# Value of Cauchy Determinant

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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 1

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.

## Proof 2

Let:

 $\ds C$ $=$ $\ds \begin {bmatrix} \dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\ \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\ \vdots & \vdots & \ddots & \vdots \\ \dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\ \end {bmatrix}$

To be proved:

 $\ds \map \det C$ $=$ $\ds \dfrac {\ds \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i} \paren {y_i - y_j} } {\ds \prod_{1 \mathop \le i, j \mathop \le n} \paren {x_i - y_j} }$ replacing $y_k \to -y_k$ in $C$ and $\map \det C$

Assume hereafter that set $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ consists of distinct values, because otherwise $\map \det C$ is undefined or zero.

Preliminaries:

Vandermonde Matrix Identity for Cauchy Matrix supplies matrix equation

$(1): \quad - C = P V_x^{-1} V_y Q^{-1}$
Definitions of symbols:
$V_x = \begin {pmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ \vdots & \vdots & \ddots & \vdots \\ {x_1}^{n - 1} & {x_2}^{n - 1} & \cdots & {x_n}^{n - 1} \\ \end {pmatrix}$
$V_y = \begin {pmatrix} 1 & 1 & \cdots & 1 \\ y_1 & y_2 & \cdots & y_n \\ \vdots & \vdots & \ddots & \vdots \\ {y_1}^{n - 1} & {y_2}^{n - 1} & \cdots & {y_n}^{n - 1} \\ \end {pmatrix}$ Definition of Vandermonde Matrix
$P = \begin {pmatrix} \map {p_1} {x_1} & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \map {p_n} {x_n} \\ \end {pmatrix}$
$Q = \begin {pmatrix} \map p {y_1} & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \map p {y_n)} \\ \end {pmatrix}$ Definition of Diagonal Matrix
$1 \mathop \le k \mathop \le n$ Polynomials
$\ds \map p x = \prod_{i \mathop = 1}^n \paren {x - x_i}$
$\ds \map {p_k} x = \prod_{i \mathop = 1, i \mathop \ne k}^n \paren {x - x_i}$

Determinant of $C$ Calculation:

 $\ds C$ $=$ $\ds \paren {-I} P V_x^{-1} V_y Q^{-1}$ Vandermonde Matrix Identity for Cauchy Matrix Symbol $I$ is the $n \times n$ identity matrix. $\text {(2)}: \quad$ $\ds \map \det C$ $=$ $\ds \map \det {-I} \map \det P \map \det {V_x^{-1} } \map \det {V_y} \map \det {Q^{-1} }$ Determinant of Matrix Product $\ds$ $=$ $\ds \paren {-1}^n \map \det I \map \det P \map \det {V_x^{-1} } \map \det {V_y} \map \det {Q^{-1} }$ Effect of Elementary Row Operations on Determinant: factoring constant $-1$ from all rows of $-I$ $\ds$ $=$ $\ds \paren {-1}^n \map \det I \dfrac {\map \det p \map \det {V_y} } {\map \det Q \map \det {V_x} }$ Matrix Product with Adjugate Matrix and Determinant of Matrix Product

Lemma: $\map \det P = \paren {-1}^m \paren {\map \det {V_x} }^2$ where $m = \frac 1 2 n \paren {n - 1}$

Details: Determinant $\map \det P$ expands to:
 $\ds \prod_{j \mathop = 1}^n \map {p_j} {x_j}$ $=$ $\ds \prod_{j \mathop = 1}^n \prod_{k \mathop = 1, k \mathop \ne j}^n \paren {x_j - x_k}$ Definition of polynomials $\map {p_j} x$
Pair factors $\paren {x_r - x_s}$ and $\paren {x_s - x_r}$ into factor $-\paren {x_s - x_r}^2$, then:
 $\ds \map \det P$ $=$ $\ds \paren {-1}^m \paren {\prod_{1 \mathop \le r \mathop < s \mathop \le n} \paren {x_s - x_r }^2}$ where $m = 1 + \cdots + \paren {n - 1} = \dfrac 1 2 n \paren {n - 1}$ $\ds$ $=$ $\ds \paren {-1}^m \paren {\map \det {V_x} }^2$ Value of Vandermonde Determinant

$\Box$

Apply the Lemma to equation (2):

 $\ds \map \det C$ $=$ $\ds \paren {-1}^{n + m} \paren {\dfrac {\map \det {V_x} \map \det {V_y} } {\map \det Q} }$ $m = \dfrac 1 2 n \paren {n - 1}$ $\ds$ $=$ $\ds \paren {-1}^{n + m} \dfrac {\ds \prod_{1 \mathop \le m \mathop < k \mathop \le n}^{\phantom n} \paren {x_k - x_m} \prod_{1 \mathop \le m \mathop < k \mathop \le n} \paren {y_k - y_m} } {\ds \prod_{k \mathop = 1}^n \map p {y_k} }$ Value of Vandermonde Determinant and Determinant of Diagonal Matrix $\ds$ $=$ $\ds \paren {-1}^{n + m} \dfrac {\ds \prod_{1 \mathop \le m \mathop < k \mathop \le n}^{\phantom n} \paren {x_k - x_m} \paren {y_k - y_m} } {\ds \prod_{k \mathop = 1}^n \prod_{j \mathop = 1}^n \paren {y_k - x_j} }$ Definition of $\map p x$ $\ds$ $=$ $\ds \paren {-1}^{n + m} \dfrac {\ds \paren {-1}^m \prod_{1 \mathop \le m \mathop < k \mathop \le n}^{\phantom n} \paren {x_k - x_m} \paren {y_m - y_k} } {\ds \paren {-1}^{n^2} \prod_{k \mathop = 1}^n \prod_{j \mathop = 1}^n \paren {x_j - y_k} }$ factoring out changed signs $\ds$ $=$ $\ds \dfrac {\ds \prod_{1 \mathop \le m \mathop < k \mathop \le n}^{\phantom n} \paren {x_k - x_m} \paren {y_m - y_k} } {\ds \prod_{k \mathop = 1}^n \prod_{j \mathop = 1}^n \paren {x_j - y_k} }$ as all signs cancel

$\blacksquare$

## Source of Name

This entry was named for Augustin Louis Cauchy.