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Let $f \in \operatorname{End}(E)$. Prove that, if $E$ has finite dimension,

$$\exists m \in \mathbb{N} \quad s.t.\quad \operatorname{Im}f^n=\operatorname{Im}f^m \quad and \quad \ker f^n = \ker f^m \quad \forall n \ge m$$

I don't know exactly how to start this proof. Could you help me? Thanks in advance!

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First, you have to see that $\operatorname{Im} f^n \subseteq \operatorname{Im} f^m$ and $\ker f^m \subseteq \ker f^n$ whenever $m \le n$. Then, because in the first case we have a sequence of subspaces that gets smaller and smaller, their dimensions also get smaller and smaller, and so eventually they have to stop, whether they reach $0$ or some other natural number. When their dimensions stop at some point $m$ in the process, then all the subspaces that come after have to have dimension equal to $\operatorname{Im} f^m$, and being subspaces of $\operatorname{Im} f^m$, they must also be equal to $\operatorname{Im} f^m$, as expected. Note that finite-dimensionality is critical here, as we needed to assume the dimensions of the subspaces were finite in order to conclude that a decreasing sequence of them would eventually terminate.

The second part proceeds almost exactly the same way. Since $\{\ker f^m\}_{m=0}^\infty$ is and increasing sequence of vector spaces, the dimension must go up, and eventually stop at or before it hits the dimension of the vector space, which is finite. Therefore there is some number $m$ for which $\ker f^n$ has the same dimension as $\ker f^m$ for all $n \ge m$, and since $\ker f^n$ contains $\ker f^m$ and has the same dimension, we must have $\ker f^n = \ker f^m$, as expected.

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  • $\begingroup$ Almost fully understood, thanks! Only one doubt: $Dim(Imf)$ and $Dim(kerf)$ do not necessarily stop at $0$ and $Dim(E)$ respectively, right? $\endgroup$ – Gibbs Nov 21 '18 at 18:43
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    $\begingroup$ @Gibbs That is correct $\endgroup$ – silvascientist Nov 21 '18 at 18:54

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