The question is simply this:
Let Q$\in$$M_n(R)$ and $\lambda$ a complex eigenvalue of Q. Show that its conjugate is also an eigenvalue of Q.
My attempt:
Let $\lambda$ denote the complex eigenvalue, and $\mu$ its complex conjugate. Since Q is orthogonal, we know that $|\lambda$|=1. Squaring, we have that $\lambda \mu$ = 1. We also know that $det(Q)=\pm1$. That is, $det(Q)=\pm \lambda \mu$. Since the determinant of Q is the product of its eigenvalues, it follows that $\mu$ is also an eigenvalue of Q.
Is it correct?
Is this result more general? That is, does it apply to any matrix?
Best Answer
If $Q \in M_n(\mathbb R)$, then the char. polynomial $p$ of $Q$ has real coefficients. Then it is easy to see that
$p(\lambda)=0$ $\quad$ iff $\quad$ $p(\overline{\lambda})=0$.
$Q$ need not to be orthogonal !