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Let $P$ be a $n\times n$ matrix, if there is $k\in \mathbb Z^+$ such that $P^k=O$, prove that $P^n=O$.

I have thought about characteristic polynomial, but it doesn't give me much information. So I think, base on the problem conclusion if there is $k$ such that $P^k=O$, then we see $\det(P)=0$. $P$ must have rank $< n$, and I have to show that every time I multiply $P$ to $P^m$, the rank of $P^{m+1}$ must less than $P^m$, how can I show this?
kelvin hong 方
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    Do you know what the characteristic polynomial of $P$ is (think about what the eigenvalues of $P^k$ are in terms of the eigenvalues of $P$)? And do you know the Cayley-Hamilton theorem? – Minus One-Twelfth Apr 17 '19 at 12:55
  • @MinusOne-Twelfth Yes, but I couldn't solve using characteristic polynomial before I look at the solution on the other post. Now I know how to do this, thanks! – kelvin hong 方 Apr 17 '19 at 12:58

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