If $P$ is an idempotent and $\langle Px,x\rangle \geq 0$, show that $P$ is a projection

Question

If $P$ is an idempotent on a Hilbert space $H$ and $\langle Px,x\rangle\geq 0$ for all $x\in H$, I’m trying to prove that P is a projection, i.e., $\text{null}P=(\text{ran}P)^{\bot}$. I have proved that $\text{null}P$ is contained in $(\text{ran}P)^{\bot}$ by showing $\langle x,y\rangle=0$ for all $x\in\text{null}P$ and $y\in\text{ran}P$. Could anyone explain why the latter is contained in the former?

Answer 1

I have solved this problem. I mentioned that I have showed $\langle x,y\rangle=0$ for all $x\in\text{null}P$ and $y\in\text{ran}P$. Then it's easy to show that $P=P_{\text{ran}P}$, where $P_{\text{ran}P}$ is the orthogonal projection of $H$ onto $\text{ran}P$. In fact, for every $x\in H$, $P_{\text{ran}P}x$ is the unique element in $\text{ran}P$ such that $(x-P_{\text{ran}P}x)\bot\text{ran}P$. It suffices to show that $(x-Px)\bot\text{ran}P$. However, $(x-Px)\in\text{null}P$ since $P$ is an idempotent. Hence $(x-Px)\bot\text{ran}P$.