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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 ?

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You need to take two different elements $h_1$ and $h_2$, but otherwise this is correct.

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  • $\begingroup$ Would be nicer, doesn't make much of a difference since both spaces span the images though (should be a comment but thanks). $\endgroup$ – Rebellos Mar 17 at 21:13
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    $\begingroup$ @Rebellos I don't think that $P(h) \perp Q(h)$ for all $h$ implies $P(H) \perp Q(H)$. $\endgroup$ – Klaus Mar 17 at 21:17
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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.

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