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:


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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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