Hrm.
If a matrix A is diagonalizable, then I claim that there exists an n such that all of the nonzero eigenvalues of A^{n} lie in the right halfplane. The requirement that Tr(A^{n}) = 0, forces all the eigenvalues to be 0, and thus A is zero... clearly nilpotent!
So the trick, then, is when the matrix is not diagonalizable.
Then again, this only works for complex valued matrices.
