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In my research of operator algebras and their connection with machine learning I of course use the well know result:

For the map $ tr:M_n \to M_n $ denoting the transpose map of matrices (meaning that $ tr(A)=A^{tr} $ we know for the completely bounded norm (cb-norm) that $ ||tr||_{cb}=n $

I keep using this result freely but to be honest it had just occurred to me that I know not a proof of this nor do I have a reference for the proof or an idea for proving it that would actually work, could operator algebraists out there please help in providing a proof or a reference to a nice (or any...) functioning proof of this useful fact?

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  • $\begingroup$ Do people actually use $\textrm{tr}$ to denote the transpose map? that conflicts with the matrix trace... $\endgroup$
    – Suvrit
    Jun 29, 2016 at 18:31
  • $\begingroup$ This is almost surely in Pisier's book, or the book of Effros and Ruan, or Paulsen's book $\endgroup$
    – Yemon Choi
    Jun 29, 2016 at 19:45

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I literally just googled "cb norm of transpose" and got a link to this math.stackexchange question which links to a proof.

Edit: the answer which contained the link disappeared, so I used Google magic again and found Tomiyama's original paper with this calculation.

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  • $\begingroup$ Author deleted his answer. It is better to give a link to the Ruan's book 'Operator spaces' $\endgroup$
    – Norbert
    Jul 10, 2016 at 21:36

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