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Is there a family of completely positive maps $L(A,B)$ depending continuously on two nonzero, symmetric positive semidefinite $n\times n$ matrices $A$ and $B$, such that $L(A,B)$ maps $A$ to $B$ and reduces to the identity when $A=B$? Can this family be chosen to be trace-preserving?

Can one give a closed formula for $L(A,B)$?

If no such family exists, what are the conditions on $A$ and $B$ for such a family to exist?

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  • $\begingroup$ For once, if $A$ and $B$ are iso-spectral (with distinct eigenvalues) such a map is $x \mapsto Ux U^\dagger$ where $ B=UAU^\dagger$ for some unitary $U$. $\endgroup$
    – lcv
    Commented Dec 11, 2019 at 16:48
  • $\begingroup$ @Icv - The same idea works even if $A$ and $B$ are just both positive definite with distinct eigenvalues. Then you can construct a matrix $C$ that sends the vectors in a spectral decomposition of $A$ to the vectors in a spectral decomposition of $B$, and the map $x \mapsto CxC^\dagger$ works. $\endgroup$
    – Nathaniel Johnston
    Commented Dec 12, 2019 at 16:10
  • $\begingroup$ @nathanielJohnston yes I believe so. If he doesn't ask for the map to be trace preserving at least. $\endgroup$
    – lcv
    Commented Dec 12, 2019 at 19:18
  • $\begingroup$ @lcv:: It is not clear to me whether $U$ or $C$ can be chosen to be continuous. $\endgroup$
    – Arnold Neumaier
    Commented Dec 13, 2019 at 10:31
  • $\begingroup$ @ArnoldNeumaier - The eigenvalues of $A$ and $B$ vary continuously with the entries of $A$ and $B$, and if those eigenvalues are distinct then so do the projections onto the eigenspaces, and thus their spectral decompositions, and thus the matrix $C$ (or $U$). Things get murky if you have repeated eigenvalues though. $\endgroup$
    – Nathaniel Johnston
    Commented Dec 13, 2019 at 12:39

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