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 the permutation $P$.
proof
If $Pv = \lambda v$, thento \(\begin{align} Cv &= c_0Iv + c_1Pv + c_2 P^2v +\cdots + c_{n-1}P^{n-1}v \\ &= c_0v + c_1\lambda v + c_2 \lambda^2 v +\cdots + c_{n-1}\lambda^{n-1} v \\ &= \big(c_0 + c_1\lambda + c_2 \lambda^2 +\cdots + c_{n-1}\lambda^{n-1}\big) v \\ \end{align}\) And the other way; note that both $C$ and $P$ simultaneously diagonalizable. Suppose $P = A^DA$, where $D$ is a diagonal matrix. \(\begin{align} C &= c_0I + c_1P + c_2 P^2 +\cdots + c_{n-1}P^{n-1} \\ &= c_0I + c_1(A^*DA) + c_2(A^*DA)(A^*DA) +\cdots + c_{n-1}(A^*DA)^{n-1} \\ &= c_0A^*IA + c_1A^*DA + c_2 A^*D^2A +\cdots + c_{n-1}A^*D^{n-1}A \\ &= A^*(c_0 I + c_1D + c_2 D^2 +\cdots + c_{n-1}D^{n-1})A \\ \end{align}\) Suppose $C = A^\Lambda A$, \(\begin{align} \Lambda &=ACA^* \\ &= A (c_0I + c_1P + c_2 P^2 +\cdots + c_{n-1}P^{n-1})A^* \\ &= c_0AA^* + c_1APA^* + c_2 AP^2A^* +\cdots + c_{n-1}AP^{n-1}A^* \\ &= c_0I + c_1(APA^*) + c_2(APA^*)(APA^*) +\cdots + c_{n-1}(APA^*)^{n-1} \\ &= c_0I + c_1 APA^* + c_2 (APA^*)^2 +\cdots + c_{n-1}(APA^*)^{n-1} \\ &= c_0 I + c_1D + c_2 D^2 +\cdots + c_{n-1}D^{n-1} \\ \end{align}\) $\square$
$Cw = \lambda w$, then it is in the span of the $n$ eigenvectors of $P$. \(\begin{align} \lambda w &= Cw \\ &= c_0w + c_1Pw + c_2 P^2w +\cdots + c_{n-1}P^{n-1}w \\ \end{align}\) Are $P^iw \perp P^j w $ for all $i \neq j$. $$
$$
\[\begin{align} (\lambda - c_0)w &= c_1w_1 + c_2 Pw_1 +\cdots + c_{n-1}P^{n-2}w_1 \\ \end{align}\]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}
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.
\[e^{i\frac{2\pi k}{n}} \ \Longleftrightarrow \ \begin{bmatrix} e^{i\frac{2\pi k(n-1)}{n}} \\ e^{i\frac{2\pi k(n-2)}{n}} \\ \vdots \\ e^{i\frac{2\pi k(2)}{n}} \\ e^{i\frac{2\pi k}{n}} \\ 1 \\ \end{bmatrix}\]