Linear map and polynomials If $f:V \to V$ a linear map, such that $\dim V =n \leq \infty$ then $\dim L(V,V) = n^2$ and $B=\{Id, f,f^2,...,f^{n^2-1}\}$ is a basis for $L(V,V)$. Therefore $B \cup \{f^{n^2}\}$ is $LD$ (because there are $n^2+1$ elements). So there are $\alpha_0, \alpha_1, ...,\alpha_{n^2}$, for some $\alpha_i$ nonzero such that $\alpha_0I_d+ \cdots+ \alpha_{n^2}f^{n^2}=0$. 
Then $P(f)(v) =0$ for all $v \in V,$ where  $P(x)=  \sum_{k=0}^{n^2}a_kx^k$.
Ok, my question is, if we don't have $\dim V < \infty$, but $\dim \operatorname{Im}(f)< \infty$, Is there a polynomial $P$ such that $P(f)=0?$
 A: Restrict $f$ to $\operatorname{Im}(f)$ to get a polynomial $Q(x)$ such that $Q(f)=0$ on $\operatorname{Im}(f)$, by the finite-dimensional case. Then $Q(f)f=0$ on $V$, so the polynomial $P(x)=Q(x)x$ does what you want.
A: Yes the statement it's true. 

Let $V$ be a vector space, $f:V\rightarrow V$ linear transformation. If $\dim \operatorname{Im}(f)<\infty$ then there exists a polynamial $P(x)$ such that $P(f) = 0$

Consider the restriction $f|_{\operatorname{Im}(f)}:\operatorname{Im}(f)\rightarrow \operatorname{Im}(f)$. This map is an endomorphism of the finite dimensional vector space $\operatorname{Im}(f)$, so there exists a polynomial $Q(x)\neq 0$ such that $Q(f|_{\operatorname{Im}(f)})=0=[Q(f)]|_{\operatorname{Im}(f)}$. Consider now the polynomial $P(x)=xQ(x)$: for all $v\in V$ we have:
\begin{gather}
[P(f)](v)=[f\circ Q(f)](v) = [Q(f)\circ f](v) = [Q(f)] (f(v)) = 0
\end{gather}
So $P(x)$ is a non trivial polynomial such that $P(f)=0$.
Also the condition that $\operatorname{Im} f$ is a finite dimensional vector space is necessary. You can take for example $V=\bigoplus_{\mathbb N}K$ with $K$ field, standard basis $\{e_i\}$ and the linear map
\begin{equation}
f:V\rightarrow V, \quad f(e_i)=e_{i+1}\quad \forall i\in \mathbb N
\end{equation}
Then you have $f^k(e_i)=e_{i+k}$ for all $i,k\in \mathbb N$. By contradiction suppose there exists a non trivial polynomial $P(x) = a_0+...+a_nx^n$ such that $P(f)=0$. But now $$[P(f)](e_1)=a_0e_1+a_1e_2+...+a_ne_{n+1}=0$$ and this implies $a_i=0$ for all $i\in \mathbb N$.
