Prove that if rank(A + B) = rank(A) + rank(B), then col(A) ∩ col(B) = {0}

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 $\operatorname{col}(A) \cap \operatorname{col}(B) = \{0\}$.