Could someone provide me with a rigorous proof as to why the derivative of the function $f:t \ni \mathbb{R} \mapsto e^{tA}\in \textrm{Mat}_n (\mathbb{R})$ is $t \mapsto A\cdot e^{tA}$ ? I didn't understand the "elementwise" arguments, as to why the above should hold and when trying to evaluate $f'$ by hand I got stuck at evaluating $$ \lim _{h\rightarrow 0} \frac{||e^{xA}\cdot e^{hA} - e^{xA} -A\cdot e^{xA} \cdot h||}{|h|},$$

where $||\cdot ||$ denotes any norm on $\textrm{Mat}_n (\mathbb{R}) $ that is multiplicative (so that $( \textrm{Mat}_n (\mathbb{R}) , || \cdot ||)$ becomes a Banach algebra) - which is what I have do to, I think (please correct me, if I'm wrong, or using an unnecessary abstract level of discourse), because in the setting of matrix-valued functions the derivative of a matrix becomes the derivative between the Banach spaces $\mathbb{R}$ and $\textrm{Mat}_n (\mathbb{R}) $ (using an isomorphism $\phi:\textrm{Mat}_n (\mathbb{R}) \rightarrow \mathbb{R}^{n^2} $ to do the derivative there and then transporting everything back to $\textrm{Mat}_n (\mathbb{R}) $ seems rather ugly to me - although I tried to do it this way and failed).

Could I replace $\mathbb{R}$ with $\mathbb{C}$ ?

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    $\begingroup$ There is a factor $h$ missing in the third term of the numerator of the limit formula. $\endgroup$ – Raskolnikov Nov 11 '11 at 19:35
  • $\begingroup$ You can just replace $e^{tA}$ by series which converge absolutely because $\|e^{tA}\|\leq e^{t\|A\|}$, that may make things easier. $\endgroup$ – Ilya Nov 11 '11 at 19:35
  • $\begingroup$ $e^{tA}$ is, by definition, just a matrix of power series in $\mathbb{R}$. To find the derivative, differentiate term-by-term. Which part of the proof do you consider in doubt? That the power series converges? That it is differentiable? $\endgroup$ – user7530 Nov 11 '11 at 19:54
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    $\begingroup$ You can make life simpler by factoring out $e^{xA}$, so $$\frac{e^{(x+h)A}-e^{xA}}{h}=\frac{e^{hA}-I_n}h\;e^{xA}.$$ Then from the power series of $e^{hA}$ only the term linear in $h$ matters. $\endgroup$ – Jyrki Lahtonen Nov 11 '11 at 20:43
  • $\begingroup$ @user7530 The problem is more of a conceptual nature I think: I have no problems believing that $e^{tA}$ converges but I can't find a way to understand how to work entrywise with the resulting matrix because the entries ar all obtained by iterating the multiplication of $tA4 with itself, so I can't see how it would help me to work entrywise since, by the multiplication process, I don't have any direct control over them. $\endgroup$ – resu Nov 12 '11 at 16:54

You can begin by recalling that there is another definition for the derivative, namely thinking of maps $\mathbb{R}^m\to\mathbb{R}^n$, one has that the derivative, if it exists, is given by $\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. Consequently, the derivative of $\text{exp}(xA)$ should be (we know it exists since $\exp(xA)$ is given by a nice power series) $\displaystyle \lim_{h\to0}\frac{\exp((x+h)A)-\exp(xA)}{h}$. Now, the important thing to notice is that $\exp((x+h)A)=\exp(xA+hA)$ which (recalling that $[X,Y]=0$ implies $\exp(X+Y)=\exp(X)\exp(Y)$) is equal to $\exp(xA)\exp(hA)$ and so the derivative of $\exp(xA)$ should be equal to

$\displaystyle \begin{aligned}\lim_{h\to0}\frac{\exp(xA)\exp(hA)-\exp(xA)}{h} &=\exp(xA)\lim_{h\to0}\frac{\exp(hA)-I}{h}\\ &= \exp(xA)\lim_{h\to0 }\frac{1}{h}\left((I+hA+O(h^2))-I\right)\\ &= \exp(xA)\lim_{h\to0}\left(A+O(h)\right)\\ &= \exp(xA)A\end{aligned}$

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    $\begingroup$ I think you can only apply your definition of the derivative if $m=1$ because otherwise either you can't divide by $h$ (since it is a vector) - or if you perceive $h$ to be a real number (so you can divide by $h$), then $x+h$ makes no sense... $\endgroup$ – resu Nov 12 '11 at 15:55
  • $\begingroup$ And does this also hold for the complex case ? I think it would, but I'm not 100% sure... $\endgroup$ – resu Nov 12 '11 at 17:34
  • $\begingroup$ @resu Notice that both $h$ and $x$ here are scalars, otherwise you cannot even define the exponential: $tA$ should be a square matrix. $\endgroup$ – awllower Feb 25 '14 at 5:17

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