# 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.

Let $C$ denote the Cauchy matrix of order $3$:

$C = \begin {pmatrix} \dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \dfrac 1 {x_1 - y_3} \\ \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \dfrac 1 {x_2 - y_3} \\ \dfrac 1 {x_3 - y_1} & \dfrac 1 {x_3 - y_2} & \dfrac 1 {x_3 - y_3} \\ \end{pmatrix}$

where the values in $\set {x_1, x_2, x_3, y_1, y_2, y_3}$ are assumed to be distinct.

Then:

 $\ds C$ $=$ $\ds -P V_x^{-1} V_y Q^{-1}$ Vandermonde Matrix Identity for Cauchy Matrix $\ds \map \det C$ $=$ $\ds \paren {-1}^3 \dfrac {\paren {x_3 - x_1} \paren {x_3 - x_2} \paren {x_2 - x_1} \paren {y_3 - y_1} \paren {y_3 - y_2} \paren {y_2 - y_1} } {\paren {x_1 - y_1} \paren {x_1 - y_2} \paren {x_1 - y_3} \paren {x_2 - y_1} \paren {x_2 - y_2} \paren {x_2 - y_3} \paren {x_3 - y_1} \paren {x_3 - y_2} \paren {x_3 - y_3} }$ Determinant of Matrix Product

## Proof

Define Vandermonde matrices

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

Define polynomials:

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

Define invertible diagonal matrices:

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

Then:

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

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 $\ds$ $=$ $\ds \paren {-1}^3 \dfrac {\map \det P \map \det {V_y} } {\map \det {V_x} \map \det Q}$ from $\det \begin {pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end {pmatrix} = \paren {-1}^3$

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

$\blacksquare$