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

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

Invariant subspace problem: 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.

Question: Do the quasi-diagonalizable operators check the invariant subspace problem ?

Remark: I post the following question on MSE : Are all the operators quasi-diagonalizable?
It's probably false, but of course, if it's true, the question here would be equivalent to the ISP, and I would be interested to know if this way is (in your opinion) promising, and ever known (reference).