I am reading an old paper by Horn (see ch.4):

And for learning convenience of users, the following is a similar discussing:
Finding the Similarity Transform of a rotation matrix

Let $R$ be a rotation matrix.

Then there exists an orthogonal matrix $U$, such that $R = U^TAU$ with

enter image description here

So, for example, if $n=3$, then $A$ should look like
$$A = \begin{bmatrix}1 & 0& 0 \\0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{bmatrix}$$

How to prove this? (WLOG, suppose the size of matrices is odd.) In fact, $A$ is also an rotation matrix.

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    $\begingroup$ it is just the real version of the spectral theorem. Each $B_i$ corresponds to a complex conjugated pair eigenvalues of $R$. $\endgroup$ – user251257 Nov 5 '16 at 2:29
  • $\begingroup$ @user251257 I see, thanks. The dimension of $B_i$ in such representation still makes me feel odd however. $\endgroup$ – sleeve chen Nov 5 '16 at 2:34
  • $\begingroup$ what do you mean? $\endgroup$ – user251257 Nov 5 '16 at 2:39
  • $\begingroup$ @user251257 I mean I am not used to the eigenvalue $e^{\pm \theta} \in R$ corresponds to the eigenvalue $B_i\in R^{2\times 2}$ $\endgroup$ – sleeve chen Nov 5 '16 at 2:46
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    $\begingroup$ $B_i$ itself is a rotation, so it's eigenvalues have the form $\exp(\pm \mathbf i \beta)$ where $\mathbf i$ is the imaginary unit and $\beta\in [0,2\pi)$. $\endgroup$ – user251257 Nov 5 '16 at 2:50

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