If $P$ is an orthogonal projector on a Hilbert space $H$, then
$\,\operatorname{im}P=(\ker P)^\perp\subset H\,$ is a closed subspace, also called the support of $P$. And vice versa: Every closed subspace determines a unique orthogonal projector, having the given subspace as support.
Let $Q$ be another orthogonal projector. Then we have
- $\operatorname{im}P\perp\operatorname{im}Q\iff PQ=0\,,$
- $\operatorname{im}P\subset\operatorname{im}Q\iff PQ=P\,.$
Both points are not hard to prove and are found in many textbooks.
My question is, if these can be condensed into one criterion as follows:
Does the equivalence
$$\operatorname{Im}P\cap\operatorname{Im}Q=\{0\}\iff\|PQ\|<1\tag{1}$$
hold?
How can it be proved/disproved?
Note that the RHS condition of $(1)$ is also symmetric since
$\|PQ\|=\|(QP)^*\|=\|QP\|$
by the orthogonality of the projectors and the isometry of the involution.
I did some textbook and online search but only came up with the "extreme" cases $\|\cdot\|\in\{0,1\}$ corresponding to the explicit points as of above.
Starting from the RHS and using as ansatz the definition of the operator norm was tried, but I couldn't see how to get through …
The validity of $(1)$ would permit to define an angle between the subspaces by $\,\cos\gamma= \|PQ\|\,$, thus measuring the "crookedness" between the projectors or their associated subspaces, respectively.
If $\,P=(\,\cdot\,| u_P)\,u_P\,$ and $\,Q=(\,\cdot\,|u_Q)\,u_Q\,$ are $1$-dimensional orthogonal projectors onto $\operatorname{span}\{u_P\}$ and $\operatorname{span}\{u_Q\}$, respectively, then
$$\|QP\| = \|PQ\| = |(u_P|u_Q)|$$
which fits well with a possible interpretation of $\|PQ\|$ as $\,\cos\gamma$.
Best Answer
I think I found a counterexample in $\ell^2$.
Let $A:=\{x\in\ell^2 : x_{2n} = n x_{2n+1}\;\forall n\in\mathbb N\}$ and let $B:=\{x\in\ell^2 : x_{2n+1}=0\;\forall n\in\mathbb N\}$. Then it can be seen that both $A$ and $B$ are closed.
Let $P$ be the orthogonal projector on $A$, and $Q$ the orthogonal projector on $B$. Clearly, $\operatorname{im} P \cap \operatorname{im} Q=\{0\}$.
Let $n\in \mathbb N$. We consider the vector $e_{2n}\in\ell^2$. It can be shown that $$ Qe_{2n} =e_{2n} $$ and $$ Pe_{2n} =(1+\frac1{n^2})^{-1}\left( e_{2n} + \frac1n e_{2n+1}\right). $$ Calculating the norm of the latter yields $$ \|PQe_{2n}\|=\|Pe_{2n}\|=(1+\frac1{n^2})^{-1/2}\to 1 \qquad (n\to\infty) $$ Thus $\|PQ\|=1$.