Rank of the difference of matrices

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  • Prove that $\operatorname{rank}(A) + \operatorname{rank}(B) \ge \operatorname{rank}(A + B)$

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Set $X=A-B$ and $Y=B$. You are asking whether $\operatorname{rank}(X) + \operatorname{rank}(Y) \ge \operatorname{rank}(X+Y)$. This is true in general. Let $W=\operatorname{Im}(X)\cap\operatorname{Im}(Y)$. Let $U$ be a complementary subspace of $W$ in $\operatorname{Im}(X)$ and $V$ be a complementary subspace of $W$ in $\operatorname{Im}(Y)$. Then we have $\operatorname{Im}(X)=U+W$ and $\operatorname{Im}(Y)=V+W$ by definition and also $\operatorname{Im}(X+Y)\subseteq U+V+W$. Therefore
$$\operatorname{rank}(X) + \operatorname{rank}(Y) = \dim U + \dim V + 2\dim W\ge \dim U+\dim V+\dim W \ge\operatorname{rank}(X+Y).$$

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 $\col A+\col B$.