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Why is the trace preserving part necessary? Is it not enough if it can take all matrices to matrices of trace 1?

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Those maps are linear, so if it preserves the trace of trace 1 matrices then it preserves the trace of any other matrix.

We just don't need to restrict ourselves by considering only trace 1 matrices (from mathematical point of view).

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    $\begingroup$ I think the question was, why can't we consider something like $$ \tilde{C}(\rho)=\frac{C(\rho)}{\mathrm{Tr}[C(\rho)]}$$ for a non trace preserving $C$ a quantum channel? (It said "[...] it can take all matrices to matrices of trace 1") $\endgroup$ – user2723984 May 8 at 13:54
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    $\begingroup$ In this case $\tilde{C}$ is not linear. $\endgroup$ – Danylo Y May 8 at 17:31

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