Hermitian Inverse: Exploring Eigenvalues of ##H^{-1}##

[ergospherical]

$H$ is an $n\times n$ Hermitian matrix with eigenvectors $\mathbf{e}_i$ and all eigenvalues negative. It's claimed that $G = \int_{0}^{\infty} e^{tH} dt$ is such that $G = H^{-1}$. I was looking at\begin{align*}
G\mathbf{e}_i &= \int_0^{\infty} \sum_{n=1}^{\infty} \frac{t^n}{n!} H^n \mathbf{e}_i dt = \mathbf{e}_i\int_0^{\infty} \sum_{n=1}^{\infty} \frac{t^n \lambda_i^n}{n!} dt = \mathbf{e}_i \int_0^{\infty} e^{\lambda_i t} dt = - \frac{1}{\lambda_i} \mathbf{e}_i
\end{align*}which is weird, because $\mathbf{e}_i = H^{-1} H\mathbf{e}_i = \lambda_i H^{-1} \mathbf{e}_i$ so $H^{-1}$ should have eigenvalues $1/\lambda_i$?

 

[PeroK]

If you try $H = -I$ you'll see you are correct. $G = -H^{-1}$.

 

[ergospherical]

Thanks, I'm surprised. It's part of an old exam question!