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Can anybody tell whether the C*-algebras $M_2(\mathcal{K}(\mathbb{H}))$ and $\mathcal{K}(\mathbb{H} \oplus \mathbb{H} )$ are isomorphic? If so, what is the isomorphism exactly?

$\mathcal{K}(\mathbb{H})$ denotes the ideal of compact operators on an infinite-dimensional Hilbert space $\mathbb{H}$.

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  • $\begingroup$ In the same way that $\mathcal{B}(H^2) = M_2(\mathcal{B}(H))$ (the algebra of bounded operators) you have $\mathcal{K}(H^2) = M_2(\mathcal{K}(H))$ $\endgroup$ – reuns Dec 2 '16 at 18:07
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Since $K(H)$ is the unique (or smallest, if $H$ is uncountably-dimensional) proper ideal, it is enough to consider the isomorphism between $M_2(B(H))$ and $B(H\oplus H)$. Let $V:H\to H\oplus H$ given by $Vx=(x,0)$; $W:H\to H\oplus H$ given by $Wy=(0,y)$. It is easy to check that $V^*(x,y)=x$, $W^*(x,y)=y$.

Define $\gamma:B(H\oplus H)\to M_2(B(H))$ by $$ \gamma(T)=\begin{bmatrix}V^*TV&V^*TW\\ W^*TV&W^*TW\end{bmatrix}. $$ It is straightforward to check that $\gamma$ is a $*$-monormorphism. So it only remains to check that $\gamma$ is onto. If $$X=\begin{bmatrix}A&B\\ C&D\end{bmatrix}\in M_2(B(H)),$$ define $T\in B(H\oplus H)$ by $$\tag{*} T(x,y)=(Ax+By,Cx+Dy). $$ Then, for all $x\in H$, $$ V^*TVx=V^*T(x,0)=V^*(Ax,Cx)=Ax, $$ so $V^*TV=A$. Similarly, $V^*TW=B$, $W^*TV=C$, $W^*TW=D$, and thus $\gamma(T)=X$. That is, $\gamma$ is onto.

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  • $\begingroup$ Could you please explain how this isomorphism transforms to isomorphism of compact operators? (I didn't understand the consequence of being unique ideal) $\endgroup$ – zionnn Nov 19 '16 at 0:44
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    $\begingroup$ If it simplifies your life, take $H$ countably dimensional. Then $B(H)$ (and $B(H\oplus H)$ and $M_2(B(H))$ all have a unique non-trivial ideal, so it has to be the same one. In any case, the way I defined $\gamma$ it is very clear that if $T\in K(H)$ then $\gamma(T)\in M_2(K(H))$. So the only exercise left would be to show that if $A,B,C,D$ are compact, then the $T$ defined in $(*)$ is compact. $\endgroup$ – Martin Argerami Nov 19 '16 at 1:03

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