Completely positive map on C*-algebra There is a quotation in a book about C*-algebra.
A positive linear functional $f$ on an operator system $E$ is completely positive map. Indeed, for $\xi=(\xi_{1}, \xi_{2},\ldots,\xi_{n})\in l_{n}^{2}$ (here, $l_{n}^{2}$ denotes $n$-dimensional Hilbert space) and $a=[a_{i, j}] \geq 0$ in $M_{n}(E)$, we have 
$$\langle f_{n}(a)\xi, \xi\rangle=f(\Sigma_{i,j=1}^{n}\bar{\xi_{i}}\xi_{j}a_{i,j})=f((\bar{\xi_{1}},\ldots,\bar{\xi_{n}}) \left(\begin{array}{ccc}
    a_{11} & \cdots &a_{1n}\\
      \vdots &  &  \vdots\\
    a_{n1}& \cdots &a_{nn}
  \end{array}\right)\left(\begin{array}{ccc}
    \xi_{1} \\
      \vdots \\
    \xi_{n}
  \end{array}\right)
) \geq 0$$
I do not know why this $f((\bar{\xi_{1}},\ldots,\bar{\xi_{n}}) \left(\begin{array}{ccc}
    a_{11} & \cdots &a_{1n}\\
      \vdots &  &  \vdots\\
    a_{n1}& \cdots &a_{nn}
  \end{array}\right)\left(\begin{array}{ccc}
    \xi_{1} \\
      \vdots \\
    \xi_{n}
  \end{array}\right)
) \geq 0$ holds? Could someone explain it to me ? Thanks.
 A: (here I'm addressing in part Tom's answer, and a mistake in the question). The vector $\xi$ is not in $l^2_n$; this makes no sense, as we need the operator system $E$ to act on each coordinate. Rather, $\xi\in \oplus_{j=1}^nH$, where $H$ is the Hilbert space such that $E\subset B(H)$. So the (bad) notation in the proof quoted in the question actually means
$$
\bar\xi\,\xi a=\langle a\xi,\xi\rangle, 
$$and
$$
(\bar{\xi_{1}},...,\bar{\xi_{n}}) \left(\begin{array}{ccc}
    a_{11} & ... &a_{1n}\\
      \vdots &  &  \vdots\\
    a_{n1}& ... &a_{nn}
  \end{array}\right)\left(\begin{array}{ccc}
    \xi_{1} \\
      \vdots \\
    \xi_{n}
  \end{array}\right)=\left\langle \left(\begin{array}{ccc}
    a_{11} & ... &a_{1n}\\
      \vdots &  &  \vdots\\
    a_{n1}& ... &a_{nn}
  \end{array}\right)
\,\left(\begin{array}{ccc}
    \xi_{1} \\
      \vdots \\
    \xi_{n}
  \end{array}\right),
\left(\begin{array}{ccc}
    \xi_{1} \\
      \vdots \\
    \xi_{n}
  \end{array}\right)
\right\rangle
$$
Now the positivity of $A$ makes the inner product non-negative.
