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I need help showing that an orthogonal matrix with all real eigenvalues is symmetric. The condition for Orthogonality is. $$O^T = O^{-1} \implies O^TO = I$$

But if O is also symmetric:

$$O^T = O = O^{-1}$$

I have tried using the similarity transform relationship:

$$S^{-1}OS = D = diag(\lambda_i) \implies (S^{-1}OS)^T = S^TO^{-1}S^{-1^T} = D$$

I don't really understand where the idea of real eigenvalues comes in other than the fact that symmetric matrices have all real eigen values. I'm not quite sure how to prove what I want to show without explicitly using it.

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Hint 1: An orthogonal matrix is normal.

Hint 2: Every normal matrix is diagonalizable via a unitary matrix (this is actually a characterization of normality).

Hint 3: Prove the result for a normal matrix.

Hint 4: The answer is somewhere here http://en.wikipedia.org/wiki/Normal_matrix in the form of a characterization of self-adjoint matrices.

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  • $\begingroup$ You're welcome. $\endgroup$ – Julien Feb 5 '13 at 19:48

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