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Why is it true that, if $f:V\to V$ is a linear map and $V$ is finite dimensional, then there must be some $n$ such that $\operatorname{Span}(\operatorname{nullspace}(f^n),\operatorname{image}(f^n))=V?$

The first thing that came to mind is the rank-nullity theorem, but I don't suppose it is of much help here. Maybe we can consider the map as a matrix, $F$, since it is a linear map? Would that help?

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    hi$n$t: rank-nullity implies that it's enough to find an $n$ such that the nullspace and the image intersect trivially.2012-05-21

2 Answers 2

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You can also work from the bottom up: $\ker f\subseteq\ker f^2\subseteq\ker f^3\subseteq\dots$ , so there must be a $k\le\dim V$ such that $\ker f^k=\ker f^{k+1}$ and hence $\ker f^k\cap f^k[V]=\{0\}$.

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We have the chain of inclusions

$\mathrm{Im}(f)\supseteq \mathrm{Im}(f^2)\supseteq \mathrm{Im}(f^3)\supseteq\cdots$

Since the dimensions are bounded from below there is some $n$ such that $\mathrm{Im}(f^i)=\mathrm{Im}(f^{i+1})$ for all $i\geq n$. Therefore $f$ induces isomorphisms $f\vert_{\mathrm{Im}(f^i)}\mathrm{Im}(f^i)\rightarrow \mathrm{Im}(f^{i+1})$. In particular $\mathrm{Im}(f^n)\cap \mathrm{Ker}(f^n)=\{\mathbf{0}\}$. The statement follows.

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    Thanks, so would it be right to say that if we take $n$ as the dimension of $V$ then the statement is always true?2012-05-21