5
$\begingroup$

Assume that $A$ and $B$ are contractions, so $I-AA^T$ and $I-BB^T$ are positive-definite matrices. Let $C=(I-AB^T)^{-1}(I-AA^{T})(I-BA^{T})^{-1}$, and show that:

$$Tr\left(\frac{1}{1-AA^T}\right)Tr\left(\frac{1}{1-BB^T}\right)-\left(Tr\left(\frac{1}{1-AB^T}\right)\right)^2 $$

$$ \ge Tr\left(\frac{1}{1-AA^T}\right)Tr\left([(A-B)^TC^2(A-B)]\right)\tag{1} $$

This is an attempt to strengthen this inequality: A similar Cauchy-Schwarz inequality with linear-algebra.

I have two questions:

(1) How can one prove this inequality?

(2) Does equality hold if and only if $A=B$?

$\endgroup$
  • 2
    $\begingroup$ Is there any motivation for this inequality? $\endgroup$ – Russel Jun 22 '14 at 1:34
  • 3
    $\begingroup$ For the scalar case, this reduces (as expected) to AM-GM; but after that, this is a Hua-type trace inequality, and seems not to easy to prove..... $\endgroup$ – Suvrit Jun 22 '14 at 2:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.