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I need your help with the differential of a matrix exponential.

A function $f$ is a mapping from a matrix to a matrix, that is, $f: \mathbb{R}^{n \times n}\rightarrow \mathbb{R}^{n \times n}$. Here $\mathbf{H} \in\mathbb{R}^{n \times n} $ is a matrix. And each element of $\mathbf{H}$ is parameterized by parameter $\theta$.

In this situation, I would like to know if it is correct $$ \frac{d e^{f(\mathbf{H})}}{d \theta} = \frac{d f(\mathbf{H})}{d\theta}e^{f(\mathbf{H})} $$

Sakurai.JJ
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    Not usually as they do not commute. See https://math.stackexchange.com/a/3872098/617446 – user619894 Jul 20 '23 at 05:26
  • Here you have the way to differentiate the exponential map : https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map#Statement – Abezhiko Jul 20 '23 at 05:56

1 Answers1

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According to this document, it holds that $$ \frac{d}{dt}\exp{\left( \mathbf{H}(t) \right)} = \exp{\left( \mathbf{H} \right)} \left\{ \mathbf{H}'-\frac{1}{2!}[\mathbf{H},\mathbf{H}']+\frac{1}{3!}[\mathbf{H},[\mathbf{H},\mathbf{H}']]-\frac{1}{4!}[\mathbf{H},[\mathbf{H},[\mathbf{H},\mathbf{H}']]]+\cdots \right\} $$

where $[\mathbf{A},\mathbf{B}] = \mathbf{A}\mathbf{B}-\mathbf{B}\mathbf{A}$ and

$$ \mathbf{H}'=\frac{d}{dt}\mathbf{H}. $$

So, the answer of the question is no.

Sakurai.JJ
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