Eigenvalues and eigenvectors of a unitary operator
I have $φ: V → V$ as a unitary operator on a complex inner product space $V$. How can I show, without using any diagonalization results, that every eigenvalue $λ$ of $φ$ satisfies $|λ|=1$ and that eigenvectors corresponding to distinct eigenvalues are orthogonal?
$\endgroup$ 11 Answer
$\begingroup$Suppose $v \neq 0$ is an eigenvector of $\phi$ with eigenvalue $\lambda$. Then$$ \langle \phi v, \phi v \rangle = \langle \lambda v, \lambda v \rangle = \lambda \bar \lambda \langle v, v \rangle = |\lambda|^2 \|v\|^2. $$Also$$ \langle \phi v, \phi v \rangle = \langle \phi^* \phi v, v \rangle = \langle v, v \rangle = \|v\|^2. $$Subtracting equations gives $0 = |\lambda|^2 \|v\|^2 - \|v\|^2 = \left( |\lambda|^2 -1 \right) \|v\|^2$. Since $v \neq 0$, $\|v\|^2 \neq 0$, and we may divide by $\|v\|^2$ to get $0 = |\lambda|^2 - 1$, as desired.
Now suppose that $u \neq 0$ is another eigenvector of $\phi$ with eigenvalue $\mu \neq \lambda$. Then$$ \langle u, \phi v \rangle = \langle u, \lambda v \rangle = \bar \lambda \langle u, v \rangle. $$Also$$ \langle u, \phi v \rangle = \langle \phi^* u, v \rangle = \langle \bar \mu u, v \rangle = \bar \mu \langle u, v \rangle $$since the eigenvalues of $\phi^*$ are the complex conjugates of the eigenvalues of $\phi$ [why?].
Since $\phi^* \phi = I$, we have $u = I u = \phi^* \phi u = \mu \phi^* u$. Since $u \neq 0$, it follows that $\mu \neq 0$, hence $\phi^* u = \frac{1}{\mu} u$. Since $|\mu| = 1$ by the above, $\mu = e^{i \theta}$ for some $\theta \in \mathbb R$, so $\frac{1}{\mu} = e^{- i \theta} = \overline{e^{i \theta}} = \bar \mu$. Thus $\phi^* u = \bar \mu u$.
Subtracting equations,$$ 0 = \bar \lambda \langle u, v \rangle - \bar \mu \langle u, v \rangle = (\bar \lambda - \bar \mu) \langle u, v \rangle. $$Hence one of the numbers $(\bar \lambda - \bar \mu)$ or $\langle u, v \rangle$ must be $0$. Since $\lambda \neq \mu$, the number $(\bar \lambda - \bar \mu)$ is not $0$, and hence $\langle u, v \rangle = 0$, as desired.
$\endgroup$
