\(\DeclarePairedDelimiterX{\Set}[2]{\{}{\}}{#1 \nonscript\;\delimsize\vert\nonscript\; #2}\) \( \DeclarePairedDelimiter{\set}{\{}{\}}\) \( \DeclarePairedDelimiter{\parens}{\left(}{\right)}\) \(\DeclarePairedDelimiterX{\innerproduct}[1]{\langle}{\rangle}{#1}\) \(\newcommand{\ip}[1]{\innerproduct{#1}}\) \(\newcommand{\bmat}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\barray}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\mat}[1]{\begin{matrix}#1\end{matrix}}\) \(\newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}}\) \(\newcommand{\mathword}[1]{\mathop{\textup{#1}}}\)
Needs:
Circulant Matrices ›
Eigenvalue Decomposition ›
Needed by:
None.
Links:
Sheet PDF ›
Graph PDF ›

Circulant Matrix Eigendecompositions

Why

It happens that all circulant matrices have the same eigenvectors.

Definition

Recall that $C \in \R ^{n \times n}$ is circulant then

\[ C = c_0I + c_1 S + c_2S^2 + \cdots + c_{n-1}S^{n-1}. \]

So $q \in \R ^n$ is an eigenvector of $C$ if and only if it is one of $S$.1

  1. Future editions will complete this development. ↩︎
 Copyright © 2023 The Bourbaki Authors — All rights reserved — Version 13a6779cc — About — Show the old page view