Hermitian inverse

  • #1
ergospherical
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##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##?
 

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  • #2
PeroK
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If you try ##H = -I## you'll see you are correct. ##G = -H^{-1}##.
 
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  • #3
ergospherical
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Thanks, I'm surprised. It's part of an old exam question!
 

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