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

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove all eigenvalues = 1 or -1 when A is circulant and satisfying
    I can think of an example, the identity matrix, but i cant think of a general case or how to set up a general case.

    2. Relevant equations

    3. The attempt at a solution
    I can only show by example for identity matrix
  2. jcsd
  3. Apr 25, 2010 #2
    remember that detA=detA^T=[detA^-1]^-1?
  4. Apr 25, 2010 #3
    I dont remember ever learning that, sorry for being clueless, but i dont 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: Apr 25, 2010
  5. Apr 26, 2010 #4
    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: Apr 26, 2010
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