Group Theory, unitary representation and positive eigenvalues

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Discussion Overview

The discussion revolves around the properties of a specific operator \( K \) defined in the context of group theory and unitary representations. Participants are exploring the implications of \( K \) having positive eigenvalues and its relationship to positive-definiteness, as well as the broader context of representations of finite groups.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents a problem regarding the operator \( K = \sum^{i=1}_{n} D^{\dagger}(g_i) D(g_i) \) and seeks to demonstrate that all eigenvalues of \( K \) are positive.
  • Another participant suggests showing that the function \( \langle x,y \rangle = y^{\dagger}Kx \) is an inner product, which would imply that \( K \) is positive-definite and thus has only positive eigenvalues.
  • A later reply expresses confusion about how to establish that \( K \) is positive-definite based on the earlier suggestion.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the proof of \( K \)'s positive-definiteness or the implications for its eigenvalues. There is ongoing clarification and exploration of the concepts involved.

Contextual Notes

The discussion includes assumptions about the properties of the operator \( K \) and the definitions of positive-definiteness and inner products, which may not be fully resolved. The mathematical steps required to demonstrate the claims remain unclear.

Aradan
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Hi, I'm new in this forum.
I have a problem i can't solve and searching on Google i couldn't find anything.
It says:


If D(g) is a representation of a finite group of order n, show that
K = \sum^{i=1}_{n} D^{\dagger} (g_i) D(g_i) has the properties:

b) All eigenvalues of K are positive


This is to prove that every representation is equivalent to a unitary representation (the problem is from the book of Matthews and Walker, problem 16-22).
I know that K = K^{\dagger} implies that all eigenvalues of K are real, but i can't figure it out how to demonstrate that they are positive also.


Thanks in advance (sorry for muy bad english)
 
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Show that the function \langle x,y \rangle = y^{\dagger}Kx is an inner product. This will imply that K is positive-definite, and hence has only positive eigenvalues.
 
Thanks for your help :)
 
Sorry. I did not understand the answer.
If ##K=\sum^n_{i=1}D^{\dagger}(g_i)D(g_i)##
How you know that ##K## is positive definite?
 

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