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

1. Apr 25, 2010

### stihl29

1. The problem statement, all variables and given/known data
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 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. Apr 25, 2010

### kof9595995

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

3. Apr 25, 2010

### stihl29

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
4. Apr 26, 2010

### 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.

Last edited: Apr 26, 2010