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I am calculating the eigenvalues and eigenvectors of a permutation matrix: $$ \mathbf{C}=\left[\begin{array}{ccccc} 0 & 0 & \cdots & 0 & 1 \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \ddots & 0 & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \end{array}\right]_{N \times N} $$

I noticed that $C^N = I$, where $I$ is an identity matrix.

Supposing the eigenvalues of $C$ is $\mathbf{\lambda}$, I got $\lambda_k = e^{-jk\frac{2\pi}{N}}$.

Now I wonder how to calculate the eigenvectors of this matrix.

Thank you for your helping~

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1 Answer 1

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Outline

Simply use the definition of eigenvectors. Suppose $v=\begin{bmatrix}v_1\\v_2\\\vdots\\v_N\end{bmatrix}$ is the eigenvector of this matrix for the eigenvalue $\lambda_k=\exp(-jk2\pi/N)$. Hence $$ Cv=\lambda_kv\implies \begin{bmatrix}v_N\\v_1\\v_2\\\vdots\\v_{N-1}\end{bmatrix} =\begin{bmatrix}\lambda_kv_1\\\lambda_kv_2\\\lambda_kv_3\\\vdots\\\lambda_kv_{N}\end{bmatrix} $$ and solve by recursive substitution.

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  • $\begingroup$ Thank you! I actually forgot the definition. $\endgroup$
    – Lemon
    Apr 22, 2021 at 8:13
  • $\begingroup$ You're welcome. Good luck! $\endgroup$ Apr 22, 2021 at 8:16

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