Does an irreducible operator generate a nuclear $C^{*}$-algebra?

Question

Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.

Definition : An operator $T \in B(H)$ is irreducible (Halmos 1968) if its commutant $\{ T\}'$ does not contain projections other than $0$ and $I$ (i.e., $T \ne T_{1} \oplus T_{2}$, or equivalently, $\{T,T^{*}\}''=B(H)$).

Definition : A $C^{*}$-algebra $\mathcal{A}$ is nuclear if and only if its enveloping von Neumann algebra $\pi_{U}(\mathcal{A})''$ (with $\pi_{U}$ its universal representation) is hyperfinite.

Let $T \in B(H)$ be an irreducible operator, and let $\mathcal{A} = C^{*}(T)$. Now $W^{*}(T) = \mathcal{A}''$ is the von Neumann algebra generated by $\mathcal{A}$ (a priori, it's not its enveloping von Neumann algebra). Now by definition of irreducibility, $W^{*}(T) = B(H)$, which is the hyperfinite $I_{\infty}$ factor. Is $\pi_{U}(\mathcal{A})''$ also hyperfinite ?

Does an irreducible operator generate a nuclear $C^{*}$-algebra ?

If no, is it nevertheless an exact $C^{*}$-algebra ?
Remark: All nuclear $C^{*}$-algebras and their $C^{*}$-subalgebras are exact.

Answer 1

See the post: Does an irreducible operator generate an exact $C^{∗}$-algebra?.

C. Eckhardt answered "no" by giving some simple, singly generated and non-exact $C^{*}$-algebras.
By simplicity, their irreducible representations are faithful.

So, there are irreducible operators $T \in B(H)$ with $C^{*}(T)$ non-exact (and a fortiori non-nuclear).