Let's consider $R_{n}$ to be an $n \times n$ real rotation matrix. $R_{n}$ can be diagonalized with a unitary matrix $U_{n}$ to $D_{n}$, which is composed of blocs of $ \left( \begin{array}{ccc} e^{j\theta} & 0 \\ 0 & e^{-j\theta} \end{array} \right) $ and eventually ones.
But what I want is to have a real rotation matrix decomposition, with blocs of $ \left( \begin{array}{ccc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array} \right)$ and ones.
How do I get the change of basis matrix that allows me to get such a decomposition ? Can I derive it from the $U_{n}$ matrix and the eigenvalues (the $\theta$s) ?
Notice that $$\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix} = \begin{pmatrix}\tfrac{1}{\sqrt{2}} & -\tfrac{j}{\sqrt{2}} \\ -\tfrac{j}{\sqrt{2}} & \tfrac{1}{\sqrt{2}}\end{pmatrix} \begin{pmatrix}e^{j\theta} & 0\\ 0 & e^{-j\theta}\end{pmatrix} \begin{pmatrix}\tfrac{1}{\sqrt{2}} & \tfrac{j}{\sqrt{2}} \\ \tfrac{j}{\sqrt{2}} & \tfrac{1}{\sqrt{2}}\end{pmatrix}.$$
So we can define $W_n$ to be a block diagonal matrix with the same block structure as $D_n$ such that for each $\begin{pmatrix}e^{j\theta} & 0\\ 0 & e^{-j\theta}\end{pmatrix}$ block in $D_n$, the corresponding block in $W_n$ is $\begin{pmatrix}\tfrac{1}{\sqrt{2}} & \tfrac{j}{\sqrt{2}} \\ \tfrac{j}{\sqrt{2}} & \tfrac{1}{\sqrt{2}}\end{pmatrix}$, and for each $\begin{pmatrix}1\end{pmatrix}$ block in $D_n$, the corresponding block in $W_n$ is also $\begin{pmatrix}1\end{pmatrix}$.
Then, $W_n^*D_nW_n = W_n^*U_n^*R_nU_nW_n$ will be block diagonal with several blocks of the form $\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}$ and then several ones.