= Matrix exponential determinant identity
{title2=$\det(e^A)=e^{\operatorname{tr}A}$}
The <determinant> is a smooth <Lie group homomorphism> from the real <general linear group> to $\mathbb R^\times$. Its derivative at the identity is the <trace>, and the flow-defined exponential on $\mathbb R^\times$ is the ordinary scalar exponential. <Naturality of the Lie group exponential> therefore proves the identity without diagonalizability assumptions. The series $\sum_{k\geq0}t^kA^k/k!$ is the flow-defined <matrix exponential> because it solves $E'=EA$, $E(0)=I$, and has inverse $E(-t)$.
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