I'm doing a course that often uses linear algebra, and it implies at a certain point that, given two symmetric real matrices A and B, both positive (semi)definite, then AB has only real and non-negative eigenvalues.
I proved it but it's too long, it's gotta be more obvious.
Here's my proof.
Unless AB=0 matrix, then AB surely has at least one non-zero complex conjugate pair of eigenvalues and complex conjugate pair of eigenvectors.
* is for conjugate, ' is for transpose.
So let c be a complex non-zero eigenvalue, and the corresponding complex eigenvector v:
Let z=Bv, z obviously is nonzero, then Az=cv and Az*=c*v*
z*'Az is real and positive because A is positive semidefinite.
z*'v=v*'Bv, and this is real and positive too because B is positive semidefinite.
Therefore c is real and positive too.