Cauchy-Binet Formula/Example/Matrix by Transpose

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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:

$\map \det {\mathbf A} = \map \det {\mathbf A^\intercal}$

$\blacksquare$


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