Let $P_1$, $P_2$ be projections such that $P_1-P_2$ is a projection.
Show that $R_{P_2} \subseteq R_{P_1}$, where $R_{P_1}$ and $R_{P_2}$ are the ranges of $P_1$ and $P_2$, respectively.
My idea so far is that since $P_1-P_2$ is a projection, then we have:
$\begin{align}(P_1-P_2)^2&=P_1-P_2\\ P_1^2-P_1P_2-P_2P_1+P_2^2&=P_1-P_2\\ P_1-P_1P_2-P_2P_1+P_2&=P_1-P_2\\ 2P_2-P_1P_2-P_2P_1&=0\end{align}$
From here, we could suppose that $x\in R_{P_2}$, then $P_2x=x$ (projection restricted to its own range), and then:
$\begin{align}2P_2x-P_1P_2x-P_2P_1x&=0\end{align}\dots$
After this last line, I tried arranging things in a few different ways but I couldn't quite see anything (although it might be staring me right in the face). Any ideas? Thank you.
From $2P_2=P_1P_2+P_2P_1 \ (*)$, we do the following:
1/ multiply by $P_1$ to the left we have $2P_1P_2=P_1P_2+P_1P_2P_1$ or $$P_1P_2=P_1P_2P_1$$
2/ multiply by $P_1$ to the right we have $2P_2P_1=P_1P_2P_1+P_2P_1$ or $$P_2P_1=P_1P_2P_1$$
Compare what we have in $1/, 2/$ and $(*)$, we get $P_1P_2=P_1P_2P_1=P_2P_1$. Therefore, $(*)$ becomes $$P_2=P_1P_2$$
It follows that $R_{P_2} \subseteq R_{P_1}$.