Prove that $rank(A+B) \geq |rank (A) - rank(B)|$ where $rank(A+B)$ of two $p \times n$ matrices $A$ and $B$.

Question

Question

Prove that the $rank(A+B) \geq |rank (A) - rank(B)|$ where $rank(A+B)$ of two $p \times n$ matrices $A$ and $B$.

My Attempt
Firstly I am going to assume $rank(A) \geq rank(B)$ so not to worry about the absolute value for now.
I have already proved that $rank(A+B) \leq rank(A) + rank (B)$. Can I not just say that the above that I want to prove is a consequence of that, i.e if $rank(A+B) \leq rank(A) + rank (B)$ ?

Then it follows that $rank(A+B) \geq rank (A) - rank(B)$. Because of properties of inequalities?

Then, we can't have a negative rank so if $rank(B) \geq rank(A)$ then we must include the absolute value?

Am I missing something?

Answer 1

Hint. This follows "as usual" from the triangle inequality $\def\r{\operatorname{rank}}\r(C+D) \le \r C + \r D$. By using it for $C = A+B$ and $D = -B$, we have $$ \r A \le \r(A+B) + \r(-B) = \r (A+B) + \r B $$ and for $C = A+B$ and $D = -A$, giving $$ \r B \le \r(A+B) + \r A $$