
Orthogonal Matrix
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 $ ?

Hi, it should be alpha^2 and not alpha^2 cause you don't take complex conjugate but rather the transpose only, and because alpha=ai a is real, alpha^2=a^2 so it cannot be equal 1.