$\newcommand{\tr}{\operatorname{tr}}$Let $A$ be a complex $n\times n$ matrix such that $\tr(A^m)=0$ for each $m=1,\ldots,n$. Show that $A$ is nilpotent.
My attempt: Since $A$ is a complex matrix, there exists a diagonal matrix $D$ and a nilpotent matrix $N$ such that $A=D+N$. Furthermore, this decomposition is unique and $D$ and $N$ commute. So we wean to show $D=0$. Since $DN=ND$, $A^2=D^2+2DN+N^2\implies \tr(A^2)=\tr(D^2)+2\tr(DN)+\tr(N^2)$. Since $N$ is nilpotent, $N^2$ is nilpotent $\implies \tr(N^2)=0$. Also, since $D$ and $N$ commute, $DN$ is nilpotent, so $\tr(DN)=0$. Hence $0=\tr(D^2)=\lambda_1^2+\cdots+\lambda_n^2$, where $\lambda_1,\ldots,\lambda_n$ are the diagonal entries of $D$. This implies each $\lambda_i=0$ and so $D=0$, which completes the proof.
I'm a little unsure though because I didn't use the fact that $\tr(A^m)=0$ for $m>2$. Is my argument correct, or does someone see a flaw?
