Let $A$ be a $p \times q$ matrix and $G$ be a $q \times p$ matrix. It is given that $AGA=A$ and $H=GA$. I need to show that $$Rank(I-H)=Rank(H-I)=q-Rank(A)$$
I proved it by this: I first showed that $Rank(H)=Rank(A)$. This is by virtue: $$GA: q\times 1 \to q\times 1$$ and $$A: q\times 1 \to q \times 1$$.
Let $x \in Null(GA)$. Then $GAx=0 \implies AGAx=0 \implies Ax=0$. Then $x \in Null(A)$. Now let $x \in Null(A)$. This gives $Ax=0 \implies GAx=0 \implies x \in Null(GA).$ Thus $Nullity(GA)=Nullity(A) \implies q-Nullity(A)=q-Nullity(GA) \implies Rank(A)=Rank(GA)$
Now I claim that $$Rank(I-H) \le Nullity(A)$$. Let $x \in Range(I-H)$. Then there exists a $y_{q \times 1}$ such that $y-Hy=x \implies Ax=Ay-AGAy=Ay-Ay=0$. Thus $x \in Nullity(A)$.
Now $Rank(I) \le Rank(I-H) + Rank(H)=Rank(I-H)+Rank(A)\implies q \le Rank(I-H)+Rank(A) \implies Nullity(A) \le Rank(I-H)$
Hence $Rank(I-H)=Nullity(A)=q-Rank(A)$.
I want to do it straight. I want to show that if $x \in Nullity(A)$ then I should be able to find a $y$ such that $(I-H)y=x$. This is , I am unable to do.
THanks for the help!!
