If $A\in C^{n\times n}$ is nilpotent, the degree of nilpotency is less than $n$. How do I prove this?
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This is about as hard as proving that the minimal polynomial of an arbitrary matrix has degree $\le n$, for which see here: https://math.stackexchange.com/questions/4949907/an-can-be-written-as-a-linear-combination-of-i-a-dots-an-1/4950053#4950053 – Qiaochu Yuan Jul 25 '24 at 18:42
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By Cayley-Hamilton, the minimal polynomial of $A$ is of degree $\le n$. If $A$ is nilpotent, this polynomial is a power of $X$. So... – Anne Bauval Jul 25 '24 at 18:46
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@QiaochuYuan I think one possible approach to the general case involves showing that the minimal polynomial of a matrix occurs as the minimal annihilator of some element. But in this case, since the monic divisors of $\lambda^m$ form a chain, that's trivial to show -- much easier than showing it in the general case. – Daniel Schepler Jul 25 '24 at 18:52
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@Daniel: the general case essentially reduces to this case via the Chinese remainder theorem (see the answer I linked to), I really don't think they're that different. – Qiaochu Yuan Jul 25 '24 at 18:57
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What do you mean by the degree of nilpotency? Is it the minimal $k$ such that $A^k=0$? – Jens Schwaiger Jul 25 '24 at 20:42