1) Consider the $n \times n$ matrix $A=(a_{ij})$ with $ a_{ij}= \begin{cases} 1,&\text{if } i
2) Suppose $A$ is an $n \times n$ matrix such that $A^2=A$. Show that $rank(A)+rank(I-A)=n$ where $I$ represents unit matrix of order $n$.
1) Consider the $n \times n$ matrix $A=(a_{ij})$ with $ a_{ij}= \begin{cases} 1,&\text{if } i
2) Suppose $A$ is an $n \times n$ matrix such that $A^2=A$. Show that $rank(A)+rank(I-A)=n$ where $I$ represents unit matrix of order $n$.
A neat argument for 2:
What formula do we know involving rank and $n$? Well, the Rank-Nullity Theorem is close ($rank+null=n$).
So we know that $rank(A)+null(A)=n$. Hence, if $ker(A)=im(A-I)$, we would be done. Well, if $v=(A-I)w$, then it is clear that $Av=A(A-I)w=(A^2-A)w=0w=0$ so $im(A-I)\subset ker(A)$. In the other direction, note that if $Av=0$, then $v=(A-I)(-v)$ so $v\in im(A-I)$. Hence $ker(A)\subset im(A-I)$ and we conclude that $ker(A)=im(A-I)$.