Cauchy-Binet Formula/Example/m greater than n

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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 right hand side of $(1)$ is a vacuous summation.

Hence the result.

$\blacksquare$


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