Let $A$ and $B$ be $n\times n$ matrices. Let $\operatorname{rank}(A)=s$ and $\operatorname{rank}(B)=t$. Then rank of $A+B\ge\cdots$

Question

$\newcommand{\rank}{\operatorname{rank}}$Suppose $A$ and $B$ are $n\times n$ matrices. Let $\rank(A)=s$ and $\rank(B)=t$. Then rank of $A+B$ is at least ..............

My attempt

• $\rank(A)=s$ $\implies$ $A$ has $s$ linearly independent $n\times 1$ column vectors. Let it be $\{x_1,x_2,\ldots,x_s\}$

• $\rank(B)=t$ $\implies$ $B$ has $t$ linearly independent $n\times 1$ column vectors. Let it be $\{y_1,y_2,\ldots,y_t\}$

How do I complete the question?

Answer 1

There is no general low bound. If $B=-A$, then $A+B=0$ with rank $0$ independently of the rank of $A$. For different $s,t$ you can probably expect $rank(A+B)\ge |s-t|$. To prove (or disprove) it, you can assume that, say, $A$ is in the RREF.

Answer 2

I find the conclusion of $R(A+B)\leq R(A)+R(B)$ on page 69 of this book. As for the lower bound, it should be $ \left|s-t \right| $.