# A similar Cauchy-Schwarz inequality with linear-algebra

Let $A$ be matrix in $M_{n}$ (i.e., $n\times n$ complex matrices), and $\|A\|\le 1$, we call it a contraction.

Assume that $A$ and $B$ are contractions such that $I-AA^*$ and $I-BB^*$ are positive-definite.

How to show that $$\text{Tr}\left(1-AA^*\right)^{-1}\cdot \text{Tr}\left(1-BB^*\right)^{-1} \ge \left(\text{Tr}(1-AB^*)^{-1}\right)^2.$$

This problem is similar to the Cauchy-Schwarz inequality. (if you don't know the contraction matrix) and I found this paper give the contraction. http://ac.els-cdn.com/S0024379507003710/1-s2.0-S0024379507003710-main.pdf?_tid=b1696360-f73a-11e3-9613-00000aab0f01&acdnat=1403131900_0ced864818647681ff916fc376a0f461

• Are you saying that the article in the link proves this? If yes, what is your question? – Michael Renardy Jun 18 '14 at 22:55
• The right hand side may be complex. I guess you may replace the bit brackets with | |. – Russel Jun 22 '14 at 1:25

Assuming $A$ and $B$ are contractions, first recall Hua's Matrix Identity \begin{equation*} (I-B^*B) + (A-B)^*(I-AA^*)^{-1}(A-B) = (I-B^*A)(I-A^*A)^{-1}(I-A^*B). \end{equation*} From this it identity immediately follows that the following matrix is (Hermitian) positive definite: \begin{equation*} \begin{pmatrix} (I-A^*A)^{-1} & (I-B^*A)^{-1}\\ (I-A^*B)^{-1} & (I-B^*B)^{-1} \end{pmatrix} \succeq 0. \end{equation*} From the operator inequality, taking trace (which preserves the positive semidefiniteness), we obtain the inequality in the OP.