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

Let $A$ be a square matrix. Invertibility of $\exp(A)$ follows easily from properties of the matrix exponential.

Is $\int_0^t \exp(A u)du$ also invertible? I believe it should be, and that the inverse should be $I - At/2 + A^2t^2/12 + ...$

This comes from expanding the real-valued function $x/(e^x - 1)$ in a power series about $x=0$. How should I approach a proof of this (or could I find it in a book somewhere?)

What about the more general case when $\Phi(t)$ is defined by

$\frac{dX}{dt} = A(t)X,\ \ X_0 = x_0 $


$X(t) = \Phi(t)x_0$? How might one show that $\int_0^t \Phi(u)du$ is invertible?

share|cite|improve this question
up vote 4 down vote accepted

Try $A = \pmatrix{0 & 1\cr -1 & 0\cr}$ for which $\exp(Au) = \pmatrix{\cos(u) & \sin(u)\cr -\sin(u) & \cos(u)\cr}$. Then $\int_0^{2\pi} \exp(Au)\ du = 0$.

More generally, if $\lambda \ne 0$ is an eigenvalue of $A$ for eigenvector $v$, then $v$ is an eigenvector of $\int_0^t \exp(Au)\ du$ with eigenvalue $(1 - e^{t\lambda})/\lambda$, so if $e^{t\lambda} = 1$, $\int_0^t \exp(Au)\ du$ is not invertible.

On the other hand, if $e^{t\lambda} \ne 1$ for all nonzero eigenvalues $\lambda$ of $A$, then $\int_0^t \exp(Au)\ du$ is invertible.

For your more general case, if $A(t)$ don't all commute, closed-form solutions may not be available, and I think it becomes harder to determine when the integral is invertible.

share|cite|improve this answer
What a great answer, thanks! – Simon May 11 '12 at 13:48

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.