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Let $\mathcal{H}_{AB}$ be a bipartite, complex Euclidean space, and let $U\colon\mathcal{H}_{AB}\to\mathcal{H}_{AB}$ be a unitary operator. Define the trace norm as $$ \lVert X\rVert_1 = \text{Tr}(\sqrt{XX^\dagger}) $$ where $X$ is a linear operator on $\mathcal{H}_{AB}$, and $X^\dagger$ denote the complex conjugate of $X$. I know $\lVert UXU^\dagger\rVert_1 = \lVert X\rVert_1$, but does the following identity hold: $$ \lVert \text{Tr}_B UXU^\dagger\rVert_1 = \lVert \text{Tr}_B X\rVert_1 $$ Assuming finite dimensionality is fine for my application.

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No. Consider $H_A=H_B=\mathbb{C}^2$ and the unitary $U(\xi\otimes \eta)=\eta\otimes \xi$. Then $U(X\otimes Y)U^\dagger=Y\otimes X$. Now if $X=\mathrm{diag}(1,1)$ and $Y=\mathrm{diag}(1,-1)$, then $\mathrm{tr}_B(X\otimes Y)=0$, while $\mathrm{tr}_B(Y\otimes X)=2Y$, which has trace norm $4$.

The inequality does hold if $X$ (in your question) is positive semi-definite, because in this case $$ \|\mathrm{tr}_B(UXU^\dagger)\|_1=\mathrm{tr}_{AB}(UXU^\dagger)=\mathrm{tr}_{AB}(X)=\|\mathrm{tr}_B (X)\|_1. $$

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