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Consider the complex $n$-by-$n$ matrices $M_n$. Suppose that $A_i$, for $i=1,\ldots,n^2$, satisfy $\mathrm{Tr}(A_i^* A_j)=\delta_{ij}$, so that together they form an orthonormal basis for $M_n$. Define a linear map $T \colon M_n \to M_n \otimes M_n$ by $T(A_i) = A_i \otimes A_i$.

Question: when is $T$ completely positive?

For example, if $A_i$ are the matrices with a single entry one and the rest zeroes in some fixed basis of $\mathbb{C}^n$, then $T$ is completely positive. In fact, I think these might be the only examples. If $T$ is completely positive, then the following are equivalent to $A_i$ being matrix units as in the above example:

  • each $A_i$ has rank one;
  • each positive semidefinite $A_i$ has trace one;
  • the set $\{0,A_1,\ldots,A_{n^2}\}$ is closed under multiplication;
  • $T(1)$ is idempotent;
  • $T^*(1) \leq 1$;
  • $T$ preserves trace.

These are sufficient conditions, but proving they are sufficient doesn't use $\mathrm{Tr}(A_i^* A_j)=\delta_{ij}$ at all. Are they necessary?

share|cite|improve this question
do you assume $A_i\geq 0$? – Kate Juschenko May 9 '11 at 17:55
We can't assume $A_i\geq 0$ because self-adjoint matrices can never span all of $M_n$ (unless $n=1$). A good example to understand the question with is the following: $A_i$ are the matrices with a single entry one and the rest zeroes in some fixed basis of $\mathbb{C}^n$. Notice that not all $A_i$ in this example are $\geq 0$, indeed, they need not even be self-adjoint. – Chris Heunen May 9 '11 at 18:11
the second condition confused me, now I see what you mean – Kate Juschenko May 9 '11 at 18:26
I'm confused. Why are these conditions equivalent? Shouldn't we be able to find sets of $A_i$ so there are no positive semidefinite $A_i$ (just take the negative of any $A_i$ which is positive semidefinite), so condition (ii) holds, but where none of the other conditions holds. – Peter Shor May 10 '11 at 0:58
The text after the question proper was meant to be helpful extra information halfway to an answer, but I see how it could be confusing. I've edited it, hopefully it is clearer now. – Chris Heunen May 10 '11 at 9:41

This is not a complete answer, but it might help.

The map $T: A_i \mapsto A_i \otimes A_i$ sends, by its very definition, the orthonormal family $(A_i)$ to an orthonormal family. It is therefore an isometry for the Hilbert-Schmidt norms.

But there are not that many completely positive maps $M_n \to M_m$ which are also isometries for the Hilbert-Schmidt-norms. Namely such a map is of the form $T(x)= D \pi(x)$, for some (not necessarily unital) $*$-homomorphism $\pi$ and some positive operator $D=T(1)\in M_m$ commuting with the range of $\pi$. This is an if-and-only-if condition provided that $\|D\|_{HS}=\sqrt n$. This statement is probably known. If you want I can expand the proof I have in mind.

This implies that such a map satisfies $Tr\circ T=c Tr$ for some positive $c=Tr(D)/n$, and more generally that for any $p>0$, $\|Tx\|_p = c_p \|x\|_p$ for $c_p=\|D\|_p/n^{1/p}$.

Coming back to your problem, I do not see how to conclude, you can already find a couple of necessary conditions on the $A_i$'s for the map $T(A_i)=A_i \otimes A_i$ to be completely positive.

Matthew asked for a proof of

A linear map $T:M_n\to M_m$ is completely positive and isometric for the Hilbert-Schmidt norm if and only if $T$ is of the form $T(x)= D \pi(x)$, for some (not necessarily unital) $*$-homomorphism $\pi$ and some positive operator $D=T(1)\in M_m$ commuting with the range of $\pi$, and such that $\|D\|_{HS}=\sqrt n$.

I only prove the "only if" direction. Assume that $T$ is cp and isometric for the Hilbert-Schmidt norm. Using the fact that $T$ is cp, by Stinespring's theorem, there is a (finite dimensional) Hilbert space $H$ and a linear map $V:\mathbb C^m \to H\otimes \mathbb C^n$ such that $T$ can be decomposed as $T(x)=V^* 1_H \otimes x V$. I claim that the assumption that $T$ is isometric implies that $VV^*$ is of the form $A \otimes 1_n$ for some positive $A \in B(H)$ (in particular $V V^*$ commutes with $1\otimes x$ for all $x \in M_n$). This will imply that $T(x) T(y) = T(1) T(xy) = T(xy) T(1)$ for all $x,y \in M_n$, and hence putting $\pi(x) = T(x) T(1)^{-1}$ (with the convention $0/0=0$) we get the proposition.

The claim is not complicated to check. By the trace property, $\langle Tx,Ty \rangle = Tr(VV^* (1\otimes x) V V^* (1\otimes y^* ))$. Writing $VV^* = \sum B_{i,j} \otimes e_{i,j}$, taking $x=e_{i,j}$, $y=e_{s,t}$, and using that $T$ preserves the scalar product, one gets $\langle e_{i,j},e_{s,t}\rangle= Tr(B_{s,i}B_{j,t})$. But $VV^*$ being self-adjoint, this becomes $\delta_{i,s}\delta_{j,t}= \langle B_{s,i},B_{t,j}\rangle$. This implies that $B_{s,i}=0$ if $s\neq i$ and that the matrices $B_{i,i}$ are all of Hilbert-Schmidt norm $1$, and that $\langle B_{i,i},B_{j,j}\rangle=1$. Thus (equality in Cauchy-Schwartz inequality), the $B_{i,i}$'s are all equal, to some matrix $U$. This proves the claim.

share|cite|improve this answer
Thanks for adding the proof! – Matthew Daws May 11 '11 at 12:12
Thanks Mikael! Isometry is necessary, but alone it is not sufficient: for example, if ($n=2$ and) $A_i$ are the normalized Pauli matrices, then $T$ is isometric but not completely positive (indeed not even positive). As you note, $T(1)$ seems to play an important role in getting a necessary condition. – Chris Heunen May 11 '11 at 20:16

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