Matrix obtained by recursive multiplication and a cyclic permutation

Question

Have you ever seen this matrix? Each row is obtained from the previous one by multiplying each element by the corresponding element of the next cyclic permutation of $(a_1,\dots, a_n)$: $$\left( \begin{array}{llllllll} 1 & 1 & 1 & \dots & 1 & 1 \\ a_1 & a_2 & a_3 & \dots & a_{n-1} & a_{n} \\ a_1 a_2 & a_2a_3 & a_3 a_4 & \dots & a_{n-1} a_n & a_{n} a_1 \\ a_1 a_2a_3 & a_2a_3 a_4 & a_3 a_4a_5 & \dots & a_{n-1} a_n a_1& a_{n} a_1 a_2 \\ a_1 a_2a_3a_4 & a_2a_3 a_4 a_5& a_3 a_4a_5 a_6& \dots & a_{n-1} a_n a_1 a_2& a_{n} a_1 a_2a_3 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_1 a_2a_3a_4 \dots a_{n-2}& a_2a_3 a_4 a_5\dots a_{n-1} & a_3 a_4a_5 a_6\dots a_n&\dots & a_{n-1} a_n a_1 a_2\dots a_{n-4}& a_{n} a_1 a_2a_3\dots a_{n-3} \\ a_1 a_2a_3a_4 \dots a_{n-1}& a_2a_3 a_4 a_5\dots a_{n} & a_3 a_4a_5 a_6\dots a_1& \dots & a_{n-1} a_n a_1 a_2\dots a_{n-3}& a_{n} a_1 a_2a_3\dots a_{n-2} \\ \end{array} \right)$$ I would like to know if there is a closed formula for the determinant; of course it is invariant (up to sign) under cyclic permutations of $(a_1,a_2,\dots,a_n)$.

Answer 1

Denote the matrix $A$, and index all $a_i$, and all rows and columns starting from $0$ for convenience. If, say, $a_0 = 0$, then $\det A = (-1)^{\lfloor (n - 1) / 2 \rfloor} \prod_{i = 1}^{n - 1} a_i^i$ by substituting and computing the remaining upper triangular determinant. Let's further assume that all $a_i$ are non-zero.

Let $p_k = \prod_{i = 0}^{k - 1} a_i$ ($p_0 = 1$ by convention). Put $a'_i = a_i p_n^{-1/n}$, and construct $A'$ similarly. Alternatively, $A'$ is $A$ with $i$-th row multiplied by $p_n^{-i/n}$. Defining $p'_k$ similarly, we now have $p'_k = p_k p_n^{-k / n}$, in particular $p'_n = 1$.

After multiplying $i$-th column of $A'$ by $p'_i$, we arrive at a Hankel matrix $$P' = \begin{pmatrix} 1 & p'_1 & \ldots & p'_{n - 1} \\ p'_1 & p'_2 & \ldots & 1 \\ \ldots & \ldots & \ldots & \ldots \\ p'_{n - 1} & 1 & \ldots & p'_{n - 2}\end{pmatrix},$$ which is a row permutation of a circulant $(1, \ldots, p'_{n - 1})$. With all substitutions in mind, we have $$\det A = (-1)^{\lfloor(n - 1) / 2\rfloor} \prod_{k = 0}^{n - 1} p_k \cdot \prod_{k = 0}^{n - 1}\left(\sum_{j = 0}^{n - 1} p'_j e^{i\frac{2\pi jk}{n}} \right).$$

One can see that the problem is as general as arbitrary circulant determinant.