Eigen values of a positive semidefinite matrix and its transpose

Question

$A\in M_n(\mathbb{C})$ is positive semi-definite so there there exists unitary matrix $U$ such that $A=U^*DU$ where $D$ is the real diagonal matrix consisting of eigen values $(\ge 0)$ of $A$, now I need to find the eigen values of $A^T$, so $A^T=U^TD^T(U^T)^*= BDB^*$ where $B=U^T$ is unitary, so $A$ and $A^T$ has same set of eigen values, Is my logic correct? Thank you.

Answer 1

Yes, your reasoning is correct.

You should probably show that $U^T$ is unitary. Note that in general, $\overline{AB} = \overline A \, \overline B$ and that $U^T = \overline{U^*}$.