Scalar Multiplication Corresponds to Multiplication by 1x1 Matrix

Theorem
Let $\map \MM 1$ denote the matrix space of square matrices of order $1$.

Let $\map \MM {1, n}$ denote the matrix space of order $1 \times n$.

Let $\mathbf A = \begin {pmatrix} a \end {pmatrix} \in \map \MM 1$ and $\mathbf B = \begin {pmatrix} b_1 & b_2 & \cdots & b_n \end{pmatrix} \in \map \MM {1, n}$.

Let $\mathbf C = \mathbf A \mathbf B$ denote the (conventional) matrix product of $\mathbf A$ with $\mathbf B$.

Let $\mathbf D = a \mathbf B$ denote the matrix scalar product of $a$ with $\mathbf B$.

Then $\mathbf C = \mathbf D$.

Proof
By definition of (conventional) matrix product, $\mathbf C$ is of order $1 \times n$.

By definition of matrix scalar product, $\mathbf D$ is also of order $1 \times n$.

Consider arbitrary elements $c_i \in \mathbf C$ and $d_i \in \mathbf D$ for some index $i$ where $1 \le i \le n$.

We have:

and: