Unitary Operators: Why is Spectrum on Unit Circle?

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Homework Help Overview

The discussion revolves around the properties of unitary operators in the context of spectral theory, specifically questioning why the spectrum of a unitary operator lies on the unit circle. The original poster expresses confusion regarding the nature of eigenvalues and their relationship to the unit circle.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between eigenvalues and the unitary operator's properties, questioning the implications of the operator being normal and the significance of the absolute value of eigenvalues. There is a focus on understanding the implications of the inner product and the lengths of vectors under the transformation of the unitary operator.

Discussion Status

The discussion is active, with participants attempting to clarify concepts and explore the mathematical relationships involved. Some guidance has been offered regarding the properties of inner products and the implications of unitarity, but there is no explicit consensus on the reasoning behind the spectrum's location.

Contextual Notes

Participants clarify that the discussion pertains to unitary operators in finite-dimensional Hilbert spaces, which may influence the interpretation of eigenvalues and the spectrum.

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Homework Statement



why is the spectrum of the unitary operator the unit circle?

Homework Equations



i know that U^(-1)=U* and i know this makes U normal
i also know that normal means UU*=U*U


The Attempt at a Solution



i know that from spectral theory there is some lambda in the spectrum
such that abs(lambda)=1, but i don't understand why ALL of them are on
the unit circle. (i understand the operator, but spectrums are confusing to me.

thanks
 
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If x is an eigenvector of U, and \lambda is its eigenvalue, then what is the length of Ux?

Ux = \lambda x
 
hmmm, i know that, so i have Ux = Lx and L is 1...so then Ux = x...? I'm just getting lost
 
Just so we're happy: you're only talkiong about operators on finite dimensional spaces, right? Because, in general, spectrum and 'set of eigenvalues' are not the same thing.
 
heh, sorry about that. a unitary operator in a Hilbert space is what I'm working with
 
Raven2816 said:
hmmm, i know that, so i have Ux = Lx and L is 1...so then Ux = x...? I'm just getting lost
L may not be 1, and you don't know what L is. But what is the length of Ux?
 
Let x be an e-vector of U with e-value L as above.What do we know?

<x,x>=<Ux,Ux>

because U is unitary. If you don't see that then consider the intermediate steps:

<x,x>=<Ix,x>=<U*Ux,x>=<Ux,Ux>

Note we've just used the unitariness of U. So now we've got to use the fact that x is an e-vector

<x,x>=<Ux,Ux>=<Lx,Lx>=...?
 
and <Lx, Lx> is the inner product of an e-vector with its e-value...so do i get one? or am i using L=1?
 
Raven2816 said:
and <Lx, Lx> is the inner product of an e-vector with its e-value...so do i get one? or am i using L=1?
L is a number. There is a formula for pulling a number multiplier out of an inner product.
 
  • #10
a formula? isn't <Lx, Lx> = ||Lx||^2?
and i know that L<x, y> = <Lx, y> ...
 
  • #11
Raven2816 said:
i know that L<x, y> = <Lx, y> ...
What about <x,Ly>?
What about <Lx,Lx>?
 
  • #12
Raven, could you satisfy my curiosity? Are you taking a course, or reading a book on your own? What is the name and level of the course or the name of the book?
 
  • #13
ahhh i see what you mean! thanks!
 

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