Unitary operator spectrum

Question

We have a unitary operator $U$ on a Hilbert space $H$ with $U^n=I$ for a $n\in \mathbb{N}$. Show, that the spectrum is subset of $A:=\{1,e^{\frac{2\pi i}{n}},\ldots,e^{\frac{2\pi(n-1)i}{n}}\}$. Attempt: Since $U$ is unitary, every eigenvalue has the form $e^{i\alpha},\alpha\in \mathbb{R}$. We have to show, that $\lambda I-U$ is invertible for every $\lambda$ except for $\lambda\in A$. For example, for $\lambda=1$ we have $I-U=U^n-U=U(U^{n-1}-I)=U(U^{-1}-I)$. I don't know if this does help to show, that an inverse doesn't exist and to show it for the other elements of $A$.

Answer 1

Spectral mapping theorem: $\sigma(U^n) = \{\lambda^n: \lambda \in \sigma(U)\}$.

Answer 2

Robert's answer, via the Spectral Mapping Theorem, is the canonical one. Here is another argument.

Since $U $ is normal, the C $^*$-algebra it generates is abelian. So, through the Gelfand transform, it is isomorphic to $ C ( \sigma (U))$ via an isomorphism that maps $U $ to the identity function. The equality $U^n=I $ is mapped to $z ^n=1$ for the identity function $z $; thus $\lambda^n=1$ for all $\lambda \in\sigma (U) $.