# $PQ = 0 \Leftrightarrow P(H) \bot Q(H)$ if $H$ is Hilbert and $P,Q$ orthogonal projections.

Exercise :

Let $$H$$ be a Hilbert space and $$P,Q \in \mathcal{L}(H)$$ orthogonal projections. Show that : $$PQ = 0 \Leftrightarrow P(H) \bot Q(H)$$

Attempt-Thoughts :

$$(\Rightarrow)$$ Let $$PQ = 0$$. Since $$P,Q$$ are orthogonal projections, $$P$$ and $$Q$$ are self-adjoint.

Let $$h \in H$$. We'll examine the operation of $$h$$ under $$P$$ and $$Q$$, as :

$$\langle P(h), Q(h) \rangle = \langle h,P^*Q(h)\rangle = \langle h, PQ(h) \rangle = 0 \implies P(h) \bot Q(h)$$ Since $$h$$ is arbitrary, this means that $$P(H) \bot Q(H)$$.

$$(\Leftarrow)$$ Let $$P(H) \bot Q(H)$$. Then, for $$h \in H$$, it is : $$\langle P(h),Q(h)\rangle = 0 \Leftrightarrow \langle h,P^*Q(h)\rangle = 0 \Leftrightarrow \langle h,PQ(h)\rangle =0 \implies PQ \equiv 0$$

Is my approach rigorous enough and correct ?

You need to take two different elements $$h_1$$ and $$h_2$$, but otherwise this is correct.
• @Rebellos I don't think that $P(h) \perp Q(h)$ for all $h$ implies $P(H) \perp Q(H)$. – Klaus Mar 17 at 21:17
If $$P$$ is a projection then $$\langle h, Ph \rangle =0$$ implies $$\langle h, P^{2}h \rangle =0$$ or $$\langle h, P^{*}Ph \rangle =0$$ or $$\langle Ph, Ph \rangle =0$$ or $$\|Ph\|^{2}=0$$ and $$Ph=0$$. So your argument is correct if you add these steps.