Show that orthogonal matrices are diagonalizable I want to prove that all orthogonal matrices are diagonalizable over $C$. I know that a matrix is orthogonal if $Q^TQ = QQ^T = I$ and $Q^T = Q^{-1}$, and that a matrix $A$ is diagonalizable if $A = PDP^{-1}$ where $D$ is a diagonal matrix. How can I start this proof? 
 A: As people have indicated, you could simply apply the spectral theorem. Here I run through a specialized argument to the orthogonal case:
Since $Q$ is orthogonal we have $\langle Qv, Qw \rangle = (Qv)^*Qw = v^* Q^T Q w = \langle v, w \rangle$. 
Given any eigenvector $v$ with eigenvalue $\lambda$, if we have some vector $w$ orthogonal to $v$ then we have $\lambda \langle v, Qw \rangle = \langle Qv, Qw \rangle = \langle v, w \rangle = 0$, so $Q$ maps $v^\perp$ into itself. We can induct on the dimension of our space to show $Q$ acts diagonalizably on $v^\perp$ so it acts diagonalizably on $v \oplus v^\perp$
We can infact say more:
Note that if $\lambda$ is an eigenvector of $Q$ then we have $|\lambda|\|v\| = \langle \lambda v, \lambda v \rangle = \langle Qv, Qv \rangle = \|v\|$. We conclude all the eigenvalues have norm $1$.
If $v,w$ are eigenvectors with different eigenvalues then we have $\langle v, w \rangle = \langle Qv, Qw \rangle = \langle \lambda v, \mu w \rangle = \lambda \mu^* \langle v, w \rangle$. Thus if $\lambda \mu^* \neq 1$ then $v$ and $w$ are orthogonal.
Combining these one can show that $Q = PRP^{-1}$ where $P$ is an orthogonal matrix and $R$ is a block diagonal matrix with $1,-1$ and $2 \times 2$ rotation matrices down the diagonal. 
A: Note that if $Q$ is orthogonal then $Q$ is normal, because
\begin{equation*}
Q Q^T = Q^T Q = I.
\end{equation*}
So the spectral theorem implies that $Q$ is diagonalizable.
