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Let $V$ be a finite-dimensional vector space over $F$ with $char(F) = 0$ and $ T: V \rightarrow V $ a linear map. Suppose that $Tr(T^n)=0$ for all $n≥1.$ Show that $T$ nilpotent.

I have seen a proof that uses the Fitting's lemma. I believe there should be some straight-forward proof from scratch. Could you help me with some suggestions? Thanks so much.

Matha Mota
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1 Answers1

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By Cayley-Hamilton theorem, there exists some coefficients $a_1$, ..., $a_n$ such that $$a_n T^n + a_{n-1}T^{n-1} + ... + a_1 T + \mathrm{det}(T) \mathrm{Id}=0$$

Taking the trace, you get $\mathrm{det}(T)=0$, so $0$ is an eigenvalue. Now you can make an induction over the dimension.

Edit : Antoher way to prove that $0$ is an eigenvalue would be to rewrite the condition on the traces with a Vandermonde system.

TheSilverDoe
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