Can an orthogonal matrix that represents a linear transformation from $\mathbb{R}^n \to \mathbb{R}^n$ have no eigenvalues? I know that if the matrix is normal and represents a transformation in a unitary space then it can be unitarily diagonalized, so it must have eigenvalues. Plus, its characteristic polynomial is never a constant, so it has roots according to the Fundamental Theorem of Algebra.
But what if we talk about orthogonal matrices that represent transformations in a Euclidean vector space? Can they have no eigenvalues?
 A: Yes, see for instance $n=2$ and the matrices in the form $\begin{pmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{pmatrix}$ for $\alpha\notin\pi\Bbb Z$.
A: For a $2n$-dimensional space this is true: just consider a generalization of the example of Gae. S.
$$T=\begin{pmatrix}J_1&&&\\
&J_2&&
\\&&\ddots&\\
&&&J_n
\end{pmatrix}\\
J_i=\begin{pmatrix}\cos(\alpha_i)&-\sin(\alpha_i)\\
\sin(\alpha_i)&\cos(\alpha_i)
\end{pmatrix};(\alpha_i\notin\pi\mathbb{Z})$$
However, the claim is not true in a $2n+1$-dimensional space; this follows trivially from the fact that a polynomial of odd degree must have at least one real root.
Actually, we can prove something more quite easily:
Claim: every $2n+1$ dimensional orthogonal matrix has an eigenvalue $\lambda\in\{-1,+1\};\lambda=\det(T)$.
Proof: Since it is orthogonal, $\det(T)=\pm1$. Let us suppose that $\det=1$ (this is not restrictive, since multiplying $T$ by $-1$ changes the sign of the eigenvalues and of the determinant)
$$\det(T-I)=\det(T^{t})\det(T-I)=\det(I-T^{t})=\\=\det(I-T)=(-1)^{2n+1}\det(T-I)=-\det(T-I)$$
Thus $\det(T-I)=0$, and $\lambda=1$ is an eigenvalue.
In $3$ dimensions, this is known as Euler's rotation theorem, since it can be phrased as "every 3-dimensional rotation has a fixed axis".
