1.5. Completely positive maps 15

Proof. Since [ei,j] E Mn(Mn(C)) is positive (it's a multiple of a rank-one

projection), the "only if" part is trivial. To prove the "if" part, assume

a= [1p(ei,j)] 2:: 0 in Mn(A) and let a^112 = [bi,j]· It follows that

n

1p(ei,j) =I:: b'k,ibk,j·

k=l

Let A c JB('h'.) be a faithful representation and define V: 1{---+ .e; ® .e; ® 'h'.

by

n

ve= I:: (j®(k®bk,je,

j,k=l

where { (j }j= 1 is the standard orthonormal basis for .e;. Then, for T =
[ti,j] E Mn(C), we have

\V*(T ® 1 ® l)V'TJ, e) \(T ® 1®1)v'TJ, ve)

n

I:: \T(j, (i) \(k, (z) \bk,j'TJ, bz,ie)

i,j,k,l=l

n n

I:: ti,j \2: bk,ibk,j'T}, e)

i,j=l k=l

i,j

for every e, 'TJ E 'h'.. Therefore, 1p(T) = V*(T ® 1 ® l)V for every TE Mn(C)
and 1p is c.p. · D

Example 1.5.13. Let ai, ... , an E A be given and define a linear map
1p: Mn(C)---+ A by 1p(ei,j) = aiaj. The previous result easily implies that 1p
is completely positive. Indeed,

I

aia* 1 aia* 2

a2ai a2a2

....
..
anai ana2

aia~ a2an I

ana~

2:: 0.