Theorem: Value of Cauchy Determinant

Let $D_n$ be a Cauchy determinant of order $n$:
:<nowiki>$\begin{vmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_2} & \cdots & \dfrac 1 {x_n + y_n} \\
\end{vmatrix}$</nowiki>
Then the value of $D_n$ is given by:
:$D_n = \dfrac {\ds \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i} \paren {y_j - y_i} } {\ds \prod_{1 \mathop \le i, \, j \mathop \le n} \paren {x_i + y_j} }$
Let $D_n$ be given by:
:<nowiki>$\begin {vmatrix}
\dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\
\dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\
\end {vmatrix}$</nowiki>
Then its determinant is given by:
:$D_n = \dfrac {\ds \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i} \paren {y_i - y_j} } {\ds \prod_{1 \mathop \le i, \, j \mathop \le n} \paren {x_i - y_j} }$

Proof:
Take the version of the Cauchy matrix defined such that $a_{i j} = \dfrac 1 {x_i + y_j}$.
Subtract column $1$ from each of columns $2$ to $n$.
Thus:
{{begin-eqn}}
{{eqn | l = a_{ij}
      | o = \gets
      | r = \frac 1 {x_i + y_j} - \frac 1 {x_i + y_1}
      | c = 
}}
{{eqn | r = \frac {\paren {x_i + y_1} - \paren {x_i + y_j} } {\paren {x_i + y_j} \paren {x_i + y_1} }
      | c = 
}}
{{eqn | r = \paren {\frac {y_1 - y_j} {x_i + y_1} } \paren {\frac 1 {x_i + y_j} }
      | c = 
}}
{{end-eqn}}
From Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant.
Now:
:$1$: extract the factor $\dfrac 1 {x_i + y_1}$ from each row $1 \le i \le n$
:$2$: extract the factor $y_1 - y_j$ from each column $2 \le j \le n$.
Thus from Determinant with Row Multiplied by Constant we have the following:
:<nowiki>$\ds D_n = \paren {\prod_{i \mathop = 1}^n \frac 1 {x_i + y_1} } \paren {\prod_{j \mathop = 2}^n y_1 - y_j} \begin {vmatrix}
1 & \dfrac 1 {x_1 + y_2} & \dfrac 1 {x_1 + y_3} & \cdots & \dfrac 1 {x_1 + y_n} \\
1 & \dfrac 1 {x_2 + y_2} & \dfrac 1 {x_2 + y_3} & \cdots & \dfrac 1 {x_2 + y_n} \\
1 & \dfrac 1 {x_3 + y_2} & \dfrac 1 {x_3 + y_3} & \cdots & \dfrac 1 {x_3 + y_n} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \dfrac 1 {x_n + y_2} & \dfrac 1 {x_n + y_3} & \cdots & \dfrac 1 {x_n + y_n} \\
\end{vmatrix}$</nowiki>
Now subtract row $1$ from each of rows $2$ to $n$.
Column $1$ will go to $0$ for all but the first row.
Columns $2$ to $n$ will become:
{{begin-eqn}}
{{eqn | l = a_{i j}
      | o = \gets
      | r = \frac 1 {x_i + y_j} - \frac 1 {x_1 + y_j}
      | c = 
}}
{{eqn | r = \frac {\paren {x_1 + y_j} - \paren {x_i + y_j} } {\paren {x_i + y_j} \paren {x_1 + y_j} }
      | c = 
}}
{{eqn | r = \paren {\frac {x_1 - x_i} {x_1 + y_j} } \paren {\frac 1 {x_i + y_j} }
      | c = 
}}
{{end-eqn}}
From Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant.
Now:
:$1$: extract the factor $x_1 - x_i$ from each row $2 \le i \le n$
:$2$: extract the factor $\dfrac 1 {x_1 + y_j}$ from each column $2 \le j \le n$.
Thus from Determinant with Row Multiplied by Constant we have the following:
:<nowiki>$\ds D_n = \paren {\prod_{i \mathop = 1}^n \frac 1 {x_i + y_1} } \paren {\prod_{j \mathop = 1}^n \frac 1 {x_1 + y_j} } \paren {\prod_{i \mathop = 2}^n x_1 - x_i} \paren {\prod_{j \mathop = 2}^n y_1 - y_j} \begin {vmatrix}
1 & 1 & 1 & \cdots & 1 \\
0 & \dfrac 1 {x_2 + y_2} & \dfrac 1 {x_2 + y_3} & \cdots & \dfrac 1 {x_2 + y_n} \\
0 & \dfrac 1 {x_3 + y_2} & \dfrac 1 {x_3 + y_3} & \cdots & \dfrac 1 {x_3 + y_n} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & \dfrac 1 {x_n + y_2} & \dfrac 1 {x_n + y_3} & \cdots & \dfrac 1 {x_n + y_n} \\
\end {vmatrix}$</nowiki>
From Determinant with Unit Element in Otherwise Zero Row, and tidying up the products, we get:
:<nowiki>$D_n = \frac {\ds \prod_{i \mathop = 2}^n \paren {x_i - x_1} \paren {y_i - y_1} } {\ds \prod_{1 \mathop \le i, j \mathop \le n} \paren {x_i + y_1} \paren {x_1 + y_j} }
\begin {vmatrix}
\dfrac 1 {x_2 + y_2} & \dfrac 1 {x_2 + y_3} & \cdots & \dfrac 1 {x_2 + y_n} \\
\dfrac 1 {x_3 + y_2} & \dfrac 1 {x_3 + y_3} & \cdots & \dfrac 1 {x_3 + y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n + y_2} & \dfrac 1 {x_n + y_3} & \cdots & \dfrac 1 {x_n + y_n} \\
\end {vmatrix}$</nowiki>
Repeat the process for the remaining rows and columns $2$ to $n$.
The result follows.
{{qed}}
A similar process obtains the result for the $a_{i j} = \dfrac 1 {x_i - y_j}$ form.