Added a handful of examples.

src/op/example/matrix.tex

@@ -0,0 +1,44 @@
+\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-ideals}
+    Let $n \in \natp$, then
+    \begin{enumerate}
+        \item $M_n(\complex)$ admits no nontrivial two-sided ideals. 
+        \item $M_n(\complex)$ admits no multiplicative functionals. 
+    \end{enumerate}
+    
+\end{proposition}
+\begin{proof}
+    Let $x = (x_{ij}) \in M_n(\complex) \setminus \bracs{0}$, then there exists $1 \le i, j \le n$ such that $x_{ij} \ne 0$. In which case, for any $1 \le k, l \le n$ and $\lambda \in \complex$, there exists $y_k, z_l \in M_n(\complex)$ such that $y_kxz_l$ is the matrix with $\lambda$ on its $(k, l)$ entry and $0$ everywhere else. Therefore every non-trivial two-sided ideal of $M_n(\complex)$ is $M_n(\complex)$. 
+\end{proof}
+
+
+