Is $\operatorname{rk}(\bar AA)=\operatorname{rk}(A^*A)$ ??

Question

I was trying to prove that $\operatorname{rank}(\bar AA)=\operatorname{rank}(A^*A)$, but I didn't know how.$\DeclareMathOperator{\rk}{rk}$

I proved that, $\rk(A)=\rk(A^T)=\rk(\bar A)=\rk(A^*)=\rk(A^*A)$

I tried $\rk(\bar AA)=\rk(A\bar A)=\rk(A^*A^T)$ but couldn't go further...

Any help is appreciated, and thanks in advance...

Answer 1

It isn't true.

Consider $A =\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. Then $\overline{A}A = A^2 = 0$ but $$A^*A = A^TA = \begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}$$