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)
spectrum of right shift operator on $\ell^2(\mathbb{Z})$
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functional-analysis
operator-theory
spectral-theory
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0related: http://math.stackexchange.com/q/617601/173147 – 2015-06-08
1 Answers
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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).
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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? – 2017-02-24