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)?
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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.
