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Let $A\subseteq B(H) $ be a $C^*$ algebra. I know that if $A$ is unital, then by von Neumann density theorem SOT closure$(A)=A''$ and $A''$ is the von Neumann algebra generated by $A$ in $B(H)$.

Is it true that if $A$ is not unital, still SOT closure $(A)=A ''$?

budi
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    Yes if $A$ is non-degenerate. For a counter example without this assumption, let $A$ be the operators on a 2-dimensional Hilbert space with matrix of the form $$\pmatrix{\lambda & 0 \cr 0 & 0}$$ – Ruy Dec 06 '20 at 13:01
  • Where Can I get a reference for this result? – budi Dec 08 '20 at 05:52

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