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  "body": "\n## Definition\n\nLet $X$ be a [[Banach space|06GZ956657]] over $\\R$ or $\\C$, and let $T \\colon X \\to X$ be a [[continuous linear map|03TM135G0N]].\nWrite $T^0 = I$, the identity map of $X$.\nThe *exponential* of $T$ is the bounded operator\n$$e^T = \\exp(T) = \\sum_{n=0}^{\\infty} \\frac{T^n}{n!},$$\nwhere the series converges in [[operator norm|06EKW1AKXF]].\n\n## Remarks\n\nThe series converges because $\\norm{T^n / n!} \\le \\norm{T}^n / n!$, and the numbers $\\norm{T}^n / n!$ have the finite sum $e^{\\norm{T}}$; see [[absolutely convergent series in a Banach space converge|082MGNWNA8]].\nThis also gives $\\norm{e^T} \\le e^{\\norm{T}}$.\n\nExample: let $D$ be the diagonal operator $D(x_n) = (\\lambda_n x_n)$ on $\\ell^2$, with $(\\lambda_n)$ a bounded sequence of complex numbers.\nThen $D^k(x_n) = (\\lambda_n^k x_n)$, so summing coordinatewise gives $e^D(x_n) = (e^{\\lambda_n} x_n)$.\n\nFor $X = \\C^n$, the definition gives the [[matrix exponential|020M04FN1H]].\nIf $ST = TS$, then $e^{S + T} = e^S e^T$; in particular $e^T$ is [[invertible|CPAD5W1CGW]] with inverse $e^{-T}$.\nThe identity $e^{S+T} = e^S e^T$ can fail when $S$ and $T$ do not commute, already for $2 \\times 2$ matrices.\nThe map $t \\mapsto e^{tT}$ from $\\R$ to bounded operators is differentiable in operator norm, with derivative $T e^{tT}$.\n",
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