Best Answer:
I'll delete this if somebody puts an answer that isn't quite as lazy as this. But for now, here's a hint:

What would you expect, in general, to be the eigenvalues of a matrix multiplied by its (conjugate) transpose? If for an orthogonal matrix, doing this gives you the identity (which has only 1 as its eigenvalues), what does that tell you?

EDIT: some elaboration

Okay, so a proof that the eigenvalues have magnitude 1:

We note that for any orthogonal matrix M, we have

<x,y> = <M^t M x, y> = (M^t M x)^t y = (M x)^t M y = <M x , M y>.

That is, all orthogonal matrices preserve the dot product of two vectors.

It follows that

||Mx||^2 = <M x, M x> = <x, x> = ||x||^2

That is, Mx has the same magnitude as x.

If v is an eigenvector with eigenvalue L, then Mv = L*v. However, we have

||v|| = ||M v|| = ||L * v|| = |L| * ||v||

That is, we deduce that ||v|| = |L| * ||v||, from which we conclude that |L| = 1.

It follows that all eigenvalues have magnitude 1.

Now, if M doesn't have complex eigenvalues, it means that M is either a reflection or the identity. If M has complex eigenvalues, then M is not purely a reflection or the identity. For example, any rotation matrix has complex eigenvalues.

Hope that helps

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