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This issue continues this question.

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

Definition : Let $(e_{n})_{n \in \mathbb{N}}$ be an orthonormal basis.
$T \in B(H)$ is banded if $\exists r \in \mathbb{N}$ such that $ (Te_{n}, e_{m})\ne 0 \Rightarrow \vert n-m \vert \leq r$.

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

Is every irreducible operator unitary equivalent to a banded operator ?

Remark: A banded operator is a thick generalization of a diagonal operator.
It's also a finite sum of finite product of weight shift operators.

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up vote 0 down vote accepted

By pooling the answers of N. Ozawa here and C. Eckhardt here we can answer "no" :

Indeed, if an operator is unitary equivalent to a banded operator, it generates an exact $C^{*}$-algebra, and there is an irreducible operator generating a non-exact $C^{*}$-algebra.

Conclusion, there is an irreducible operator which is not unitary equivalent to a banded operator.

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