# spectrum of right shift operator on $\ell^2(\mathbb{Z})$

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)

## 1 Answer

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).

• 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
• I suppose not, since that conclusion can be drawn about any |z| = k. – A.A May 13 '11 at 16:36
• @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? – user359396 Feb 24 '17 at 5:53