Let $E$ a finite dimensional linear space with $\dim E=n$ and $u\in\mathcal L(E)$. I need to prove that $u$ is nilpotent iff $u$ is trigonalizable with unique eigenvalue $0$. I know that the proof would be easy if we use JCF and the minimal polynomial or Cayley-Hamilton theorem but the problem is that this question is asked before these theorems have been introduced. Is there an elementary proof?
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If you know the cyclic decomposition theorem, that also makes it easy. (And off the top of my head, the "direct" proofs of the fact I can think of all pretty much look like a special case of the proof of the cyclic decomposition theorem.) – Daniel Schepler Oct 10 '17 at 21:03
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this? https://math.stackexchange.com/questions/518038 – Bart Michels Oct 10 '17 at 21:04
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Oh, wait, the result only requires an upper triangular matrix representation, not necessarily one with 1 only on the +1 off-diagonal. That makes things significantly easier. (Just show the kernel is nontrivial, pick a nonzero element $x$ of the kernel, then use induction on the dimension and apply inductive hypothesis to $\bar u \in \mathcal{L}(E / \langle x \rangle)$.) – Daniel Schepler Oct 10 '17 at 21:06