Matrix Space is Module

Theorem
Let $\struct {R, +, \circ}$ be a ring.

Let $\mathbf A = \sqbrk a_{m n}$ be an $m \times n$ matrix over $\struct {R, +, \circ}$.

Then the matrix space $\map {\MM_R} {m, n}$ of all $m \times n$ matrices over $R$ is a module.

Proof
This follows as $\map {\MM_R} {m, n}$ is a direct instance of the module given in the module of all mappings, where $\map {\MM_R} {m, n}$ is the $R$-module $R^{\closedint 1 m \times \closedint 1 n}$.

The $S$ of that example is the set $\closedint 1 m \times \closedint 1 n$, while the $G$ of that example is the $R$-module $R$.

Let $\sqbrk a_{i j}$ and $\sqbrk b_{i j}$ be the $\tuple{i, j}$th element of $m \times n$ matrices $\mathbf A$ and $\mathbf B$ respectively.

Let $\lambda$ and $\mu$ be arbitrary elements of $R$.

We have for all $i \in \closedint 1 m$, $j \in \closedint 1 n$, by egregious abuse of notation: