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  "body": "\n## Statement\n\nLet $A$ be a real $n \\times n$ matrix, and for $t \\in \\mathbb{R}$ let $e^{t A} = \\sum_{k=0}^{\\infty} \\frac{t^k}{k!} A^k$, a series that converges for each $t$.\nThen the map $t \\mapsto e^{t A}$ is differentiable on $\\mathbb{R}$, and\n$$\\frac{d}{dt}\\, e^{t A} = A\\, e^{t A} = e^{t A} A.$$\nConsequently, for each $x_0 \\in \\mathbb{R}^n$, the function $x(t) = e^{t A} x_0$ is the unique solution on $\\mathbb{R}$ of the [[initial value problem|075VBY4K4X]] $x' = A x$, $x(0) = x_0$.\n\n## Remarks\n\nUniqueness: if $x$ is a solution, then $\\frac{d}{dt}\\left(e^{-tA} x(t)\\right) = 0$, so $e^{-tA} x(t) = x_0$.\nIf $A = P D P^{-1}$ with $D$ diagonal, then $e^{tA} = P e^{tD} P^{-1}$, and $e^{tD}$ is diagonal with the entries $e^{t \\lambda_k}$.\nThe matrix $e^{tA}$ is the [[fundamental matrix|0MQ24RBG93]] of $x' = A x$ with value $I$ at $t = 0$.\n\n## Explanations\n\n### L1 human\n\nFor one equation $y' = a y$, the solution is $e^{a t} y_0$.\nThe matrix exponential is the same formula for a system, with the matrix $A$ in place of the number $a$.\nThe series is the same as the series of $e^x$.\n",
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