3
$\begingroup$

In a Hilbert space, let $U$ be a continuous operator which it unitary. Prove $\sigma (U)\subseteq \Bbb{S}^1$.

It is important for me to know how I am doing, and I didn't come by a clear explanation so it would be appreciated to have your correcting and guiding.

Attempt: $U^{*}U=UU^{*}=I$ and therefore $U^{-1}=U$. I have $||Ux||^2=<Ux,Ux>=<U*Ux,x>=<x,x>=||x||^2$ and the same goes for $U^*$ and therefore $||U||=||U^{-1}||=1$. Looking at $U-\lambda$ I can take $-\lambda U(U^{-1}-{1\over \lambda}I)=U-\lambda I$. Clearly, for $\lambda\ne 0$, $-\lambda U$ is defined and it a linear operator. $(U^{-1}-{1\over \lambda}I)$ is invertible if $|\lambda|\ge 1$ and therefore $\sigma (A^{-1})=\sigma (A)\subseteq \{|z|\le1\}$. But from the RHS, $U-\lambda I$ is invertable if $|\lambda|\le 1$ and therefore $\sigma(A)\subseteq \Bbb{S}^1$. I have noticed my claims tend to have holes in them many times which I cannot spot. If it is substantially true, how can I make sure it is completely formal?

$\endgroup$
  • 1
    $\begingroup$ It's true in general that $\Vert Ax \Vert = \Vert A^* x \Vert = \Vert x \Vert$ for $A$ a unitary operator on a given Hilbert space $H$. Even the converse is true! Maybe the formula $(U-\lambda I)^* =(U^* - \lambda ^* I)$ may help, but otherwise it's seems to be fine. $\endgroup$ – noctusraid Jan 26 '16 at 12:25
3
$\begingroup$

In your proof slight modification is needed.

We have, $UU^* =U^*U =I$. So $U$ is invertible with $U^{-1}= U^*$.

Also as you have already shown that $||U || = 1 = ||U^{-1}||$, we must have $$\sigma (U) \subset \{\lambda \in \mathbb{C} : |\lambda| \leq 1\}\;\; \text{and}\;\; \sigma (U^{-1}) \subset \{\lambda \in \mathbb{C} : |\lambda| \leq 1\}.$$

Now to show $\sigma (U) \subset \{\lambda \in \mathbb{C} : |\lambda| = 1\}, $ it is sufficient show that for $(U-\lambda I)$ is invertible for every $\lambda$ satisfying $|\lambda| <1$. As $U$ is already invertible, sufficient to show that $(U-\lambda I)$ is invertible for every $\lambda$ satisfying $0 <|\lambda| <1$.

Now from the relation $-\lambda U (U^{-1} - \frac{1}{\lambda}I) = (U- \lambda I)$, It is clear that, for $\lambda \neq 0$,

if $(U^{-1} - \frac{1}{\lambda}I)$ is invertible then $(U - \lambda I)$ becomes invertible.

Now note that If $0 <|\lambda| < 1,$ then $|\frac{1}{\lambda}| > 1$, then in that case $(U^{-1} - \frac{1}{\lambda}I)$ invertible and hence $(U- \lambda I)$ invertble. This gives us $\sigma (U) \subset \{\lambda \in \mathbb{C} : |\lambda| = 1\}. $

$\endgroup$
  • $\begingroup$ Why does $|\frac{1}{\lambda}| > 1$ imply that $(U^{-1}-\frac{1}{\lambda})$ is invertible? $\endgroup$ – Dan Mar 26 at 10:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.