Let $A,B\in M_n$ be complex square matrices. Let $C=ABA^{-1}B^{-1}$ be the multiplicative commutator of $A, B$ and assume that both $A$ and $C$ are normal. Now assume that $AC=CA$ and $0\notin\operatorname{conv}\sigma(B)$, and diagonalize $A = U\Lambda U^*, C = UMU^*$ for some $U$ unitary and $\Lambda, M$ diagonal. Let $\mathcal{B} = U^*BU = [\beta_{ij}]_{i,j=1}^n$, then is it true that $\beta_{ii} \neq 0$ for all $i=1,\dots,n$ ?
I've tried to use the following: if $B$ is also normal, then since $0\notin\operatorname{conv}\sigma(B)$, we have $$ \beta_{ii} = e_i^T\mathcal{B}e_i=u_i^*Bu_i=u_i^*Q^*\operatorname{diag}(b_1,\dots,b_n)Qu_i=\sum_{i=1}^n \lVert\xi_i\rVert^2b_i, $$ where $u_i$ is the $i$-th column of $U$, $Q$ is unitary, and $\xi=Qu_i=[\xi_i]$. Since $\xi$ has norm $1$, it follows that $\beta_{ii}$ is a convex combination of the eigenvalues of $B$, and so it can't be zero. But, I fail to establish that $B$ is normal.