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  "body": "\n## Proof\n\nWe compare partial sums and bound the difference with a matrix norm. The binomial theorem is where $A B = B A$ enters.\n\n1. For an $n \\times n$ matrix $X = (x_{ij})$ over $\\mathbb{K}$, let $\\norm{X} = n \\max_{i,j} \\abs{x_{ij}}$. Then $\\abs{x_{ij}} \\le \\norm{X}$, $\\norm{X + Y} \\le \\norm{X} + \\norm{Y}$ and $\\norm{c X} = \\abs{c} \\norm{X}$ for $c \\in \\mathbb{K}$.\n2. For matrices $X, Y$, each entry of $X Y$ satisfies $\\abs{\\sum_l x_{il} y_{lj}} \\le n \\max \\abs{x_{il}} \\max \\abs{y_{lj}}$. Multiplying by $n$ gives $\\norm{X Y} \\le \\norm{X} \\norm{Y}$. By induction, $\\norm{X^k} \\le \\norm{X}^k$ for $k \\ge 1$.\n3. Let $S_N(X) = \\sum_{k=0}^{N} X^k / k!$. By the [[definition of the matrix exponential|0K665H5V1S]], each entry of $S_N(X)$ tends to the matching entry of $e^X$ as $N \\to \\infty$.\n4. The finite sums $\\sum c_{jk} A^j B^k$ with $c_{jk} \\in \\mathbb{K}$ form a subring of the ring of matrices. It is commutative, because $A B = B A$. So the [[binomial theorem|0EHWPWGHB2]] applies: $(A + B)^m = \\sum_{k=0}^{m} \\binom{m}{k} A^k B^{m-k}$.\n5. Dividing by $m!$ and summing over $m \\le N$ gives $S_N(A + B) = \\sum_{k + l \\le N} \\frac{A^k B^l}{k! \\, l!}$, where $k, l \\ge 0$.\n6. Expanding the product gives $S_N(A) S_N(B) = \\sum_{k \\le N, \\, l \\le N} \\frac{A^k B^l}{k! \\, l!}$.\n7. Let $T_N$ be the set of pairs $(k, l)$ with $k \\le N$, $l \\le N$ and $k + l \u003e N$. By steps 5 and 6, $\\Delta_N = S_N(A) S_N(B) - S_N(A + B) = \\sum_{(k,l) \\in T_N} \\frac{A^k B^l}{k! \\, l!}$.\n8. Let $a = \\norm{A}$ and $b = \\norm{B}$. For $(k, l) \\in T_N$, $k$ and $l$ are not both $0$. Steps 1 and 2 then give $\\norm{A^k B^l} \\le a^k b^l$. When $k = 0$ or $l = 0$, the factor $I_n$ is simply left out.\n9. By steps 1, 7 and 8, $\\norm{\\Delta_N} \\le \\sum_{(k,l) \\in T_N} \\frac{a^k b^l}{k! \\, l!}$.\n10. Let $s_N(x) = \\sum_{k=0}^{N} x^k / k!$ for real $x$. Steps 5 to 7 hold for the commuting real numbers $a$ and $b$ too. So the right side of step 9 equals $s_N(a) s_N(b) - s_N(a + b)$.\n11. By the [[power series of the exponential function|06JZGMYTYD]], $s_N(x) \\to e^x$ for each real $x$. So the right side of step 9 tends to $e^a e^b - e^{a+b}$, which is $0$ by the [[exponential of a sum|0E8HNVBCSP]].\n12. By steps 1, 9 and 11, each entry of $\\Delta_N$ tends to $0$.\n13. Entry $(i, j)$ of $S_N(A) S_N(B)$ is $\\sum_l S_N(A)_{il} S_N(B)_{lj}$, a finite sum of products. By step 3 and the [[algebra of limits|07XBM3A9J5]], it tends to $\\sum_l (e^A)_{il} (e^B)_{lj}$, the entry $(i, j)$ of $e^A e^B$.\n14. By step 3, entry $(i, j)$ of $S_N(A + B)$ tends to entry $(i, j)$ of $e^{A+B}$.\n15. By steps 7, 13 and 14, entry $(i, j)$ of $\\Delta_N$ tends to entry $(i, j)$ of $e^A e^B - e^{A+B}$. By step 12, this entry is $0$.\n16. Hence $e^{A+B} = e^A e^B$.\n\n## Idea\n\nBecause $A$ and $B$ commute, $(A + B)^m$ expands by the binomial theorem, exactly as for numbers. Then the partial sums of $e^{A+B}$ and of $e^A e^B$ differ only by the terms $A^k B^l / (k! \\, l!)$ with $k + l \u003e N$. A submultiplicative norm bounds this tail by the same tail for the real numbers $\\norm{A}$ and $\\norm{B}$, which tends to $0$.\n",
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