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I was wondering about something. Say $A_{n \times n}$ is a matrix and it's characteristic polynomial is $P(x)=x^n$ (all eigenvalues are $0$), can you say that $A$ is a nilpotent matrix?

I really don't know how to start proving this. I don't know how to identify if a matrix is nilpotent except for multiplying it hoping to get the zero matrix.

Thank you :)


I think I know the answer. It's correct according to Hemilton-Caley right ? P(A)=0 means A^n=0 which means A is nilpotent, right ?

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  • $\begingroup$ Do you know about jordan normal form? $\endgroup$ – Git Gud Apr 27 '14 at 13:43
  • $\begingroup$ yes I do. *BTW I'm working under algebric-closed field (C) $\endgroup$ – neta Apr 27 '14 at 13:45
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Indeed the Cayley-Hamilton theorem says, if the characteristic polynomial of a matrix$~a$ is $X^n$, that $A^n=0$, so that $A$ is nilpotent of order at most$~n$. Conversely if some power of $A$ is $0$, then the minimal polynomial is $X^k$ where $k$ is the exponent of the first power of $A$ that is$~0$, and the characteristic polynomial will be $X^n$ where $n$ is the size of the matrix. Thus one can recognise nilpotent matrices from their characteristic polynomial (but one cannot distinguish their Jordan types).

All this is independent of the field used, since over any field $X^n$ has $0$ as its unique root.

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