A proposition about Voiculescu's Theorem in C*-algebra It is the quotation below:
Exploiting the duality between completely positive map $A \rightarrow M_{n}(C)$ and states on $M_{n}(A)$, it is not too hard to deduce the next result from Glimm's lemma.
(Glimm's lemma) Let $A\subset B(H)$ be a separable C*-algebra containing no nonzero compact operators on $H$. If $\phi$ is a state on $A$, then there exist orthonormal vectors $\{\xi_{n}\}$ such that $\langle a\xi_{n}, \xi_{n} \rangle \rightarrow \phi(a)$ for all $a\in A$.
Proposition 1.7.1. Let $H$ be separable Hilbert space, $1\in A\subset B(H)$ be a separable C*-algebra and $\phi: A\rightarrow M_{n}(C)$ be a unital completely positive map such that $\phi|_{A\cap K(H)}=0$. Then there exist isometries $V_{k}:l_{n}^{2}\rightarrow H$ with the following properties: 
(1) the ranges of the $V_{k}$ 's are pairwise orthogonal;
(2) $||\phi(a)-V_{k}^{*}aV_{k}||\rightarrow 0$ for every $a\in A$.
(Here, $K(H)$ denotes the compacts and $l_{n}^{2}$ denotes the n-dimensional Hilbert space.)
I do not know how to utilise the duality to deduce the proposition, could someone show me more details?
 A: You can read about such duality in the first pages of Chapter 6 in Paulsen. Given $\phi\in\text{UCP}(A,M_n(\mathbb C))$, it can be identified with the state $s_\phi$ on $M_n(A)$ by 
$$
s_\phi([a_{kj}])=\frac1n\,\sum_{k,j}\phi(a_{kj})_{kj}.
$$
The hypothesis that $\phi$ is zero on the compacts implies that so is $s_\phi$. So we can apply Glimm's Lemma to $M_n(A)\subset B(H^n)$ to obtain orthonormal vectors $\xi_k=(\xi_k^1,\ldots,\xi_k^n)\in H^n$ with $s_\phi(A)=\lim_k\langle B\xi_k,\xi_k\rangle$ for all $B\in M_n(A)$. 
(this construction does not yield that the ranges of the $V_k$ are pairwise orthogonal; I have some doubts that such a thing is possible).
Now define $W_k:l_n^2\to H$ by $W_ke_j\mapsto \sqrt n\xi_k^j$. We have, for any $a\in A$,
$$
\phi(a)_{ij}=n\,s_\phi(a\otimes E_{ij})=n\,\lim_k\,\langle(a\otimes E_{ij})\xi_k,\xi_k\rangle=n\,\lim_k\langle a\,\xi_k^j,\xi_k^i\rangle\\
=\lim_k\,\langle a\,W_ke_j,W_ke_i\rangle=\lim_k\, (W_k^*aW_k)_{ij}.
$$
As these are limits of matrices, an entry-wise limit implies norm limit. So $\|\phi(a)-W_k^*aW_k\|\to0$. 
Note also that
$$
\|W_k(\sum_{s=1}^nc_se_s)\|^2=n\,\|\sum_{s=1}^nc_s\xi_k^s\|^2
$$
doesn't look like an isometry. But
$$
\lim_k\,\langle W_ke_s,W_ke_t\rangle=
\lim_k\,\langle \xi_k^s,\xi_k^t\rangle=\lim_k\,\langle E_{ts}\xi_k,\xi_k\rangle=\frac1n\,s_\phi(I\otimes E_{st})=\phi(I)_{st}=\delta_{st}.
$$
So the components of $\xi_k$ are almost orthonormal and, more importantly, $W_k^*W_k\to I$. Again, the weak convergence implies norm convergence, as $W_k^*W_k\in M_n(\mathbb C)$; so $\|W_k^*W_k-I\|\to0$. So, for big enough $k$, $W_k^*W_k$ is invertible.
Write $W_k=V_k\,|W_k|$ the polar decomposition. From the previous paragraph, for big $k$ we have that $V_k$ is an isometry, and $\|V_k-W_k\|\to0$ (because $W_k^*W_k\to I$). So we get
$$
\|\phi(a)-V_k^*aV_k\|\to0
$$
for isometries $V_k$. 

Edit: here is an example that shows that the condition of being zero on the compacts is essential. Let $n=1$, and $\{e_j\}$ an orthonormal basis for $H$. Let $A=B(H)$ (or $K(H)$, which would be enough). Define a ucp map $\phi:B(H)\to\mathbb C$ by 
$$\tag1
\phi(a)=\tfrac12\,\langle ae_1,e_1\rangle+\tfrac12\,\langle ae_2,e_2\rangle. 
$$
Suppose that $\phi(a)=\lim_k V_k^*aV_k$ for isometries $V_k:\mathbb C\to H$. It is easy to check that $V_k\lambda=\lambda v_k$ for some $v_k\in H$ with $\|v_k\|=1$, and that $V_k^*x=\langle x,v_k\rangle$. Thus
$$
\phi(a)=\lim_k\langle av_k,v_k\rangle.
$$
With $P=E_{11}+E_{22}$, we have that from $(1)$ that $\phi(a)=\phi(aP)=\phi(Pa)$. So 
$$
\phi(a)=\lim_k\langle aPv_k,Pv_k\rangle. 
$$
As $Pv_k=\lambda_{k,1}e_1+\lambda_{k,2}e_2$ with $1\geq\|Pv_k\|^2=|\lambda_{k,1}|^2+|\lambda_{k,2}|^2$ the bounded sequences $\{\lambda_{k,j}\}_k$ admit a convergent subsequence for each $j=1,2$. So we obtain a subsequence of $\{Pv_k\}$ that converges to a certain $v$, and thus
$$\tag2
\phi(a)=\langle av,v\rangle.
$$
Say $v=\lambda_1e_1+\lambda_2e_2$. Comparing $(1)$ and $(2)$ we get:
$$
\tfrac12=\phi(E_{11})=\langle E_{11}v,v\rangle=\alpha,
$$
and similarly $\beta=\tfrac12$. But then
$$
0=\phi(E_{21})=\langle E_{21}v,v\rangle=\tfrac12,
$$
a contradiction. 
