Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb{C}$ ?

share|improve this question
1  
There is a factor $h$ missing in the third term of the numerator of the limit formula. –  Raskolnikov Nov 11 '11 at 19:35
    
You can just replace $e^{tA}$ by series which converge absolutely because $\|e^{tA}\|\leq e^{t\|A\|}$, that may make things easier. –  Ilya Nov 11 '11 at 19:35
    
$e^{tA}$ 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? –  user7530 Nov 11 '11 at 19:54
    
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. –  Jyrki Lahtonen Nov 11 '11 at 20:43
    
@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. –  resu Nov 12 '11 at 16:54
add comment

1 Answer

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

$\displaystyle \begin{aligned}\lim_{h\to0}\frac{\exp(xA)\exp(hA)-\exp(xA)}{h} &=\exp(xA)\lim_{h\to0}\frac{\exp(hA)-I}{h}\\ &= \exp(xA)\lim_{h\to0 }\frac{1}{h}\left((I+hA+h^2A^2+O(h^3))-I\right)\\ &= \exp(xA)\lim_{h\to0}\left(A+hA^2+O(h^2)\right)\\ &= \exp(xA)A\end{aligned}$

share|improve this answer
1  
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... –  resu Nov 12 '11 at 15:55
    
And does this also hold for the complex case ? I think it would, but I'm not 100% sure... –  resu Nov 12 '11 at 17:34
    
@resu Notice that both $h$ and $x$ here are scalars, otherwise you cannot even define the exponential: $tA$ should be a square matrix. –  awllower Feb 25 at 5:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.