If $\mathrm{rank}(A)+\mathrm{rank}(I_n-A)=n$, then $A^2=A.$

Question

Let $A$ be a $n \times n$ matrix with complex elements. Show that, if $\mathrm{rank}(A)+\mathrm{rank}(I_n-A)=n$, then $A^2=A.$

My try: I used the inequalities $\mathrm{rank}(A+B) \le \mathrm{rank}(A)+\mathrm{rank}(B)$ and $\mathrm{rank}(AB) \ge \mathrm{rank}(A)+\mathrm{rank}(B)-n$, but I didn't get to any result.

Answer 1

Hint: let $v\in\mathbb{C}^n$ such that $Av=0$ and $(I_n-A)v=0$. Then $v=0$.

Answer 2

Hint #1 (from @egreg): what can you say about $\ker A\cap\ker(I-A)$?

Since $Ax=0$ and $(I-A)x=0$ $\Rightarrow$ $x=0$ then $\ker A\cap\ker(I-A)=\{0\}$.

Hint #2: Use the rank-nullity theorem.

Since $\operatorname{rank}A+\operatorname{rank}(I-A)=n$ the rank-nullity theorem gives that $\dim\ker A+\dim\ker(I-A)=n$. Together with Hint #1 we can conclude that $\Bbb C^n=\ker(A)\oplus\ker(I-A)$, i.e. any $x\in\Bbb C^n$ can be decomposed as $x=x_1+x_2$ where $x_1\in\ker A$ and $x_2\in\ker(I-A)$.

Hint #3: show that $A(I-A)x=0$ for every $x\in\Bbb C^n$.

For any $x\in\Bbb C^n$ we have $x=x_1+x_2$ as above where $Ax_1=0$ and $(I-A)x_2=0$. Then $$A(I-A)x=A(I-A)x_1+A(I-A)x_2=(I-A)\underbrace{Ax_1}_{=0}+A\underbrace{(I-A)x_2}_{=0}=0.$$ Hence, $A(I-A)=0$.