Why is $\mathrm{Ran}(P) = \mathrm{Ker}(I - P)$ for a projection mapping $P$?

Question

Let $P$ be a projection mapping in a Banach space $X$. I am trying to understand why $\mathrm{Ran}(P) = \mathrm{Ker}(I - P)$: If $u \in \mathrm{Ker}(I - P)$ then $(I - P)u = 0 \Longleftrightarrow Iu = u = P(u)$ and so $u\in \mathrm{Ran}(P)$. But if $w \in \mathrm{Ran}(P)$ then we know only that $w = P(u)$ for some $u\in X$. Knowing that $P^2 = P$ we may then deduce $P(w) = P^2(u) = P(u)\Longleftrightarrow P(w - u) = 0 \Longleftrightarrow w - u \in \mathrm{Ker}(P)$. But I don't see how this would imply that $(I - P)u = 0$.

Edit: It just hit me: $(I - P)(P(u)) = P(u) - P^2(u) = P(u) - P(u) = 0$. So yeah, the claim holds.

Answer 1

I think a projection means $P^2=P$.

Then $y∈ \text{Im}(P) \iff \exists x\in X$, s.t. $Px=y$, which implies $$Py=P^2x=Px=y\implies y∈ \text{Ker}(I-P).$$

Conversely, if $y∈ \text{Ker} (I-P)$, then $y=Py∈ \text{Im}(P)$.