Assume values $\set {x_1, \ldots, x_n, -y_1, \ldots, -y_n}$ are distinct, then replace in the preceding equation $y_i$ by $-y_i$, $1 \le i \le n$:

$\map \det {\begin{smallmatrix} \dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\ \dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\ \vdots & \vdots & \cdots & \vdots \\ \dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_2} & \cdots & \dfrac 1 {x_n + y_n} \\ \end{smallmatrix} } = \paren {-1}^n \dfrac {\ds \prod_{1 \mathop \le j \mathop < i \mathop \le n} \paren {x_i - x_j} \quad \prod_{1 \mathop \le j \mathop < i \mathop \le n} \paren {y_j - y_i} } {\ds \prod_{i \mathop = 1}^n \prod_{j \mathop = 1}^n \paren {x_i + y_j} }$ Value of Cauchy Determinant

$\blacksquare$