Determinant of a matrix exponential

Statement

Let $\mathbb{K}$ be $\R$ or $\C$, and let $A$ be an $n \times n$ matrix over $\mathbb{K}$. Then $$\det(e^A) = e^{\tr A},$$ where $e^A$ is the matrix exponential, $\det$ is the determinant and $\tr$ is the trace.

Remarks

If $A$ is triangular over $\C$ with diagonal entries $\lambda_i$, then $e^A$ is triangular with diagonal entries $e^{\lambda_i}$, and both sides equal $e^{\lambda_1 + \cdots + \lambda_n}$. The general case follows by triangularisation over $\C$. It shows again that $e^A$ is invertible, because $e^{\tr A} \ne 0$. If $\tr A = 0$, then $e^A$ lies in the special linear group.

Explanations

L1 human

The determinant of the exponential of a matrix is the exponential of its trace. This links the product of the eigenvalues of $e^A$ with the sum of the eigenvalues of $A$. It is the matrix form of the rule $e^{a} e^{b} = e^{a + b}$.