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.
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $46 \ \text{(iv)}$