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Let $H$ be a Hilbert space, on which $P,Q$ be projection operators. Let $S:=\mathcal{R}(P)\cap\mathcal{R}(Q)$ be the intersection of ranges, then it is easy to show the orthogonal complement $S^{\bot}$ is an invariant subspace of $PQP$. The question is how to show $||PQP|_{S^{\bot}}||<1$ , if $S^{\bot}$ is finite-dimensional?

Sorry to have modified the problem. I made mistake and found out that if $PQ=QP$, which means $PQ$ is also projection on $H$, then $PQP$ shall vanish, but generally it does not, and should have the above norm estimate but I am not sure how to get it anyway.

Roy Han
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    If $PQ = QP$, then $PQP = PPQ = PQ$, which need not vanish. – Branimir Ćaćić Aug 08 '13 at 08:26
  • @BranimirĆaćić Thank you for your comment! I was working in the context that $S^{\bot}$ is finite dimensional, so in this case if $PQ=QP$, then $\mathcal{R}(P|{S^{\bot}})\cap\mathcal{R}(Q|{S^{\bot}})={0}$, and this means $PQP|{S^{\bot}}=0$ since the norm can be calculated by $||PQP|{S^{\bot}}||=\sup_{||y||=1,y\in\S^{\bot}}<PQPy,y>=\sup<QPy,Py>=0$ by the observation that $S^{\bot}$ is an invariant subspace of $P$ and $Q$ under the commutative assumption. – Roy Han Aug 08 '13 at 10:29
  • @BranimirĆaćić But I am not clear about the case when commutative assumption is dropped. It's quite clear that $||PQP||=1$, and its restriction on $S^{\bot}$ has $||PQP|_{S^{\bot}}||\leq 1$, but I am not sure if the equality can be reached. – Roy Han Aug 08 '13 at 10:33

1 Answers1

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The following classification of pairs of projections is a result by Halmos ("Two subspaces", Trans. AMS, 1969):

Given two orthogonal projections $P,Q$ on a Hilbert space $H$ there exist an orthogonal sum decomposition $H=H_{0,0}\oplus H_{1,0}\oplus H_{0,1}\oplus H_{1,1}\oplus K\oplus K$ having the following properties. $H_{i,j}$ are invariant under $P$ and $Q,$ restriction of $(P,Q)$ onto $H_{i,j}$ is $(i\cdot Id,j\cdot Id),$ and $$P|_{K\oplus K}=\begin{pmatrix}I & 0\\ 0 & 0 \end{pmatrix},\ Q|_{K\oplus K}=\begin{pmatrix}B & (B-B^2)^{1/2}\\ (B-B^2)^{1/2} & I-B \end{pmatrix},$$ where $B$ is a selfadjoint operator on $K$ such that $0<B<I.$ (Strict inequalities mean that $0,1$ are not eigenvalues of $B.$)

The space $S$ is $H_{1,1}.$ Hence $S^\perp=H_{0,0}\oplus H_{1,0}\oplus H_{0,1}\oplus K\oplus K.$ Operator $PQP$ restricted to $S^\perp$ equals to $0\oplus 0\oplus 0\oplus B\oplus 0.$ Since $K$ is finite-dimensional, the strict inequalities $0<B<I$ mean that $\varepsilon I<B<(1-\varepsilon)I$ for some $\varepsilon>0.$ Hence $||PQP|_{S^\perp}||=1-\varepsilon.$

Yurii Savchuk
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