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Let $H$ be a separable Hilbert space with a fixed orthonormal basis $\{e_n\}_n$. For a bounded operator $T$ on $H$, the diagonal of $T$ is the unique operator $D_T$ on $H$ which is diagonal with respect to the above basis, and whose diagonal entries are given by $d_n=\langle T(e_n),e_n\rangle$. It is well know that if $T$ is positive and $D_T=0$, then necessarily $T=0$.

Question: If $T$ is positive and $D_T$ is compact, is $T$ necessarily compact?

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2 Answers 2

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Nope. For each $n$ let $T_n$ be the $n\times n$ matrix all of whose entries are $\frac{1}{n}$. This is a rank $1$ projection. So $T = \bigoplus T_n$ is a projection with infinite dimensional range, and hence is not compact. But its diagonal entries go to zero as $n \to \infty$, which means that $D_T$ is compact.

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  • $\begingroup$ Thanks Nik. Very nice example! This means that the natural conditional expectation from the Calkin algebra onto $\ell^\infty/c_0$ is not faithful. I wonder if there exists one that is faithful... $\endgroup$
    – Ruy
    Mar 18, 2017 at 22:23
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    $\begingroup$ Hmm ... I don't think so. In fact I think any such conditional expectation $\Phi$ will take my operator $T$ to $0$. To see this fix $k$ and for $1 \leq i \leq k$ let $P_i$ be the orthogonal projection onto the basis vectors $e_j$ with $j \equiv i$ (mod $k$). Then $\Phi(T) = \Phi(P_1TP_1 + \cdots + P_kTP_k)$, but each $P_iTP_i$ is a compact operator plus something that has norm at most $\frac{1}{k}$, so the same is true of $P_1TP_1 + \cdots + P_kTP_k$. Since $k$ was arbitrary, this shows that $\Phi(T) = 0$. $\endgroup$
    – Nik Weaver
    Mar 19, 2017 at 1:38
  • $\begingroup$ This is now beyond my original motivation but I got curious: the diagonal of your operator is compact alright but the diagonal sequence goes to zero very very slowly. Can you find examples with faster convergence? Say $\ell^1$? Another related question is faithfulness of the conditional expectation on the quotient of $B(H)$ by the trace class operators or other Schatten ideals. $\endgroup$
    – Ruy
    Mar 19, 2017 at 12:09
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    $\begingroup$ I think these are good research questions for you! $\endgroup$
    – Nik Weaver
    Mar 19, 2017 at 14:38
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    $\begingroup$ @Ruy: if the $d_n$ are $l^1$, this means that $T$ is trace class and hence it is in particular compact. $\endgroup$
    – Simon Henry
    Mar 20, 2017 at 15:31
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If T also is sparse, i.e. for some number N then for each i <Te_i, e_j> is non-zero for at most N values of j’s, (and for each j it is non-zero for at most N values of i’s) then the claim is true.

I have a proof, but I guess this is well known. My own literature search has been unsuccessful, so anyone having a reference will make me happy.

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    $\begingroup$ Hi @Magnus, this must be another well known fact that I haven't heard of (among many) so feel free to scratch a proof since I'll certainly benefit from it! $\endgroup$
    – Ruy
    Oct 8, 2020 at 2:44

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