Definition:Cauchy Determinant

Theorem
Let $C_n$ be a square Cauchy matrix of order $n$ given by:


 * $\begin{bmatrix}

\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_m + y_1} & \dfrac 1 {x_m + y_2} & \cdots & \dfrac 1 {x_m + y_n} \\ \end{bmatrix}$

Then the determinant of $C_n$ is given by:


 * $\det \left({C_n}\right) = \dfrac {\displaystyle \prod_{1 \le i < j \le n} \left({x_j - x_i}\right) \left({y_j - y_i}\right)} {\displaystyle \prod_{1 \le i, j \le n} \left({x_i + y_j}\right)}$

If $C_n$ is given by:


 * $\begin{bmatrix}

\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_m - y_1} & \dfrac 1 {x_m - y_2} & \cdots & \dfrac 1 {x_m - y_n} \\ \end{bmatrix}$

then its determinant is given by:


 * $\det \left({C_n}\right) = \dfrac {\displaystyle \prod_{1 \le i < j \le n} \left({x_j - x_i}\right) \left({y_j - y_i}\right)} {\displaystyle \prod_{1 \le i, j \le n} \left({x_i - y_j}\right)}$

Proof
Take the version of the Cauchy matrix defined such that $a_{ij} = \dfrac 1 {x_i + y_j}$.

Subtract column 1 from each of columns 2 to $n$.

Thus:

From Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant.

Now:
 * extract the factor $\dfrac 1 {x_i + y_1}$ from each row $1 \le i \le n$;
 * 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:


 * $\displaystyle \det \left({C_n}\right) = \left({\prod_{i = 1}^n \frac 1 {x_i + y_1}}\right) \left({\prod_{j = 2}^n y_1 - y_j}\right) \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_m + y_2} & \dfrac 1 {x_m + y_3} & \cdots & \dfrac 1 {x_m + y_n} \\ \end{vmatrix}$

Now subtract row 1 from each of rows 2 to $n$.

Column 1 will go to zero for all but the first row.

Columns 2 to $n$ will become:

From Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant.

Now:
 * extract the factor $x_1 - x_i$ from each row $2 \le i \le n$;
 * 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:


 * $\displaystyle \det \left({C_n}\right) = \left({\prod_{i = 1}^n \frac 1 {x_i + y_1}}\right) \left({\prod_{j = 1}^n \frac 1 {x_1 + y_j}}\right) \left({\prod_{i = 2}^n x_1 - x_i}\right) \left({\prod_{j = 2}^n y_1 - y_j}\right) \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_m + y_2} & \dfrac 1 {x_m + y_3} & \cdots & \dfrac 1 {x_m + y_n} \\ \end{vmatrix}$

From Determinant with Unit Element in Otherwise Zero Row, and tidying up the products, we get:


 * $\det \left({C_n}\right) = \frac {\displaystyle \prod_{i = 2}^n \left({x_i - x_1}\right) \left({y_i - y_1}\right)} {\displaystyle \prod_{1 \le i, j \le n} \left({x_i + y_1}\right) \left({x_1 + y_j}\right)}

\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_m + y_2} & \dfrac 1 {x_m + y_3} & \cdots & \dfrac 1 {x_m + y_n} \\ \end{vmatrix}$

Repeat the process for the remaining rows and columns $2$ to $n$.

The result follows.

A similar process obtains the result for the $a_{ij} = \dfrac 1 {x_i - y_j}$ form.