Suppose we have a right circulant matrix ($n \times n$) $$ C= \begin{bmatrix} c_0 & c_{n-1} & c_{n-2} & \cdots & c_1\\ c_1 & c_0 & c_{n-1} & \cdots & c_2\\ c_2 & c_1 & c_0 &\cdots & c_3\\ \vdots & \vdots & \vdots &\ddots&\vdots \\ c_{n-1} & c_{n-2} & c_{n-3} &\cdots & c_0 \end{bmatrix} $$ and the first $\ell$, where $\ell \ll n$, elements of the first column of $C$ is non-zero, namely, $$ \begin{bmatrix} c_0 & c_1 & c_2 & \cdots & c_{n-1}\\ \end{bmatrix}^T = \begin{bmatrix} c_0 & c_1 & c_2 & \cdots & c_{l-1} & 0 & 0 & \cdots & 0 \\ \end{bmatrix}^T $$
It's well known that circulant matrix can be diagonalized by discrete Fourier transform, namely, $$ FCF^H= \begin{bmatrix} \lambda_0 & 0 & \cdots & 0\\ 0 & \lambda_1 & \cdots & 0\\ \vdots & \vdots &\ddots&\vdots \\ 0 & 0 &\cdots & \lambda_{n-1} \end{bmatrix} $$
where $F$ is Fourier matrix ($F_{pq}=e^{-j2\pi\frac{pq}{n}}$), $F^H$ is conjugate matrix of $F$, and $\lambda_i$ is the eigenvalues of $C$
If we only have $m$, where $\ell < m < n$, eigenvalues of sparse circulant matrix $C$, how can we recover $C$?