# A stronger Cauchy-Schwarz inequality for traces of compression matrices

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 $A=B$?

• Is there any motivation for this inequality? – Russel Jun 22 '14 at 1:34
• 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..... – Suvrit Jun 22 '14 at 2:33