Let A and B be (complex matrices) positive definite with trace 1.
Given A < B, (B-A is pos def )
then
A^2 < AB (AB-A^2 is pos def)
CompuChip
Aug27-08, 04:53 AM
Maybe you can use that
A B - A^2 = A(B - A)
and from the result that
One thing that is true is this: if A and B are hermitian (or real
symmetric) with all their eigenvalues in [0, a] and [0, b]
respectively, then A B has all its eigenvalues in [0, a b].
source with proof (http://sci.tech-archive.net/Archive/sci.math/2006-03/msg02721.html)
Dragonfall
Aug27-08, 04:18 PM
A or B might not be real symmetric.
Dragonfall
Aug28-08, 02:37 PM
This is not homework! But I guess this section will get more viewers.
morphism
Aug28-08, 05:33 PM
Usually self-adjointness is included in any notion of positivity for complex operators. How are you defining "positive definite" for a complex matrix A?
Dragonfall
Aug29-08, 03:05 AM
A matrix M such that for all vectors v, <v, Mv> (inner product, the usual one for complex vector spaces) is a real, positive number.
morphism
Aug29-08, 11:43 AM
But if <v, Mv> is real for all v, then M is self-adjoint.