Simultaneous similarity of two complex matrices and their hermitian conjugates implies their unitary similarity

Question

Let $A$ and $B$ be two similar complex $n \times n$ matrices, i.e., there exists $P \in GL(\mathbb{C},n)$ such that $A=P B P^{-1}$. Furthermore, suppose that $A^{*} = P B^{*} P^{-1}$, where $M^{*}$ designates the conjugate transpose of a complex matrix $M$. I would like to show that $A$ and $B$ are unitarily similar. I know one can easily get relations between real matrices by addition and subtraction of the two equations, but I don't see where unitarity would come from. I also tried to us the polar decomposition ($P=U H$, with $U$ unitary and $H$ Hermitian positive definite) but I don't see how to get rid of $H$ then. Only basic linear algebra if possible and, if necessary, only basic topology, please. I hope I interpreted my textbook problem correctly : it says "$A$ and $A^{*}$ are simultaneously similar to $B$ and $B^*$". I am originally a physicist but now also an amateur mathematician (using * instead of + :-).

Answer 1

As announced in my answer to Bruno B's comment, the second part of Specht's original proof to his theorem contains the answer to my question. So I just adapt it here.

By taking the Hermitian conjugate of $A=PBP^{-1}$, we obtain $P^{*-1} B^{*} P^{*}=A^{*}$. Therefore, $P B^{*} P^{-1} = P^{*-1} B^{*} P^{*} $ or $P^{*} P B^{*} = B^{*} P^{*} P $. Since $P^{*} P$ is similar to a diagonal matrix with only strictly positive eigenvalues, $\lambda_1,\dots,\lambda_n$, there exists a matrix $H$, similar to a diagonal matrix with elements $\sqrt{\lambda_1},\dots,\sqrt{\lambda_n}$, and satisfying $H^2 = P^{*} P$.

Since there exists a real polynomial $p$ such that $p(\lambda_i) = \sqrt{\lambda_i}$, for all $i=1,\dots, n$, $H$ is a polynomial of $P^{*} P$ and therefore commutes with $B$.

Now, we define $U=H^{-1} P^{*}$, which is unitary, because $U U^{*} = H^{-1} P^{*} P H^{-1} = H^{-1} H^2 H^{-1} = 1$, to obtain finally, $A=PBP^{-1}=U^{*} H B H^{-1} U = U^{*} B U$.