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?

I assume your matrix is an $n\times n$ matrix with, say, complex coefficients.
Since $A^k=0$, the spectrum of $A$ is $\{0\}$ (or the characteristic polynomial of $A$ is $X^n$). 
Next we can find an invertible matrix $P$ such that $PAP^{-1}$ is upper-triangular with $0$'s on the diagonal. 
So
$$
\mbox{trace}A=\mbox{trace}(PAP^{-1})=0
$$
where we use the fact that $\mbox{trace} (AB)=\mbox{trace}(BA)$ in general.
