Let $\mathcal{S}$ be an operator system such that every unital, positive map $\phi:\mathcal{S}\to M_n$ is completely positive. We have to prove that every positive map $\phi:\mathcal{S}\to M_n$ is completely positive.
There is a sequence of hints given in the book Completely Bounded Maps and Operator Algebras by Vern Paulsen (Exc 6.2) to prove the preceding statement:
Let $\phi:\mathcal{S}\to M_n$ be positive and set $\phi(1)=P\ge0$. Let $Q$ be the projection onto range of $P$, and let $R\ge 0$ be such that $(I-Q)R=0$ and $RPR=Q$. The construction of $R$ is as follows: Since $P$ is positive, $P$ is unitarily diagonalizable. So if $P$ is of the form $\text{diag}(p_1,\ldots,p_k,0,\ldots,0)$ with $p_i>0$, then $Q$ is of the form $\text{diag}(1,\ldots,1,0,\ldots,0)$. Then take $R$ to be $\text{diag}({p_1}^{-1/2},\ldots,{p_k}^{-1/2},0,\ldots,0)$. Let $\psi:\mathcal{S}\to M_n$ be any positive, unital map, and set $\phi'(a)=R\phi(a)R+(I-Q)\psi(a)(I-Q)$. Then
(i) Show that, $\phi'$ is unital, positive. This one is straight forward. $\phi'(1)=R\phi(1)R+(I-Q)\psi(1)(I-Q)=RPR+(I-Q)^2=Q+(I-Q)=I$.
Again since $a\mapsto R*\phi(a)R$ and $a\mapsto (I-Q)^*\psi(a)(I-Q)$ are positive, $\phi'$ is positive.
(ii) Show that if $(a_{ij})$ is positive in $M_k(S)$, but $\phi_k((a_{ij}))$ is not positive, then $\phi_k'((a_{ij}))$ is not positive either. This is the only part I cannot figure out.
(iii) Deduce the statement. If we assume our hypothesis of the statement. Here $\phi'$ will be completely positive i.e. $\phi_k'$ is positive for all $k$. By part (ii) above, $\phi_k$ has to be positive for all $k$. Hence, $\phi$ is completely positive. This completes the proof of our statement.
I'm stuck with the part (ii). I have tried even with $k=1$ case, but get nothing. Can anyone help me with this part? Thanks for your help in advance.
Part (ii) is phrased in a way that sounds weird to me, as it seems that one has a rather direct way of checking things.
Once you know that $\phi'$ is positive and unital, you know it is completely positive by hypothesis. Let $R'=\operatorname{diag}(p_1^{1/2},\ldots,p_k^{1/2},0,\ldots,0)$. Then $R'R=Q$. Note that $R'$ is positive. Then, since $R'(I-Q)=0$, \begin{align} R'\phi'(a)R'&=R'R\phi(a)RR'+R'(I-Q)\psi(a)(I-Q)R'\\[0.3cm] &=R'R\phi(a)RR'=Q\phi(a)Q=\phi(a). \end{align} That is, $\phi$ is equal to the completely positive map $a\longmapsto R'\phi'(a)R'$.
Edit: justification for $Q\phi(a)Q=\phi(a)$.
Let $a\in S$ be positive. We have $$ \phi(a)\leq\phi(\|a\|\,1)=\|a\|\,P. $$ Then $$ (I-Q)\phi(a)(I-Q)\leq\|a\|\,(I-Q)P(I-Q)=0. $$ We can rewrite this as $$ [\phi(a)^{1/2}(I-Q)]^*\phi(a)^{1/2}(I-Q)=0. $$ Then $\phi(a)(I-Q)=0$. That is, $\phi(a)=\phi(a)Q$. As the positive elements span $S$, we get that $\phi(a)=\phi(a)Q$ for all $a\in S$.