Rank of a Matrix Sum [duplicate]

Solution 1:

The rank does not behave well under sum

Example $0=A+(-A)$ for any $A$.

On the other hand, the rank is subadditive:

$\operatorname{rank}(A+B)\leq\operatorname{rank}(A)+\operatorname{rank}(B)$

Proof. I denote by "span of a set" the vector space generated by that set.

The rank of a matrix is the dimension of the span of the set of its columns. The span of the columns of $A+B$ is contained in the span of {columns of $A$ and columns of $B$}.

Edit. From a comment:

Let $C_A$ be the span of the columns of $A$ and $C_B$ the span of the columns of $B$. Let $c=\dim(C_A\cap C_B)$. The span of the columns of $A+B$ is contained in the span of $C_A\cup C_B$.

Then $$\operatorname{rank}(A+B)\leq \dim(C_A)+\dim(C_B)-\dim(C_A\cap C_B)=\operatorname{rank}(A)+\operatorname{rank}(B)-c.$$

Now, let $R_A$ be the span of the rows of $A$ and $R_B$ the span of the rows of $B$. Let $d=\dim(R_A\cap R_B)$.

Since the rank by columns equals the rank by row we have $$\operatorname{rank}(A+B)\leq \dim(R_A)+\dim(R_B)-\dim(R_A\cap R_B)=\operatorname{rank}(A)+\operatorname{rank}(B)-d.$$

In conclusion $$\operatorname{rank}(A+B)\leq \operatorname{rank}(A)+\operatorname{rank}(B)-\max\{c,d\}.$$