# 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) Consider now a more general result, and let $$A,B$$ be generic matrices of type $$A:\mathbb{C}^n\to\mathbb{C}^m$$ and $$B:\mathbb{C}^m\to\mathbb{C}^n$$. Note that we need this constraint on domains and codomains only because otherwise $$\operatorname{tr}(AB)$$ isn't well defined. We want to prove that $$\lvert \operatorname{tr}(AB)\rvert \le \|A\|_{\rm op} \operatorname{tr}|B|$$. To this end, write their SVDs as $$A = \sum_i a_i |a_i^L\rangle\!\langle a_i^R|, \qquad U = \sum_i u_i |u_i^L\rangle\!\langle u_i^R|.$$ Thus $$\lvert\operatorname{tr}(AB)\rvert = \lvert\sum_{i} a_i \langle a_i^R|B|a_i^L\rangle\rvert \le \sum_i a_i \lvert \langle a_i^R|B|a_i^L\rangle\rvert \le \|B\|_{\rm op}\operatorname{tr}|A|,$$ where we used $$\operatorname{tr}|A|=\sum_i a_i$$ and $$\lvert \langle v|B|w\rangle\rvert\le \|B\|_{\rm op}$$ for any pair of normal vectors $$|v\rangle,|w\rangle$$. One could further characterise the $$B$$ that maximises $$|\operatorname{tr}(AB)|$$ among those operators with fixed operator norm $$\|B\|_{\rm op}$$, and from the above equation it's not surprising that this maximum is achieved with $$B=\|B\|_{\rm op} \sum_i |a_i^R\rangle\!\langle a_i^L|$$.
In fact, this can be generalised even further, and seen as a special case of a "tracial matrix Holder inequality": $$|\langle A,B\rangle|\le \|A\|_p\|B\|_p$$ for $$1/p+1/q=1$$. See e.g. this MO answer. See also this other answer of mine for more rambling about operator norms.