Proposition 35.1.5.label Let $n \in \natp$. For each $y \in M_{n}(\complex)$, let
then the following are equivalent:
- (1)
$\phi_{y}$ is a pure state of $M_{n}(\complex)$.
- (2)
$y$ is a projection operator with rank $1$.
- (3)
There exists $v \in \complex^{n}$ with $\norm{v}_{\complex^n}= 1$ such that $\dpn{x, \phi_y}{M_n(\complex)}= \dpn{xv, v}{\complex^n}$ for all $x \in M_{n}(\complex)$.
Proof. (1) $\Leftrightarrow$ (2): By Proposition 35.1.4, $y \ge 0$ with $\text{tr}(y) = 1$. Via an orthogonal change of coordinates, assume without loss of generality that $y$ is diagonal. In which case, $y$ corresponds to an extreme point of $S(M_{n}(\complex))$ if and only if it is of rank $1$. As $\text{tr}(y) = 1$, $y$ is a projection.
(2) $\Rightarrow$ (3): Let $v \in \complex^{n}$ be a unit eigenvector of $y$, then for each $x \in M_{n}(\complex)$,
$\square$
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