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Let $H$ be a complex infinite dimensional separable Hilbert space. There are various extensions of the following well known result:

Theorem (Lomonosov): Every nonscalar $T \in B(H)$ which commutes with a nonzero compact operator $K$ has a nontrivial hyperinvariant subspace.

It has been shown that there exist operators $T$ which do not commute with any nonzero compact $K$. This led to the following two generalisations; the first one can be found, for instance, in the book Kubrusly, C. S. Hilbert space operators. Birkhauser, Boston, 2003 (Problem and Solution 12.4), while the second one is obtained in Lauric, V. (1997). Operators $\alpha$-Commuting with a Compact Operator. Proceedings of the American Mathematical Society, 125(8), 2379-2384.

Theorem: Let $T \in B(H)$ be nonscalar. If there exists a nonzero compact $K$ such that $\operatorname{rank} (TK-KT) \leq 1$, then $T$ has a nontrivial hyperinvariant subspace.

Theorem: Let $T \in B(H)$ be nonscalar. If there exists a nonzero compact $K$ such that $TK= \alpha KT$ for some $\alpha \in \mathbb{C}$, then $T$ has a nontrivial hyperinvariant subspace.

I was wondering if the following natural generalisation is true: if there exists a nonzero compact $K$ such that $\operatorname{rank}(TK - \alpha KT) \leq 1$ for some $\alpha \in \mathbb{C}$, then there exists a nontrivial hyperinvariant subspace.

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Looking again at Solution 12.4 in Kubrusly's book, I have noticed that the proof can be used even to prove the statement above, with some small changes. I will briefly sketch such small modifications: instead of considering the operator $TS-ST$, we will consider $C:=TS-\alpha ST$ ($\alpha \neq 0$). The claim:

If $\operatorname{rank}(C) =1$, then $\mathcal{R}(C) \subseteq \mathcal{R}(S)$.

(where $\mathcal{R}(\cdot)$ denotes the range) still holds, and it can be proved as in the cited book. The remaining part of the proof of (a) does not need any modification. The next step in Kubrusly's book (part (b)) is still applicable, with $C:=TS-\alpha ST$ and not as $TS-ST$ (as before). This time, the case $LC=0$ leads to a contradiction because of Lauric extension of Lomonosov Theorem (see the reference above). The case with $\operatorname{rank}(LC)=1$ is proved via part (a). We obtain again a contradiction and thus we conclude the proof of the above statement. I refer to Kubrusly, C. S. Hilbert space operators. Birkhauser, Boston, 2003 (Problem and Solution 12.4) for all the missing details and notations.

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