Cauchy-Binet Formula/Example/Matrix by Transpose

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

Let $\mathbf A^\intercal$ be the transpose $\mathbf A$.

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 $\mathbf A^\intercal_{j_1 j_2 \ldots j_m}$ denote the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf A^\intercal$.

Then:
 * $\ds \map \det {\mathbf A \mathbf A^\intercal} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \paren {\map \det {\mathbf A_{j_1 j_2 \ldots j_m} } }^2$

where $\det$ denotes the determinant.

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

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 $\mathbf B$.

From the definition of transpose $\mathbf A^\intercal$ is an $n \times m$ matrix.

Hence the Cauchy-Binet Formula can be applied directly:
 * $\ds \map \det {\mathbf A \mathbf A^\intercal} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf A^\intercal_{j_1 j_2 \ldots j_m} }$

Note that by construction:
 * $\mathbf A_{j_1 j_2 \ldots j_m}$ is a square matrix

Also, by definition of transpose:
 * $\mathbf A^\intercal_{j_1 j_2 \ldots j_m} = \paren {\mathbf A_{j_1 j_2 \ldots j_m} }^\intercal$

The result follows from Determinant of Transpose:
 * $\map \det {\mathbf A} = \map \det {\mathbf A^\intercal}$