Prove/Disprove: AB-A^2 is Positive Definite

[Dragonfall]

URGENT: Can you prove or disprove:

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]

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[/size]​

 

[Dragonfall]

A or B might not be real symmetric.

 

[Dragonfall]

This is not homework! But I guess this section will get more viewers.

 

[morphism]

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]

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]

But if <v, Mv> is real for all v, then M is self-adjoint.

 

[Dragonfall]

M is indeed self-adjoint.