Suppose that is an eigenpair of (and
then if
will only have real eigenvalues if is real and positive.
Let A be a Matrix with no real eigenvalues. It is possible that the following Matrix B has real eigenvalues?
B=A-A*A^{T}+A^{T}
If I multiply a Matrix with it's transponse I get a symmetric matrix, which has two real roots so A*A^{T} has real eigenvalues. If I subtract A and add A^{T} it is not symmetric anymore... but how can I derive if it still has no real eigenvalues?
Profuse and sincere apologies for those two responses, I suspect it was those magic mushrooms.
The idea of post multipling by isn't correct because even though and have the same eigenvalues, they will (usually) have different eigenvectors.
The basic property that answers both of your questions is that the eigenvalues of a real symmetric matrix are real.
So, even if the matrix (with all real elements) has complex eigenvalues, the matrices and will both have real eigenvalues since they are both symmetric.
For your earlier example this is also symmetric, (consider it as and will also have real eigenvalues irrespective of whether the eigenvalues of are real or complex.