Added explicit descriptions of states in matrix algebras.

src/op/example/matrix.tex

@@ -26,3 +26,61 @@
 \begin{proof}
     (1), (2): By \autoref{proposition:matrix-algebra-spectrum}, $\sigma(x)$ is finite. By \autoref{proposition:log-identity-component}, there exists $y \in M_n(\complex)$ such that $x = \exp(y)$. In which case, $x \in G_0(M_n(\complex))$. 
 \end{proof}
+
+
+\begin{proposition}
+\label{proposition:matrix-algebra-state-space}
+    Let $n \in \natp$. For each $y \in M_n(\complex)$, let
+    \[
+    \phi_y: M_n(\complex) \to \complex \quad x \mapsto \dpn{x, y}{F} = \text{tr}(y^*x)
+    \]
+
+    then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi_y \in S(M_n(\complex))$. 
+        \item $y \ge 0$ and $\text{tr}(y) = 1$.
+    \end{enumerate}
+\end{proposition}
+% "Obvious" so won't prove. 
+
+\begin{proposition}
+\label{proposition:matrix-algebra-pure-state}
+    Let $n \in \natp$. For each $y \in M_n(\complex)$, let
+    \[
+    \phi_y: M_n(\complex) \to \complex \quad x \mapsto \dpn{x, y}{F} = \text{tr}(y^*x)
+    \]
+
+    then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi_y$ is a pure state of $M_n(\complex)$. 
+        \item $y$ is a projection operator with rank $1$. 
+        \item 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 $x \in M_n(\complex)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Leftrightarrow$ (2): By \autoref{proposition:matrix-algebra-state-space}, $y \ge 0$ 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 $x \in M_n(\complex)$,
+    \[
+        \dpn{x, \phi_y}{M_n(\complex)} = \text{tr}(y^*x) = \text{tr}(yx) = \dpn{xv, v}{\complex^n}
+    \]
+\end{proof}
+
+\begin{example}
+\label{proposition:spectrum-pure-state-counterexample}
+    Let
+    \[
+    A = \begin{bmatrix}
+        1 & 0 \\
+        0 & -1
+    \end{bmatrix}
+    \]
+
+    then $A$ is a self-adjoint element of $M_2(\complex)$. By \autoref{proposition:matrix-algebra-pure-state}, the mapping $T \mapsto \dpn{Tv, v}{\complex^2}$ is a pure state on $M_2(\complex)$ for every unit vector $v \in \complex^2$. In particular, if $v = (\sqrt{2}, \sqrt{2})/2$, then $\dpn{Tv, v}{\complex^2} = 0 \not\in \sigma_{M_2(\complex)}(A)$. Therefore
+    \[
+    \sigma_{M_2(\complex)}(A) \subsetneq \bracsn{\dpn{T, \phi}{M_2(\complex)}|\phi \in P(M_2(\complex))}
+    \]
+\end{example}
+
+
+