Let $\mathcal H$ be a Hilbert space and $\mathcal B(\mathcal H)$ be the space of all bounded linear operators on $\mathcal H.$ For $A, B\in \mathcal B(\mathcal H), $ define the linear map $\tau_{A, B}: \mathcal B(\mathcal H)\to\mathcal B(\mathcal H) $ by $$\tau_{A, B}(X)=AX-XB,~ X\in \mathcal B(\mathcal H).$$
The famous Kleinecke-Shirokov's theorem says that if $A=B$ and $X\in \ker \tau_{A, A}\cap \text{ran} \tau_{A, A},$ then $X$ is quasinilpotent, that is , the spectral radius $r(X)$ is $0.$
My question is the following:
Is it true for $\tau_{A, B}$ also. If not, is it true if we assume $AB=BA.$
Any help or reference will be appreciated.
I found the following counterexample (after posting this question):
Take $A=\begin{bmatrix} 0&1\\ 0&0 \end{bmatrix}$ and $B=\begin{bmatrix} 0&0\\ 0&0 \end{bmatrix}.$ Then $X= \begin{bmatrix} 1&0\\ 0&0 \end{bmatrix}=A\begin{bmatrix} 0&0\\ 1&0 \end{bmatrix}$ and also $AX=0.$