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I was wondering about the following: $m>1$ integer, and $A$ real matrix. $A^m=0$. Is $t=0$ the only eigenvalue of A?

Is it true?

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    Yes! ${}{}{}{}$2011-09-18
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    I tried to prove, but found it complex. any suggestions?2011-09-18
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    If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $Av = \lambda v$, so $A^2v = \lambda^2 v$ .... $A^m v = \lambda^m v$.2011-09-18
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    Seen [this](http://en.wikipedia.org/wiki/Nilpotent_matrix)?2011-09-18
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    @user16268: Prove the more general implication: if $\lambda$ is an eigenvalue of $A$, then $\lambda^n$ is an eigenvalue of $A^n$; this holds for all $n\gt 0$, and also for all negative $n$ if $A$ is invertible. (Careful: The implication is not reversible).2011-09-18

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