Part of a problem in Roman's Advanced Linear Algebra asks to show that if $\rho$ and $\sigma$ are projections over a field not of characteristic $2$, then $\rho-\sigma$ is a projection implies $\rho\sigma=\sigma\rho=\sigma$.
There is a hint which states that $\rho$ is a projection iff $\iota-\rho$ is a projection, so $\rho-\sigma$ is a projection iff $\iota-(\rho-\sigma)=(\iota-\rho)+\sigma$ is a projection.
I can prove the hint, but I don't know how to use it to prove $\rho\sigma=\sigma\rho=\sigma$. I suppose $(\iota-\rho)+\sigma$ is a projection, hence idempotent, and use the equality to determine that $2\sigma=\rho\sigma+\sigma\rho$, but that's all. How can the conclusion be made?