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?)



Powers of orthogonal matrices is closed
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.


