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?
 A: 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.
A: Do you know that the trace of a matrix is the sum of it's eigenvalues counted with multiplicity?  Think about the Jordan Canonical Form of your matrix to see why this is so.
Now for a matrix that satisfies $A^k = 0$ prove that $0$ is its only eigenvalue.
A: 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, $f(x)=x^n$,  $\forall i$     $a_i=0$  $$ trac A=\sum_{i=1}^n {a_i}=0$$
