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Let $V$ be a vector space finitely generated over $\Bbb{C}$ and let $\alpha \in \operatorname{End}(V)$ satisfy $\operatorname{spec}(\alpha) = \{0\}$. Show that $\alpha$ is nilpotent.

I'm not sure how to do this. Any solutions/hints are greatly appreciated.

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  • $\begingroup$ Are you familiar with the Cayley-Hamilton theorem, or the minimal polynomial of a linear transformation? $\endgroup$ – Eric Wofsey Oct 25 '15 at 6:56
  • $\begingroup$ I know the CH theorem, that's it though. $\endgroup$ – 1233dfv Oct 25 '15 at 7:13
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By finite dimensionality of$~V$, and therefore that of$~\operatorname{End}(V)$, there must be some nontrivial relation between powers of$~\alpha$. This gives a nonzero polynomial $P$ such that $P[\alpha]=0$. Since $\Bbb C$ is algebraically closed we can factor $P=c\prod_{i=1}^d(X-r_i)$ with $c\neq0$ the leading coefficient of$~P$, and one will have $\prod_{i=1}^d(\alpha-r_i\mathbf I)=0\in\operatorname{End}(V)$. Since $\alpha$ has no other eigenvalues than$~0$, the factors $ \alpha-r_i\mathbf I$ with $r_i\neq0$ are invertible, and (by multiplying by their inverses) we can remove them from the identity. What remains is that $\alpha^m=0$ where $m$ is the number of factors with $r_i=0$. So $\alpha$ is nilpotent.

Note that I did not use the Cayley-Hamilton theorem, nor mentioned minimal polynomials.

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