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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.

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  • $\begingroup$ what does the asterisk mean? is it just the transpose? $\endgroup$ Mar 12, 2022 at 7:02
  • $\begingroup$ @ChrisSanders It is either the induced map on dual spaces, the adjoint with respect to the natural Hermitian inner products, or the conjugate transpose. However, these ideas are all isomorphic, so you can take your pick. $\endgroup$
    – Aaron
    Mar 12, 2022 at 7:11
  • $\begingroup$ @ChrisSanders Affirming what Aaron has said: * means conjugate transpose, which is akin to the adjoint wrt to Hermitian inner product. $\endgroup$ Mar 12, 2022 at 7:37

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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 $$Av+Bw =A(v_b)+B(w_a)=(A+B)(w_a+v_b).$$

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  • $\begingroup$ Thanks so much! This is great. I'm wondering: is it that the hypothesis $B^*A = 0$ is necessary to prove $\text{im}(A) \oplus \text{im}(B)$ but that one can prove that $\text{rank}(A) + \text{rank}(B) = \text{rank}(A + B)$ without using that hypothesis? (I'm asking because I was eventually able to `prove' this result using a different method, but my proof never used the hypothesis that $B^*A = 0$, or its consequence that $\text{im}(A) \perp \text{im}B$.) $\endgroup$ Mar 12, 2022 at 19:39
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    $\begingroup$ The hypothesis is necessary. Otherwise you get the image of the sum is the sum (but not necessarily direct) of the images, in which case the dimension of the sum will not necessarily be the sum of the dimensions. The other condition will not give it, unfortunately. $\endgroup$
    – Aaron
    Mar 12, 2022 at 20:15
  • $\begingroup$ Thanks again. I see the flaw in my proof now. You rock! $\endgroup$ Mar 12, 2022 at 21:36

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