Let $A$ be a square matrix of order n such that ∃m ∈ N such that $A^m$ = 0. Then... Then.
1) The only eigenvalue (complex) of $A$ is $0$ and then $\text{tr}(A)=0$.
2) $\text{tr} (A) = 0$ but $A$ can have complex eigenvalues $ \ne 0$.
3) We can deduce that $\det(A)=0$ but not $\text{tr}(A)=0$.
Only one of those is true but I really don't know how to start, I've tried to write a generic matrix $3 \times 3$ and imposing $A^2=0$ but this method is obviously long, dumb and not generic... 
 A: This question may be answered quite simply from first principles without the machinery of characteristic or minimal polynomials, to wit:
Let $\lambda$ be an eigenvalue of $A$; then
$A\vec v = \lambda \vec v \tag 1$
for some vector $\vec v \ne 0$.  It follows that
$A^2 \vec v = A(A\vec v) = A(\lambda v) = \lambda A\vec v = \lambda^2 \vec v; \tag 2$
if
$A^k \vec v = \lambda^k \vec v, \tag 3$
then
$A^{k + 1} \vec v = A(A^k \vec v) = A\lambda^k \vec v = \lambda^k A \vec v = \lambda^k \lambda \vec v = \lambda^{k + 1} \vec v; \tag 4$
we have just shown by induction that
$A^q \vec v = \lambda^q \vec v \tag 5$
for every integer $q \ge 1$; therefore, since
$A^m = 0, \tag 6$
it follows that
$\lambda^m \vec v = A^m \vec v = 0, \tag 7$
and since $\vec v \ne 0$, we find
$\lambda^m = 0, \tag 8$
forcing
$\lambda = 0. \tag{9}$
We see that every eigenvalue of $A$ is zero, whence
$\text{tr}(A) = \sum_i \lambda_i = 0, \tag{10}$
where $\lambda_i$ ranges over the complete set of eigenvalues of $A$.  Thus we see that (1) binds but that (2) and (3) are false.
Incidentally, we also have
$(\det A)^m = \det A^m = 0, \tag{11}$
so
$\det A = 0 \tag{12}$
as well.
