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.


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.