$\DeclareMathOperator{\End}{End}$ I'm trying to show that:
Show that the homomorphism $\lambda: k[X] \to \End_k(V) : p \mapsto p(A)$ corresponding to the $k[X]$-module strucutre of $V$ as in (see below) has a nontrivial kernel.
I've shown that:
Let $V$ be a finite dimensional $k$-vector space and let $A \in \End_k(V)$. Show that $V$ admits a unique $k[X]$-module structure such that $X.v = A(v)$ for all $v\in V$
This is my try:
So we need to show that there exists a polynomial $f \neq 0$ such that $f(A) = 0$. In other words, $f(A).v$ must be $0$ for all $v\in V$. So $\sum_{i=0}^n a_i (A(v))^i$ must be $0$ where some $a_i$ are non-zero. If all $(A(v))^i$ where L.I. this would be impossible, so apperently, we need to show that $(A(v))^i$ are linear dependent and that we can choose $a_i$ so that $\sum_{i=0}^n a_i (A(v))^i=0$. How can we show this ?
$\DeclareMathOperator{\End}{End}$Let the dimension of $V$ be $n$. Then the dimension of $\End(V)$ is $n^2$. Hence the endomorphisms $I, A, A^2, \dots A^{n^2}$ must be linearly dependent. The coefficients of the dependence relation are the coefficients of a polynomial $f$ for which $f(A) = 0$.
As pointed out in Voldemort's answer, by the Cayley-Hamilton theorem, you can actually stop at $n$ rather than $n^2$.