Added the Riesz decomposition.
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@@ -26,19 +26,3 @@
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\begin{proof}
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(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, $x \in G_0(M_n(\complex))$.
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\end{proof}
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\begin{proposition}
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\label{proposition:matrix-algebra-ideals}
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Let $n \in \natp$, then
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\begin{enumerate}
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\item $M_n(\complex)$ admits no nontrivial two-sided ideals.
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\item $M_n(\complex)$ admits no multiplicative functionals.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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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 $M_n(\complex)$.
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\end{proof}
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