Let $A$ and $B$ be two square Hermitian matrices such that $AB = 0 = BA$. Prove that $\mbox{rank} (A + B) = \mbox{rank}(A) + \mbox{rank} (B)$.
This is what I have so far. It is clear that rank($A + B$) $\leq$ rank($A$) + rank($B$). So, it only suffices to show that rank($A + B$) $\geq$ rank($A$) + rank($B$).
Thanks in advance!
Claim 1: $A$ and $B$ can be simultaneously unitarily diagonalized.
By the Schur Unitary Theorem for commuting matrices, we have that there exists $U$ unitary such that \begin{equation}\label{1}\tag{1} U^* A U = T_A \quad \text{and} \quad U^* B U = T_B \end{equation} where $T_A$ and $T_B$ are upper triangular matrix not necessarily equal to each other. Since $A$ and $B$ are Hermitian and using \eqref{1}, we have $$T_A^* = T_A \quad \text{and} \quad T_B^* = T_B$$ The above implies that $T_A$ and $T_B$ are diagonal matrices. Thus, $A$ and $B$ can be simultaneously unitarily diagonalized. Thus, by Claim 1, we know that $A$ and $B$ can be simultaneously unitarily diagonalized. That is $$A = UD_AU^* \quad \text{and} \quad B = UD_BU^*$$ where $U$ is unitary and $D_A$ and $D_B$ are diagonal matrices containing the non-zeros eigenvalues of $A$ and $B$, respectively. Thus $$A + B = UD_AU^* + UD_BU^* \quad \Rightarrow \quad \text{rank}(A+B) = \text{rank }A + \text{rank }B$$