If $A$ is a subalgebra of the algebra of $n \times n$-matrices $M_n(K)$, or a subalgebra of the algebra $\operatorname{End}_K(V)$ of $K$-linear maps on a vector space $V$ (see Example 1.3), then $A$ has a natural module, which we will now describe.

Let $A$ be a subalgebra of $M_n(K)$, and let $V=K^n$, the space of column vectors, that is, of $n \times 1$-matrices. By properties of matrix multiplication, multiplying an $n \times n$-matrix by an $n \times 1$-matrix gives an $n \times 1$-matrix, and this satisfies axioms (i) to (iv). Hence $V$ is an $A$-module, the natural $A$-module. Here $A$ could be all of $M_n(K)$, or the algebra of upper triangular $n \times n$-matrices, or any other subalgebra of $M_n(K)$.

Let $V$ be a vector space over the field $K$. Assume that $A$ is a subalgebra of the algebra $\operatorname{End}_K(V)$ of all $K$-linear maps on $V$ (see Example 1.3). Then $V$ becomes an $A$-module, where the action of $A$ is just applying the linear maps to the vectors, that is, we set
$$
A \times V \rightarrow V,(\varphi, v) \mapsto \varphi \cdot v:=\varphi(v)
$$
To check the axioms, let $\varphi, \psi \in A$ and $v, w \in V$, then we have
$$
(\varphi+\psi) \cdot v=(\varphi+\psi)(v)=\varphi(v)+\psi(v)=\varphi \cdot v+\psi \cdot v
$$
by the definition of the sum of two maps, and similarly
$$
\varphi \cdot(v+w)=\varphi(v+w)=\varphi(v)+\varphi(w)=\varphi \cdot v+\varphi \cdot w
$$
since $\varphi$ is $K$-linear. Moreover,
$$
\varphi \cdot(\psi \cdot v)=\varphi(\psi(v))=(\varphi \psi) \cdot v
$$
since the multiplication in $\operatorname{End}_K(V)$ is given by composition of maps, and clearly we have $1_A \cdot v=\operatorname{id}_V(v)=v$

Let $B$ be an algebra and $A$ a subalgebra of $B$. Then every $B$-module $M$ can be viewed as an $A$-module with respect to the given action. The axioms are then satisfied since they even hold for elements in the larger algebra $B$. We have already used this, when describing the natural module for subalgebras of $M_n(K)$, or of $\operatorname{End}_K(V)$.