Return to Answer

I'll give a couple of methods to do this:

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 $B=\|B\|_{\rm op} \sum_i |a_i^R\rangle\!\langle a_i^L|$.

I'll give a couple of methods to do this:

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.

I'll give a couple of methods to do this:

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.

I'll give a couple of methods to do this:

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 $B=\|B\|_{\rm op} \sum_i |a_i^R\rangle\!\langle a_i^L|$.

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, $U=\sum_i |a_i^R\rangle\!\langle a_i^L\rvert$.

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

I'll give a couple of methods to do this:

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.