I've known that for an $f$ orthogonal operator to be diagonalizable, the eigenvalues have to be real, but what doesn't make sense to me, is that they have to be $-1$ or $1$ in order to be diagonalizable. I've learnt that those eigenvalues correspond to rotations and reflections, but why $-1$ and $1$? What makes them special? My thoughts are that since it is called "orthogonal" their bases have to be orthogonal, and by doing a rotation or a reflection, their base remain to be orthogonal after the transformation. What is it?
2026-03-28 13:11:06.1774703466
Why are $-1,1$ the only diagonalizable eigenvalues for orthogonal operators?
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