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Question:

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:

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

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

I have two questions:

(1) How can one prove this inequality?

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

where $$\dfrac{1}{1-AA^T}=(1-AA^T)^{-1}$$

I know this inequality is stronger the Cauchy-Schwarz inequality.But I can't prove it.

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    $\begingroup$ Does $\displaystyle \frac{1}{1-AA^{\top}}$ denote the inverse of the matrix $\mathrm{I} - AA^{\top}$ ? I don't think you notations are standard. $\endgroup$ – jibounet Jul 9 '14 at 8:58
  • $\begingroup$ @jibounet,Yes,that's your mean $\endgroup$ – math110 Jul 11 '14 at 2:58

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