Let $K$ be a field, $n \in \mathbb N$, $V$ an $n$-dimensional $K$-Vector space and $f: V \to V$.
Show: $\exists$ Basis $B$ of $V$ such that $M_{f}=J(\lambda,n)$ $\iff$ $V \cong K[X]/(X-\lambda)^{n}$ with $V$ a $K[X]$-Module
Ideas:
"$\rightarrow$" Let $B$ be Basis for $V$ such that $M_{f}=J(\lambda,n)$
By definition of the Jordan Matrix it means that:
for $B:=\{v_{1},...,v_{n}\}$ we've got:
$f(v_{1})=\lambda v_{1}$,
$f(v_{2})=\lambda v_{2} + v_{1}$ and $f(v_{i})=\lambda v_{i}+v_{i-1}$ $\forall i \in \{2,...n\}$
Now looking at a map:
$\varphi:V\to K[X]/(X-\lambda)^{n}$,
$\varphi(p(f)(v))=[p(f)(v)]$
I know that $dim_{K}V=n=dim_{K}K[X]/(X-\lambda)^{n}$. So all I need to do is that the map is injective. I am unsure of how I should define $[p(f)(v)]$ as an equivalence class in $\{[0],..,[(X-\lambda)^{n-1}]\}$
How do I show that $ker\varphi=0$?
"$\leftarrow$" Main idea, we find the equivalence classes of $[p(f)(v_{i})]$
$\forall i$ $\in\{1,...,n\}$ and show that they are not the same and then somehow lead it back to $f(v_{i})=\lambda v_{i}+v_{i-1}$ $\forall i \in \{2,...,n\}$
Any help would be greatly appreciated.
For the forward implication, the idea is to pick bases of $V$ and of $K[X]/(X-\lambda)^n$ such that the endomorphism $f$ has the same matrix with respect to the former basis as $x$ does with the latter.
Consider the basis $\{ m_k = (x-\lambda)^{n-k}: 0 < k \leq n\}$. Then $$xm_1 = x(x-\lambda)^{n-1} = \lambda(x-\lambda)^{n-1}=\lambda m_1$$ and for $k > 1$, $$xm_k = x(x-\lambda)^{n-k} = (x-\lambda)^{n-k+1} + \lambda(x-\lambda)^{n-k} = \lambda m_k + m_{k-1}.$$ Then the matrix of $x$ with respect to this basis is exactly the Jordan block $J_n(\lambda)$. So mapping your $v_i$ to $m_i$ for all $i$ provides a $K[X]$-module isomorphism.
The reverse follows since the image of this basis for $K[X]/(X-\lambda)^n$ will be a Jordan basis for $V$.