TITLE: Is there a proof that the $C^{*}$-algebras don't see the invariant subspace problem?
QUESTION [12 upvotes]: This post is an appendix of this one.  
Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.  
Invariant subspace problem: Let $T \in B(H)$. Is there a non-trivial closed $T$-invariant subspace?  
Hypothesis : The ISP admits a negative answer, i.e., there are ISP counter-examples.  
Definition : A category $\mathcal{S}$ of operator algebras see the ISP if $ \forall T, T' \in B(H)$ with  $\mathcal{S}(T) \simeq \mathcal{S}(T')$:
$$
T \text{ is an ISP counter-example} \Leftrightarrow T' \text{ is an ISP counter-example }
$$
Proposition: The category $W^{*}$ of von Neumann algebras, doesn't see the ISP.
proof: Under the previous hypothesis,  let $T \in B(H)$ be an ISP counter-example. Then $T$ is irreducible,  i.e.,  $W^{*}(T) = B(H)$.  But there are many irreducible operators checking the ISP, for example, the unilateral shift $S$. So $W^{*}(T) \simeq W^{*}(S)$, $S$ checks the ISP and $T$ not. $\square$
This post asks about an equivalent result for the category of $C^{*}$-algebras : 

Is there a proof that the category of $C^{*}$-algebras doesn't see the
  ISP ?

REPLY [17 votes]: C*-algebras don't see the ISP. The operators $T\in B(H)$ and $T\oplus T\in B(H\oplus H)$
generate isomorphic C*-algebras, but the latter clearly has non-trivial invariant subspaces.
To have both operators in the same Hilbert space, pick isometries $v_1,v_2\in B(H)$ with orthogonal ranges that add up to $H$. Then
$$
T\mapsto v_1Tv_1^*+v_2Tv_2^*
$$ 
is an injective *-endomorphism of $B(H)$ that maps $T$ to an operator with non-trivial invariant subspaces.