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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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    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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The eigenvalues of $ \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} $ are $\cos\theta \pm i\sin\theta= e^{\pm i\theta}$. This is an orthogonal matrix.

If a matrix with real entries is symmetric (equal to its own transpose) then its eigenvalues are real (and its eigenvectors are orthogonal). Every $n\times n$ matrix whose entries are real has at least one real eigenvalue if $n$ is odd. That is because the characteristic polynomial has real coefficients so the complex conjugate of a root is another root, and you can't have an odd number of roots if they come in pairs of distinct entries.

But generally, the eigenvalues of matrices with real entries need not be real.

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    @CarlMummert : But in that case, this matrix would still not have real eigenvalues. $\qquad$2016-11-01
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No, a real matrix does not necessarily have real eigenvalues; an example is $\pmatrix{0&1\\-1&0}$.

On the other hand, since this matrix happens to be orthogonal and has the eigenvalues $\pm i$ -- for eigenvectors $(1\mp i, 1\pm i)$ -- I think you're supposed to consider only real eigenvalues in the first place.

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    @anon, otherwise what he's trying to prove is not true, as shown by my counterexample.2011-09-24
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I guess it depends whether you are working with vector spaces over the real numbers or vector spaces over the complex numbers.In the latter case the answer is no, however in the former the answer has to be yes.Is it not guys ?

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    this answer is technically true but not at all helpful, and obviously only confuses things given the discussion following the answer2017-11-03