Implications on structure of $B$ when $rank(A-B) = s$ for a fixed $A$

Question

Consider the case where $A \in \mathbb{R}^{n \times K}$ where $n > K$ and $\text{rank}(A) = K$. Suppose we know $$ \text{rank}(A - B) = s$$ for some $s < K$. What does this say about the relationship between the column space of $B$ and $A$?

Answer 1

It says you get $B$ from $A$ (or vice versa) by adding a matrix of rank $s$. What else do you want it to say?

EDIT: The column space of $A$ has dimension $K = \text{rank}(A)$. The column space of $B$ could be anything from $K - s$ to $K$.

Answer 2

It has failed because:

$\dim (A+(-B)) \ge \dim A$