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The set of completely bounded (CB) maps forms can be considered as a complex span of the set of completely positive (CP) maps. Can we find a basis for this complex linear space of CB maps such that every element of this is a CP map. What I am looking for is an explicit construction method for such a basis (if exists).

In same spirit we may consider the CB Hermiticity preserving maps which can be written as $\psi_1-\psi_2$, where $\psi_i\in$ CP maps and ask the similar question.

Even a partial answer is also helpful for me. (Example: Characterisations of all such maps between $\mathcal{B}(\mathbb{C}^n)\longrightarrow\mathcal{B}(\mathbb{C}^n)$, and similar partial cases). I hope, I explained my question correctly. I am ready to explain any ambiguity (if exists) and give further explanation if required (also re-edit my question to make it lucid and self-explanatory). Advanced thanks for all helps.

EDIT: As pointed by Tom Cooney, the above statement is not true for non-injective von Neumann algebras. Now the question is for injective $C^*$ algebras, does there exists a method by which we can actually construct a basis. For time being, we can consider only CB maps from $\mathcal{B(H)}$ to itself and ask for an explicit example of such basis.

Re-edit: Actually I tried to construct such basis for the maps from $\mathcal{B}(\mathbb{C}^n)$ to itself. However, structure of such maps, when CP, is well understood (Choi-Krauss representation), but does explicitly give a basis explicitly. At the best there is a result of Choi on the extremal points of such CP maps, but I failed to use it for the construction.

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    [This](http://en.wikipedia.org/wiki/Choi%27s_theorem_on_completely_positive_maps) and [this](http://en.wikipedia.org/wiki/Stinespring_factorization_theorem) may help.2012-07-09
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    Some comments: 1. What do you mean by convex span? Convex combinations of cp maps will again be cp and will not include cb maps which are not cp. 2. Yes, cb maps $B(\mathbb C^n) \to B(\mathbb C^n)$ are linear combinations of cp maps. However there are cb maps between operator algebras that cannot be decomposed as a linear combination of cp maps. See U. Haagerup's "Injectivity and decomposition of completely bounded maps" for further details.2012-07-09
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    @TomCooney Sorry, my stupid mistake. I wanted to write 'complex'. I have changed the question accordingly. I am trying to get a copy of the above mentioned paper. As per I know, by Wittstock’s decomposition theorem any cb map $\phi:\mathcal{A}\longrightarrow\mathcal{B(H)}$ can be written as $\psi_1\pm \psi_2\pm(\psi_3+\psi_4)$, where $\mathcal{A}$ is a $C^*$-algebra, $\psi_i$ are cp maps. Hence I am slightly uncomfortable about your statement, unless you want to say 'real linear'. That question I have asked in 2nd paragraph. Can you please explain it a bit more?2012-07-09
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    Yes, as you say, by Wittstock's theorem, any cb map from a $C^*$-algebra into an injective $C^*$-algebra (for example, $B(H)$) is a linear combination of completely positive maps. However if $N$ is any non-injective von Neumann algebra, then $CB(N,N) \neq \textrm{span} \ CP(N,N)$. This is shown in the paper of Haagerup's mentioned in my earlier comment.2012-07-09
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    Continuing on from above: A simpler example (due to R. Smith) is that not every completely bounded map from $C([0,1])$ to itself is in the span of the completely positive maps on this space.2012-07-09
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    @TomCooney Thanks a lot. I did not know the above result. I am trying to get a copy of the paper. My original intention was the maps on $C^*$ algebra. So I am again modifying question. Is the above question valid for $C^*$ algebras then?2012-07-09
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    The two counterexamples mentioned above are $C^*$-algebras,so not in general. However, for injective $C^*$-algebras (including the important example $B(H)$), the answer is yes: $CB(B(H),B(H))= \textrm{span} \ CP(B(H),B(H))$.2012-07-09
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    @TomCooney Thanks a lot. For that case, is there a method to construct a basis where each element is a cp map? You see I want an explicit method for getting such basis.2012-07-09
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    I think that you should follow Norbert's suggestion above and work with the Kraus operators in the finite-dimensional situation. Also, about your edit, it should be "the statement is not true for NON-injective von Neumann algebras" (and some other noninjective $C^*$-algebras).2012-07-09
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/4048/discussion-between-rsg-and-tom-cooney)2012-07-09
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    One fundamental point that seems not to have been mentioned yet: in infinite dimensions, you need to be very careful about what you mean by a basis (Hamel? Schauder? Unconditional? Biorthogonal?) So if you are primarily interested in the finite-dimensional case, I suggest you clarify this in the question2012-07-10

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