If Q is an orthogonal matrix and $\lambda$ is a complex eigenvalue, show that its conjugate is also an eigenvalue of Q. The question is simply this:
Let Q$\in$$M_n(R)$ and $\lambda$ a complex eigenvalue of Q. Show that its conjugate is also an eigenvalue of Q.
My attempt:
Let $\lambda$ denote the complex eigenvalue, and $\mu$ its complex conjugate. Since Q is orthogonal, we know that $|\lambda$|=1. Squaring, we have that $\lambda \mu$ = 1. We also know that $det(Q)=\pm1$. That is, $det(Q)=\pm \lambda \mu$.  Since the determinant of Q is the product of its eigenvalues, it follows that $\mu$ is also an eigenvalue of Q.
Is it correct?
Is this result more general? That is, does it apply to any matrix?
 A: If $Q \in M_n(\mathbb R)$, then the char. polynomial $p$ of $Q$ has real coefficients. Then it is easy to see that
$p(\lambda)=0$ $\quad$ iff $\quad$  $p(\overline{\lambda})=0$.
$Q$ need not to be orthogonal !
A: Let $Q$ be orthogonal and $(\lambda,v)$ is an eigenpair then:
$$Qv=  \lambda v$$
and assume $\lambda$ is complex. Apply conjugation: 
$$Q^{*} v^{*} = \lambda^{*} v^{*}$$
Since $Q$ is real then
$$Q v^{*} = \lambda^{*} v^{*}$$
Thus $(\lambda^{*},v^{*})$ is an eigenpair. As @Fred points out below, $Q$ need not be orthogonal.
A: No, that reasoning is not correct. One can very well have a set of complex numbers whose product is $\pm1$ but which is not closed under complex conjugation. For instance $\{\exp(\frac{\pi\mathbf i}6),\exp(\frac{\pi\mathbf i}3)\}$ whose product is $-1$. This is just a simple example; in general there can be more than two eigenvalues, leaving a lot of freedom to make counterexamples.
A good argument would be to observe that orthogonal matrices have real entries, therefore a real characteristic polynomial, and the for every root$~z$ of a real polynomial, $\overline z$ is also a root of that polynomial.
