Show that if $AB=BA=0$ then $rank(A+B)=rank(A)+rank(B)$ We are supposed to show that if $AB=BA=0$ then $rank(A+B)=rank(A)+rank(B)$
Maybe this fact help to answer the question:
$rank(A+B)\le rank(A)+rank(B)$ 
and
$rank(A)+rank(B)=dim(Im(B))$.
But I got stuck when trying to prove that $rank(A+B)\ge rank(A)+rank(B)$

Take
$$
A=B=\begin{pmatrix} 0 & 1 \cr 0 & 0 \end{pmatrix}.
$$
Then $AB=BA=0$, but 
$$
1=rank(A+B)\neq rank(A)+rank(B)=2.
$$
So Daniel is right.
