Proposition 30.1.3.label Let $n \in \natp$, then
- (1)
For each $x \in G(M_{n}(\complex))$, there exists $y \in M_{n}(\complex)$ such that $x = \exp(y)$.
- (2)
$G(M_{n}(\complex)) = G_{0}(M_{n}(\complex))$.
- (3)
$I(M_{n}(\complex))$ is trivial.
Proof. (1), (2): By Proposition 30.1.2, $\sigma(x)$ is finite. By Proposition 29.3.2, there exists $y \in M_{n}(\complex)$ such that $x = \exp(y)$. In which case, $x \in G_{0}(M_{n}(\complex))$.$\square$
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