Let $A\in M_{n\times n}(R).$ If $rank(A) +rank(A-I_{n})=n$, show that $trace(A)=rank(A)$
I have already known that an idempotent $A^2=A$ implies...$$(1)rank(A) +rank(A-I_{n})=n\qquad(2)trace(A)=rank(A)$$ But the original question is as above.
Does $rank(A) + rank(A-I_n)=n$ implies $A^2=A$ so that I can consequently prove that $trace(A)=rank(A)$ ? Or $rank(A) + rank(A-I_n)=n$ can directly implies that $trace(A)=rank(A)$ even without the condition that $A^2=A$???