Infinite dimensional space of solutions Let $E = \mathbb{C}[X]$. I consider the equation 
$$(f+\mathrm{Id}_{E})^{2n} = \mathrm{Id}_{E}\tag{$\star$}
$$ 
(with $f \in \mathcal{L}(E)$, $\mathcal{L}(E)$ being the vector space of linear maps from $E$ to itself.) If $S$ denotes the set of all the solutions to $(\star)$, then I would like to prove that $\mathrm{Span}(S)$ is an infinite dimensional vector space. 
My idea would be to find a family of $n$ linearly independent linear maps from $E$ to itself, for $n$ as large as I want. But I'm not sure that I could easily find such a family. Would there be any other method ? I was actually thinking about defining $s_{k}$ ($0 \leq k \leq n$) as follows : $s_{k} \, : \, \mathbb{C}[X] \, \rightarrow \, \mathbb{C}[X]$ and : $\forall P = \displaystyle \sum_{j=0}^{N} a_{j}X^{j}, \; s_{k}(P) = \sum_{j \neq k} a_{j}X^{j} - a_{k}X^{k}$. Because, for all $k$, $s_{k}^{2} = \mathrm{Id}_{E}$. Therefore, $s_{0},\ldots,s_{n}$ are solutions (which, at the first glance, seem linearly independent) to the equation $f^{2n} = \mathrm{Id}_{E}$. 
 A: You could define, for every nonnegative integer $p$, a linear map $f_p$ by the formula $$f_p(X^j)=\begin{cases}0&\text{if }j\neq p\\\left(e^{\frac{2i\pi} {2n}}-1\right)X^p & \text{if }j=p\end{cases}$$
These maps are clearly linearly independent, and you'll find $$(f_p+\mathrm{id}_E)^{2n}=\mathrm{id}_E$$

Another solution althogether, that will work over any coefficient field is to define, for every $p\geq 0$, and endomorphism $c_p$ by the formula
$$c_p(X^j)=\begin{cases}X^j&\text{if}\quad j < p\\ X^{j+1} & \text{if}\quad p\leq j<p+2n-1\\ X^{p} & \text{if}\quad j=p+2n-1\\ X^j&\text{if}\quad j\geq p+2n \end{cases}$$
then $c_p^{2n}=\mathrm{id}_E$ and $f'_p=c_p-\mathrm{id}_E$ is a linearly independent family of solutions.
A: Here's a nice way to go about it: we can rewrite $\star$ as
$$
(f + \operatorname{id})^{2n} - \operatorname{id} = 0
$$
That is, letting $p$ denote the polynomial $p(z) = (z + 1)^{2n} - 1$, we are simply looking for the endomorphisms satisfying $p(f) = 0$.
Let $E',E''$ be subspaces of $E$ such that $\dim(E') < \infty$ and $E = E' \oplus E''$.  Define $f|_{E''} = \operatorname{id}_{E''}$.  On the other hand, $f|_{E'}$ is simply a matrix.  We can define, for example, $f|_{E'}$ to be a diagonal matrix whose (diagonal) entries $\lambda_j$ satisfy $p(\lambda_j) = 0$ (note that $\lambda_j \neq 1$).
By this formulation, you should have no trouble finding an infinite family.
