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This might be a basic question, nonetheless I cannot give a proof.

Given an orthogonal matrix $A$ with eigendecomposition $A = Q \Lambda Q^{-1}$ with only non-real eigenvalues. Given also a diagonal real matrix $\Phi$ with $\Phi_{ii} = \Phi_{jj}$ if $\Lambda_{ii} = \overline{\Lambda_{jj}}$. The following matrix power is defined as $[\Lambda^\Phi]_{ii} := \Lambda_{ii}^{\Phi_{ii}}$.

Why is $Q \Lambda^\Phi Q^{-1}$ orthogonal? (Unitarity is simple, but why is it real?)

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    $\begingroup$ Have you tried what happens if $A=-I$ and $\Phi=\frac12I$? $\endgroup$ Commented Nov 25, 2014 at 15:15
  • $\begingroup$ @Joonas: Sorry, I just realized that I was quite loose on the restrictions. I'll edit the question. $\endgroup$
    – Jiro
    Commented Nov 25, 2014 at 15:18

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The complex eigenvalues of a real matrix come in conjugate-complex pairs, with conjugate-complex pairs of eigenvectors. But the converse is true: If you associate to conjugate-complex pairs of eigenvectors any conjugate-complex pairs of eigenvalues, the result will be real because it is the sum of two terms which are conjugate-complex. You can take any diagonal matrix $\Psi$ with $\Psi_{ii} =\overline{\Psi_{jj}}$ if $\Lambda_{ii} = \overline{\Lambda_{jj}}$, and $Q\Psi Q^{-1}$ will be orthogonal.

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  • $\begingroup$ Thanks for the answer. I guess this is also true for any real matrix, right? Is there a short argument for this? Or textbook reference. Sorry, this is definitely no research level question. $\endgroup$
    – Jiro
    Commented Nov 26, 2014 at 11:55
  • $\begingroup$ @Sebastian, What do you mean, "any real matrix", for which matrix? Was my argument not short enough? $\endgroup$ Commented Nov 26, 2014 at 13:28
  • $\begingroup$ All fine. I was just asking whether the restriction on $Q$ matters, but I already realized that it does not. So that the fact is true for any paris of conjugate-complex pairs eigenvalues and eigenvectors. And also your answer was short enough. Thanks $\endgroup$
    – Jiro
    Commented Nov 26, 2014 at 13:51

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