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So I'm trying to find the eigenvalues and eigenvectors of the backwards shift operator $T:F^\infty \to F^\infty$ defined as $Tv=T(z_1,z_2,....)=(z_2,z_3,....)$.

Arriving at the system of equations from $Tv = \lambda v$: $$z_2 = \lambda z_1$$ $$z_3 = \lambda z_2$$ $$\vdots$$ $$z_{n+1} = \lambda z_n$$ $$\vdots$$

(1) If we fix $z_1$ we arrive at the set of eigenvectors, $$ \{ (z_1,\lambda z_1, \lambda^2 z_1, ....) : \lambda \epsilon \mathbb{F} \}$$

(2) If $\lambda = 1$, then our eigenvectors are then $$ \{ (z,z,...) : z \epsilon \mathbb{F} \}$$

(3) If $\lambda = 0$, then clearly $(z,0,0,...)$ describes the eigenvectors.

If $\lambda \ne 0$, then we have

$$z_1 = \frac{z_2}{\lambda}$$ $$z_2 = \frac{z_3}{\lambda}$$ $$\vdots$$ $$z_{n} = \frac{z_{n+1}}{\lambda}$$ $$\vdots$$

But this system of equations also implies that the 1st element $z_1$ is a function of $z_2$, which is a function of $z_3$, and so on. So they are all "chained together" to infinity. That is to say,
$$ z_1 = \frac{1}{\lambda} \frac{z_3}{\lambda} $$
$$ = \frac{1}{\lambda^2} \frac{z_4}{\lambda} $$ $$ \vdots $$
$$ = \lim_{n \to \infty} \frac{z_n} {\lambda^{n-1}} $$

And we can repeat the argument for any $z_i$ since the vector has infinite length. Now if $\lambda = 1$, we have the previous case (set 2). If $\lambda = -1$, then then the limit is not necessarily convergent, e.g. $z_n = (-1)^n$ could be a constraint, but $z_n=1$, $n=1,2...$ would not give a convergent limit. But if $|\lambda| \lt 1$, then the limit implies $z_1 \to \infty$, ... ,$z_n \to \infty$. Similarly if $|\lambda| \gt 1$, then we must get $z_1 \to 0,z_2 \to 0,...)$.

Doesn't this contradict the result that set 1 describes eigenvalues of T? In comparing to similar posts such as this thread, I see that set 1 should be the correct answer, but I'm not sure where my logic has lead me astray to believe otherwise. I suspect that " we can repeat the argument for any $z_i$" might not be true, but I'm not sure why. Any ideas?

chris fritz
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    What is the definition of $F^{\infty}$? – Kavi Rama Murthy Nov 23 '18 at 07:21
  • "if $\lambda = -1$, then the limit is not defined" - how do you conclude this? What if $z_n = (-1)^n$? –  Nov 23 '18 at 07:27
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    Your limit argument in general seems to be assuming that $z_n$ is constant as $n \to \infty$. I'm not sure why you need a limit argument at all. Plainly for any nonzero $z$, the definition shows that $(z, \lambda z, \lambda^2 z, \lambda^3 z , \ldots)$ works as an eigenvector corresponding to the eigenvalue $\lambda$. This is true for any real or even complex $\lambda$, or indeed in any field. –  Nov 23 '18 at 07:31
  • I'm using the limit argument because the system of equations imply that each entry depends on the next (i.e $z_n = f(z_{n+1}))$. Since the list has infinite length, this should mean that the limiting behavior of the entries is relevant to at least the first entry. The only instance I could think of where an entry doesn't depend on the next ad infinitum is if the next entry is zero, but this would imply all entries are zero (for nonzero $\lambda$). – chris fritz Nov 24 '18 at 07:59
  • Still unclear why $(z,\lambda z, \lambda^2 z, \ldots)\in F^{\infty}$ does not (?) hold for any $z$ and $\lambda$. Just rephrasing things asked already: what is your $F^{\infty}$ ($\implies$ why does the limit behavior bother you)? – metamorphy Dec 01 '18 at 23:00

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