Value of Cauchy Determinant/Proof 1

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Theorem

Let $D_n$ be a Cauchy determinant of order $n$:

$\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}$


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:

$\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}$


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:

\(\ds a_{ij}\) \(\gets\) \(\ds \frac 1 {x_i + y_j} - \frac 1 {x_i + y_1}\)
\(\ds \) \(=\) \(\ds \frac {\paren {x_i + y_1} - \paren {x_i + y_j} } {\paren {x_i + y_j} \paren {x_i + y_1} }\)
\(\ds \) \(=\) \(\ds \paren {\frac {y_1 - y_j} {x_i + y_1} } \paren {\frac 1 {x_i + y_j} }\)

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:

$\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}$


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:

\(\ds a_{i j}\) \(\gets\) \(\ds \frac 1 {x_i + y_j} - \frac 1 {x_1 + y_j}\)
\(\ds \) \(=\) \(\ds \frac {\paren {x_1 + y_j} - \paren {x_i + y_j} } {\paren {x_i + y_j} \paren {x_1 + y_j} }\)
\(\ds \) \(=\) \(\ds \paren {\frac {x_1 - x_i} {x_1 + y_j} } \paren {\frac 1 {x_i + y_j} }\)

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:

$\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}$

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

$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}$


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

The result follows.

$\blacksquare$


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


Source of Name

This entry was named for Augustin Louis Cauchy.


Sources

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