Hi,

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:

ABv=cv

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*'Az=cz*'v=(Az*)'v=c*v*'z

z*'v=v*'Bv, and this is real and positive too because B is positive semidefinite.

Therefore c is real and positive too.