# Binet-Cauchy Identity

## Theorem

$\displaystyle \left({\sum_{i \mathop = 1}^n a_i c_i}\right) \left({\sum_{j \mathop = 1}^n b_j d_j}\right) = \left({\sum_{i \mathop = 1}^n a_i d_i}\right) \left({\sum_{j \mathop = 1}^n b_j c_j}\right) + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \left({a_i b_j - a_j b_i}\right) \left({c_i d_j - c_j d_i}\right)$

where all of the $a, b, c, d$ are elements of a commutative ring.

Thus the identity holds for $\Z, \Q, \R, \C$.

## Proof 1

Expanding the last term:

 $\displaystyle$  $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop \le n} \left({a_i b_j - a_j b_i}\right) \left({c_i d_j - c_j d_i}\right)$ $\displaystyle$ $=$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop \le n} \left({a_i c_i b_j d_j + a_j c_j b_i d_i}\right)$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop \le n} \left({a_i d_i b_j c_j + a_j d_j b_i c_i}\right)$ $\displaystyle$ $=$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop \le n} \left({a_i c_i b_j d_j + a_j c_j b_i d_i}\right) + \sum_{i \mathop = 1}^n a_i c_i b_i d_i$ These new terms are the same $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop \le n} \left({a_i d_i b_j c_j + a_j d_j b_i c_i}\right) - \sum_{i \mathop = 1}^n a_i d_i b_i c_i$ $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n a_i c_i b_j d_j - \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n a_i d_i b_j c_j$ Completing the sums $\displaystyle$ $=$ $\displaystyle \left({\sum_{i \mathop = 1}^n a_i c_i}\right) \left({\sum_{j \mathop = 1}^n b_j d_j}\right) - \left({\sum_{i \mathop = 1}^n a_i d_i}\right) \left({\sum_{j \mathop = 1}^n b_j c_j}\right)$ Factoring terms indexed by $i$ and $j$

Hence the result.

$\blacksquare$

## Proof 2

This is a special case of the Cauchy-Binet Formula:

$\displaystyle \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \, \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$

where:

$\mathbf A$ is an $m \times n$ matrix
$\mathbf B$ is an $n \times m$ matrix.
For $1 \le j_1, j_2, \ldots, j_m \le n$:
$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$.

In this case $m = 2$, giving:

$\displaystyle \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop \le n} \map \det {\mathbf A_{j_1 j_2} } \, \map \det {\mathbf B_{j_1 j_2} }$

## Also known as

The Binet-Cauchy Identity is also known as Binet's formula.

## Source of Name

This entry was named for Jacques Philippe Marie Binet and Augustin Louis Cauchy.

## Historical Note

The Binet-Cauchy Identity is a special case of the Cauchy-Binet Formula, which was presented by Jacques Philippe Marie Binet and Augustin Louis Cauchy on the same day in $1812$.