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We investigate the Hilbert space $\ell^2(\mathbb{N}_0)$ with standard orthonormal basis vectors $e_n:=(0,...,0,1,0,...).$

Consider the family of self-adjoint rank $1$ projections $P_n\bullet:= \langle \bullet,e_n \rangle e_n.$

Take any $n\in\mathbb{N}_0$. My question is this: Does there exist a bounded linear operator $T$ on $\ell^2(\mathbb{N}_0)$ that commutes with all $P_m$ for $m \neq n$ but not with $P_n$?

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2 Answers 2

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Maybe a quicker way to see this is, if $TP_m = P_mT$ for all $m \neq n$ then $TP = PT$ where $P =\sum_{m\neq n} P_m = I - P_n$. Since $T$ commutes with $I$, it must therefore commute with $P_n$.

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  • $\begingroup$ I realized this just a few minutes ago, right before seeing your answer. $\endgroup$ Mar 25, 2018 at 18:00
  • $\begingroup$ I guess if you work with von Neumann algebras this is the natural thing to do. $\endgroup$
    – Nik Weaver
    Mar 25, 2018 at 20:22
  • $\begingroup$ I have not actually worked with von Neumann algebras, and the previous version of my answer was pretty clunky, even though with the same "no" conclusion. Just a bit later after that, it occurred to me that actually the matter is quite simple, and I posted the new version of the answer, and then saw your answer. $\endgroup$ Mar 26, 2018 at 0:18
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The answer is no. Indeed, without loss of generality (wlog), $n=0$, so that $TP_m=P_mT$ for all $m=1,2,\dots$ (assuming here that $\mathbb{N}_0=\{0,1,\dots\}$). Since $P_0=I-\sum_{m\ge1}P_m$ and $T I=IT$, it follows that $TP_0=P_0T$. (Here, of course, $I$ is the identity operator and the operator $S:=\sum_{m\ge1}P_m$ is defined by the condition that $Sx:=\sum_{m\ge1}P_m x$ for all $x\in\ell^2(\mathbb{N}_0)$.)

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