A question about to determinant of matrix

Question

Find the determinant of the matrix $\begin{vmatrix} 1& 2 &3 &... & n\\2 &3 &4 &... &1 \\3 &4 &5 &... &2 \\. &. &. &... .&. \\n &1 &2 &... &n-1 \end{vmatrix}$

We can find for in the case of $n=2,3..$

but how to find $n \times n$

Answer 1

Before we start, let us recall a circulant matrix is a $n \times n$ matrix of the form

$$C = \begin{bmatrix} c_0 & c_1 & c_2 & \cdots & c_{n-1}\\ c_{n-1} & c_0 & c_1 & & c_{n-2}\\ \vdots & c_{n-1} & c_0 & \ddots & \vdots\\ c_2 & & \ddots & \ddots &c_1\\ c_1 & c_2 & \cdots & c_{n-1} & c_0 \end{bmatrix}$$ whose entries at row $i$, column $j$ depends only on $i - j \pmod n$.

For each circulant matrix, we can associate with it a polynomial $$f(x) = c_0 + c_1 x + c_2 x^2 + \cdots + c_{n-1} x^{n-1}$$

In terms of this polynomial, the eigenvalues of $C$ are simply $f(1)$, $f(\omega), \cdots, f(\omega^{n-1})$ where $\omega = e^{\frac{2\pi i}{n}}$ is the primitive $n^{th}$ root of unity.

For the matrix at hand, if we reverse the order from $2^{nd}$ to $n^{th}$ row, we obtain a circulant matrix with $(c_0,c_1,\ldots,c_{n-1}) = (1,2,\ldots,n)$. This means the determinant we seek equals to

$$(-1)^{\frac{(n-1)(n-2)}{2}} \prod_{j=0}^{n-1} f(\omega^j)\quad\text{ where }\quad f(x) = 1 + 2x+\cdots + nx^{n-1}$$

It is easy to see $f(\omega^0) = f(1) = \sum\limits_{k=0}^{n-1} (k+1) = \frac{n(n+1)}{2}$.

For the other $f(\omega^j)$ where $j \ne 0$, we use the identity

$$f(x) = (x+\cdots +x^n)' = \left(\frac{x}{1-x}(1-x^n)\right)' = \left(\frac{x}{1-x}\right)'(1-x^n) - n \frac{x^n}{1-x}$$

Since $x^n = 1$ at such $\omega^j$, we find

$$f(\omega^j) = -\frac{n}{1-\omega^j}\quad\text{ for }\quad j = 1,\ldots,n-1$$

Recall $1 + x + \cdots + x^{n-1} = \frac{x^n-1}{x-1} = \prod_{j=1}^{n-1}(x - \omega^j)$, we obtain

$$\prod_{j=1}^{n-1} f(\omega^j) = (-n)^{n-1}\prod_{j=1}^{n-1}\frac{1}{1-\omega^j} = \frac{(-n)^{n-1}}{1+ 1^1 + 1^2+\cdots + 1^{n-1}} = \frac{(-n)^{n-1}}{n}$$

Combine all these, the determinant we seek is

$$(-1)^{\frac{(n-1)(n-2)}{2}}\times \underbrace{\frac{n(n+1)}{2}}_{f(1)}\times \underbrace{\frac{(-n)^{n-1}}{n}}_{\prod_{j=1}^{n-1}f(\omega^j)} = (-1)^{\frac{n(n-1)}{2}}\frac{(n+1)n^{n-1}}{2}$$