Consider the right shift operator on $\ell^2(\mathbb{Z})$. Is there a way of calculating (well, showing what it is since I already know it's $z$ s.t $|z| = 1$) its spectrum without reference to it being unitary and with just basic linear operator and spectral theory? How about if I assume that it exists and use the vector with zero everywhere except the 0th position, where it is 1? (If you don't understand that, ignore it)
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4You should use the search function: http://math.stackexchange.com/questions/29219/spectrum-of-a-linear-operator http://math.stackexchange.com/questions/32417/spectrum-of-right-shift-operator – Plop May 12 '11 at 20:21
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related: http://math.stackexchange.com/q/617601/173147 – glS Jun 08 '15 at 07:10
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If $\delta$ is that vector you mentioned and $S$ is your shift, then for any $z$ with $|z| = 1$ and positive integer $n$, let $v = \sum_{j=0}^n z^j S^{-j} \delta$. Compare $\|S v - z v\|$ to $\|v\|$ to see that $z$ is in the spectrum. On the other hand, if $|z| > 1$ or $|z| < 1$ you can construct $(S - z I)^{-1}$ using geometric series (different ones in those two cases).
Robert Israel
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OK, I guess this requires you to know that the approximate spectrum is set of values where that expression is not bounded below. If I had only the basic defn. of spectrum (i.e. $S-zI$ has no inverse), would a valid tactic be: suppose that $(S-zI)^{-1}$ exists, which means for some $x$, $(S-zI)x = \delta$ (as defined by your answer). Then this would imply that $x$ is the vector with elements say $c/z^i$ going one way and $cz^i$ going the other way, which is not in $\ell^2$ when $|z| = 1$.
I guess this must related to your suggest method above, but for I can't see it as a beginner.
– A.A May 13 '11 at 16:27 -
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1@Robert Is it possible to, instead of constructing the inverse, show that there is a vector in $\ell^2$ that is orthogonal to the range of $S - zI$? If so, how would you do this? – Feb 24 '17 at 05:53