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?



The simplest approach is to relate the equation $Ax = \lambda x$ to $(-A)y = \mu y$.

  • $\begingroup$ I don't really understand how to go from here. μ is an eigenvalue for -A, and y is an eigenvector... Now what? $\endgroup$
    – kmahar
    Apr 16 '14 at 20:55
  • $\begingroup$ @kmahar: You relate the two equations. And through that relationship, your knowledge of the full solution space to $Ax = \lambda x$ for $(\lambda, x)$ will be related to the full solution space to $(-A)y = \mu y$ for $(\mu, y)$. $\endgroup$
    – user14972
    Apr 17 '14 at 4:49

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