8
$\begingroup$

On the standard space $l^2$ let us consider the left shift operator $$ L(c_1,c_2,c_3,\ldots)=(c_2,c_3,c_4,\ldots). $$ It is well known that the spectrum of $L$ is the whole unit disk in the complex plane. I would like to approximate $L$ by some sequence of finite-dimensional operators $L_n$. A naive way to do this is to set $L_n$ as follows $$ L_n=\left(\begin{array}{ccccc} 0 & 1 & 0 & \ldots & 0 \\ 0 & 0 & 1 & \ldots & 0 \\ 0 & 0 & 0 & \ddots & 0 \\ 0 & 0 & 0 & \ldots & 1 \\ 0 & 0 & 0 & \ldots & 0 \end{array}\right) $$

However the spectrum of $L_n$ consists only of $0$. Could one suggest more reasonable finite-dimensional approximation sequence $L_n$ such that spectrum of operator $L_n$ gradually fills the unit disk? References are welcome.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ What kind of approximation do you have in mind? Certainly it is not possible in the norm topology. $\endgroup$
    – Tomasz Kania
    Nov 19, 2016 at 14:14
  • $\begingroup$ Google Berg's method. It's used for approximating the generators of irrational rotation algebras, one of which can be the shift. $\endgroup$
    – David Handelman
    Nov 19, 2016 at 14:14
  • $\begingroup$ I am not sure what topology is suitable for this task, but would like to understand how to find finite-dimensional approximations which do not ignore continuous spectrum. $\endgroup$
    – Anton
    Nov 19, 2016 at 20:11
  • 1
    $\begingroup$ Every $|z|<1$ is an eigenvalue with eigenvector $z^n$, so you could just take finitely many of these, truncate them, and make them eigenvectors of a finite-dimensional approximation. It's not so obvious though (to me) if these "approximations" still converge in the strong operator topology; perhaps this will depend on a suitable choice of the eigenvalues. $\endgroup$
    – Christian Remling
    Nov 20, 2016 at 20:21
  • $\begingroup$ Could you explain why you have accepted an answer which does not provide a "good" approximation of the shift? More generally, perhaps you should look at work of Steffen Roch and his coauthors which has a systematic look at "finite section methods" for approximating operators $\endgroup$
    – Yemon Choi
    Nov 27, 2016 at 14:08

3 Answers 3

Reset to default
6
$\begingroup$

If you replace it with the cyclic shift operator, you get a circulant matrix (the same as your $L_n$ except that the bottom-left entry is $1$). The eigenvalues of that matrix are the $n$th roots of unity. So as $n$ grows, the spectrum fills the unit circle (it does not fill the unit disk, though).

Your $L_n$ is a highly non-normal matrix; the circulant version is normal. If you want to understand this better, read Chapter 7 of Trefethen & Embree's Spectra and Pseudospectra, which deals specifically with your example.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Although this example has been accepted: I feel I should point out that (as David remarks) there are senses in which this is NOT a good approximation of the unilateral backward shift. This approximation does not "see" most of the spectrum of that operator! $\endgroup$
    – Yemon Choi
    Nov 27, 2016 at 14:07
  • 1
    $\begingroup$ Agreed; I was a bit surprised to see it accepted. It may be an acceptable substitute in some situations. $\endgroup$
    – David Ketcheson
    Nov 27, 2016 at 14:39
3
$\begingroup$

I think that the numerical range is an appropriate tool for your question. Your naive approximations $L_n$ of the shift operator are nilpotent. For such matrices $M$ (nilpotent of size $n$), the numerical range ${\cal H}(M)$ is a disk $D(0;r_n)$ with radius $$r_n=\|M\|\cos\frac\pi{n+1}\,$$ where $\|M\|$ is the standard operator norm. In your situation, $\|L_n\|=1$, so that $$r_n=\cos\frac\pi{n+1}\rightarrow1^-.$$ I suspect that for reasonnable operators $L$, the finite dimensional approximations $L_n=P_n^*LP_n$ ($P_n$ the orthogonal projection on an increasing sequence of subspaces) has the property that the union ${\cal H}(L_n)$, which is a non-decreasing sequence for inclusion, contains the spectrum of $L$, exactly as ${\cal H}(M)$ contains the spectrum of $M$. This would be true if $L\mapsto {\cal H}(L)$ is lcs for a rather weak topology on operators.

Improve this answer
$\endgroup$
1
$\begingroup$

You may probably benefit from the analysis provided in this paper which is intimately related to your question and similar ones.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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