Is every irreducible operator unitary equivalent to a banded operator? 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.

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.