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:

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

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