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

  • Thread starter Thread starter Dragonfall
  • Start date Start date
  • Tags Tags
    Positive
Click For Summary

Homework Help Overview

The discussion revolves around the properties of complex matrices, specifically focusing on the positive definiteness of the expression AB - A^2 given certain conditions on matrices A and B. The original poster presents a conjecture involving positive definite matrices with a trace of 1.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between matrices A and B, particularly considering the implications of A < B and the expression A B - A^2. There is a discussion on the definitions of positive definiteness in the context of complex matrices and the conditions under which these properties hold.

Discussion Status

The discussion is active, with participants questioning the definitions and properties of positive definite matrices, particularly in the complex case. Some guidance has been offered regarding the implications of self-adjointness and the nature of eigenvalues in relation to the matrices involved.

Contextual Notes

There is a mention of the matrices not necessarily being real symmetric, which raises questions about the assumptions being made in the original conjecture. Additionally, the definition of positive definiteness for complex matrices is under scrutiny, indicating potential gaps in understanding or agreement among participants.

Dragonfall
Messages
1,023
Reaction score
5
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)
 
Physics news on Phys.org
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].
 
A or B might not be real symmetric.
 
This is not homework! But I guess this section will get more viewers.
 
Usually self-adjointness is included in any notion of positivity for complex operators. How are you defining "positive definite" for a complex matrix A?
 
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.
 
But if <v, Mv> is real for all v, then M is self-adjoint.
 
M is indeed self-adjoint.
 

Similar threads

Replies
1
Views
2K
Replies
14
Views
4K
  • · Replies 40 ·
2
Replies
40
Views
6K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K