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Let us define the operator $T$ over $\ell_2$ $$T({x_1},{x_2},{x_3}, \ldots ) = ({x_1},\frac{{{x_2}}}{2},\frac{{{x_2}}}{3}, \ldots )$$

what is the point spectrum of this operator and the spectrum?

attempt:

what is puzzling me that when I try to get the point spectrum by setting $T(x)=\lambda x$, I end up getting no solution since $\lambda x_n=x_n/n$ for all $n$. if I solve using $x_1$, I get 1, with $x_2$ I get $1/2$ and so on. Should I deduce the spectrum is empty?

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    $\begingroup$ Similar to math.stackexchange.com/q/1355228 $\endgroup$
    – Jean Marie
    Jan 4, 2022 at 17:43
  • $\begingroup$ what is puzzling me that when I try to get the point spectrum by setting $T(x)=\lambda x$, I end up getting no solution since $\lambda x_n=x_n/n$ for all $n$ $\endgroup$
    – user1011087
    Jan 4, 2022 at 17:46
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    $\begingroup$ No contradiction in fact : take $(x_1,x_2, ... x_n, \cdots)$ of the form $(0,0, \cdots, 0, 1, 0, \cdots)$ with a single $1$ $\endgroup$
    – Jean Marie
    Jan 4, 2022 at 17:53

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For your additional question, consider the "canonical" base vectors of $\ell_2$. This gives you a lot of eigenvalues. Actually, your argument shows that there are no other eigenvalues. If you know a theorem about the spectrum of compact operators, you can now show that your operator is compact and are done. Otherwise, you have to calculate the resolvent.

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  • $\begingroup$ makes perfect sense, and so apparent, how could I miss that!! $\endgroup$
    – user1011087
    Jan 4, 2022 at 17:55
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    $\begingroup$ See for self adjointness here. $\endgroup$
    – Jean Marie
    Jan 4, 2022 at 17:55
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    $\begingroup$ Yes, there is a theorem which says that all nonzero spectral values are eigenvalues. Zero may or may not be an eigenvalue, but it belongs to the spectrum if the space has infinite dimension. I do not think that the theorem has a particular name. $\endgroup$
    – Martin Väth
    Jan 4, 2022 at 18:26
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    $\begingroup$ The theorem I mentioned does not necessarily hold for self-adjoint operators unless they are compact. And self-adjointness usually does not help in proving compactness. You can still go the alternative way and calculate the resolvent: If $\lambda\ne\frac1n$ and $\lambda\ne0$, consider $Ax-\lambda x=y$ coordinate-wise and show that it implies $x=R(\lambda)y$ for some bounded operator $R(\lambda)$ in $\ell_2$ (and that the corresponding $x$ satisfies the equation). $\endgroup$
    – Martin Väth
    Jan 4, 2022 at 18:35
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    $\begingroup$ By "$R(\lambda)$ is a bounded operator in $\ell_2$" I mean that the domain is $\ell_2$, that is, your calculation (hence definition of $R(\lambda)y$) works for every $y\in\ell_2$. Just try to calculate $R(\lambda)y$ by writing $Ax-\lambda x=y$ in coordinates, and you should be able understand what I mean. $\endgroup$
    – Martin Väth
    Jan 4, 2022 at 21:25

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