LA11

From Exampleproblems

Let $A\,$ be an $n\times n$ matrix such that $A^2=A\,$ , and let $I\,$ be the $n\times n$ identity matrix. Prove that ${\rm rank}(A)+{\rm rank}(A-I)=n\,$ .

Let $x\in{\rm Col}(A)$ , so there exists a vector $y\in\mathbb{R}^n$ such that $Ay=x\,$ , or $A^2y=Ax=x\,$ , so $(A-I)x=0\,$ ; that is, ${\rm Col}(A)\subseteq{\rm Null}(A-I)$ . Further, if $x\in{\rm Null}(A-I)$ , $(A-I)x=0\,$ implies $Ax=x\in{\rm Col}(A)$ , so ${\rm Col}(A)={\rm Null}(A-I)\,$ . Then, by the Rank Theorem, $n={\rm rank}(A-I)+{\rm dim\ Null}(A-I)={\rm rank}(A-I)+{\rm dim\ Col}A={\rm rank}(A)+{\rm rank}(A-I)$ .