# Cauchy-Binet Formula

## Theorem

Let $\mathbf A$ be an $m \times n$ matrix.

Let $\mathbf B$ be an $n \times m$ matrix.

Let $1 \le j_1, j_2, \ldots, j_m \le n$.

Let $\mathbf A_{j_1 j_2 \ldots j_m}$ denote the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.

Let $\mathbf B_{j_1 j_2 \ldots j_m}$ denote the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$.

Then:

$\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m} }\right) \det \left({\mathbf B_{j_1 j_2 \ldots j_m} }\right)$

where $\det$ denotes the determinant.

## Proof

Let $\left({k_1, k_2, \ldots, k_m}\right)$ be an ordered $m$-tuple of integers.

Let $\epsilon \left({k_1, k_2, \ldots, k_m}\right)$ denote the sign of $\left({k_1, k_2, \ldots, k_m}\right)$.

Let $\left({l_1, l_2, \ldots, l_m}\right)$ be the same as $\left({k_1, k_2, \ldots, k_m}\right)$ except for $k_i$ and $k_j$ having been transposed.

Then from Transposition is of Odd Parity:

$\epsilon \left({l_1, l_2, \ldots, l_m}\right) = - \epsilon \left({k_1, k_2, \ldots, k_m}\right)$

Let $\left({j_1, j_2, \ldots, j_m}\right)$ be the same as $\left({k_1, k_2, \ldots, k_m}\right)$ by arranged into non-decreasing order.

That is:

$j_1 \le j_2 \le \cdots \le j_m$

Then it follows that:

$\det \left({\mathbf B_{k_1 \cdots k_m} }\right) = \epsilon \left({k_1, k_2, \ldots, k_m}\right) \det \left({\mathbf B_{j_1 \cdots j_m} }\right)$

Hence:

 $\displaystyle \det \left({\mathbf A \mathbf B}\right)$ $=$ $\displaystyle \sum_{1 \mathop \le l_1, \mathop \ldots \mathop , l_m \mathop \le m} \epsilon \left({l_1, \ldots, l_m}\right) \left({\sum_{k \mathop = 1}^n a_{1 k} b_{k l_1} }\right) \cdots \left({\sum_{k \mathop = 1}^n a_{m k} b_{k l_m} }\right)$ $\quad$ Definition of Determinant $\quad$ $\displaystyle$ $=$ $\displaystyle \sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \cdots a_{m k_m} \sum_{1 \mathop \le l_1, \mathop \ldots \mathop , l_m \mathop \le m} \epsilon \left({l_1, \ldots, l_m}\right) b_{k_1 l_1} \cdots b_{k_m l_m}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \cdots a_{m k_m} \det \left({\mathbf B_{k_1 \cdots k_m} }\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \epsilon \left({k_1, \ldots, k_m}\right) \cdots a_{m k_m} \det \left({\mathbf B_{j_1 \cdots j_m} }\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sum_{1 \mathop \le j_1 \mathop \le j_2 \mathop \le \cdots \mathop \le j_m \le n} \det \left({\mathbf A_{j_1 \cdots j_m} }\right) \det \left({\mathbf B_{j_1 \cdots j_m} }\right)$ $\quad$ $\quad$

If two $j$s are equal, $\det \left({\mathbf A_{j_1 \cdots j_m} }\right) = 0$.

$\blacksquare$

## Examples

### Cauchy-Binet Formula: $m = n$

Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be a square matrices of order $n$.

Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.

Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$.

Then:

$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$

That is, the determinant of the product is equal to the product of the determinants.

### Cauchy-Binet Formula: $m = 1$

Let $\mathbf A = \left[{a}\right]_{1 n}$ be a row matrix with $n$ columns.

and $\mathbf B = \left[{b}\right]_{n 1}$ be a column matrix with $n$ rows.

Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$.

Then:

$\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{j \mathop = 1}^n a_j b_j$

where:

$a_j$ is element $a_{1 j}$ of $\mathbf A$
$b_j$ is element $b_{j 1}$ of $\mathbf B$.

### Cauchy-Binet Formula: Matrix by Transpose

Let $\mathbf A$ be an $m \times n$ matrix.

Let $\mathbf A^\intercal$ be the transpose $\mathbf A$.

Let $1 \le j_1, j_2, \ldots, j_m \le n$.

Let $\mathbf A_{j_1 j_2 \ldots j_m}$ denote the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.

Let $\mathbf A^\intercal_{j_1 j_2 \ldots j_m}$ denote the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf A^\intercal$.

Then:

$\displaystyle \det \left({\mathbf A \mathbf A^\intercal}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \left({\det \left({\mathbf A_{j_1 j_2 \ldots j_m} }\right)}\right)^2$

where $\det$ denotes the determinant.

### Cauchy-Binet Formula: $m > n$

Let $\mathbf A$ be an $m \times n$ matrix.

Let $\mathbf B$ be an $n \times m$ matrix.

Let $m > n$.

Then:

$\displaystyle \det \left({\mathbf A \mathbf B}\right) = 0$

### Cauchy-Binet Formula: $m = 2$

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

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

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

## Also known as

The Cauchy-Binet Formula is also known, confusingly, as the Binet-Cauchy Identity, which is a direct consequence of this.

## Source of Name

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

## Historical Note

The Cauchy-Binet Formula was presented by Augustin Louis Cauchy and Jacques Philippe Marie Binet to Journal de l'École Polytechnique on the same day in $1812$.

## Sources

• 1813: J. BinetMémoire sur un système du formules analytiques, et leur application à des considérations géométriques (J. l'École Polytechnique Vol. 9: 280 – 354)
• 1815: CauchyMémoire sur les fonctions qui ne peuvent obtenir que deux valeurs égales et de signes contraires par suite des transpositions operees entre les variables qu'elles renferment (J. l'École Polytechnique Vol. 10: 29 – 112)