Elements of SO(n) is block-diagonalizable I am not able to show that elements of SO(n) are conjugate to a block-diagonal matrix with 2x2 blocs that are rotation matrices, and a 1x1 bloc 1 if n is odd. Can someone help me please?
 A: Orthogonal matrices are normal, that is, $A^*A=AA^*$ and, therefore they are diagonalizable. In other words, every eigenvalue has algebraic multiplicity equals to its geometric multiplicity. Therefore, its real Jordan form has only the diagonal blocs.
Fix $A\in SO(n)$. Take $\lambda\in \operatorname{spec}(A)$. Then there is a eigenvector $v$ such that
$$Av=\lambda v.$$
Thus,
$$v^*A^*=\overline{\lambda}v^*.$$
Then,
$$v^*A^*Av=\overline{\lambda}v^*\lambda v=|\lambda|\operatorname{||}v\operatorname{||}.$$
By the other side,
$$ v^*A^*Av=v^*Iv=\operatorname{||}v\operatorname{||}.$$
Therefore,
$$|\lambda|\operatorname{||}v\operatorname{||}=\operatorname{||}v\operatorname{||}.$$
As $v$ is an egeinvector and, therefore, non-null, $|\lambda|=1$. Thus $\lambda=\cos(\theta)+i\sin(\theta).$
Then, its Jordan form is of the form
$$\left(
\begin{array}{cccccc}
R_{t_1} & &  &  &  &\\ 
& \ddots &  &  &  &\\
 &  & R_{t_p} &  & &\\
 &  &         & 1  & &\\
 &  &         &   & \ddots&\\
 &  &         &   & &1
\end{array}\right)$$
In which
$$R_{t}=
\left(\begin{array}{cc} 
\cos(t) & \sin(t)\\
-\sin(t) & \cos(t) 
\end{array}\right).$$
Note that as $\det(A)=1$, then one has that
$$\lambda\in\operatorname{A}\Leftrightarrow \overline{\lambda}\in\operatorname{A}.$$
And, if $-1\in\operatorname{A}$, then it has an even multiplicity.
Therefore, if $n$ is odd, 1 must be an eigenvalue.
