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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 $K$ ?

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    $K$ doesn't contain vectors, it contains sets of vectors – 2017-01-13

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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.

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    Since $d$ is a factor of $m$, there are only finitely many possibilities for $kerp(u)$ , please sir can you explain this part ? because I can't understand why this set has to be finit , if $x$ is a vector form $Ker(d(u))$ than for every $a$ in $\mathbb C$ we have $ax$ in $Ker(d(u))$ – 2017-01-13
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    @Seginus, $K$ is a set of *subspaces*, not a subspace. – 2017-01-13