Let A be a square matrix of order n. Prove that if $A^2 = A$, then $\operatorname{rank}(A) + \operatorname{rank}(I - A) = n$.

Question

$A(I-A) = 0\implies\operatorname{rank}(A) + \operatorname{rank}(I-A)\le n$.

I managed to get this but wasn't able to go further. Any help would be appreciated.

Answer 1

$dim(kerA) +rank A =n$, $A(I-A) =0$ implies $rank(I-A)\leq dim(kerA)$

Answer 2

Let $x\in \Bbb R^n$ then $$x=Ax+(x-Ax)\in \operatorname{Im}A+\operatorname{Im}(I-A)$$ and then let $y\in \operatorname{Im}A\cap \operatorname{Im}(I-A)$ so $y=Ax=z-Az$ for some $x,z\in \Bbb R^n$ so applying $A$ we get $$y=Ax=A^2x=Az-A^2z=Az-Az=0$$ so $$\operatorname{Im}A\cap \operatorname{Im}(I-A)=\{0\}$$

hence we get $$\Bbb R^n=\operatorname{Im}A\oplus \operatorname{Im}(I-A)$$ and the result follows by taking the dimension.

Answer 3

$A^2=A$, thus $A$'s minimal polynomial must divide $x(x-1)$, which indicates:
1). $A$'s all possible eigenvalues are $1,0$.
2). $A$ is diagonalizable. That's to say, exists invertible $P$ such that $$A=P\text{diag}(1,1,\cdots,1,0,0,\cdots,0)P^{-1}=P\Lambda P^{-1}$$

Note that $$\text{rank}A=\text{rank}\Lambda$$ And that $$\text{rank}(I-A)=\text{rank}(P(I-A)P^{-1})=\text{rank}(I-\Lambda)$$ Can you take it from here?

Answer 4

As you have proved $\operatorname{rank}(A) + \operatorname{rank}(I-A)\le n$, the opposite inequality comes from the subadditivity of rank $$ n=\operatorname{rank}(I)=\operatorname{rank}(A+I-A)\le\operatorname{rank}(A) + \operatorname{rank}(I-A). $$