Cauchy-Binet Formula/Example/m greater than n

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

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

Let $m > n$.

Then:
 * $\map \det {\mathbf A \mathbf B} = 0$

Proof
The Cauchy-Binet Formula gives:
 * $(1): \quad \ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$

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 $\mathbf B$.

But here $m > n$.

Therefore the set $\set {j_1, j_2, \ldots, j_m}$ such that:
 * $1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n$

is the empty set.

Thus the of $(1)$ is a vacuous summation.

Hence the result.