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

where $\det$ denotes the determinant.

Proof
Let $\tuple {k_1, k_2, \ldots, k_m}$ be an ordered $m$-tuple of integers.

Let $\map \epsilon {k_1, k_2, \ldots, k_m}$ denote the sign of $\tuple {k_1, k_2, \ldots, k_m}$.

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

Then from Transposition is of Odd Parity:
 * $\map \epsilon {l_1, l_2, \ldots, l_m} = -\map \epsilon {k_1, k_2, \ldots, k_m}$

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

That is:
 * $j_1 \le j_2 \le \cdots \le j_m$

Then it follows that:
 * $\map \det {\mathbf B_{k_1 \cdots k_m} } = \map \epsilon {k_1, k_2, \ldots, k_m} \map \det {\mathbf B_{j_1 \cdots j_m} }$

Hence:

If two $j$s are equal:
 * $\map \det {\mathbf A_{j_1 \cdots j_m} } = 0$

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