I am interested in computing the determinant of the following circulant matrices: let $n=p^k$ for $p$ a prime and $k\in \mathbb{N}$, take $a\in \mathbb{N}$ to be such that $a<p$ and $(a,p)=1$. Consider the matrix
$$A = \left[ \begin{array}{ccccccccc} 1&1&1&...&1&0 & ... & 0 & 0\\ 0&1&1&& & ...& & 0 &0& \\0&0&1&& & ...& & 0&0\\&&&&\vdots\\1&&&&&...&&0&1\end{array}\right]$$
where $A$ is the circulant matrix with the first row given by the vector with the first $a$ coordinates equal to $1$ and the rest equal to $0$.
I am aware that the eigenvalues of this matrix are given by $$\lambda_j = 1+\omega^j + \omega^{2j} + \dots + \omega^{(a-1)j},\quad j = 0, 1, \dots, n-1,$$
where $\omega$ is primitive $n$-th root of unity.
I have done several computations of the determinant of some specific matrices of this form and it seems like $\det(A)=a$.
I would like to know if there is some general result already known about circulant matrices of this form that I could cite or I would like to be pointed in the right direction of how to prove this. Thank you!
Disclaimer: I am a graduate student and this came up for a problem I am working on. It is not a homework problem or part of any class.
Since you already have an expression for the eigenvalues, you can prove it yourself. Note that we have $$ \lambda_0 = a, \qquad \lambda_j = \frac{1 - \omega^{aj}}{1 - \omega^j} \quad j = 1,2,\dots,n-1. $$ Because $\gcd(a,n) = 1$, the map $x \mapsto ax$ is bijective over $\{1,2,\dots,n-1\} \subset \Bbb Z/n\Bbb Z$. It follows that $\prod_{j=1}^{n-1} (1 - \omega^j) = \prod_{j=1}^{n-1} (1 - \omega^{aj})$, so that $$ \prod_{j=1}^{n-1} \lambda_j = \prod_{j=1}^{n-1}\left( \frac{1 - \omega^{aj}}{1 - \omega^j}\right) = \frac{\prod_{j=1}^{n-1} (1 - \omega^{aj})}{\prod_{j=1}^{n-1} (1 - \omega^j)} = 1. $$ Thus, we come to the conclusion that $\det(A) = \prod_{j=0}^{n-1} \lambda_j = a \cdot 1 = a$, which was what we wanted.