# 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. (Using matrix inequalities) 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.
2. (Via SVD) Another approach is to leverage the SVD of the operators. Notice that, if the SVD of $$A$$ reads $$A = \sum_i a_i |a_i^L\rangle\!\langle a_i^R|,$$ then, for any unitary $$U$$, $$AU$$ is an operator which has the same singular values and left singular vectors as $$A$$. Similarly, $$A$$ and $$UA$$ have the same singular values and right singular vectors. The question to maximise $$\lvert {\rm Tr}(AU)\rvert$$ thus amounts to maximising $$\lvert{\rm Tr}(AU)\rvert = \left\lvert \sum_i a_i \langle u_i,a_i^L\rangle\right\rvert$$ over the sets of orthonormal bases $$\{\lvert u_i\rangle\}_i$$. The connection with the above is via $$|u_i\rangle=U^\dagger \lvert a_i^R\rangle$$. From the above expression, it is clear that Cauchy-Schwarz implies $$\lvert{\rm Tr}(AU)\rvert \le \sum_i a_i = {\rm Tr}|A|,$$ and that the maximum is achieved when $$|u_i\rangle=|a_i^L\rangle$$, that is, $$U=\sum_i |a_i^R\rangle\!\langle a_i^L\rvert$$.
3. (Via SVD, generalised result) Let us try now for a more general result. Relax the constraint of $$U$$ being unitary, and let it by an arbitrary (squared) matrix. The only difference compared to the previous case is that we can now have non-unit singular values, call these $$u_i$$. The trace we are interested in now reads $$\lvert{\rm Tr}(AU)\rvert = \left\lvert\sum_i a_i u_i \langle u_i,a_i^L\rangle\right\rvert.$$ We can again use Cauchy-Schwarz to get an upper bound, but this time the singular values $$u_i$$ are in the way. That is not that big of a problem though: observe that $$\|U\|_{\rm op}=\max_i u_i$$, and thus we can write $$\lvert{\rm Tr}(AU)\rvert \le \|U\|_{\rm op}{\rm Tr}|A|.$$ We can also, if we want, characterise the $$U$$ that achieves the maximum: this will be the (or a) unitary operator connecting right and left singular vectors of $$A$$, as in the previous case.