How to prove the complement $P^\perp$ of a projection matrix $P$ have relation $I-P=P^\perp$

Question

I want to know how to prove that for a projection matrix $P$ and its complement matrix $P^\perp$.

We have $$I-P=P^\perp$$

I do know the intuition that $P$ and $P^\perp$ project a vector into two different subspaces.

But can we prove it in a algebra way?

P can be represented by $$P=U_1U_1^T;P^\perp=U_2U_2^T$$, in which $$\begin{bmatrix}U_1&U_2\end{bmatrix}$$ is an orthogonal matrix.

Answer 1

Suppose $P$ is an orthogonal projection onto $U$ then for

$$v=v_U+v_{U_\perp} \implies Pv=v_U$$

and $$(I-P)v=v-v_U=v_{U_\perp}$$ thus $(I-P)$ is an orthogonal projection onto $U_\perp$.

Answer 2

Let $U = Im(P) = P(V)$, then for $x \in V$ you have the following orthogonal sum: $$x = Px \oplus (x-Px) = Px \oplus (I-P)x \mbox{ with } Px \in U, \; (I-P)x \in U^{\perp} $$ So, $I-P$ is the orthogoonal projector onto $U^{\perp}$.

Answer 3

Let $V$ be a vector space and $x\in V$. Suppose $U$ is a subspace of $V$. Let $P$ be the projection onto $U$ and $P^{\perp}$ be the projection on the orthogonal complement $U^{\perp}$ of $U$. Then we have the following decomposition $$x=y+z\hspace{0.2cm}\text{with}\hspace{0.2cm}y:=P(x),z=P^{\perp}(x)$$ From the above relation we obtain $$P^{\perp}(x)=x-P(x)=(I-P)(x)$$ Since it holds for any vector $x\in V$ it must then be the case that $$P^{\perp}=I-P$$