Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The matrix exponential is a well know thing but when I see online it is provided for matrices. Does it the same expansion for a linear operator? That is if $A$ is a linear operator then $$e^A=I+A+\frac{1}{2}A^2+\cdots+\frac{1}{k!}A^k+\cdots$$

share|cite|improve this question
What is your native language? As it stands, it is not very clear what is being asked. Maybe you can post in your native language and someone can translate. – Pedro Tamaroff Sep 29 '12 at 22:01
@PeterTamaroff I would say that Medan is asking if the matrix definition of the exponential extends to linear maps, and the answer would be yes. – Fly by Night Sep 29 '12 at 22:05
@FlybyNight Oh, well. Good then. But this is still in risk of uncalled downvotes. – Pedro Tamaroff Sep 29 '12 at 22:07
sorry, I had to read it before posting. Fly by Night had a right guess. I wanted to make sure I can extend matrix exponential to the case of linear operators. – Medan Sep 30 '12 at 3:26
up vote 1 down vote accepted

Yes, you can define an exponential of any linear BOUNDED operator by this series. If the operator is unbounded then it is not always possible.

share|cite|improve this answer
Yes, this is important. My operator looks like $A:=\frac{\partial}{\partial x}+\frac{\partial^2}{\partial x^2}$. If I remember correctly it can be bounded or unbounded depending what is the space of functions. However, I am looking at the "splitting methods" for the odes and pde and in order to prove the splitting error they just do expansions for the exponential. – Medan Sep 30 '12 at 13:17
And what is your space? I am not sure about your note, differential operators tend to by unbounded... Anyway, you can not use the series, you will loose differentiability in the process. – kalvotom Sep 30 '12 at 16:14
Assume the space is such the linear operator $A$ is unbounded. And assume it is a Laplacian operator on the interval for $x$. Then, I can't represent $e^A$, however, I can discretize the operator and get the approximation matrix of it $A_h$(using three points, for example). Then I can do $e^{A_h}$. Is that correct? Sounds like cheating, just by doing a consistent discretization it avoids all unboundedness problems? – Medan Sep 30 '12 at 19:26
I do not think so. It depends on what you are trying to do. Even if you are on an interval, than there are many operators which act as a Laplacian but with a different domain. Their exponential is then different also. Your discretiazation might correspond to one of those operators, but I am not sure about that. On the other hand, I am positive that you can write down an explicit expression for the exponential of any of those operators. It will act as some integral operator. – kalvotom Sep 30 '12 at 20:41
Let me tell you where my question comes from. Consider the pde $$u_t=Au$$ with $A:=\frac{\partial}{\partial x},u(0)=u_0$. Then the solution can be written as $$u(t)=u_0e^{At}$$So this is where I get stuck, if $A$ is a matrix I can do the matrix expansion and get $u(t)$ as a series, otherwise I can't. I want to keep it as an operator and not specify how I would discretize that, but I run into trouble working with unbounded operators? – Medan Sep 30 '12 at 21:50

As you have suggested, if $A$ is a linear operator then:

$$\exp A = I + A + \frac{1}{2}A^2 + \cdots + \frac{1}{k!}A^k + \cdots \, . $$

These are very common in physics. Here is a link to a PDF file.

share|cite|improve this answer

The exponential series has a remarkably "ubiquitiuos" convergence. As soon as you have a $\mathbb Q$-algebra $M$ with a norm such that $||X Y||\le c\cdot ||X||\cdot ||Y||$ for some $c$, then $\exp(A)$ converges for all $A$ with respect to this norm. Hence if $M$ is complete, you indeed obtain an element of $M$. Moreover, if $AB=BA$ then $\exp(A+B)=\exp(A)\exp(B)$ holds.

There are even cases when the exponential series is useful even when division by $k!$ is undefined. One just has to be careful that $A$ must be nilpotent enough (i.e. $A^k=0$ for all $k$ for which divison by $k!$ is undefined)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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