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How can I compute the derivative of the exponential of a sum of matrices, for example the generators in the algebra of a Lie Group?

I think about a matrix of the form

$$M=e^{\sum{\lambda_i*A_i}}$$

where the $\lambda$s are scalars and the $A$s are matrices, for example the generators of the algebra. Is there a simple way to express $\frac{dM}{d\lambda_j}$ in terms of these generators?

Thanks in advance

Shaun
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Piano Piano a Tratti Forte
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  • Don't you just get $\frac{\partial M}{\partial \lambda_j}=A_je^{\sum_i\lambda_iA_i}$ – drC1Ron Jun 01 '22 at 11:13
  • https://math.stackexchange.com/questions/4454084/higher-order-derivative-of-exponential-map – user619894 Jun 01 '22 at 11:29
  • @RonnyLandsverk this is what I get using Lie-Trotter formula and taking the derivative inside of the limit for n to infinity but I am not sure if I can do this.The problem is that the result you mentioned trivially holds for only one matrix but if I have more than one generator in the sum and they do not commute ,does it still hold? Can I still rearrange the matrix product in the Lie -Trotter formula? – Piano Piano a Tratti Forte Jun 01 '22 at 11:57
  • @RonnyLandsverk furthermore, do I have to multiply Aj from the left or the right because I don't think it commutes with the exponential of the sum, since the generators do not commute generally. – Piano Piano a Tratti Forte Jun 01 '22 at 12:00
  • @PianoPianoaTrattiForte Perhaps I misinterpret the notation. Is it a matrix exponential where the matrix argument is a sum of matrices $A_i$ scaled by $\lambda_i$? – drC1Ron Jun 01 '22 at 12:18
  • @RonnyLandsverk exactly, the exponent is a sum of matrices ,which do not necessarily commute , weighted by the scalar coefficients lambda and the derivative is taken with respect to one of them – Piano Piano a Tratti Forte Jun 01 '22 at 13:26
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    See https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map – Eero Hakavuori Jun 01 '22 at 13:41

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