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?

  • 4
    $\begingroup$ 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. $\endgroup$ – Tom Stephens Aug 10 '10 at 12:45
  • $\begingroup$ @Tom Stephens: That is the answer already, thanks. Please feel free to add it as such (maybe quoting $\frac{d}{dt}e^{X(t)} = \int_0^1 e^{\alpha X(t)} \frac{dX(t)}{dt} e^{(1-\alpha) X(t)}\,d\alpha$) $\endgroup$ – Tobias Kienzler Aug 10 '10 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. $\endgroup$ – Tom Stephens Aug 10 '10 at 12:53
  • 1
    $\begingroup$ I've no access to dx.doi.org/10.1137/S0895479895283409 , but it might have what you need I suppose. $\endgroup$ – J. M. isn't a mathematician Aug 10 '10 at 23:07
  • 1
    $\begingroup$ 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. $\endgroup$ – J. M. isn't a mathematician Aug 11 '10 at 11:51

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)

| cite | improve this answer | |

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:


| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Tobias Kienzler Aug 11 '10 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. $\endgroup$ – Pierre-Yves Gaillard Aug 11 '10 at 12:56
  • $\begingroup$ Are the $a_*$ and $a'$ equivalent in this answer? $\endgroup$ – kennytm Aug 11 '10 at 13:27
  • $\begingroup$ Dear KennyTM: Yes, you're perfectly right! Thanks and congratulation for having solved my LaTeX problem. You're a wizard! $\endgroup$ – Pierre-Yves Gaillard Aug 11 '10 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 $A' e^{A(x)} =e^{A(x)}A'$

| cite | improve this answer | |

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.