eigenvalues of a matrix with zero $k^{th}$ power For a matrix $A$, where $A^k=0$, $k\ge1$, need prove that $trace(A)=0$; i.e sum of eigenvalues is zero. How do you approach this problem?

hint: $A$ is a real matrix but you assume $A$ is a complex matrix and $f(x)=(x-a_1)(x-a_2)...(x-a_n)$ is its characteristic polynomial in the complex field. By induction you can prove that trac($A^k$)=$\sum_{i=1}^n {a_i^k}$ , and if  $\forall k\in\mathbb N$ trac($A^k$)=$\sum_{i=1}^n {a_i^k}=0$  then   $a_i=0$. Hence, $f(x)=x^n$,  $\forall i$     $a_i=0$  $$ trac A=\sum_{i=1}^n {a_i}=0$$
