Suppose, $A,B\in \mathbb{C}^{d\times d}$, $A$ is normal, and $\text{dim ker}(AB-BA)\ge (1-\delta)d$. Does then follow that $\text{dim ker}(A^*B-BA^*)\ge (1-\delta)d$, possibly for different $\delta$?
This is known to hold for $\delta=0$ because there is polynomial $P$ such that $A^*=P(A)$. But here, if $v\in \text{ker}(AB-BA)$, then $P(A)Bv=BP(A)v$ need not hold as $B$ might not commute with higher powers of $A$, so one could assume that $v\in \bigcap_{i=0}^{d-1}\text{ker}((AB-BA)A^i)$, which would certainly imply $P(A)Bv=BP(A)v$ but then the dimension of this intersection is perhaps too small.
I'm not sure if the statement is true.