Vandermonde Matrix Identity for Cauchy Matrix/Examples/3x3

EXAMPLE 3x3
Illustrate $3\times 3$ case for Vandermonde Matrix Identity for Cauchy Matrix and Value of Cauchy Determinant:

Then:

Details:

Define Vandermonde matrices


 * $\displaystyle V_x = \paren {\begin{smallmatrix}

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

Define polynomials:


 * $\displaystyle \map p x = \paren {x-x_1} \paren {x-x_2} \paren {x-x_3}$


 * $\displaystyle \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:


 * $\displaystyle P = \paren {\begin{smallmatrix}

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

Then:
 * $\displaystyle P = \paren {\begin{smallmatrix}

\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{smallmatrix} }, \quad Q = \paren {\begin{smallmatrix} \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{smallmatrix} }$

Determinant of Diagonal Matrix implies


 * $\displaystyle \det\paren { P } = \paren {x_1 - x_2} \paren {x_1 - x_3} \paren {x_2 - x_1}\paren {x_2 - x_3} \paren {x_3 - x_1}\paren {x_3 - x_2}$


 * $\displaystyle \det\paren { 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}$

Vandermonde Determinant implies


 * $\displaystyle \det\paren {V_x} =

\paren {x_3 - x_2} \paren {x_3 - x_1} \paren {x_2 - x_1}$


 * $\displaystyle \det\paren {V_y } = \paren {y_3 - y_2} \paren {y_3 - y_1} \paren {y_2 - y_1}$

Determinant of Matrix Product and Definition:Inverse Matrix imply


 * $\displaystyle \det\paren {V_x^{-1} } =\dfrac{1}{\det\paren {V_x} },\quad

\det\paren {Q^{-1} } =\dfrac{1}{\det\paren {Q} }$

Then:

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

Case $n\times n$
The methods of the $3\times 3$ example apply unchanged:

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


 * $\det \paren {\begin{smallmatrix}

\frac {1} {x_1 - y_1} & \frac {1} {x_1 - y_2} & \cdots & \frac {1} {x_1 - y_n} \\ \frac {1} {x_2 - y_1} & \frac 1 {x_2 - y_2}  & \cdots & \frac {1} {x_2 - y_n} \\ \vdots               & \vdots                & \cdots & \vdots \\ \frac {1} {x_n - y_1} & \frac {1} {x_n - y_2} & \cdots & \frac {1} {x_n - y_n} \\ \end{smallmatrix} } = (-1)^n \dfrac {\prod_{1 \mathop \le j < 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} } {\prod_{i \mathop = 1}^n \prod_{j \mathop = 1}^n \paren {x_i - y_j} }$ Value of Cauchy Determinant

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


 * $\det \paren {\begin{smallmatrix}

\frac {1} {x_1 + y_1} & \frac {1} {x_1 + y_2} & \cdots & \frac {1} {x_1 + y_n} \\ \frac {1} {x_2 + y_1} & \frac 1 {x_2 + y_2} & \cdots & \frac {1} {x_2 + y_n} \\ \vdots & \vdots & \cdots & \vdots \\ \frac {1} {x_n + y_1} & \frac {1} {x_n + y_2} & \cdots & \frac {1} {x_n + y_n} \\ \end{smallmatrix} } = (-1)^n \dfrac {\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} } {\prod_{i \mathop = 1}^n \prod_{j \mathop = 1}^n \paren {x_i + y_j} }$ Value of Cauchy Determinant