# Why states are completely positive?

A is C*-algebra. Let $\phi: A \to \mathbb C$ be a state (positive linear functional with $\mid \mid \phi \mid \mid =1$). Why any state is a completely positive map? Another quetion, what can I say, for example, about norm of $\phi_2$? Is it equal to the norm of $\phi$?

(Here $\phi_n$ is a map between $M_n(A) \to M_n(\mathbb C), \phi_n (a_{ij})=[\phi(a_{ij})]$.)

• What is $\phi_2$ ? – Bumblebee Mar 21 '16 at 7:43
• I'm sorry for missing explanations. Here $\phi_n$ is a map between $M_n(A) \to M_n(\mathbb C), \phi_n (a_{ij})=[\phi(a_{ij})]$. – Maria Gerasimova Mar 21 '16 at 7:46

Let $a = (a_{ij}) \in \def\Mat{\operatorname{Mat}}\Mat_n(A) \ge 0$, $x \in \mathbf C^n$, we have \begin{align*} \def\<#1>{\left<#1\right>} \<\phi_n(a)x, x> &= \<\left(\sum_{j=1}^n \phi(a_{ij})x_j\right)_i, x>\\ &= \sum_{i,j=1}^n \phi(a_{ij})x_i\bar x_j\\ &= \phi\left(\sum_{i,j} x_i\bar x_j a_{ij} \right) \end{align*} So, it suffices to show that $\sum_{i,j} x_i \bar x_j a_{ij}$ is positive (as $\phi$ is positive). We may assume that $A$ has a unit, otherwise consider $A^+$, $A$'s unitization. We have $$\sum_{i,j} x_i \bar x_j a_{ij} =\begin{pmatrix} x_1\cdot 1 \\ \vdots \\ x_n \cdot 1\end{pmatrix}^* a \begin{pmatrix} x_1\cdot 1 \\ \vdots \\ x_n \cdot 1\end{pmatrix}$$ So, as $a \ge 0$, we have $\sum_{i,j} x_i \bar x_j a_{ij} \ge 0$, this proves that $\phi$ is completely positive.