Let $\displaystyle \alpha $ be an imaginary eigenvalue of an orthogonal matrix Q adn let X be a corresponding eigenvector. Prove that $\displaystyle X^T X = 0$

Ok here is what I did:

$\displaystyle X^TQ^{-1}QX = X^T Q^{-1} \alpha X = \alpha X^T Q^{-1}X = \alpha X^T Q^T X = \alpha (QX)^T X $

$\displaystyle = \alpha ( \alpha X)^T X = a X \overline{\alpha} X = \alpha \overline{\alpha} X^T X = {|\alpha|}^2 X^T X $

Also $\displaystyle X^TQ^{-1}QX = X^T X $

So $\displaystyle X^T X = {|\alpha|}^2 X^T X $

i.e. $\displaystyle (1-{|\alpha|}^2) X^T X = 0 $

i.e. $\displaystyle X^T X = 0 $ or $\displaystyle (1- {|\alpha|}^2) = 0 $

Why can't $\displaystyle (1- {|\alpha|}^2) = 0 $ ?