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I want to find the eigenvalue of the differential operator $D(f)=f'=λf$. By solving the differential equation $f'=λf$ I get the eigenfunction ${e^{\lambda t}}$ which means $D(e^{\lambda t})=\lambda e^{\lambda t}=λf$.

What I understand by this is that ${e^{\lambda t}}$ is the eigenvector but what I'm not sure if the eigenvalue can be $\lambda$ itself.

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  • $\begingroup$ Basically all $\lambda \in \Bbb R$ (or $\Bbb C$) are eigenvalues. $\endgroup$ – user137731 Jul 10 '16 at 0:08
  • $\begingroup$ What you want to answer is: for what values of $\lambda$ is there an eigenfunction? Based on your computation, it's clear that there was no restriction you must place on $\lambda$ for there to be an eigenfunction and so all $\lambda\in\Bbb R$ (or $\lambda\in\Bbb C$) are eigenvalues. $\endgroup$ – Cameron Williams Jul 10 '16 at 0:09
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You should read it as "let $\lambda \in \mathbb{C}$, let us try to find an eigenfunction of $D$ with eigenvalue $\lambda$". Then you do some work and find $e^{\lambda t}$ is such an eigenfunction. So any $\lambda \in \mathbb{C}$ is an eigenvalue. (Switch $\mathbb{C}$ for $\mathbb{R}$ if you are just working over $\mathbb{R}$.)

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