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

[stihl29]

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

 

[kof9595995]

remember that detA=detA^T=[detA^-1]^-1?

 

[stihl29]

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?

 

[kof9595995]

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.