Let \(A\in M_n(F)\), where \(F\) has characteristic \(0\). Suppose \[ \operatorname{tr}(A^k)=0 \] for every \(k\ge1\). What can you conclude about \(A\)?
Solution
The matrix \(A\) is nilpotent.
Pass to an algebraic closure of \(F\) and let the eigenvalues of \(A\) be \[ \lambda_1,\ldots,\lambda_n \] with algebraic multiplicity. Then \[ \operatorname{tr}(A^k)=\lambda_1^k+\cdots+\lambda_n^k. \] Thus every power sum \[ p_k:=\sum_i\lambda_i^k \] vanishes.
Write the characteristic polynomial as \[ \chi_A(x)=x^n-e_1x^{n-1}+e_2x^{n-2}-\cdots+(-1)^ne_n, \] where \(e_j\) are the elementary symmetric functions of the eigenvalues. Newton’s identities give, for \(1\le m\le n\), \[ me_m=\sum_{i=1}^m(-1)^{i-1}e_{m-i}p_i. \] Since every \(p_i=0\) and \(m\) is invertible in characteristic \(0\), induction gives \[ e_1=e_2=\cdots=e_n=0. \] Hence \[ \chi_A(x)=x^n. \] By Cayley–Hamilton, \[ A^n=0. \] Therefore \(A\) is nilpotent.