Turning the Ring of Polynomials, $\mathbb{F}[\lambda]$, into a Module. Modules associated to a linear operator. Suppose that $\mathbb{F}$ is a field and $V$ a vector
space over $\mathbb{F}$( i.e. an $\mathbb{F}$-module). Let $T:V\rightarrow V$ be a linear operator on $V$ 
(i.e. an $\mathbb{F}$-module
homomorphism). Let $\mathbb{F}[\lambda]$ be the ring of polynomials over $\mathbb{F}$. For
\begin{equation*} f(\lambda)=a_n\lambda^n+....+a_0\in \mathbb{F}[\lambda] \end{equation*}
we define
\begin{equation*} f(T)=a_nT^n+...+a_1T+a_0I. \end{equation*}
We can make $V$ into an $\mathbb{F}[\lambda]$-module using $T$ by defining the action of $f(\lambda)\in \mathbb{F}[\lambda]$
to be
\begin{equation*} f(\lambda)T\cdot v:= f(T)v=a_nT^n(v)+...+a_1T(v)+a_0I(v) \end{equation*}
We call $V$ the $\mathbb{F}[\lambda]$-module associated to the linear operator $T$.

So this is a concept thats been introduced in my homework. It wasn't something explored in class. My questions regarding the above is the following:


*

*So far, actions by Rings on Modules taught to me looked mainly like scalar multiplication from linear algebra. By the looks of this definition, I'm guessing its fair to say that
actions can be much more varied then this?

*If an action by $R$ is a map $M\times R\rightarrow M$. How is
this an action? Is $f(T)$ an element of $\mathbb{F}[\lambda]$?

*Is $T^n$ just the composition of $T$ with itself $n$ times?

*Is $I$ just an identity map? i.e. I(v)=v?


Thank you for your time. 
 A: To answer your questions in order:



*

*Modules can be far more varied in structure than vector spaces — the example you’re exploring is a good example.





*

*To define the module multiplication $R \times M \to M$ in your example, we take our ring $R$ to be $\mathbb{F}[\lambda]$. The definition of the action is what brings $T$ into the picture. We define the action $f\cdot v :=f(T)(v)$, where we note that $f(T)$ is a linear transformation in its own right.





*

*Yes, the notation $T^n$ means $T\circ T\circ\dots\circ T$ a total of $n$ times.





*

*Yes, the $I$ in the definition is the identity linear transformation on $V$.



To give a concrete example, let’s take $V = \mathbb{R}^2$, and 
$$T = \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}.$$
If $f(\lambda) = \lambda^2 + 2$, then we have 
$$f\cdot\begin{bmatrix} 1\\0\end{bmatrix} = \left(\begin{bmatrix} 1&1\\0&1\end{bmatrix}^2 + \begin{bmatrix} 2&0\\0&2\end{bmatrix}\right)\cdot \begin{bmatrix} 1 \\0\end{bmatrix}=\begin{bmatrix} 3&2\\0&3\end{bmatrix}\begin{bmatrix} 1 \\0\end{bmatrix} = \begin{bmatrix} 3 \\0\end{bmatrix}.$$
