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  "body": "\n## Proof\n\nWe differentiate the series entry by entry, and prove uniqueness with the factor $e^{-tA}$.\n\n1. Let $c = n \\max_{i, j} \\abs{A_{ij}}$. We claim $\\abs{(A^k)_{ij}} \\le c^k$ for all $i, j$ and $k \\ge 0$. For $k = 0$, the entries of $A^0 = I$ are $0$ or $1$.\n2. If the claim holds for $k$, then $\\abs{(A^{k+1})_{ij}} = \\abs{\\sum_l (A^k)_{il} A_{lj}} \\le n\\, c^k \\max_{l, j} \\abs{A_{lj}} = c^{k+1}$. This proves the claim by induction.\n3. Entry $(i, j)$ of the [[matrix exponential|0K665H5V1S]] $e^{tA}$ is the power series $\\varphi_{ij}(t) = \\sum_{k \\ge 0} (A^k)_{ij}\\, t^k / k!$. By step 1, its terms are bounded in absolute value by $(c \\abs{t})^k / k!$.\n4. The series $\\sum_k (c \\abs{t})^k / k!$ [[converges for every $t$|0PVJRACB9H]]. So $\\varphi_{ij}$ converges absolutely for every $t \\in \\R$, and its [[radius of convergence|0660QX1VPR]] is $\\infty$.\n5. By [[termwise differentiation of power series|002X7EK91H]], $\\varphi_{ij}$ is differentiable on $\\R$ with $\\varphi_{ij}'(t) = \\sum_{k \\ge 1} (A^k)_{ij}\\, t^{k-1}/(k-1)! = \\sum_{k \\ge 0} (A^{k+1})_{ij}\\, t^k / k!$.\n6. We have $(A^{k+1})_{ij} = \\sum_l A_{il} (A^k)_{lj} = \\sum_l (A^k)_{il} A_{lj}$. Finite sums of convergent series may be summed termwise. So $\\varphi_{ij}' = \\sum_l A_{il} \\varphi_{lj} = \\sum_l \\varphi_{il} A_{lj}$.\n7. In matrix form, $\\frac{d}{dt} e^{tA} = A\\, e^{tA} = e^{tA} A$ for each $t \\in \\R$.\n8. Existence. Let $x(t) = e^{tA} x_0$. Each entry of $x$ is a fixed combination of the $\\varphi_{ij}$, so by step 7 $x'(t) = A e^{tA} x_0 = A x(t)$.\n9. At $t = 0$ only the term $k = 0$ remains, so $e^{0 A} = I$ and $x(0) = x_0$. So $x$ solves the [[initial value problem|075VBY4K4X]] for the [[system|0VJF99CZ0Z]] $x' = A x$ on $\\R$.\n10. Since $(-t)^k A^k = t^k (-A)^k$, we have $e^{-tA} = e^{t(-A)}$. Step 7 for the matrix $-A$ gives $\\frac{d}{dt} e^{-tA} = -A e^{-tA} = -e^{-tA} A$.\n11. For differentiable matrix functions $U, V$, the entrywise [[product rule|0V9A43H7WV]] and [[linearity of differentiation|0M0TKFRX2A]] give $(U V)' = U' V + U V'$.\n12. Let $Z(t) = e^{tA} e^{-tA}$. By steps 7, 10 and 11, $Z' = A e^{tA} e^{-tA} - e^{tA} A e^{-tA}$. Since $e^{tA} A = A e^{tA}$, this is $0$.\n13. So each entry of $Z$ [[is constant|003QDZ4T16]], and $Z(t) = Z(0) = I$ for all $t$.\n14. Uniqueness. Let $y$ be any solution on $\\R$ with $y(0) = x_0$, and let $w(t) = e^{-tA} y(t)$. By steps 10 and 11, $w' = -e^{-tA} A y + e^{-tA} A y = 0$.\n15. So $w$ is constant, and $w(t) = w(0) = y(0) = x_0$.\n16. By step 13, $y(t) = e^{tA} e^{-tA} y(t) = e^{tA} w(t) = e^{tA} x_0$. So $x(t) = e^{tA} x_0$ is the unique solution.\n\n## Idea\n\nEach entry of $e^{tA}$ is a power series in $t$ with infinite radius, so it can be differentiated term by term. This shifts $A^k$ to $A^{k+1}$, which is $A e^{tA}$. For uniqueness, $e^{-tA}$ undoes the flow: it turns any solution into a constant.\n",
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