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I was trying to show that orthogonal matrices have eigenvalues $1$ or $-1$.

Let $u$ be an eigenvector of $A$ (orthogonal) corresponding to eigenvalue $\lambda$. Since orthogonal matrices preserve length, $ \|Au\|=|\lambda|\cdot\|u\|=\|u\|$. Since $\|u\|\ne0$, $|\lambda|=1$.

Now I am stuck to show that lambda is only a real number. Can any one help with this?

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    Real symmetric matrices do, but otherwise not necessarily.2011-09-24
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    You need to specify over what field you are working; in some contexts, "orthogonal matrices" is reserved to matrices operating on *real* vector spaces, and as such can have only real eigenvalues (though their characteristic polynomial may have complex roots). In other contexts, "orthogonal matrices" does not restrict the field of the underlying vector space.2011-09-25
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    This isn't true as shown below. Orthogonal matrices do have two differences, whether they have a determinant of 1 or -1, not the eigenvalues being 1 or -1. The ones that have a determinant of 1 form the Special Orthogonal Group...2013-09-10

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