3
$\begingroup$

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)

$\endgroup$
4
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – A.A May 13 '11 at 16:27
  • $\begingroup$ I suppose not, since that conclusion can be drawn about any |z| = k. $\endgroup$ – A.A May 13 '11 at 16:36
  • 1
    $\begingroup$ @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? $\endgroup$ – user359396 Feb 24 '17 at 5:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.