If $A $ is a square matrix of size $n$ with complex entries such that $Tr(A^k)=0 , \forall k \ge 1$ , then is it true that $A$ is nilpotent ?

Question

If $A$ is a square matrix of size $n$ with complex entries and is nilpotent , then I can show that all the eigenvalues of $A^k$ , for any $k$ , is $0$ , so $Tr(A^k)=0 , \forall k \ge 1$ . Now conversely if $A $ is a square matrix of size $n$ with complex entries such that $Tr(A^k)=0 , \forall k \ge 1$ , then is it true that $A$ is nilpotent ?

Answer 1

Yes, it is true. Let $\lambda_i, i=1,\ldots, n$ denote the eigenvalues of your matrix. Then $\sum \lambda_i^k=0, k\in \mathbb{N}^*.$ This implies that $\lambda_i=0$ for all $i=1,\ldots, n$. Just found that it is a duplicate: Traces of all positive powers of a matrix are zero implies it is nilpotent