I'm working on a problem out of Roman's Advanced Linear Algebra.
Let $V$ be a vector space over a field $F$ ($\operatorname{char}(F)\neq 2$), with $\rho$ and $\sigma$ projections. Prove:
The difference $\rho-\sigma$ is a projection iff $\rho\sigma=\sigma\rho=\sigma$, in which case $ \operatorname{im}(\rho-\sigma)=\operatorname{im}(\rho)\cap\ker(\sigma),\quad \ker(\rho-\sigma)=\ker(\rho)\oplus \operatorname{im}(\sigma). $
I've essentially proven everything except that $\ker(\rho-\sigma)\subseteq\ker(\rho)\oplus\operatorname{im}(\sigma)$. From the relations $\rho\sigma=\sigma\rho=\sigma$ I see $\ker(\rho)\subseteq\ker(\sigma)$ and $\operatorname{im}(\sigma)\subseteq\operatorname{im}(\rho)$. I take $v\in\ker(\rho-\sigma)$, and conclude that $\rho(v)=\sigma(v)$, but I don't see way to write $v$ as a sum of elements in $\ker(\rho)$ and $\operatorname{im}(\sigma)$. Does anyone know how to show this containment? Thanks.