Is this a circulant matrix?

Let $\mathbf A = \left\{a_{ij}\right\}_{i,j=0}^n \in \mathbb C^{(n+1) \times (n+1)}$ be defined as $$ a_{ij} = c_{i+j \operatorname{mod} (n+1)}, $$ so $$ \mathbf A = \begin{pmatrix} c_0 & c_1 & \dots & c_n\\ c_1 & c_2 & \dots & c_0\\ \vdots & \vdots & \ddots & \vdots\\ c_n & c_0 & \dots & c_{n-1} \end{pmatrix}. $$ Let's assume every index is taken modulo $n+1$. Then for $\mathbf y = \mathbf A \mathbf x$ $$ y_m = \sum_{j=0}^n a_{mj} x_j = \sum_{j = 0}^n c_{m+j} x_j $$

Let's use the following discrete Fourier transform $$ \mathcal{F}[\mathbf y]_k = \sum_{m=0}^n y_m \exp\left\{-2\pi i\frac{k m}{n+1}\right\} $$ Then $$ \mathcal{F}[\mathbf y]_k = \sum_{m=0}^n \sum_{j=0}^n c_{m+j}x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\}\exp\left\{-2\pi i\frac{k (m+j)}{n+1}\right\} = \\ = \sum_{j=0}^n x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\} \sum_{m=0}^n c_{m+j} \exp\left\{-2\pi i\frac{k (m+j)}{n+1}\right\} = \\ =\sum_{j=0}^n x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\} \sum_{m=-j}^{n-j} c_{m+j} \exp\left\{-2\pi i\frac{k (m+j)}{n+1}\right\} =\\ =\sum_{j=0}^n x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\} \sum_{m'=0}^{n} c_{m'} \exp\left\{2\pi i\frac{k m')}{n+1}\right\} = \\ =\sum_{j=0}^n x_j \exp\left\{2\pi i\frac{k j}{n+1}\right\}\mathcal{F}[\mathbf c]_{k} = \mathcal{F}^{*}[\mathbf x^*]_k \mathcal{F}[\mathbf c]_{k} $$ Thus for $\mathbf A\mathbf x = \lambda \mathbf x$ $$ \mathcal{F}[\mathbf c]_{k} \mathcal{F}^{*}[\mathbf x^*]_k = \lambda\mathcal{F}[\mathbf x]_k $$ Let for simplicity $C_k = \mathcal{F}[\mathbf c]_{k}$. $$ C_{k} \mathcal{F}^{*}[\mathbf x^*]_k = \lambda\mathcal{F}[\mathbf x]_k $$ Recall that discrete Fourier transform can be written in form $$ \mathcal{F}[\mathbf x]_k = (\mathbf F\mathbf x)_k $$ with $\mathbf F^{-1} = \frac{1}{n+1}\mathbf F^H$. Using $\mathbf C = \operatorname{diag} C_k$ one can write the eigenproblem as $$ \mathbf C \mathbf F^* \mathbf x = \lambda \mathbf F \mathbf x $$

The eigenvalues satisfy $$ \operatorname{det}(\mathbf C \mathbf F^* - \lambda \mathbf F) = 0\\ \operatorname{det}(\mathbf C - \lambda\frac{1}{n+1} \mathbf F\mathbf F^\top) = 0 $$ The $\frac{1}{n+1} \mathbf F\mathbf F^\top$ has the following form $$ \frac{1}{n+1} \mathbf F\mathbf F^\top = \begin{pmatrix} 1 & 0 & \dots & 0 & 0\\ 0 & 0 & \dots & 0 & 1\\ 0 & 0 & \dots & 1 & 0\\ \vdots & \vdots & \cdot & \vdots & \vdots\\ 0 & 1 & \dots & 0 & 0 \end{pmatrix} $$ and the eigenvalues (except for $\lambda = C_0$) are the solutions to the problem $$ \left| \mathbf C_\text{AC} - \lambda \mathbf J_n \right| = 0\\ \left| \mathbf J_n \mathbf C_\text{AC} - \lambda \mathbf I_n \right| = 0 $$ where $\mathbf J_n$ is an exchange matrix of size $n$. We arrived to a problem of finging the eigenvalues of the antidiagonal matrix $\mathbf J_n \mathbf C_\text{AC}$.

Next, following this post one can note that $$ (\mathbf J_n \mathbf C_\text{AC})^2 = \operatorname{diag} \begin{pmatrix}C_1 C_n & C_2 C_{n-1} & \dots & C_n C_1\end{pmatrix} $$ so eigenvalues of the original problem are $$ \lambda_0 = C_0\\ \lambda_{i, n-i+1} = \pm \sqrt{C_i C_{n-i+1}},\quad i = 1, 2, \dots $$

This is called an anticirculant matrix, which is a special case of Hankel matrix. The eigenvalue/eigenvector formula for circulant matrix does not apply.