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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.

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$Av=\lambda v\implies||Av||=||\lambda v||\implies\langle Av,Av\rangle=\langle\lambda v,\lambda v\rangle$ 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 $\langle v,v\rangle\neq 0$

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    For the pairs of complex eigenvalues you have that their multiplication is $1$ since for every $z\in\mathbb{C}$ you have $z\overline{z}=|z|^{2}$ and we already noted that if $z$ is an eigenvalue then $|z|=1$. We know that there are real roots since the degree of the char.poly is odd, and we have even number of complex roots (that are not real) so there are odd number of real roots, if all of them were$-1$ them since we have an odd number of them the determinant would of been $-1$. OK ?2012-08-30
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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 $\pm 1$.