Proving eigenvalues = 1 or -1 when A = A transpose = A inverse A is circulant

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Homework Help Overview

The discussion revolves around proving that all eigenvalues of a circulant matrix \( A \) are either 1 or -1, given that \( A \) is symmetric (\( A = A^T \)) and orthogonal (\( A = A^{-1} \)). The original poster reflects on specific examples, particularly the identity matrix, while seeking a more general approach.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between determinants and eigenvalues, with one suggesting that since the determinant is the product of eigenvalues, it may imply that eigenvalues must be 1 or -1. Others discuss the properties of determinants in relation to matrix inverses.

Discussion Status

The discussion is active, with participants sharing insights about eigenvalues and determinants. Some guidance has been offered regarding the properties of determinants, but there is no explicit consensus on the general case or a complete solution yet.

Contextual Notes

Participants note a lack of familiarity with certain determinant properties and express uncertainty about their relevance to the problem at hand. The original poster is seeking a general case beyond specific examples.

stihl29
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Homework Statement


Prove all eigenvalues = 1 or -1 when A is circulant and satisfying
A=A^T=A^-1
I can think of an example, the identity matrix, but i can't think of a general case or how to set up a general case.

Homework Equations





The Attempt at a Solution


I can only show by example for identity matrix
 
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remember that detA=detA^T=[detA^-1]^-1?
 
I don't remember ever learning that, sorry for being clueless, but i don't see the relation.

OH, maybe since det A is the product of eigenvalues, and because 1 or -1 is the only number ^-1 that stays the same ?
is that right?
 
Last edited:
You can justify that by eigenvalues, but the more straightforward method is, to consider the relation detA*detB=detAB, so detA*det(A^-1)=detI, I think now you can see where it's going.
 
Last edited:

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