Orthogonal matrix with eigenvalues $\in \mathbb{R}$ is symmetric

Question

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.

You can actually take $S$ to be unitary. – Julien Feb 05 '13 at 19:21 — Julien, Feb 05 '13 at 19:21
Related to: https://math.stackexchange.com/q/913093/497335 – Bach Aug 02 '19 at 02:32 — Bach, Aug 02 '19 at 02:32

score 5 · Accepted Answer · answered Feb 05 '13 at 19:21

5

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.

answered Feb 05 '13 at 19:21

Julien

45,674

You're welcome. – Julien Feb 05 '13 at 19:48

Orthogonal matrix with eigenvalues $\in \mathbb{R}$ is symmetric

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