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 $T=0$.



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> Question: If $T$ is positive and $D\_T$ is compact, is $T$ necessarily compact?
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If the diagonal of a positive operator is compact, is the operator itself compact?
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.


