# Why are $-1,1$ the only diagonalizable eigenvalues for orthogonal operators?

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?

• An orthogonal operator, by definition, preserves the inner product. If $v$ is an eigenvector, we necessarily have $\Vert f(v)\Vert=\Vert v\Vert$. Which eigenvalues are possible for this equation to hold? Commented Apr 28, 2023 at 14:06
• @Levent is there a more geometrical way to visualise this? Commented Apr 30, 2023 at 11:56
• @Levent I've reasked my thoughts on a new, and better written question: math.stackexchange.com/questions/4689455/… Commented Apr 30, 2023 at 12:11