cardinality / linear algebra Let $E$ be a $\mathbb C$-vector space with a finite dimension , and $u$ an endomorphism of $E$
denote $K=\{KerP(u) ; P\in\mathbb C[X]\}$  and $=\{ImP(u) ; P\in\mathbb C[X]\}$
show that $K$ and $I$ are finite and has same cardinality 
the confusion I has for this problem is that those two sets are infinite ?? because if a vector is in $K$ every translation of the vector is also in $K$ ?

Let's prove that $K$ is finite.
Let $m\in\mathbb C[X]$ be the minimal polynomial of $u$. (Any polynomial $m$ such that $m(u)=0$ works.)
Let $p\in\mathbb C[X]$.
Write $d = \gcd(m,p) = rm+sp$ with $r,s\in\mathbb C[X]$.
Then $d(u)=s(u)p(u)$ and $\ker d(u) \supseteq \ker p(u)$.
Write $p=qd$ with $q\in\mathbb C[X]$.
Then $p(u)=q(u)d(u)$ and $\ker d(u) \subseteq \ker p(u)$.
Therefore, $\ker p(u) = \ker d(u)$.
Since $m$ has only finitely many (monic) factors,
there are only finitely many possibilities for $d$ and so 
there are only finitely many possibilities for $\ker d(u)$.
The same argument proves that $I$ is finite.
$\def\im{\operatorname{im}}$
Indeed, $d(u)=p(u)s(u)$ implies $\im d(u) \subseteq \im p(u)$ and $p(u)=d(u)q(u)$ implies $\im d(u) \supseteq \im p(u)$.
Therefore, $\im p(u) = \im d(u)$.
It remains to prove that $K$ and $I$ have the same cardinality.
