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.

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    What kind of approximation do you have in mind? Certainly it is not possible in the norm topology. – Tomek Kania Nov 19 '16 at 14:14
  • Google Berg's method. It's used for approximating the generators of irrational rotation algebras, one of which can be the shift. – David Handelman Nov 19 '16 at 14:14
  • 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. – Anton Nov 19 '16 at 20:11
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    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. – Christian Remling Nov 20 '16 at 20:21
  • 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 – Yemon Choi Nov 27 '16 at 14:08

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.

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    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! – Yemon Choi Nov 27 '16 at 14:07
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    Agreed; I was a bit surprised to see it accepted. It may be an acceptable substitute in some situations. – David Ketcheson Nov 27 '16 at 14:39

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.

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

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