It's easy for scalars, $(\exp(a(x)))' = a' e^a$. But can anything be said about matrices? Do $A(x)$ and $A'(x)$ commute such that $(\exp(A(x)))' = A' e^A = e^A A'$ or is this only a special case?

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    $\begingroup$ Have a look at this wikipedia article: the exponential map, although I don't know if this article directly addresses the commutative part of your question. $\endgroup$ Aug 10, 2010 at 12:45
  • $\begingroup$ @Tom Stephens: That is the answer already, thanks. Please feel free to add it as such (maybe quoting $\frac{d}{dt}e^{X(t)} = \int_0^1 e^{\alpha X(t)} \frac{dX(t)}{dt} e^{(1-\alpha) X(t)}\,d\alpha$) $\endgroup$ Aug 10, 2010 at 12:48
  • $\begingroup$ I don't have anything intelligent to say about this. I would like to see someone else post an answer who is more familiar with these operations. In particular, I would like to see the issue of commutativity addressed since it is not at all obvious to me. $\endgroup$ Aug 10, 2010 at 12:53
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    $\begingroup$ I've no access to dx.doi.org/10.1137/S0895479895283409 , but it might have what you need I suppose. $\endgroup$ Aug 10, 2010 at 23:07
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    $\begingroup$ The reason for specializing to (block) triangular IIRC is that one does a preliminary Schur decomposition A=QTQ* of the input matrix first, evaluates the function at the resulting (block) triangular matrix T, and then multiplies back the unitary matrices Q. Now that I think about it, maybe you should take a look at maths.manchester.ac.uk/~higham/fm/index.php also. $\endgroup$ Aug 11, 2010 at 11:51

3 Answers 3


Tom Stephens linked to the exponential map, which states that

$ \frac{d}{dt}e^{X(t)} = \int\limits_0^1 e^{\alpha X(t)} \frac{dX(t)}{dt} e^{(1-\alpha) X(t)} d\alpha $

If $X(t)$ and $\frac{d}{dt}X(t)$ commute, the latter also commutes with $\exp(X(t))$ and then it simplifies into $ \frac{d}{dt}e^{X(t)} = \frac{d X(t)}{dt} e^{X(t)}$.

A counter-example is $$X(t) = \begin{pmatrix} \cos(t) & \sin(t) \\\ \sin(t) & -\cos(t) \end{pmatrix}$$ at $t=0$, i.e. $X(0) = \sigma_3, \dot X(0) = \sigma_1$ ($\sigma_i$ are the non-commuting Pauli Matrices)


The question comes down to computing what's called the "derivative of", the "differential of", or the "tangent map to", the exponential map from $M_n(\mathbb R)$ into itself at a given matrix $A$ (not necessarily the zero matrix). There is a classical formula for this. Here is the first reference I found: pages 1 and 2 of


by Peter Woit. Here it is (with Peter Woit's notation) $$\exp_*(X)\ Y=\exp(X)\ \frac{1-e^{-ad(X)}}{ad(X)}\ Y.$$ Here is a reference for the Chain Rule:


It reads, in Peter Woit's notation and under appropriate assumptions,

$$(f\circ g)\_*(x)=f_*(g(x))\circ g_*(x).$$

[Thank you to KennyTM for having edited this formula.]

EDIT 1. Here are the two formulas written in another notation:

$$\exp'(X)=\exp(X)\ \frac{1-e^{-ad(X)}}{ad(X)}\quad,$$

$$(f\circ g)'(x)=f'(g(x))\circ g'(x).$$

EDIT 2. Here is another reference. This is a post by Akhil Mathew:


  • $\begingroup$ Does $ad(X)$ mean $ad(X) Y = [X,Y], ad(X)^2 Y = [X, [X,Y]]$ etc.? In that case I'm missing a Matrix it's acting on, and also what $ad(X)^{-1}$ would mean. $\endgroup$ Aug 11, 2010 at 11:19
  • $\begingroup$ Yes, ad(X) means what you said. If T is an indeterminate, then the expression (1-exp(-T))/T defines a formal power series with infinite radius of convergence. Hence, it can be evaluated on any endomorphism of any finite dimensional complex vector space. $\endgroup$ Aug 11, 2010 at 12:56
  • $\begingroup$ Are the $a_*$ and $a'$ equivalent in this answer? $\endgroup$
    – kennytm
    Aug 11, 2010 at 13:27
  • $\begingroup$ Dear KennyTM: Yes, you're perfectly right! Thanks and congratulation for having solved my LaTeX problem. You're a wizard! $\endgroup$ Aug 11, 2010 at 13:44

Only I khnow is that if your matrix $A$ and $A\in GL(n,\mathbb{R})$ is invertible we can say that $A'(X)=-AXA^{-1}$. So $(e^{A(x)})' =A(x)'e^ {A(x)} =-AXA^{-1}e^{A(x)}$ , since $A$ is not commutative you can't say $A' e^{A(x)} =e^{A(x)}A'$


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