# Trace norm of partial trace of unitary transformation

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.

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 $$\|\mathrm{tr}_B(UXU^\dagger)\|_1=\mathrm{tr}_{AB}(UXU^\dagger)=\mathrm{tr}_{AB}(X)=\|\mathrm{tr}_B (X)\|_1.$$