How to prove that $P\mathcal H \perp Q \mathcal H$ if $P,Q$ and $P + Q$ are all projections?

Question

Let $\mathcal H$ be a Hilbert space and $P,Q \in B(\mathcal H)$ are projections. Then $(P+Q)$ is a projection iff $P \mathcal H \perp Q \mathcal H.$

"$\impliedby$" part is easy. How do I prove "$\implies$" part?

Let $(P + Q)$ be an orthogonal projection. Then $$P + Q = (P + Q)^2 = P^2 + PQ + QP + Q^2 = P + Q + PQ + QP.$$

So $PQ + QP = 0.$ Let $x,y \in \mathcal H.$ Then $$\left \langle Px, Qy \right \rangle = \left \langle QPx, y \right \rangle = \left \langle -PQx, y \right \rangle = -\left \langle Qx, Py \right \rangle$$ From here how do I conclude that $\left \langle Px, Qy \right \rangle = 0\ $? Any help in this regard will be warmly appreciated.

Thanks for your time.

Answer 1

If you multiply $PQ+QP=0$ from the left with $(1-P)$, you get $(1-P)QP=0$. If you multiply the same equation with $P$ from the left and the right, you get $2PQP=0$. Thus $QP=(1-P)QP+PQP=0$, which implies $$ \langle Px,Qy\rangle=\langle QP x,y\rangle=0 $$ for all $x,y\in\mathcal{H}$.