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I'm studying Mark Wilde's "Quantum Information Theory" and the author sometimes use the inequality $\mathrm{Tr}(\prod_\mathcal{H'}Y) \leq \mathrm{Tr}(Y)$ where $Y\in \mathcal{H}'$ is a density matrix and $\prod_\mathcal{H'}=V^{\dagger}V$ with isometry $V: \mathcal{H} \rightarrow \mathcal{H}'$. As far as I know, this inequality does not hold for general matrix $Y$. So I tried to prove the inequality using the positive semi-definite condition of $Y$, but I cannot grasp any clue. Does the inequality really holds? And if it does, how can I prove it? I appreciate any help.

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Hint: Write $\Pi_{\mathcal H'}=\sum \lambda_i|i\rangle\langle i|$ in an eigenbasis $|i\rangle$, with $\lambda_i=0,1$. Then use that the trace is cyclic, and $\langle i|Y|i\rangle\ge0$.

(Note that the inequality cannot hold for a general matrix: If it holds for $Y=Y_0$, it cannot hold for $Y=-Y_0$.)

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    $\begingroup$ Oh, it was a simple task. Thanks for your answer! $\endgroup$
    – asdf
    May 2 '20 at 16:52

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