Linear Algebra – Solving Circular Determinant Problems

Question

I'm stuck in this question:
How calculate this determinant ?
$$\Delta=\left|\begin{array}{cccccc}
1&2&3&\cdots&\cdots&n\\
n&1&2&\cdots&\cdots& n-1\\
n-1&n&1&\cdots&\cdots&n-2\\
\vdots &\ddots & \ddots&\ddots&&\vdots\\
\vdots &\ddots & \ddots&\ddots&\ddots&\vdots\\
2&3&4&\cdots&\cdots&1
\end{array}\right|$$
Thanks a lot.

Answer 1

Since the matrix with determinant $\Delta$ is circulant, its eigenvalues $\{\lambda_j\}_{j=1}^n$ are given by

$$\lambda_j=1+2\omega_j+3\omega_j^2+\ldots+n\omega_j^{n-1},$$

where $\omega_j=\exp\left(\frac{2\pi ij}{n}\right)$. The determinant is the product of the eigenvalues.

For an explicit formula not involving products and sums, see sequence A052182 in OEIS. It can be shown that

$$\Delta(n)=(-1)^{n-1} \cdot n^{n-2} \cdot \frac{n^2+n}{2}$$