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

**[Invariant subspace problem][1]**: Let $T \in B(H)$. Is there a non-trivial closed $T$-invariant subspace? 


**Remark**: This problem is known for the Banach spaces in general, but still open for an Hilbert space.
  
Let $(e_{n})_{n \in \mathbb{N}}$ be an *orthonormal basis*.  
**Definition** : $T \in B(H)$ is **banded** if $\exists r \in \mathbb{N}$ such that $(Te_{n}, e_{m})\ne 0 \Rightarrow \vert n-m \vert \leq r$.   

**Remark**: A *banded* operator is a thick generalization of a diagonal operator.  It's also a finite sum of finite product of (*orthonormal*) weighted shift operators (which check obviously the ISP).


>**Question**: Do the banded operators check the invariant subspace problem ?

**Remarks**: 

- MO post : [Is there an operator algebraic reformulation of the invariant subspace problem?][5]

- Here is a question on MSE  : [Is every operator unitary equivalent to a banded operator ?][2]  
 N. Ozawa answered "no" in its comment below (see also sections 16.3 and 16.4 of its book [here][3]), because if an operator is unitary equivalent to a banded operator, it generates an exact $C^{∗}$-algebra, however D. Voiculescu gave examples of quasidiagonal operators which don't generate exact $C^{*}$-algebras:  *A note on quasidiagonal operators*, Operator Theory, 1988, 265-274.   

- MSE post: [Does an irreducible operator generate an exact $C^{∗}$-algebra?](https://math.stackexchange.com/questions/455949/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).   

- An ISP counter-example is necessarily irreducible, but the Voiculescu examples are not :   
MSE post : [Is every irreducible operator unitary equivalent to a banded operator?][4]   
By pooling the answers of N. Ozawa and C. Eckhardt we can also answer "no". 
  


 


  [1]: http://en.wikipedia.org/wiki/Invariant_subspace_problem
  [2]: https://math.stackexchange.com/questions/452908/are-all-the-operators-thick-diagonalizable
  [3]: http://www.ams.org/bookstore-getitem/item=gsm-88
  [4]: https://math.stackexchange.com/questions/454135/is-every-irreducible-operator-unitary-equivalent-to-a-banded-operator
  [5]: https://mathoverflow.net/questions/138070/is-there-an-operator-algebraic-reformulation-of-the-invariant-subspace-problem