Let $H$ be a projection matrix onto a subspace $V$ i.e. $Im(H)=V$. Show that $I - H$ is a projection matrix onto the subspace $V^\perp$.
I already have shown that $(I-H)^T=(I-H)=(I-H)^2$. I still need to show that $Im(I-H)=Im(H)^\perp$. I actually already have a proof of this statement using eigenvalues but I really want to do it the brutal way
I managed to do this inclusion $Im(I-H)\subseteq Im(H)^\perp$, but I struggle to show the reverse inclusion. I tried to pick $y \in Im(H)^\perp$, so that $\langle y,Hx \rangle$ $\forall x$ and I need to show that there exists $z \in \Bbb R^n$ such that $y=(I-H)z$. I squeezed the inner product equality I have in many ways but still cannot find something interesting. I also tried to pick different choices of $x$ but nothing came out. Any ideas to solve this ?
If $\langle y,Hx\rangle=0$ for all $x$, then $$ \langle Hy,Hy\rangle=\langle y,H^\ast Hy\rangle=\langle y,H^2y\rangle=\langle y,Hy\rangle=0. $$ Therefore $Hy=0$. In turn, $y=Hy+(I-H)y=(I-H)y\in\operatorname{Im}(I-H)$.