Finite-dimensional approximations of the shift operator 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.  
 A: 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.
A: 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.
A: You may probably benefit from the analysis provided in this paper which is intimately related to your question and similar ones.
