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Let $A$ be a $C^\ast$-algebra, $\mathcal{K}$ the algebra of all compact operators on a separable Hilbert space. Is $A$ (isomorphic to) a hereditary subalgebra of $A\otimes\mathcal{K}$?

Thanks.

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If $A$ is $\sigma$-unital you can do the following. Let $h_A \in A$ be a strictly positive element. Then you can consider the corner $$ \overline{h(A \otimes \mathcal K)h} \ , $$ where $h = h_A \otimes e_{11}$.

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    $\begingroup$ Btw, is it true that if $1$ is the unit in the unitization of $A$ then $h=1\otimes e_{11}$ is in the unitization of $A\otimes K$, so that $h(A\otimes K)h$ is isomorphic to $A$ ? (if yes, then this is a way to view $A$ as an hereditary sub-algebra). I know that it holds if I take $1_{M(A)}\otimes e_{11}\in M(A)\otimes M(K)\subseteq M(A\otimes K)$, but I'm not sure that hereditary sub algebras can be viewed by conjugating with an element from the multiplier algebra. $\endgroup$ – Shirly Geffen Aug 28 '17 at 8:23

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