Scalar Multiplication Corresponds to Multiplication by 1x1 Matrix

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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:

\(\ds c_i\) \(=\) \(\ds \sum_{j \mathop = 1}^i a_{j j} b_j\) Definition of Matrix Product (Conventional)
\(\ds \) \(=\) \(\ds a b_j\) Definition of $\mathbf A$

and:

\(\ds d_i\) \(=\) \(\ds a b_j\) Definition of Matrix Scalar Product
\(\ds \) \(=\) \(\ds c_i\) from above

$\blacksquare$


Sources

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