# Spectrum and operators [closed]

Hwo can help me to prove that if the spectrum of a normal operator lies on a circle {z∈C:∣z∣=1}, then this operator is unitary.

I'm very thankful for any ideas and help

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## closed as off-topic by abx, Robert Israel, Yemon Choi, Piotr Hajlasz, Neil HoffmanJan 9 at 20:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – abx, Robert Israel, Piotr Hajlasz, Neil Hoffman
If this question can be reworded to fit the rules in the help center, please edit the question.

• – Martin Sleziak Jan 9 at 19:33
• @MartinSleziak The result stated there is only valid on finite-dimensional spaces, and in general the answers there seem more of a distraction than a help. – Yemon Choi Jan 9 at 19:46
• I think this question would be more appropriate on MSE - it is not a duplicate of the one that Martin has linked to. That said, here is a hint: do you know the result that whenever $T\in B(H)$ is normal one has $\Vert T \Vert$ = spectral radius of $T$? In particular, if $T$ is normal and has spectrum contained in the unit circle, then consider $\Vert T\Vert$ and $\Vert T^{-1} \Vert$ – Yemon Choi Jan 9 at 19:47
• In fact, it seems you have previously asked the same question on Mathematics. Since it is your own post, you should still see both the question and the answers posted there: Spectrum of a normal operator . Unitary operator.. (And in case it is useful to people who cannot see deleted posts, here is link to the cached version.) – Martin Sleziak Jan 9 at 20:09