How to find a matrix $P$ invertible such that $PA=BP$ where,
$$A= \begin{pmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \\ \end{pmatrix} $$ and $$B= \begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \\ \end{pmatrix} $$ Is there any operation to find $P$.If I try to find the roots of $λ^2−2λcosθ+1=0$ I get $e^{iθ}$ and $e^{−iθ}$ then I try to find characteristic vector by trying to find $v$ such that $Av=v\operatorname{exp}(iθ)$ which ultimately leads me to calculation similar to finding $P$ such that $PA=BP$ but I want to avoid that bad calculation and want to tackle this using operation on matrices if possible.
Knowing eigenvectors it's straightforward.
To save our time, we can reason as follows
All matrices $A_{\theta}$ pairwise commute and are diagonalizable. Then there is a common matrix $P$ (that does not depend on $\theta$) that diagonalizes all these matrices. Then, to find such a matrix $P$, it suffices to diagonalize
$A_{\pi/2}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\rightarrow diag(i,-i)$.
EDIT. Answer to the OP's comment. OK, I understand..
Then put $P=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and solve $P\begin{pmatrix}0&-1\\1&0\end{pmatrix}=diag(i,-i)P$.
Then $P$ depends on 2 parameters; randomly choose them and (except if you are particularly unlucky) the obtained $P$ is invertible and works for any $A_{\theta}$..