While doing some research to help me on my homework, I read that if $P$ projects onto the column space of $A$, then $I - P$ projects onto the left nullspace.
I'm trying to prove this for my own purposes because I don't understand where this came from. Can anyone help me understand why this is? Thank you!
First you I want to remind what it means that $P: V \to V$ is a projection. $$ P \text{ is a Projection} \Leftrightarrow P^2 = P$$ So you need to check this condition for $I-P$: $$(I-P)^2 = I^2 - PI -IP - P^2 =I - P - P + P = I-P$$ now you want to check $\ker P = \text{ran} (I-P)$.
We start with $\subseteq$:
take $x \in \ker P$ then you know $Px = 0$ and $( I - P)x = x$ which implies that $x \in \text{ran}(I-P)$.
at last $\supseteq$:
take $x \in \text{ran}(I-P)$ then you know there is a $y\in V$ with $(I-P)y = x$. if you take the image of $(I-P)$ on both sides you get $$(I-P)^2y = (I-P)x$$ now you can use the characteristic of a projection and you get $$(I-P)y = (I-P)x \\ x = x - Px \quad| -x\\ 0 = -Px$$ which means nothing else than $x\in\ker P$.