$A\in M_{2n+1}(\mathbb{R})$, 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. 

Matrices for which $X^{\top}\! X = E$ are orthogonal, and they preserve the length of every vector. If an orthogonal matrix has a real eigenvector then it can only fix it or reflect it, i.e. the eigenvalue can only be $+1$ or $-1$ respectively. Some orthogonal matrices have no real eigenvalues, e.g. a rotation of $\pi/2$ in the plane, but that's because all of their eigenvalues appear in complex conjugate pairs. In the case where $X$ acts on an odd-dimensional space, there must be at least one real eigenvalue: you can't pair an odd number of eigenvalues off into complex conjugates. But we've already seen that if an orthogonal matrix has a real eigenvalue then it must be $\pm 1$.
