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.


4 Answers 4


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


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