Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Digital systems, linear algebra, introduction to formal proof / 5 / e

Prove both inclusions. If $v$ belongs to the column space of $I-P$, then $v=(I-P)u$ for some $u$ and $Pv=(P-P^2)u=0$. Thus $v\in \mathcal{N}(P)$.

Conversely, if $Pv=0$, then $v=(I-P)v$, which explicitly writes $v$ as an image of a linear map under $I-P$. Hence $v\in \mathcal{C}(I-P)$. The two inclusions prove $\mathcal{N}(P)=\mathcal{C}(I-P)$. No orthogonality assumption was used.

Set $Q=I-P$. Since $Q^2=I-2P+P^2=I-P=Q$, the previous result applies to this projector and gives $\mathcal{N}(Q)=\mathcal{C}(I-Q)$. Substituting $Q=I-P$ proves $\mathcal{N}(I-P)=\mathcal{C}(P)$. These are the kernel-image identities for complementary projectors.