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  "body": "\n## Statement\n\nLet $\\mathbb{K}$ be $\\R$ or $\\C$, and let $A$ and $B$ be $n \\times n$ matrices over $\\mathbb{K}$ with $AB = BA$.\nThen\n$$e^{A + B} = e^A e^B,$$\nwhere $e^A$ is the [[matrix exponential|020M04FN1H]].\n\n## Remarks\n\nThe proof multiplies the two series and uses the binomial formula, which needs $AB = BA$.\nIt gives $e^A e^{-A} = I_n$, so $e^A$ is always invertible.\nWithout $AB = BA$, the identity can fail, for example for $A = \\begin{pmatrix} 0 \u0026 1 \\\\ 0 \u0026 0 \\end{pmatrix}$ and $B = \\begin{pmatrix} 0 \u0026 0 \\\\ 1 \u0026 0 \\end{pmatrix}$.\n\n## Explanations\n\n### L1 human\n\nFor numbers, $e^{a + b} = e^a e^b$.\nFor matrices, this rule holds only when the two matrices commute.\nIn particular, $e^{sA} e^{tA} = e^{(s + t)A}$ for all scalars $s$ and $t$.\n",
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