$\operatorname{rank}(M)=\operatorname{rank}(A)+\operatorname{rank}(B-A)$

Question

Let $A, B \in M_n,$ and
$M= \begin{pmatrix} A & A \\ A& B \end{pmatrix}\\$

I have to prove that $$ \operatorname{rank}(M)=\operatorname{rank}(A)+\operatorname{rank}(B-A)$$

Any ideas?

Answer 1

Hint: Note that $$ \pmatrix{I&0\\-I & I}\pmatrix{A&A\\A&B} = \pmatrix{A & A\\0 & B-A}, \\ \pmatrix{A & A\\0 & B-A} \pmatrix{I & -I\\0 & I} = \pmatrix{A & 0\\0&B-A}. $$