# Proving the inequality $|\mathrm{tr}(AU)|\le \mathrm{tr}|A|$ in Uhlmann's theorem

In Nielsen and Chuang, in the Fidelity section, (Lemma 9.5, page 410 in the 2002 edition), they prove the following.

$$\mathrm{tr}(AU) = |\mathrm{tr}(|A|VU)| = |\mathrm{tr}(|A|^{1/2}|A|^{1/2}VU)|$$ $$|\mathrm{tr}(AU)| \leq \sqrt{\mathrm{tr}|A| \mathrm{tr}(U^{\dagger}V^{\dagger}|A|VU)} = \mathrm{tr}|A|$$

First, is $$|A|$$ the positive matrix in the polar decomposition of $$A$$? And second, apparently, the second equation comes from the first due to Cauchy Schwartz inequality of Hilbert-Schmidt.

How does the first expression lead to the second?

## 1 Answer

The idea is to use CS inequality in the form $$\newcommand{\tr}{\operatorname{Tr}}\lvert \sum_{ij}A_{ij}^* B_{ij}\rvert\le\sqrt{\sum_{ij} \left\lvert A_{ij}\right\rvert^2}\sqrt{\sum_{ij}\left\lvert B_{ij}\right\rvert^2}$$, which in matrix formalism reads $$\lvert\tr(A^\dagger B)\rvert\le\sqrt{\tr(A^\dagger A})\sqrt{\tr(B^\dagger B)}$$.

Therefore, $$\lvert\tr(AU)\rvert=\lvert\tr(\lvert A\rvert VU)\rvert =\lvert\tr(\lvert A\rvert^{1/2} \underbrace{\lvert A\rvert^{1/2} VU}_{D})\rvert \le\sqrt{\tr(\lvert A\rvert^{1/2}\lvert A\rvert^{1/2})}\sqrt{\tr(D^\dagger D)},$$ and putting in the explicit expression for $$D$$ you get the result.

You can also have a look at this other answer for more details on how to prove this result.