Problem Statement.
Let $A, B \in \mathcal{M}^{m \times n}(\mathbb{C})$ such that $\text{range }A^* \cap \text{range }B^* = \{0\}$ and $B^*A = 0$. Prove that $\text{rank}(A + B) = \text{rank } A + \text{rank }B$.
My Question.
I asked about this question here, and a kind person graciously offered a proof. Using a key insight from their proof, I developed a different proof. The linked proof is cleaner, proves something even stronger than the desired result, and is probably more in line with what the question writers intended. But I find my proof more personally intuitive, which is important to me only because I have a qualifying exam in a couple days and I'd rather use techniques that I am more personally comfortable with. But, for that same reason, I want to be 100% sure that I'm not making an error!
My Proof.
First, $\text{null }A + \text{null }B = V$ because $$(\text{null }A + \text{null }B)^\perp \subseteq (\text{null }A)^\perp \cap (\text{null }B)^\perp = \text{range }A^* \cap \text{range B}^* = \{0\}.$$ So now, let $\Gamma = \{\gamma_1, \dots, \gamma_i\}$ be a basis for $\text{null }A \cap \text{null }B$. Extend $\Gamma$ by $\{\alpha_1, \dots, \alpha_j\}$ to a basis for $\text{null }A$. Also extend $\Gamma$ by $\{\beta_1, \dots, \beta_k\}$ to a basis for $\text{null }B$. Because $\text{null }A + \text{null B} = V$, we have that $$\{\gamma_1, \dots, \gamma_i\} \cup \{\alpha_1, \dots, \alpha_j\} \cup \{\beta_1, \dots, \beta_k\}$$ is a basis for $V$. For any $\alpha \in \{\alpha_1, \dots, \alpha_j\}$, we have $(A + B)\alpha = A\alpha + B\alpha = B\alpha \neq 0$, and similarly $(A + B)\beta \neq 0$ for any $\beta \in \{\beta_1, \dots, \beta_k\}$. Furthermore, because $B^*A = \{0\}$, we conclude that $\text{range }A \perp \text{range }B$, meaning that there is no way for $T\alpha$ to cancel $T\beta$. On the other hand, for any $\gamma \in \{\gamma_1, \dots, \gamma_i\}$, we have $(A + B)\gamma = 0$. Thus, $\text{nullity }(A + B) = i$, the number of vectors in $\{\gamma_1, \dots, \gamma_i\}$. So $\text{rank }(A + B) = j + k$. But $\text{rank }A = k$ and $\text{rank }B = j$. So $\text{rank }(A + B) = \text{rank }A + \text{rank }B$.
Thanks for taking a read!