Inverse of a quasipositive matrix with negative spectral bound

Question

A square matrix is quasipositive if all off-diagonal elements are nonnegative. The spectral bound of a square matrix is defined as $$s(A) = \max\{\Re (\lambda) : \lambda \mbox{ is an eigenvalue of } A\}.$$ I would like to show that if $s(A) < 0$ and if $A$ is quasipositive then $$A^{-1} = -\int_{0}^{\infty} e^{At}dt.$$ Well, since $s(A) < 0$, then $0$ is not an eigenvalue of $A$, so $A$ is invertible. I know that for a quasipositive matrix $A$ the spectral bound $s(A)$ is an eigenvalue associated with a nonnegative eigenvector (i.e. all coordinates are nonnegative) $v$ of $A$ and a nonnegative eigenvector $u$ of the transposed matrix $A^{T}$. Also, I know that $$e^{At} = \sum_{n=0}^{\infty} \frac{1}{n!}(At)^{n},$$ but I don't know how to relate these concepts. Any answers or hints will be really appreciated.

Answer 1

Hint: We can apply the fundamental theorem of calculus for any function of one variable. In particular, we have $$ A \left(-\int_{0}^{\infty} e^{At}dt\right) = -\int_0^\infty Ae^{At} = -\int_0^\infty \left(\left.\frac{d}{d\tau} e^{A\tau}\right|_{\tau = t}\right)\,dt = -\left. e^{At}\right|_{t=0}^\infty $$