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  "body": "\n## Statement\n\nLet $\\mathbb{K}$ be $\\R$ or $\\C$, and let $A$ be an $n \\times n$ matrix over $\\mathbb{K}$.\nThen\n$$\\det(e^A) = e^{\\tr A},$$\nwhere $e^A$ is the [[matrix exponential|020M04FN1H]], $\\det$ is the [[determinant|00KDZRW4C0]] and $\\tr$ is the [[trace|0XC1RDBY7A]].\n\n## Remarks\n\nIf $A$ is triangular over $\\C$ with diagonal entries $\\lambda_i$, then $e^A$ is triangular with diagonal entries $e^{\\lambda_i}$, and both sides equal $e^{\\lambda_1 + \\cdots + \\lambda_n}$.\nThe general case follows by [[triangularisation|0363JMQ42Y]] over $\\C$.\nIt shows again that $e^A$ is invertible, because $e^{\\tr A} \\ne 0$.\nIf $\\tr A = 0$, then $e^A$ lies in the special linear group.\n\n## Explanations\n\n### L1 human\n\nThe determinant of the exponential of a matrix is the exponential of its trace.\nThis links the product of the eigenvalues of $e^A$ with the sum of the eigenvalues of $A$.\nIt is the matrix form of the rule $e^{a} e^{b} = e^{a + b}$.\n",
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