Orthogonal Vector Space with Identity Matrix and Orthogonal Projection

Question

I'm doing some homework and I'm having some trouble trying to prove something about orthogonal vector spaces. Given a vector space $S \subseteq \mathbb{R}^m$ and $P$ an orthogonal projection on $S$ it asks me to prove that $(\mathbb{I} - P)x \in S^\perp \ \forall x \in \mathbb{R}^m$. This is what I did:

First I took $x,y \in \mathbb{R}^m$ and defined $s=Px$ and $v=(\mathbb{I} - P)y$. Because $P$ is an orthogonal projection on $S$ I know that $s \in S$. So, I want to see that the inner product of $s$ and $v$ is zero for any $x$ and $y$ to prove that $v \in S^\perp$. So:

$$ s^\dagger v = (Px)^\dagger (\mathbb{I} - P)y \\ = x^\dagger P^\dagger y - x^\dagger P^\dagger P y \\ = x^\dagger P^\dagger y - x^\dagger y $$

I got this far. I tried to look for some identity that proves this but I couldn't find any. Any help is appreciated.

Answer 1

Edit:

Since $P$ is orthogonal, we have $P^T = P$ by definition. Take $x$ in $\mathbb R^n$ and $y$ in $S$. Since $P$ is a projection on $S$, $S$ is the image of $P$ so we have $y = Py$ (because $P^2 = P$). Now

$$ \langle x - Px, y \rangle = \langle x, y \rangle - \langle Px, y \rangle = x^Ty - x^T \underbrace{P^T}_{=P} y = x^Ty - x^T(P y) = x^Ty - x^Ty = 0. $$

Hence the result.

Answer 2

As $P$ is an orthogonal projection we know P is self adjoint, i.e. $\langle P x, y \rangle = \langle x, P y \rangle$. Taking $x\in \mathbb{R}^m$ and $y \in S$ we have $$ \begin{aligned} \langle (I-P) x, y \rangle &= \langle x, y \rangle - \langle P x, y \rangle \\ &= \langle x, y \rangle - \langle x, P y \rangle \\ &= \langle x, y \rangle - \langle x, y \rangle \\ &=0 \end{aligned} $$ where on the penultimate line we used the fact that $P y = y$.