Determinant of Matrix Product/Proof 3

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Theorem

Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be a square matrices of order $n$.

Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.

Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$.


Then:

$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$


That is, the determinant of the product is equal to the product of the determinants.


Proof

The Cauchy-Binet Formula gives:

$\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m}}\right) \det \left({\mathbf B_{j_1 j_2 \ldots j_m}}\right)$

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 $\mathbf B$.

When $m = n$, the only set $j_1, j_2, \ldots, j_m$ that fulfils $1 \le j_1 < j_2 < \cdots < j_m \le n$ is $\left\{ {1, 2, \ldots, n}\right\}$.

Hence the result.

$\blacksquare$


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