Spectrum of unitary operators lie in the unit circle

Question

$U$ - unitary operator on complex Hilbert space. Show that $\sigma(U) \subset \{z \in\mathbb C : |z| = 1\}.$

So $U$ is unitary operator if is is surjective and it preserves a scalar product and a unitary operator is a bounded linear operator $U : H → H$ on a Hilbert space H that satisfies $U^*U = UU^* = I$, where $U^*$ is the adjoint of $U$, and $I : H \to H$ is the identity operator.

But how do I show that $\sigma(U) \subseteq \{z \in C : |z| = 1\}.$ ?

Thanks in advance for any help

Answer 1

Here's a few useful observations that should help you, and that should be proven if you're not familiar with them. All of them have easy proofs.

For an operator $T\in\mathcal B(H)$, we have the following

• $\sigma(T^*)=\{\overline{\lambda}:\lambda\in\sigma(T)\}$
• If $T$ is invertible, then $\sigma(T^{-1})=\{\lambda^{-1}:\lambda\in\sigma(T)\}$
• $\|T^*T\|=\|T\|^2$
• $\sup\{|\lambda|:\lambda\in\sigma(T)\}\leq\|T\|.$

Answer 2

A unitary operator $U$ is invertible with inverse $U^{-1}=U^*$. So $\|U\|=\|U^{-1}\|=1$. Any operator $B$ with $\|B\| < 1$ is such that $I-B$ is invertible with $(I-B)^{-1}=\sum_{n=0}^{\infty}B^{n}$. Therefore, if $|\lambda| < 1$, the following is invertible: $$ U-\lambda I = U(I-\lambda U^{-1}) $$ And, for $|\lambda| > 1$, the following is also invertible: $$ U-\lambda I=-\lambda\left(I-\frac{1}{\lambda}U\right) $$ So the spectrum of a unitary $U$ is a subset of the unit circle in the complex plane.