For invertible matrix $B$, does always apply $\operatorname{rank}(AC)=\operatorname{rank}(ABB^{\dagger}C)$?

Question

For invertible complex matrix $B$, for $B^{\dagger}$ the Hermitian conjugate, does for all compatible complex matrices $A,C$ necessarily apply $\operatorname{rank}(AC)=\operatorname{rank}(ABB^{\dagger}C)$?

Answer 1

Let $$A=\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} ;C= \begin{pmatrix} 0 & 0\\ 1 & 0\\ \end{pmatrix} ;B= \begin{pmatrix} i & 0\\ 1 & -i\\ \end{pmatrix}. $$ Then $$B^*=\begin{pmatrix} -i & 1 \\ 0 & i \\ \end{pmatrix} ;AC= \begin{pmatrix} 0 & 0\\ 0 & 0\\ \end{pmatrix} ;ABB^*C= \begin{pmatrix} i & 0\\ 0 & 0\\ \end{pmatrix}. $$

It is clear, that $rank(B)=2$ thus $B$ is invertible, but $0 = rank(AC) \ne rank(ABB^*C) = 1$, so we constructed a counterexample.