Let A,B be in $M_{mxn}(\mathbb{R})$
Prove that if rank(A + B) = rank(A) + rank(B), then col(A) ∩ col(B) = {0}
I started with a proof by contradiction, since we know that rank(A + B) ≤ rank(A) + rank(B). I assume that if rank(A + B) < rank(A) + rank(B), then col(A) ∩ col(B) = {0}. I realize that we don't have to do the > case. However I am a little confused on how to show col from rank.
Note
$$\dim (\operatorname{col}(A) + \operatorname{col}(B)) = \dim \operatorname{col}(A) + \dim \operatorname{col}(B) - \dim (\operatorname{col}(A) \cap \operatorname{col}(B)).$$
Since $\operatorname{col}(A + B)$ is a subspace of $\operatorname{col}(A) + \operatorname{col}(B)$, $\dim \operatorname{col}(A + B) \le \dim (\operatorname{col}(A) + \operatorname{col}(B))$. Also, $$\dim \operatorname{col}(A) + \dim \operatorname{col}(B) = \operatorname{rank}(A) + \operatorname{rank}(B) = \operatorname{rank}(A + B).$$ Thus
$$\dim \operatorname{col}(A + B) \ge \operatorname{rank}(A + B) + \dim(\operatorname{col}(A) \cap \operatorname{col}(B)),$$
or
$$\operatorname{rank}(A + B) \ge \operatorname{rank}(A + B) + \dim(\operatorname{col}(A) \cap \operatorname{col}(B)).$$
Therefore $\dim (\operatorname{col}(A) \cap \operatorname{col}(B)) \le 0$, which forces $\operatorname{col}(A) \cap \operatorname{col}(B) = \{0\}$.
Your proof by contradiction is wrong on a conceptual level. If you want to prove the statement by contradiction, then you must, from a negation of the statement, prove a contradiction. In your case, the statement is$\newcommand{\rank}{\operatorname{rank}}$
For all matrices $A,B$, if $\rank(A+B)=\rank A + \rank B$, then $\mathrm{col} A\cap \mathrm{col} B = \{0\}$.
The negation of this statement is
There exists such two matrices $A,B$ that $\rank (A+B) = \rank A + \rank B$ and $\mathrm{col} A \cap \mathrm{col} B\neq \{0\}$
and I do not recommend you proving your statement by contradiction. Instead, try to prove the statement
For all matrices $A,B$, if $\mathrm{col} A + \mathrm{col} B \neq \{0\}$, then $\rank (A+B) \neq \rank A + \rank B$.
