Home » Subject » Algebra

trace inequality

Importance: Medium ✭✭
Author(s):
Subject: Algebra
Keywords:
Recomm. for undergrads: no
Posted by: Miwa Lin
on: October 11th, 2008

Let $ A,B $ be positive semidefinite, by Jensen's inequality, it is easy to see $ [tr(A^s+B^s)]^{\frac{1}{s}}\leq [tr(A^r+B^r)]^{\frac{1}{r}} $, whenever $ s>r>0 $.

What about the $ tr(A^s+B^s)^{\frac{1}{s}}\leq tr(A^r+B^r)^{\frac{1}{r}} $, is it still valid?


Bibliography



* indicates original appearance(s) of problem.
add new comment

clarification

On August 28th, 2010 ojs says:

I suppose that A and B are square hermitian matrixes and that s and r are real numbers. But what are the brackets supposed to represent? Sorry for my ignorance if it is obvious.

reply

It is open.

On October 12th, 2008 Miwa Lin says:

It is open.

reply

this inequality is not true.

On December 20th, 2010 Anonymous says:

this inequality is not true. It was denied some years ago. See X.Z Zhan's

reply

Comment viewing options

Select your preferred way to display the comments and click "Save settings" to activate your changes.