How does a matrix define a module? For example, $$A=\begin{bmatrix}
    1 & 0 & 0\\
    0 & 1 & 0  \\
    0 & 0& -1
\end{bmatrix}
$$
How does $A$ define a $\mathbb{R}[x]-\text{module}$?
 A: Consider a $V$ - three dimensional real vector space. We have obvious action of $A$ on V: $v \to Av, v \in V.$ Now if we suppose to act a $x \in \mathbb{R}[x]$ on V like $xv \to Av$ then this action extends to action of $\mathbb{R}[x]: (\sum a_n x^n)v \to \sum a_n A^n v$. It sure is a module structure of V.
A: The module is $\mathbb{R}^3$ with the action
$$
(a_0+a_1x+\dots+a_nx^n)v=
a_0v+a_1Av+\dots+a_nA^nv
$$
for $v\in\mathbb{R}^3$.
More generally, if $T\colon V\to V$ is a linear map, where $V$ is a vector space over $\mathbb{R}$, we get a structure of $\mathbb{R}[x]$-module on $V$ by defining, for $v\in V$,
$$
(a_0+a_1x+\dots+a_nx^n)v=
a_0v+a_1T(v)+\dots+a_nT^n(v)
$$
where $T^k=\underbrace{T\circ\dots\circ T}_{k\text{ times}}$.
The key is that we get an $\mathbb{R}$-algebra homomorphism $\mathbb{R}[x]\to S$, where $S$ is the algebra of endomorphisms of $V$, by sending $x$ to $T$. Such a homomorphism obviously defines an $\mathbb{R}[x]$-module structure on $V$.
A: One has the following proposition:

Proposition. Let $k$ be a field, then the $k[X]$-modules are exactly the $k$-vector spaces endowed with an endomorphism.

Proof. Let us proceed by inclusions.


*

*Let $E$ be a vector space and $\varphi\in\textrm{End}_k(E)$, then one can define an external multiplication of $k[X]$ on $E$ in the following manner:
$$\cdot\colon\begin{cases}k[X]\times E&\rightarrow&E\\
(f,x)&\mapsto&f(\varphi)(x)\end{cases}.$$
For any $\displaystyle f:=\sum_{i=0}^na_nX^n\in k[X]$, $f(\varphi)\colon E\rightarrow E$ is define by:
$$f(\varphi)(x):=\sum_{i=0}^na_n\varphi^i(x),$$
where $\varphi^i$ is $\varphi$ composed $i$ times with itself. It is easy to check that $(E,+,\cdot)$ is a $k[X]$-module.

*Let $(M,+,\cdot)$ be a $k[X]$-module, by restriction to $k$ $\cdot$ induces a $k$-vector space structure on $M$. Furthermore, let define $\varphi\colon M\rightarrow M$ by:
$$\varphi(x):=X\cdot x.$$
You are interested in the forward inclusion.
Remark. This is a key observation in order to deal with the reduction of endomorphisms, in particular to establish Frobenius decomposition theorem.
