\section{The Matrix Algebra}
\label{section:matrix-algebra}

\begin{definition}[Matrix Algebra]
\label{definition:matrix-algebra}
    Let $n \in \natp$ and $M_n(\complex)$ be the set of all $n \times n$ matrices with entries in $\complex$, then $M_n(\complex)$ equipped with the operator norm is the \textbf{matrix algebra} over $\complex$. 
\end{definition}

\begin{proposition}
\label{proposition:matrix-algebra-spectrum}
    Let $n \in \natp$ and $x \in M_n(\complex)$, then $\sigma(x)$ is the set of eigenvalues of $x$. 
\end{proposition}
\begin{proof}
    By the rank-nullity theorem, $\lambda - x$ is invertible if and only if $\lambda$ is not an eigenvalue of $x$. 
\end{proof}

\begin{proposition}
\label{proposition:matrix-algebra-index}
    Let $n \in \natp$, then
    \begin{enumerate}
        \item For each $x \in G(M_n(\complex))$, there exists $y \in M_n(\complex)$ such that $x = \exp(y)$. 
        \item $G(M_n(\complex)) = G_0(M_n(\complex))$. 
        \item $I(M_n(\complex))$ is trivial. 
    \end{enumerate}
\end{proposition}
\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}