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David Ketcheson
David Ketcheson
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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.

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.

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.

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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David Ketcheson
David Ketcheson
  • 1.3k
  • 7
  • 16

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.

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.