Suppose $A \in \mathcal{O}(n,\mathbb{R})$. Then $A^{T}A=AA^{T}=A^{*}A=AA^{*}=1$ (where * denotes the Hermitian conjugate). Thus $A$ is normal and hence by spectral theorem it has a decomposition such that $A=SDS^{*}$ where $S \in \textbf{U}(n,\textbf{C})$ with orthonormal eigenvectors as columns and $D$ is complex diagonal. Then $$ A^{T}A=(SDS^{*})^{T}SDS^{*}=\bar{S}DS^{T}SDS^{*}=I $$
Where $I$ is the identity as usual.
I'm stuck at this point. Typically I would want to use the fact that S is unitary to simplify the issue, but I don't see how I can get the eigenvalues out of this. I also tried another method and got the following:
Suppose $A$ is as before. Let $v$ be an eigenvector of $A$ with eigenvalue $\lambda$, i.e. $Av=\lambda v$. Then $$ (Av)^T Av=(\lambda v)^T \lambda v \implies v^T A^T Av=v^T \lambda^{2} v $$ hence by orthogonality of A, $$ v^T v=\lambda^{2} v^T v \implies \lambda^2=1 $$
Thus $\lambda=\pm 1$.
Now this is all fine and good, but my class notes suggest that $|\lambda|=1$ and that the eigenvalues are either real or complex conjugate pairs. How do I get the conjugate pairs condition? It doesn't seem apparent from either of these methods. Thanks for listening!
If $A \in \mathcal{O}(n, \mathbb{R})$, then $A$ is also in $\mathcal{O}(n, \mathbb{C})$, and satisfies the property $\langle Av, Aw \rangle = \langle v, w \rangle$ for any two vectors $v, w \in \mathbb{C}^n$, where $\langle \cdot, \cdot \rangle$ is the standard inner product given on column vectors by $\langle v, w \rangle = \sum_i v_i \overline{w_i}$.
If $\lambda$ is an eigenvalue of $A$, then since we are over $\mathbb{C}$ it has an eigenvector $v$. Then $$\langle v, v \rangle = \langle Av, Av \rangle = \langle \lambda v, \lambda v \rangle = \lambda \overline{\lambda} \langle v, v \rangle$$ And since $\langle v, v \rangle \neq 0$, we have $|\lambda| = 1$. The fact that all eigenvalues are real or complex conjugate pairs comes from the fact that the characteristic polynomial has real coefficients.