Eigenvalues of circulant matrices

A circulant matrix is defined as
$$C=\begin{bmatrix}c_{0}&c_{n-1}&\dots &c_{2}&c_{1}\\c_{1}&c_{0}&c_{n-1}&&c_{2}\\\vdots &c_{1}&c_{0}&\ddots &\vdots \\c_{n-2}&&\ddots &\ddots &c_{n-1}\\c_{n-1}&c_{n-2}&\dots &c_{1}&c_{0}\\\end{bmatrix}$$
where $C_{j, k}=c_{j-k \mod n}$, the $k$-th eigenvalue $\lambda_k$ and eigenvector $x_k$ satisfy $C\cdot x_k=\lambda_k x_k$, or, equivalently, $n$ equations as:
$$\sum_{j=0}^{m-1}c_{m-j}x_j+\sum_{j=m}^{n-1}c_{n-j+m}x_j=\lambda_k x_m\quad m=0,1,\dots,n-1$$
with $c_n=c_0$, $x_m$ is the $m$-th compoent of an eigenvector $x_k$. Changing the dummy summing ($j\to m-j$ and $j\to n-j+m$) variables yields
$$\sum_{j=1}^{m}c_j x_{m-j} +\sum_{j=m+1}^{n}c_j x_{n+m-j}=\lambda_k x_m$$
with $m=0,1,2,\dots,n-1$. One can "guess" a solution that $x_j=\omega^j$, therefore the equation above turns into
$$\begin{align}&\sum_{j=1}^{m}c_j \omega^{m-j} +\sum_{j=m+1}^{n}c_j \omega^{n+m-j}=\lambda_k \omega^m\\ \leftrightarrow &\sum_{j=1}^{m}c_j \omega^{-j} +\omega^{n}\sum_{j=m+1}^{n}c_j \omega^{-j}=\lambda_k\end{align}$$
Let $\omega$ be one of the $n$-th square root of unity, i.e., $\omega = \exp(-2\pi\mathbb{i}/n)$,  then $\omega^{n}=1$; we have an eigenvalue
$$\lambda = \sum_{j=0}^{n-1}\omega^{-j} c_j$$
with corresponding eigenvector
$$x= (1, \exp(2\pi\mathbb{i}/n), \exp(2\pi\mathbb{i}/n)^2,\dots,\exp(2\pi\mathbb{i}/n)^{n-1})^T$$
The $k$-th eigenvalue is generated from $\omega^{-k} = \exp(2\pi k\mathbb{i}/n)$ with $k\in [0,n-1]$, yields the $k$-th eigenvalue and eigenvector:
$$\lambda_k = \sum_{j=0}^{n-1}c_j \exp(2\pi k\mathbb{i}/n)$$
$$x_k = (1, \exp(2\pi k\mathbb{i}/n), \exp(2\pi k\mathbb{i}/n)^2,\dots,\exp(2\pi k\mathbb{i}/n)^{n-1})^T$$
The eigenspace is just the DFT matrix, and ALL circulant matrices share same eigenspace, it is easily to verify that circulant matrices have following properties:
If $A$ and $B$ are circulant matrices, then
  1. $AB=BA=W^\ast\Gamma W$ with $W$ is the DFT matrix and $\Gamma=\Gamma_A\Gamma_B$, $\Gamma_i$ represents diagonal matrix consists of eigenvalues of $i$; $\Gamma_A = WAW^\ast$;
  2. $B+A=A+B=W^\ast\Omega W$, $\Omega=\Gamma_A+\Gamma_B$;
  3. If $\mathrm{det}(A)\ne 0$, then $A^{-1}=W^\ast \Gamma_A^{-1}W$;
The proof is straightforward:
  1. $AB=W^\ast \Gamma_AW W^\ast\Gamma_BW= W^\ast \Gamma_A\Gamma_BW=W^\ast \Gamma_B\Gamma_AW=BA$
  2. $W(A+B)W^\ast=\Gamma_A+\Gamma_B$; 
  3. $AW^\ast \Gamma_A^{-1}W=W^\ast \Gamma_AW W^\ast \Gamma_A^{-1}W=\mathbb{I}$


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