For example, $$A=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0& -1 \end{bmatrix} $$
How does $A$ define a $\mathbb{R}[x]-\text{module}$?
How does a matrix define a module?
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$\begingroup$
abstract-algebra
ring-theory
modules
3 Answers
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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.
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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.
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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 $V$.