Each row of $C$ is the same sequence permuted. Specifically, circular shifted in discrete time. Such a matrix is called circulant. In DSP circular convolution of a finite discrete time input signals ${x_i }$ with a system with response ${c_i }$ is represented as matrix multiplication
\[C x = \begin{bmatrix} c_0 & c_{n-1} & \cdots & c_2 & c_{1}\\ c_1 & c_0 & \cdots & c_3 & c_{2}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ c_{n-2} & c_{n-3} & \ddots & c_0 & c_{n-1}\\ c_{n-1} & c_{n-2} & \cdots & c_{1} & c_0\\ \end{bmatrix} \begin{bmatrix} x_0\\ x_1\\ x_2\\ \vdots\\ x_{n-2}\\ x_{n-1} \end{bmatrix}\\\]When writing out the multiplication the convolution becomes clear \(y_k = \sum^{n-1}_{i=0} c_{[(k-i) \mod n]}x_{i}\) For example,
\[\begin{align} y_1 &= \sum^{n-1}_{i=0} c_{[(1-i) \mod n]}x_{i} \\ &= c_{[1 \mod n]}x_0 + c_{[0 \mod n]}x_1 + c_{[-1 \mod n]}x_2 + c_{[-2 \mod n]}x_3 + \cdots + c_{[2-n \mod n]}x_{n-1}\\ &= c_1x_0 + c_0x_1 + c_{n-1}x_2 + c_{n-2}x_2 +\cdots + c_{2}x_{n-1} \\ \end{align}\]Since each row of $C$ is a permutation of ${c_0, c_1, \cdots, c_{n-1}}$, $C$ can be written as linear combination of permutation matrices.
\[C = c_0I + c_1P + c_2 P^2 +\cdots + c_{n-1}P^{n-1}\]Where $P$ is the permutation $\big((n-1),(n-2), \cdots,1,0\big)$.
\[P = \begin{bmatrix} e_1 & e_2 & \cdots & e_{n-1} & e_0\\ \end{bmatrix} = \begin{bmatrix} 0 & \cdots & \cdots & 0 & 1\\ 1 & 0 & & \huge0 & 0\\ & 1 & \ddots & & \vdots\\ & \huge0 & \ddots &\ddots & \vdots \\ & & & 1 & 0\\ \end{bmatrix}\]From the decomposition of $C$ into a linear composition, we see that their is a 1-to-1 relationship between eigenvalues and eigenvectors of $P$ and $C$.
Lemma
The spectrum of $C$ is the same as permutation $P$. Moreover, they are simultaneously diagonalizable.
proof
First note that $CP = PC$, which follows from $C$s representation as a linear combination of permutation matrices.
\[\begin{align} CP &= c_0IP + c_1P^2 + c_2 P^2 +\cdots + c_{n-1}P^{n} \\ &= P\big(c_0I + c_1P + c_2 P^2 +\cdots + c_{n-1}P^{n-1}\big) \\ &= PC \\ \end{align}\]Suppose $Pv = \lambda v$, then
\[PCv = CPv = C\lambda v \quad \Longrightarrow \quad P\underbrace{Cv}{} = \lambda \underbrace{Cv}{}.\]Which means that $Cv$ and $v$ are both eigenvectors of $P$ corresponding to the same eigenvalue, $\lambda$. In the next section we learn that $P$ has $n$ distinct eigenvalues, implying that $Cv$ is a multiple of $v$. Specifically, there is $\exists \alpha$ s.t. $Cv = \alpha v$. Therefore,
To compute the eigenvalues of $P$, solve $\det(P-\lambda I)=0$.
\[\require{color} \begin{align} \det(P - \lambda I) &= \det\begin{bmatrix} -\lambda & \cdots & \cdots & 0 & 1\\ 1 & -\lambda & & \huge0 & 0\\ & 1 & \ddots & & \vdots\\ & \huge0 & \ddots &\ddots & \vdots \\ & & & 1 & -\lambda\\ \end{bmatrix}\\ &= -\lambda \det\big(P_{0,0}\big) + (-1)^{n-1}\det\big(P_{0,n-1}\big)\\ &= -\lambda \det\begin{bmatrix} -\lambda & & & \huge0 \\ 1 & -\lambda & & \\ & \ddots &\ddots & \\ \huge0 & & 1 & -\lambda\\ \end{bmatrix} + (-1)^{n-1} \det\begin{bmatrix} 1 & -\lambda & & \huge0 \\ & 1 & \ddots & \\ & & \ddots & -\lambda \\ \huge0 & & & 1 \\ \end{bmatrix}\\ \end{align}\]The sub-matrices $P_{0,0}$ and $P_{i,n-1}$ are lower triangular and upper triangular, respectively; so their determinants are easy to compute–just multiply the diagonal elements. Continuing the calculation,
\(\begin{gather} 0 = \det(P - \lambda I) = (-1)^{n}\lambda^n + (-1)^{n-1}\\ \end{gather}\) Therefore, any eigenvalue of $P$ satisfies $1 - \lambda^n$= 0, which means the eigenvalues are the roots of unity:
\[1, \ \omega, \ \omega^2,\ \cdots,\ \omega^{n-1} \text{ where } \omega = e^{i\frac{2\pi}{n}}.\]The associated eigenvectors can be solved by determining the null space of $P - \lambda I$. \(\begin{align} (P - \lambda I) v &= \begin{bmatrix} -\lambda & & & && 1\\ 1 & -\lambda & & \Huge0 & & \\ & 1 & -\lambda & & & \\ & & \ddots & \ddots& & \\ & \Huge0 & & \ddots & \ddots & \\ & & & & 1 & -\lambda \\ \end{bmatrix} \begin{bmatrix} \lambda^{n-1} \\ \lambda^{n-2} \\ \vdots \\ \lambda^2 \\ \lambda \\ 1 \\ \end{bmatrix} &= \begin{bmatrix} 1 - \cancelto{1}{\lambda^{n}} \\ \lambda^{n-1} - \lambda^{n-1} \\ \vdots \\ \lambda^{3} - \lambda^{3} \\ \lambda^{2} - \lambda^{2} \\ \lambda - \lambda \\ \end{bmatrix}\\ &= \vec{0} \end{align}\)
Finally, each eigenvalue, $e^{i\frac{2\pi k}{n}}$, corresponds to a specific eigenvector.
\[e^{i\frac{2\pi k}{n}} \ \Longleftrightarrow \ \begin{bmatrix} e^{i\frac{2\pi k(n-1)}{n}} \\ e^{i\frac{2\pi k(n-2)}{n}} \\ \vdots \\ e^{i\frac{2\pi k(2)}{n}} \\ e^{i\frac{2\pi k}{n}} \\ 1 \\ \end{bmatrix}\]Because $C$ can be written as linear combination of permutation matrices, its eigenvalues are of the form
\[\begin{equation} \lambda_C = c_0 + c_1 \lambda_P + c_2 \lambda_P^2 +\cdots + c_{n-1}\lambda_P^{n-1} \\ \end{equation}\]where $\lambda_P = e^{i\frac{2\pi k}{n}}$ is an eigenvalue of $P$. Substituting further,
\[\begin{equation} \lambda_C = c_0 + c_1 e^{i\frac{2\pi k}{n}} + c_2 e^{i\frac{4\pi k}{n}} +\cdots + c_{n-1}e^{i\frac{2\pi k(n-1)}{n}} \\ \end{equation}\]