I am going through Functional analysis text by J.Conway, and encountered with next problem (2.3.4) :
Let $P$ and $Q$ be projections. Show $P+Q$ is a projection if and only if $\text{ran}P\perp \text{ran}Q$.
Here projection is defined as idempotent $P$ such that $\text{ker} P=(\text{ran}P)^\perp$, on given Hilbert space $\mathcal{H}$.
My solution : ($\Rightarrow$) If $\text{ran}P\not\perp\text{ran}Q$, then there exists $h\in\text{ran}P$ s.t. $h\not\perp \text{ran}Q$, i.e. $Qh\neq 0$. Then by idempotent condition, $$P+Q=(P+Q)(P+Q)=P^2+PQ+QP+Q^2=P+PQ+QP+Q$$ which gives $$PQ=-QP.$$ Therefore, since $h\in\text{ran}P$ and $Qh\neq0$, $$\tilde{h}:=PQh=-QPh=-Qh\neq 0$$ and $\tilde{h}\in \text{ran}P\cap\text{ran}Q$. Finally $$\begin{align} \Bigg|\Bigg|(P+Q)\frac{\tilde{h}}{||\tilde{h}||}\Bigg|\Bigg|^2&=\Bigg\langle(P+Q)\frac{\tilde{h}}{||\tilde{h}||},(P+Q)\frac{\tilde{h}}{||\tilde{h}||}\Bigg\rangle\\ (\because P+Q\text{ is hermitian})&=\Bigg\langle(P+Q)^2\frac{\tilde{h}}{||\tilde{h}||},\frac{\tilde{h}}{||\tilde{h}||}\Bigg\rangle\\(\because P+Q\text{ is idempotent})&=\Bigg\langle(P+Q)\frac{\tilde{h}}{||\tilde{h}||},\frac{\tilde{h}}{||\tilde{h}||}\Bigg\rangle\\&=\frac{1}{||\tilde{h}||^2}(\langle P\tilde{h},\tilde{h}\rangle+\langle Q\tilde{h},\tilde{h}\rangle)\\(\because\tilde{h}\in\text{ran}P\cap \text{ran}Q)&=\frac{2||\tilde{h}||^2}{||\tilde{h}||^2}=2>1=||P+Q||\end{align}$$ a contradiction, by using the theorem that $||P||=1$ if $P$ is a projection.
($\Leftarrow$) By $PQh=QPh=0$ for all $h\in\mathcal{H}$, we have idempotent condition : $$(P+Q)^2=P+Q.$$ By the theorem (idempotent $P$ is projection iff $P$ is hermitian) we can conclude $(P+Q)^\ast=P^\ast+Q^\ast=P+Q$ is hermitian, hence projection.
Question : Is my solution correct? Is there any simpler solutions?