# Closed form expression for matrix exponential derivative with respect to scalars

I'd like to evaluate generic expressions of the following form:

$$\frac{d}{da}\exp\left[aX + bY\right]$$

where $$a,b$$ are scalars and $$X,Y$$ are arbitrary complex matrices. Replacing the exponential with its defining series gives us an infinite series of terms which are each symmetrized by the derivative:

$$\sum_{n=0}^{\infty}\frac{1}{n!}\frac{d}{da}\left(aX+bY\right)^{n}$$

For say $$n=3$$ this gives a term of the form

$$X(aX+bY)^{2}+(aX+bY)X(aX+bY)+(aX+bY)^{2}X$$

I don't see a clean way to resum such a series into a nice analytic form, but I thought maybe there was a clean result for this up to commutators. Any insights or literature suggestions are very welcome.

In general, the derivative of the exponential map is given by $$\frac{d}{dt}e^{Z(t)} = e^{Z}\frac{1 - e^{-\mathrm{ad}_{Z}}}{\mathrm{ad}_{Z}}\frac{dZ(t)}{dt}$$ Thus, for your case of $$Z(a) = aX + bY$$, we have \begin{align*} \frac{d}{da}e^{Z} &= e^{Z}\frac{1 - e^{-\mathrm{ad}_{Z}}}{\mathrm{ad}_{Z}}(X) \\ & = e^Z \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!} \mathrm{ad}_Z^k(X) \end{align*} Where $$\mathrm{ad}_Z^k(X)$$ denotes $$[\overbrace{Z,[Z,\cdots,[Z}^{k \text{ times}},X]\cdots]]$$.

From the other answer based on Hall's text, we also have $$\left. \frac{d}{da} e^Z \right|_{a = 0} = e^{bY}\left\{ X - \frac 1{2!}b[Y,X] + \frac 1{3!}b^2[Y,[Y,X]] - \cdots \right\}$$

The number $$b$$ is really irrelevant to your question.

For all $$X,Y\in M_n(\mathbb{C})$$, we have

$$\frac{d}{dt}e^{X+tY}\big|_{t=0}=e^X\left\{ Y-\frac{[X,Y]}{2!}+\frac{[X,[X,Y]]}{3!}-\cdots \right\}.$$

More generally, if $$X(t)$$ is a smooth matrix-valued function, then $$\frac{d}{dt}e^{X(t)}=e^{X(t)}\left\{ \frac{I-e^{-\operatorname{ad}_{X(t)}}}{\operatorname{ad}_{X(t)}}\bigg(\frac{dX}{dt}\bigg) \right\}$$

See Theorem 5.4 (Derivative of Exponential) and its proof in Brian Hall's Lie Groups, Lie Algebras, and Representations.

If $$X$$ and $$Y$$ commute, then $$\frac{d}{da}e^{aX+bY}=\frac{d}{da}e^{aX}e^{bY}=Xe^{aX+bY}.$$

Let me add a different way of writing the result. If $$X(t)$$ is an operator-valued function, then $$\frac{\mathrm d}{\mathrm dt} \mathrm e^{X(t)} = \int_0^1 \mathrm e^{(1-\lambda) X(t)} \frac{\mathrm dX}{\mathrm dt} \mathrm e^{\lambda X(t)}\; \mathrm d\lambda . \tag{*}$$

Proof. We begin with the identity $$\mathrm e^{-\Lambda X} \frac{\mathrm d}{\mathrm dt} \mathrm e^{\Lambda X} = \int_0^\Lambda \mathrm e^{-\lambda X} \frac{\mathrm dX}{\mathrm dt} \mathrm e^{\lambda X}\; \mathrm d\lambda .$$ These two expressions are equal because they satisfy the same initial value problem as functions of $$\Lambda$$. Setting $$\Lambda = 1$$ gives the desired result (*).

Applied to your example: $$\frac{\mathrm d}{\mathrm da} \mathrm e^{aX + bY} = \int_0^1 \mathrm e^{(1-\lambda)(aX+bY)}\, X\, \mathrm e^{\lambda(aX+bY)}\; \mathrm d\lambda .$$