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Let $A$ be an $n\times n$ complex nilpotent matrix. Then we know that because all eigenvalues of $A$ must be $0$, it follows that $\text{tr}(A^n)=0$ for all positive integers $n$.

What I would like to show is the converse, that is,

if $\text{tr}(A^n)=0$ for all positive integers $n$, then $A$ is nilpotent.

I tried to show that $0$ must be an eigenvalue of $A$, then try to show that all other eigenvalues must be equal to 0. However, I am stuck at the point where I need to show that $\det(A)=0$.

May I know of the approach to show that $A$ is nilpotent?

  • 2
    What are the eigenvalues of $A^2, A^3 ...$? [as values, and in terms of the eigenvalues of $A$ in general, whether or not nilpotent]. Cayley Hamilton might help.2012-06-16
  • 0
    For the record, here is [a related more general question](http://math.stackexchange.com/q/88424).2015-05-18
  • 0
    See http://math.stackexchange.com/questions/1229233 for a slightly stronger claim, and http://math.stackexchange.com/questions/1798703/traces-of-powers-of-a-matrix-a-over-an-algebra-are-zero-implies-a-nilpotent/1845002#1845002 for a generalization (distinct from that referenced by Marc van Leeuwen).2017-04-03

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