# 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.

• I changed $< f_{n}(a)\xi, \xi >$ to $\langle f_{n}(a)\xi, \xi\rangle$. That is standard usage. – Michael Hardy Mar 11 '14 at 16:26
• @Michael Hardy Ok thanks~ – Yan kai Mar 11 '14 at 16:32

(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.
• @Martin Argerami:But, I suppose, in you equation above, $(\xi_{1},\ldots,\xi_{n}) \left(\begin{array}{ccc} a_{11} & \cdots &a_{1n}\\ \vdots & & \vdots\\ a_{n1}& \cdots &a_{nn} \end{array}\right)\left(\begin{array}{ccc} \xi_{1} \\ \vdots \\ \xi_{n} \end{array}\right)$ is an element in operator system $E$ – Yan kai Mar 12 '14 at 1:45
• @Martin Argerami And $\langle \left(\begin{array}{ccc} a_{11} & \cdots &a_{1n}\\ \vdots & & \vdots\\ a_{n1}& \cdots &a_{nn} \end{array}\right) \left(\begin{array}{ccc} \xi_{1} \\ \vdots \\ \xi_{n} \end{array}\right), \left(\begin{array}{ccc} \xi_{1} \\ \vdots \\ \xi_{n} \end{array}\right) \rangle$ is a complex number, So what does your equal sign mean? – Yan kai Mar 12 '14 at 1:49
• Because $\xi_j$ is vector in a Hilbert space. So there's no obvious meaning for $\bar \xi_j$ nor $\bar\xi_j\,\xi_j$ – Martin Argerami Mar 12 '14 at 3:12