A fact about linear transformations on a finite dimensional vector space

Let F be a field and let V be a finite dimensional vector space over F. Let \varphi : V \rightarrow V be a linear transformation. Prove that there exists a positive natural number m such that \mathsf{im}\ \varphi^m \cap \mathsf{ker}\ \varphi^m = 0, where \varphi^m denotes the mth iterate of \varphi.

Note that if \varphi^a(x) = 0, then \varphi^{a+1}(x) = 0. Similarly, if x = \varphi^{a+1}(y), then x = \varphi^a(\varphi(y)). This gives us two chains of subspaces in V, one ascending and one descending: \mathsf{ker}\ \varphi^1 \subseteq \mathsf{ker}\ \varphi^2 \subseteq \cdots \subseteq \mathsf{ker}\ \varphi^s \subseteq \mathsf{ker}\ \varphi^{s+1} \subseteq \cdots and \mathsf{im}\ \varphi^1 \supseteq \mathsf{im}\ \varphi^2 \supseteq \cdots \supseteq \mathsf{im}\ \varphi^t \supseteq \mathsf{im}\ \varphi^{t+1} \supseteq \cdots.

Note that because V is finite dimensional, these chains must stabilize. Let s and t be minimal such that \mathsf{ker}\ \varphi^s =\mathsf{ker}\ \varphi^{s+1} and \mathsf{im}\ \varphi^t = \mathsf{im}\ \varphi^{t+1}, respectively, and let m = \mathsf{max}(s,t). We claim that \mathsf{ker}\ \varphi^m \cap \mathsf{im}\ \varphi^m = 0.

Let x \in \mathsf{ker}\ \varphi^m \cap \mathsf{im}\ \varphi^m. Now x = \varphi^m(y) for some y \in V, and so 0 = \varphi^{m}(x) = \varphi^{2m}(y). In particular, y \in \mathsf{ker}\ \varphi^{2m} = \mathsf{ker}\ \varphi^m. So x = \varphi^m(y) = 0. Thus \mathsf{ker}\ \varphi^m \cap \mathsf{im}\ \varphi^m = 0, as desired.

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  • Nate  On June 25, 2012 at 5:38 pm

    I don’t understand the string of equalities on the very last line, particularly the equation of phi^2m(y) = phi^m(y). Could someone please explain that step to me?

    • nbloomf  On August 17, 2012 at 11:29 pm

      It should be more clear now. Thanks!

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