need to calculate the smallest singular value of $C_m^d$, and show that it is non-zero.

The eigendecomposition of a circulant matrix is well-known (see [@DavisP-94; @Arbenz-91]). A circulant $C$ of order $m$, defined by the symbol $(c_0, \ldots, c_{m - 1})$, can be written as $$C = \sum_{j = 0}^{m - 1} c_j \Pi^j,$$ where $$\Pi = \begin{bmatrix}
    \; 0 \;\; & 1              &                &   \\[3pt]
              & \smash{\ddots} & \smash{\ddots} &   \\
              &                & \smash{\ddots} & \;\; 1 \;\; \\
    \; 1 \;\; &                &                & \;\; 0 \;\; \\
  \end{bmatrix}.$$

The spectral decomposition of $\Pi$ is $\Pi = F \Omega F^{\ast}$, where $$\Omega = {\operatorname{diag}}(1, \omega, \omega^2, \ldots, \omega^{m - 1}), \quad
  \omega = \frac{2\pi i}{m}, \quad i = \sqrt{-1},$$ while $$F_{j, k} = \frac{1}{\sqrt{m}} \omega^{kj}, \quad
  0 \leq k, j \leq m - 1.$$ Hence, $C$ can be decomposed as $$C = F \Lambda F^{\ast}, \quad
  \Lambda = {\operatorname{diag}}(\lambda_0, \ldots, \lambda_{m - 1})
  = \sum_{j = 0}^{m - 1} c_j \Omega^j.$$ The eigenvalues of a real symmetric circulant $C$ are real, and given by $$\lambda_k(C) = c_0 + \sum_{j = 1}^{m - 1} c_j \cos \frac{2\pi k j}{m},
  \quad k = 0, \ldots, m - 1.
  \label{4.4}$$ They 