Added the Riesz decomposition.
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Bokuan Li
2026-06-05 21:57:02 -04:00
parent 09a94756ea
commit 99d772d1c8
3 changed files with 79 additions and 16 deletions

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@@ -26,19 +26,3 @@
\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}