if range $A^* \cap$ range $B^* = 0$ and $B^*A = 0$ then $\text{rank }(A + B) = \text{rank }A + \text{rank }B$ Problem. 
Let $A, B \in \mathcal{M}_{m,n}(\mathbb{C})$ 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 Attempt. 
Taking inspiration from the proof of the rank-nullity theorem, I start by defining four sets of vectors that will, together, be a basis for $\mathbb{C}^n$. Let $\Gamma = \{\gamma_1, \dots, \gamma_i\}$ be a basis for $\text{null }A \cap \text{null }B$. Use $\{\alpha_1, \dots, \alpha_j\}$ to extend $\Gamma$ to a basis for $\text{null }A$ and $\{\beta_1, \dots, \beta_k\}$ to extend $\Gamma$ to a basis for $\text{null B}$. Let $\{\delta_1, \dots, \delta_\ell\}$ be such that $$\{\gamma_1, \dots, \gamma_i\} \cup \{\alpha_1, \dots, \alpha_j\} \cup \{\beta_1, \dots, \beta_k\} \cup \{\delta_1, \dots, \delta_\ell\}$$ is a basis for $\mathbb{C}^n$. I won't write out the rest of my proof because it is long, long, long. But what it amounts to is: if $\text{null }A + \text{null }B = \mathbb{C}^n$, i.e. if $\{\delta_1, \dots, \delta_\ell\}$ is empty, then the result I want follows. But, I'm having a hard time proving that $\text{null }A + \text{null }B = \mathbb{C}^n$. Maybe it isn't even true? Maybe there is a really easy, slick way to do the proof that I am totally overlooking?
I've spent literally two days on this! Please help! Thanks so much.
 A: Note, I will use the terminology $\operatorname{im}(A)$ (image of $A$) instead of range, and $\ker(A)$ instead of null..
We will prove the stronger statement that $\operatorname{im}(A+B)=\operatorname{im}(A)\oplus\operatorname{im}(B)$.  Since $\operatorname{im}(A+B)\subset \operatorname{im}(A)+\operatorname{im}(B)$ always holds because $(A+B)v=Av+Bv$, we need to show the reverse inclusion and that the intersection is trivial.
With the Hermetian inner product on $\mathbb C^n$ and $\mathbb C^m$, we have the fundamental relations $\operatorname{im}(A)^{\perp}=\ker(A^*)$ and other similar relations (see here for example, although this only works over $\mathbb R$, but the result is true over $\mathbb C$ too).
Since $B^*A=0$, the image of $A$ is contained in the kernel of $B^*$, which is the orthogonal complement of $\operatorname{im}(B)$, so $\operatorname{im}(A)\perp \operatorname{im}(B)$.  In particular, their intersection is trivial.
Since $\{0\}=\operatorname{im}(A^*)\cap \operatorname{im}(B^*) = \ker(A)^{\perp} \cap \ker(B)^{\perp}=(\ker(A)+\ker(B))^{\perp}$, we have that $\ker(A)+\ker(B)$ must be all of $\mathbb C^n$
Given $v, w$, we wish to show that $Av+Bw\in\operatorname{im}(A+B)$.  Write $v=v_a+v_b, w=w_a+w_b$ where $v_a, w_a\in \ker(A), v_b, w_b\in \ker B$.  Then $$Av+Bw =A(v_b)+B(w_a)=(A+B)(w_a+v_b).$$
