I came across this question in a study guide:
Let $A$ and $B$ be two real $5 \times 5$ matrices, such that $A^{2} = A$, $B^2 = B$, and $I – (A+B)$ is invertible. Prove that rank$(A)$ = rank$(B)$.
Any ideas on where to start? I know that for projection matrices, $Range(P) = Kernel(I-P)$. Is it enough to say that for $x \in Range(A)$, $(I – (A + B))x = -Bx \in Range(B)$, and since $I – (A+B)$ is a bijection, the dimensions of the two ranges are the same? Thanks in advance.
Best Answer
Note $I-(A+B)$ is invertible, hence when composed with another matrix $M$, the rank will not change. Now consider $(I-(A+B))A$ as well as $B(I-(A+B))$. What can you conclude along with the conditions of $A^2=A, B^2=B$?