Is it true that if $A=BA^{*}A$ then $A^{*}=B^{*}AA^{*}$

Question

I wonder is it true that for any $n \times n$ matrices $A$, and $B$.// If $A=BA^{*}A$ is true, does it imply that $A^{*}=B^{*}AA^{*}$?// I used mathematica to check the condition in 3 by 3 case, and it works, but could anyone have an idea to prove that this is true for n dimensions？// Thanks! P.S. I do have a proof for this when A and B are partial isometries, but it confuses me that whether it was true for any general matrices.

Answer 1

If $A^*$ denotes the adjoint matrix (complex conjugate) of $A$ that is not true. Take:

$$A = \left( \begin{array}{c c} 1 & 0 \\ 0 & 0 \end{array} \right) \qquad B = \left( \begin{array}{c c} 1 & 1 \\ 0 & 0 \end{array} \right) $$

Then:

$$A^* = \left( \begin{array}{c c} 1 & 0 \\ 0 & 0 \end{array} \right) \qquad B^* = \left( \begin{array}{c c} 1 & 0 \\ 1 & 0 \end{array} \right)$$

But:

$$BA^*A= \left( \begin{array}{c c} 1 & 0 \\ 0 & 0 \end{array} \right) \qquad B^*AA^* = \left( \begin{array}{c c} 1 & 0 \\ 1 & 0 \end{array} \right) $$

You need to impose more conditions if you want that result to be true or $*$ does not mean adjoint matrix.

EDIT: I changed $B$ and $B^*$ following Jonas Meyer's advice so now it is a complete counterexample.