TITLE: Rank of the difference of matrices
QUESTION [10 upvotes]: Let $A$ and $B$ be to $n \times n$ matrices. My question is:
Is $\operatorname{rank}(A-B) \geq \operatorname{rank}(A) - \operatorname{rank}(B)$ true in general? Or maybe under certain assumptions?

REPLY [4 votes]: It is somewhat more insightful to write your inequality as $\def\rk{\operatorname{rank}}\rk(A)\leq\rk(B)+\rk(A-B)$, and then substituting $A+B$ for $A$ as
$$
  \rk(A+B)\leq\rk A+\rk B.
$$
Since by definition the rank of a matrix$~M$ is the dimension of the space $\def\col{\operatorname{col}}\col M$ spanned by its columns, this follows from the inclusion of subspaces
$$
  \col(A+B)\subseteq\col A+\col B,
$$
since from this one gets
$$
   \rk(A+B)=\dim\col(A+B)\leq\dim(\col A+\col B)\leq\dim\col A+\dim\col B
  =\rk A+\rk B.
$$
The mentioned inclusion of subspaces is simply because every column of $A+B$ is the sum of the corresponding columns of $A$ and $B$, both of which lie in the subspace $\col A+\col B$. The other inequality used, $\dim(\col A+\col B)\leq\dim\col A+\dim\col B$, is also a generality about subspaces: combining bases of $\col A$ and of $\col B$ gives a sequence of $\dim\col A+\dim\col B$ vectors that span $\col A+\col B$.