Cauchy-Binet Formula/Example/m equals 1

Theorem

Let $\mathbf A = \left[{a}\right]_{1 n}$ be a row matrix with $n$ columns.

and $\mathbf B = \left[{b}\right]_{n 1}$ be a column matrix with $n$ rows.

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

Then:

$\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{j \mathop = 1}^n a_j b_j$

where:

$a_j$ is element $a_{1 j}$ of $\mathbf A$
$b_j$ is element $b_{j 1}$ of $\mathbf B$.

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 $m = 1$, the relation:

$1 \le j_1 < j_2 < \cdots < j_m \le n$

degenerates to:

$1 \le j \le n$

By definition of order $1$ determinant:

$\det \left({\left[{a_j}\right]}\right) = a_j$
$\det \left({\left[{b_j}\right]}\right) = b_j$

Hence the result.

$\blacksquare$