I'm trying to prove the following statement :
If $A$ is a real unitary matrix then $A$ is similar to $A^*$.
Here is what I have so far:
$A$ is unitary and therefore is unitary diagonalizable. Thus, there is unitary $P$ and diagonal $D$ such that : $P^{*}AP = D$.
In order to show that $A$ is similar to $A^*$ I should show that : $P^{*}A^*P = D$ (That is, both $A$ and $A^*$ have the same eigenvalues).
But all I know about $A^*$ is that $A^* \overbrace{=}^{A \ is \ unitary} A^{-1} \overbrace {=}^{A \ is \ real} A^t$.
Can you please give an hint how can I continue from here?
Thank you
First of all not all real unitary matrices can be diagonalized over the reals, for example a 45 degree rotational operator in $\mathbb{R}^2$--it has no real eigenvalue. They are always unitarily diagonalized over the complex field though.
Now if you assume $A$ has only real eigenvalues, $P^*AP=D$, then clearly $(P^*AP)^* =P^*A^*P = D^*=D$
If however $D$ is not real, you don't have $D^*=D$ in general. But since the eigenvalues are the roots of the characteristic polynomial, they come in conjugate pairs. This means that $D$ and $D^*$ have the same spectrum, hence is similar by a permutation matrix.