Solution
= Solution
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 \b[$\mathcal N(P)=\mathcal C(I-P)$]. No <orthogonality> assumption was used.