Cauchy-Binet Formula/Example/m equals 1

Theorem
Let $\mathbf A = \sqbrk a_{1 n}$ be a row matrix with $n$ columns.

and $\mathbf B = \sqbrk b_{n 1}$ be a column matrix with $n$ rows.

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

Then:
 * $\ds \map \det {\mathbf A \mathbf B} = \sum_{j \mathop = 1}^n a_j b_j$

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:
 * $\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \mathop \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$

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 $\mathbf B$.

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:
 * $\det \sqbrk {a_j} = a_j$
 * $\det \sqbrk {b_j} = b_j$

Hence the result.