I'm working on the following problem:
"If $Ax = \lambda x$, find an eigenvalue and an eigenvector of $e^{At}$ and also of $-e^{-At}$."
So far, I have figured that $e^{\lambda t}$ will be an eigenvalue of $e^{At}$, and that $A$ has the same eigenvector matrix as $e^{At}$, so $x$ will be an eigenvector for $e^{At}$.
However, I am a bit stuck on the $-e^{-At}$ part. I know that $e^{-At}$ is the inverse of $e^{At}$, so I think that means $\frac{1}{e^{\lambda t}}$ would be an eigenvalue for simply $e^{-At}$... but how do I find the eigenvalue for $-e^{-At}$? And how would I find an eigenvector?
Thanks!
The simplest approach is to relate the equation $Ax = \lambda x$ to $(-A)y = \mu y$.