$A\in M_{2n+1}(\mathbb{R})$, with $AA^t=I$, prove that either $1$ or $-1$ is an eigenvalue. So basically the question is in the title. I did it this way:
Use the fact that if $\lambda$ is an eigenvalue (with eigen vector $x$) of a normal operator $T$, then $x$ is an eigenvector corresponding to $\bar{\lambda}$ of $T^*$. 
Since $AA^t=I$, and our entries are in the reals, we know that if $\lambda$ is an eigenvalue of $A$, then it is also an eigenvalue of $A^t$ (with the same eigenvector). 
To prove the existence of such an eigenvalue we see that the char poly of $A$ is going to have odd degree (here we use the $2n+1$ assumption), and hence a root, so at least one eigenvalue exists. 
Hence, say $P$ is the change of coordinate matrix such that $PAP^{-1}=B$ where $B$ has as a first column $(\lambda,0,....,0)^t$, by the above remarks we know that this is also the first column of $B^t$, so $$BB^t=PAP^{-1}PA^tP^{-1}=PIP^{-1}=I$$the $(1,1)$ entry of the LHS is $\lambda^2$ and the RHS is $1$, so $\lambda^2=1$, and we are done. 
Im trying to find other ways to do this problem. I ask because I missed the session and I saw that this was one of the problems done, but they havent learned the first claim (the one of normal operators) that I used. (I believe they were talking about minimal polynomials that day). If anyone knows about another way it would be appreciated. 
Thanks. 

$Av=\lambda v\implies||Av||=||\lambda v||\implies\langle Av,Av\rangle=\langle\lambda v,\lambda v\rangle$
now using the fact the over $\mathbb{R}$ it holds that $A^{t}=A^{*}$
and since $A$ is orthogonal we have $\langle v,v\rangle=|\lambda|^{2}\langle v,v\rangle\implies|\lambda|=1$.
Since, as you stated, the char poly of $A$ is of odd degree it has
a real root $\lambda$ and the claims follow since $|\lambda|=1$
and $\lambda\in\mathbb{R}$ imply $\lambda\in\{-1,1\}$.
Note: I used that $v\neq 0$ since it is an eigenvector to deduce $\langle v,v\rangle\neq 0$
