What would the tranpose of $Y$ look like $$Y ={A}^{*} \operatorname{diag} \left( b \right) {A} $$
where $*$ is the conjugate tranpose, i.e., hermitian.
We have $\operatorname{diag}(b)^*=\operatorname{diag}(\overline{b})$, hence
$Y^*=A^*\operatorname{diag}(\overline{b})A^{**}=A^*\operatorname{diag}(\overline{b})A$.
$Y^T=A^T\operatorname{diag}(\overline{b})(A^{*})^T=A^T\operatorname{diag}(\overline{b})conj(A)$?
– abina shr Jan 17 '19 at 11:27