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Suppose I have $f\in L(V,V)$ such that $f^n=0$ but $f^k\neq 0\,\,\forall k. Would I be right in thinking that there exists $v\in V$ such that $\{f^k(v)|0\leq k\leq n-1\}$ is linearly independent? If I am right, how could I find this $v$?

Thanks

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    Since $f^{n-1}\neq 0$, you can find $v\in V$ such that $f^{n-1}(v)\neq 0$. Now, show that $\{f^k(v)\mid 0\leq k\leq n-1\}$ is linearly independent.2011-11-20
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    Thanks, @DavideGiraudo. Do you mean that for any $v\in V$ such that $f^{n-1}(v)\neq 0$ the set is linearly independent?2011-11-20
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    Yes, and you can try to prove it, for example assuming that $\sum_{j=0}^{n-1}\alpha_j f^j(v)=0$ for $(\alpha_0,\ldots,\alpha_{n-1})\neq (0,\ldots,0)$ and getting a contradiction.2011-11-20
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    @DavideGiraudo: Thanks, got it now.2011-11-20

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