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Let $T$ be the backward shift operator: $Tv = T(v_1,v_2,....) = (v_2,v_3,....)$. I would like to determine all the eigenvectors and eigenvalues. So far I have the following:

  1. It is evident that $(\alpha, 0,0,....)$ is an eigenvector for the eigenvalue $0$.

  2. It is also easy to see that $(\alpha, \alpha, \alpha, ...)$ is an eigenvector for the eigenvalue $1$.

  3. Finally I observed that if $\lambda$ is any scalar in the underlying field then $(1,\lambda, \lambda^2, \lambda^3,...) , (\lambda, \lambda^2, \lambda^3,...)$ etc. are all eigenvectors for the eigenvalue $\lambda$.

Now I need to either argue why these are all eigenvectors or find more. But although I think these are all I don't know how to prove it. How to proceed from here?

newb
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    Suppose $Tv = \lambda v$, that is $0 = Tv - \lambda v = (v_2 - \lambda v_1, , v_3 - \lambda v_2,, \dotsc)$. – Daniel Fischer Aug 14 '13 at 12:39
  • @DanielFischer Oh right. So there are three different eigenvalues each with one corresponding eigenvector? Sort of hard to believe given that the space has infinitely many dimensions. – newb Aug 14 '13 at 14:14
  • Duh. That was a stupid comment: there is in fact one eigenvector for every $\lambda \in \mathbf F$. Though, if $\mathbf F$ is a finite field there are still only finitely many eigenvectors while the whole space has infinite dimension. – newb Aug 14 '13 at 14:25
  • @DanielFischer Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Oct 05 '13 at 08:08

1 Answers1

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If $v$ satisfies

$$0 = Tv - \lambda v = (v_2 - \lambda v_1, v_3 - \lambda v_2, \dotsc, v_{n+1} - \lambda v_n, \dotsc),$$

then $v_{n+1} = \lambda v_n$ for all $n$, and hence $v_n = \lambda^{n-1}\cdot v_1$. Conversely, if we have $v_n = \lambda^{n-1}\cdot v_1$ for $n \geqslant 2$, then $Tv = \lambda v$, and either $v = 0$ (if and only if $v_1 = 0$) or $v$ is an eigenvector to the eigenvalue $\lambda$.

Therefore all eigenspaces are one-dimensional, a basis for $E_\lambda$ is $\{g^{(\lambda)}\}$ with $g^{(\lambda)}_n = \lambda^{n-1}$.

Daniel Fischer
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