Let $a_0, a_1, a_2, \dots, a_n \in \mathbb{C}$. How can I compute the following determinant?
$$\begin{vmatrix} a_0 & a_1 & a_2 & \dots & a_{n-1} \\ -a_{n-1} & a_0 & a_1 & \dots & a_{n-2} \\ -a_{n-2} & -a_{n-1} & a_0 & \dots & a_{n-3} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -a_1 & -a_2 & a_3& \dots & a_{0} \\ \end{vmatrix}_{(n)}$$
Let $\zeta\in\mathbb{C}$ with $\zeta^{2n}=1=-\zeta^n$. I think, $v=(1,\zeta,\ldots,\zeta^{n-1})$ is an eigenvector with eigenvalue $\sum_{k=0}^{n-1}a_k\zeta^k$. Now let $\zeta:=e^{\pi i/n}$. This should give $$\det A=\prod_{k=1}^n\sum_{l=0}^{n-1}a_l\zeta^{l(2k-1)}.$$
Added on request: Argument without eigenvalues adapted from Determinant of cyclic matrix, proof without eigenvectors.. Let $f(\omega)=\sum_{k=0}^{n-1}a_k\omega^k$ and $M=(\zeta^{(k-1)(2l-1)})_{k,l=1}^n$. Then $$AM=\begin{pmatrix} f(\zeta)&f(\zeta^3)&\cdots&f(\zeta^{2n-1})\\ \zeta f(\zeta)&\zeta^3f(\zeta^3)&\cdots&\zeta^{2n-1}f(\zeta^{2n-1})\\ \vdots&\vdots&&\vdots\\ \zeta^{n-1}f(\zeta)&\zeta^{3(n-1)}f(\zeta^3)&\cdots&\zeta^{(n-1)(2n-1)}f(\zeta^{2n-1}). \end{pmatrix}$$ Since the determinant is linear in every column, we can factor out $f(\zeta)$, $f(\zeta^3)$ and so on, $$\det(A)\det(M)=\det(AM)=f(\zeta)f(\zeta^3)\ldots f(\zeta^{2n-1})\det(M).$$ Since $M$ is an invertible Vandermonde matrix, $\det(M)\ne 0$ and $\det(A)=f(\zeta)f(\zeta^3)\ldots f(\zeta^{2n-1})$.
