The most usual definition of the completely positive map (c.p.) between two C\*-algebras (say, unital) is the following: $\sigma: A \to B$ should satisfy $\sigma(1)=1$ and for each $n \in \mathbb{N}$ the map $\sigma\_n: M\_n(A) \to M\_n(B)$ defined ``cordinatewise''should be positive.  

However I met also the following definition: $\sigma(1)=1$ and for each $n \in \mathbb{N}, a\_1,...,a\_n \in A, b\_1,...,b\_n \in B$ we have $\sum\_{i,j}b\_i^\*\sigma(a\_i^\*a\_j)b\_j \geq 0$. I was able to show that the first definition implies the second: how to prove that the converse is also true (if it is so)? 



Completely positive maps-equivalent definition
The converse follows from two observations:


1) If $a=[a\_{ij}]$ is a positive element in $M\_n(A)$, then $a=x^\*x$ for some $x=M\_n(A)$, and if we write this out in matrix entries we have
$$
a\_{ij} = \sum\_k x\_{ki}^\*x\_{kj}.
$$


2) The matrix $[a\_{ij}]$ is positive in $M\_n(A)$ if and only if it is positive in every GNS representation $\pi$ of $A$; and this is implied by the condition
\begin{equation}
\sum\_{ij}c\_i^\*a\_{ij}c\_j\geq 0
\end{equation}
for all $c\_1,\dots c\_n\in A$. To see this fix a state $\rho$ on $A$, let $H$ denote the GNS space associated to $\rho$ and $\pi:A\to B(H)$ the GNS representation. $H$ is spanned by equivalence classes $[c]$ with $c\in A$. Note that since $A$ is unital, by the GNS construction we have $[c]=\pi(c)[1]$. To check that $[a\_{ij}]$ is positive in the representation $\pi$, fix vectors $[c\_1], \dots [c\_n]$ in $H$ and complute:
\begin{align}
\sum\_{ij}\langle \pi([a\_{ij}])[c\_j],[c\_i]\rangle\_H &= \sum\_{ij}\langle \pi(c\_i^\*)\pi(a\_{ij})\pi(c\_j)[1],[1]\rangle\_H \\
&=\sum\_{ij}\langle \pi(c\_i^\*a\_{ij}c\_j)[1].[1]\rangle\_H\\
&= \rho \left(\sum\_{i,j} c\_i^\*a\_{ij}c\_j\right)\\
&\geq 0
\end{align}
by assumption.


Now, for the proof: suppose $\sum\_{ij} b\_i^\*\sigma(a\_i^\*a\_j)b\_j\geq 0$ for all $a's$ and $b's$. Fix a positive element $[a\_{ij}]\in M\_n(A)$, factor it as in the first observation. To check the positivity of $\sigma([a\_{ij}])$, we check the condition of the second observation; so fix $b\_1, \dots b\_n$; we have
\begin{equation}
\sum\_{ij}b\_i^\*\sigma(a\_{ij})b\_j = \sum\_k\sum\_{ij}b\_i^\*\sigma(x\_{ki}^\*x\_{kj})b\_j\geq 0,
\end{equation}
where the last inequality follows by hypothesis.


