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Consider an $n\times n$ nilpotent complex matrix $A$. Is it true that $A^n=0$ always? If not, is there a minimum $d$ (in terms of $n$) such that $A^d=0$ always?

JJ Beck
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  • @EwanDelanoy I put "nilpotent" in the title and forgot to put it in the actual question. Sorry.. – JJ Beck Jan 30 '14 at 06:06

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$A^n=0$ is true for $n \times n$ nilpotent matrices. Proof: All eigenvalues of a nilpotent matrix are $0$. Hence the characteristic polynomial is $x^d=0$ for some $d$. However, the characteristic polynomial has degree $n$. So, $d=n$. By Cayley-Hamilton Theorem, $A^n=0.$

jim
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voldemort
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