# Matrix exponential and rank

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

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

and

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

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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.