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I'm working on the following question:

"Suppose that $A$ is a complex square matrix such that the trace of $A^k$ is zero for every $k \in \mathbb{N}$. Show that all the eigenvalues of $A$ are zero."

I think perhaps I should use the fact that the trace of a matrix is equal to the sum of its eigenvalues? Thoughts?

walter595738
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  • https://math.stackexchange.com/questions/656742/nilpotent-matrices-condition-with-traces shows that $A$ is nilpotent. I assume you know how to go from there? – darij grinberg Mar 29 '19 at 06:16

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Let $A$ be $n \times n$. $\sum \lambda_i^{k} =0$ for all $k\geq 1$. Hence $\sum f(\lambda_i) =f(0) n$ for every polynomial $f$, hence for all continuous functions $f$ (by Weierstarss Theorem). If $\lambda_i \neq 0$ we can find a continuous function $f$ such that $f(\lambda_i)=1$ and $f(0)=f(\lambda_j)=0$ for all $j \neq i$ leading to a contradiction.

Kavi Rama Murthy
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