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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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    @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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Recall the definition of an eigenvalue. $\lambda $ is an eigenvalue of $A$ if there exists a vector $v$ such that $Av=\lambda v$. Hence $A^m v=\lambda^m v$. But what is $A^m v$ if $A^m$ is the all zeros matrix? What does that tell us about the other side $\lambda^m v$? Can you solve it from here?