can also be viewed as the discrete Fourier transform (DFT) of the symbol $(c_0, \ldots, c_{m - 1})$.

For real and symmetric $C$, i.e., when $c_k = c_{m - k}$, for $k = 1, \ldots, m - 1$, from (\[4.4\]) it also follows that $$\lambda_k(C) = c_0 + \sum_{j = 1}^{m - 1} c_j \cos \frac{2\pi k j}{m}
  = c_0 + \sum_{j = 1}^{m - 1} c_{m - j} \cos \frac{2\pi k (m - j)}{m}
  = \lambda_{m - k}(C).$$ So, all the eigenvalues, except $\lambda_0(C)$, and possibly $\lambda_{\frac{m}{2}}(C)$, for even $m$, are multiple.

Therefore, the eigenvalues of the circulant $C_m^d$ from (\[4.1\]) are $$\lambda_k(C_m^d) = t_0^d + 2 \sum_{j = 1}^{r} t_j^d \cos \frac{2\pi k j}{m},
  \quad k = 0, \ldots, m - 1.
  \label{4.6}$$

For prime orders $m$, the nonsingularity of $C_m^d$ is a consequence of the following theorem from [@Geller-Kra-Popescu-Simanca-2004].

Let $m$ be a prime number. Assume that the circulant $C$ of order $m$ has entries in $\mathbb{Q}$. Then $\det C = 0$ if and only if $$\lambda_0 = \sum_{j = 0}^{m - 1} c_i = 0,$$ or all the symbol entries $c_i$ are equal. \[thmGKPS\]

If $m$ is prime, then we must have $\det C_m^d \neq 0$, 