Suppose that $A,B \in M_n(\mathbb{C})$ are nilpotent matrices such that $AB = BA$. Prove that $AB$ and $A + B$ are nilpotent matrices.
Thoughts:
1) Since $0$ is the only eigenvalue of a nilpotent matrix. $$ABx = A(\lambda_Bx) = \lambda_A(\lambda_Bx) = \lambda_B(\lambda_Ax) = B(\lambda_Ax) = BAx = 0$$ I don't think this is correct because $x$ needs to be a eigenvector for both $A$ and $B$.
2) If $n$ is the nilpotent index for $A$ and $m$ is the nilpotent index for $B$. Let $n<m$, then $(AB)^n = (AB)(AB)\dots(AB)$ n times. Switch around the $AB$ with $BA$ and vice versa until all the $A$s are grouped together. since $A^n = 0$, $(AB)^n = 0$ and $n$ is the nilpotent index fo $AB$?
No idea about proving this for $A + B$
Hint : Look at $(A+B)^{(m+n)}$