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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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    related: http://math.stackexchange.com/q/617601/1731472015-06-08

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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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    @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