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We define $M \in \mathcal{M}_{n}(\mathbb{R}) \mapsto \exp(M)=\sum _{n=0}^{+\infty}\frac{1}{n!}M^n \in \mathcal{M}_{n}(\mathbb{R})$. Few ideas and though: Is it correct that $\exp(M)$ is a polynomial? (Because of the sum and $M^n$). Usually when we want to prove that a function is $\mathcal{C}^{1}$. We start by showing that is continuous ? (how?) and the derivative of that function is continuous and I struggle here. (Is it the same process, but with the differential in this case?)

Gary
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Loca
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  • Suppose that it is a polynomial in $M$. What is the degree of that polynomial? Do you know about power series? – Gary Jun 07 '22 at 11:26
  • I assume it's $+\infty$ but such as Formal power series – Loca Jun 07 '22 at 11:28
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    The question doesn't really make so much sense. What $\exp(M)$ be $\mathcal C^1$ means ? Do you mean $M\mapsto \exp(M)$ is $\mathcal C^1(\mathcal M_n(\mathbb R),\mathcal M_n(\mathbb R))$ ? – Surb Jun 07 '22 at 11:29
  • Yes it's true, I'm sorry for missleading you. I edit it now. – Loca Jun 07 '22 at 11:29
  • A hint to show that map is continuous is to show for M, N in $\mathcal{M}_{n}(\mathbb{R}), |||exp(M+H)-exp(M)|||\leqslant exp(|||M|||+|||H|||)-exp(|||M|||)$ but I struggle to prove it. – Loca Jun 07 '22 at 11:36
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    Check this: https://math.stackexchange.com/q/1361636/42969, or this: https://math.stackexchange.com/q/2043/42969. – Martin R Jun 07 '22 at 11:38
  • @Loca The degree of a non-zero polynomial is a non-negative integer. $+\infty$ is not a non-negative integer (or a number for that matter). – Gary Jun 07 '22 at 11:42
  • Thank you for the links, If I understand, you assume that in order to show that the function is C1, We have to prove that Dexp(M)(H) is continuous ? – Loca Jun 07 '22 at 11:46
  • @Gary Thank you for that information ! – Loca Jun 07 '22 at 11:49
  • $exp(M+H)=\sum_{k=0}^{+\infty}\frac{(M+H)^{k}}{k!}=\ \sum_{k=0}^{+\infty}\frac{M^{k}}{k!}+\sum_{k=0}^{+\infty}\frac{1}{k!}\sum_{j=0}^{k-1}M^{k-j-1}HM^{j}+o(|||H|||)=\exp(M)+Dexp(M)(H)+o(|||H|||)$ \ In order to prove that $M \mapsto exp(M)$ is $\mathcal{C}^{1}$ let show that $Dexp(M)(H):M \mapsto \sum_{k=0}^{+\infty}\frac{1}{k!}\sum_{j=0}^{k-1}M^{k-j-1}HM^{j}$ converge uniformly.

    $|||\frac{1}{k!}\sum_{j=0}^{k-1}M^{k-j-1}HM^{j}|||\leqslant \frac{1}{k!}\sum_{j=0}^{k-1}|||M^{k-1}||| \times|||H|||=\frac{1}{(k-1)!}|||M^{k-1}||| \times|||H|||$

     – Loca Jun 08 '22 at 08:36
  • and $\sum_{k=1}^{+\infty}\frac{1}{(k-1)!}|||M^{k-1}||| \times|||H|||$ converge to $exp(|||M|||)\times|||H|||$ – Loca Jun 08 '22 at 08:37

1 Answers1

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Each component of $e^M$ is a power series on $\mathbb{R}^{n \times n}$ that converges absolutely everywhere because $\sum_{n}\|M^n/n!\| \leq e^{\|M\|} < \infty$. It is well known that power series are smooth. Hence each component of $e^M$ is smooth. Hence $e^M$ is smooth.

Mason
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