Positive matrices I: Positive linear functionals

Let $\mathbb{M}_n$ denotes the algebra of n by n complex matrices. A matrix $A \in \mathbb{M}_n$ is said to be positive semi-definite or just positive if any one of the following equivalent conditions holds:

1. $x^*Ax \geq 0$ for any $x \in \mathbb{C}^n$;
2. $A = X^*X$ for some $X \in \mathbb{M}_n$;
3. $A$ is Hermitian (i.e. $A = A^*$) and all its eigenvalues are nonnegative (i.e. $\sigma(A) \subseteq [0, \infty)$).

We write $A \geq O$ if $A$ is a positive matrix.

A linear subspace $\mathcal{S}$ of $\mathbb{M}_n$ is an operator system if it contains $I$ and closed under the adjoint operation. 

Let $\mathcal{S} \subseteq \mathbb{M}_n$ be an operator system. A linear map $\Phi: \mathcal{S} \rightarrow \mathbb{M}_k$ is said to be positive if $\Phi(A) \geq O$ whenever $A \geq O$.

Note that we can write any $X \in \mathcal{S}$ as $X = A + iB$ where $A, B \in \mathcal{S}$ are Hermitian matrices given by $A = \frac{1}{2}(X + X^*)$ and $B = \frac{1}{2i}(X - X^*)$. Further, if $Y \in \mathbb{M}_n$ is Hermitian, then $Y = P - N$ for some positive matrices $P, N \in \mathcal{S}$. Here $P = \frac{1}{2}(\|Y\|I + Y)$ and $N = \frac{1}{2}(\|Y\|I - Y)$. The above decomposition gives the following result.

Proposition. If $\Phi: \mathcal{S} \rightarrow \mathbb{M}_k$ is a positive linear map, then $\Phi(X^*) = \Phi(X)^*$ for any $X \in \mathcal{S}$.

We now consider positive linear functionals on an operator system $\mathcal{S}$. It turns out in this case positivity is equivalent to that the functional attains its norm at the identity.

Theorem. Let $\varphi$ be a linear functional on $\mathcal{S}$. Then $\varphi$ is positive if and only if $\|\varphi\| = \varphi(I)$.

Proof: (=>) Let $X \in \mathcal{S}$. It is enough to show that $|\varphi(X)| \leq \|X\|\varphi(I)$. By replace $X$ by $e^{-i\theta}X$ for some suitable $\theta \in \mathbb{R}$, we may assume that $\varphi(X) \geq 0$. Write $X = A + iB$ where Hermitian $A, B \in \mathcal{S}$ are given by $A = \frac{1}{2}(X + X^*)$ and $B = \frac{1}{2i}(X - X^*)$. Since $\varphi(A), \varphi(B) \in \mathbb{R}$, we have $\varphi(X) = \varphi(A)$. Now write $A = P - N$ where $P, N \in \mathcal{S}$ are positive and given by $P = \frac{1}{2}(\|A\|I + A)$ and $N = \frac{1}{2}(\|A\|I - A)$. Note that $O \leq P \leq \|P\|I$, so $\varphi(P) \leq \|P\|\varphi(I)$. Now we have $\varphi(X) = \varphi(A) = \varphi(P) - \varphi(N) \leq \varphi(P) \leq \|P\|\varphi(I) \leq \|A\|\varphi(I) \leq \|X\|\varphi(I)$.
(<=) WLOG, assume $\|\varphi\| = \varphi(I) = 1$. Let $X \in \mathcal{S}$ be positive. Let $a = \min \sigma(X), b = \max \sigma(X)$ and $J = [a, b] \supseteq \sigma(X)$. We are going to show that $\varphi(X) \in J$, hence $\varphi(X) \geq 0$. Suppose not. Take a closed disk $D$ centered at $z_0$ with radius $r$ such that $J \subseteq D$ and $\varphi(X) \notin D$. It follows that $\sigma(X - z_0I) \subseteq  \{z: |z| \leq r\}$. Now $|\varphi(X) - z_0| = |\varphi(X - z_0I)| \leq \|\varphi\|\|X_0 - z_0I\| \leq r$. This contradicts the fact that $\varphi(X) \notin D$.

The above characterization immediately gives the following extension theorem for positive linear functionals.

Theorem (Krein Extension Theorem). Every positive linear functional on $\mathcal{S}$ extends to a positive  linear functional on $\mathbb{M}_n$.

Proof: Suppose $\varphi$ is a positive linear functional on $\mathcal{S}$. Then $\|\varphi\| = \varphi(I)$. By Hahn-Banach Theorem, $\varphi$ extends to a linear functional $\tilde{\varphi}$ on $\mathbb{M}_n$ with $\|\tilde{\varphi}\| = \|\varphi\| = \varphi(I) = \tilde{\varphi}(I)$, so $\tilde{\varphi}$ is positive.

Positive linear functionals on $\mathbb{M}_n$ have the following characterization.

Proposition. Let $\varphi$ be a positive linear functonal on $\mathbb{M}_n$. Then there exists a positive matrix $X \in \mathbb{M}_n$ such that $\varphi(A) = \mathrm{tr} AX$ for all $A \in \mathbb{M}_n$.

Proof: For $1 \leq i, j \leq n$, let $E_{ij} \in \mathbb{M}_n$ be the matrix with entry 1 at i-th row and j-th column and other entries zero and let $p_{ij} = \varphi(E_{ij})$. Take any $A = [a_{ij}] \in \mathbb{M}_n$. Then $\varphi(A) = \sum_{i, j} a_{ij}p_{ij} = \mathrm{tr}(AX)$ where $X = [x_{ij}]$ with $x_{ij} = p_{ji}$. Let $P = [p_{ij}]$, so that $X = P^*$. For any $x \in \mathbb{C}^n$, $x^*Xx = \sum_{i, j} \bar{x_i} x_{ij} x_j = \varphi(\sum_{i, j} \bar{x_i} E_{ji} x_j) = \varphi([\bar{x_j}x_i]) \geq 0$ since $\varphi$ and $[\bar{x_j}x_i] = xx^*$ are positive. Hence $X$ is positive.

Reference:
Bhatia - Positive Definite Matrices