# Vandermonde Matrix Identity for Cauchy Matrix/Examples/3x3

## Example of Vandermonde Matrix Identity for Cauchy 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.

Then:

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

Details:

Define Vandermonde matrices

$V_x = \begin {pmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ x_1^2 & x_2^2 & x_3^2 \\ \end {pmatrix}, \quad V_y = \begin {pmatrix} 1 & 1 & 1 \\ y_1 & y_2 & y_3 \\ y_1^2 & y_2^2 & y_3^2 \\ \end {pmatrix}$

Define polynomials:

$\map p x = \paren {x - x_1} \paren {x - x_2} \paren {x - x_3}$
$\map {p_1} x = \paren {x - x_2} \paren {x - x_3}, \quad \map {p_2} x = \paren {x - x_1} \paren {x - x_3}, \quad \map {p_3} x = \paren {x - x_1} \paren {x - x_2}$

Define invertible diagonal matrices:

$P = \begin {pmatrix} \map {p_1} {x_1} & 0 & 0 \\ 0 & \map {p_2} {x_2} & 0 \\ 0 & 0 & \map {p_3} {x_3} \\ \end {pmatrix}, \quad Q = \begin {pmatrix} \map p {y_1} & 0 & 0 \\ 0 & \map p {y_2} & 0 \\ 0 & 0 & \map p {y_3} \\ \end {pmatrix}$

Then:

$P = \begin {pmatrix} \paren {x_1 - x_2} \paren {x_1 - x_3} & 0 & 0 \\ 0 & \paren {x_2 - x_1}\paren {x_2 - x_3} & 0 \\ 0 & 0 & \paren {x_3 - x_1}\paren {x_3 - x_2} \\ \end {pmatrix}, \quad Q = \begin {pmatrix} \paren {y_1 - x_1}\paren {y_1 - x_2}\paren {y_1 - x_3} & 0 & 0 \\ 0 & \paren {y_2 - x_1}\paren {y_2 - x_2}\paren {y_2 - x_3} & 0 \\ 0 & 0 & \paren {y_3 - x_1}\paren {y_3 - x_2}\paren {y_3 - x_3} \\ \end {pmatrix}$
$\map \det P = \paren {x_1 - x_2} \paren {x_1 - x_3} \paren {x_2 - x_1} \paren {x_2 - x_3} \paren {x_3 - x_1} \paren {x_3 - x_2}$
$\map \det Q = \paren {y_1 - x_1} \paren {y_1 - x_2} \paren {y_1 - x_3} \paren {y_2 - x_1} \paren {y_2 - x_2} \paren {y_2 - x_3} \paren {y_3 - x_1} \paren {y_3 - x_2} \paren {y_3 - x_3}$
$\map \det {V_x} = \paren {x_3 - x_2} \paren {x_3 - x_1} \paren {x_2 - x_1}$
$\map \det {V_y } = \paren {y_3 - y_2} \paren {y_3 - y_1} \paren {y_2 - y_1}$
$\map \det {V_x^{-1} } = \dfrac 1 {\map \det {V_x} }, \quad \map \det {Q^{-1} } = \dfrac 1 {\map \det Q}$

Then:

 $\ds \map \det C$ $=$ $\ds \map \det {-I} \map \det P \map \det {V_x^{-1} } \map \det {V_y} \map \det {Q^{-1} }$ Determinant of Matrix Product $\ds$ $=$ $\ds \paren {-1}^3 \dfrac {\map \det P \map \det {V_y} } {\map \det {V_x} \map \det Q}$ $\det \begin {pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end {pmatrix} = \paren {-1}^3$

Insert the four determinant equations and simplify to obtain the equation for $3 \times 3$ $\map \det C$.

$\blacksquare$