Simple formula for determinant of circulant matrix built on an arithmetic sequence.

Question

Let $a$ be an arithmetic sequence: $$a_i=a_1+\lambda(i-1),\tag1$$ and consider a $n\times n$ circulant matrix $M_{n}(a)$ "built" on rotational shifts of the sequence $a$, i.e. with elements:

$$M_{ij}=a_{1+(j−i)\operatorname{mod}n}.\tag2$$

Prove: $$ \det M_n(a)=\frac{a_1+a_n}2(-n\lambda)^{(n-1)}.\tag3 $$

An example: $$ \det\begin{pmatrix} 1&2&3&4&5\\ 5&1&2&3&4\\ 4&5&1&2&3\\ 3&4&5&1&2\\ 2&3&4&5&1\\ \end{pmatrix}=\frac{1+5}2(-5)^4=1875 $$

Answer 1

Write $a$ for $a_0$ and $\newcommand{\ze}{\zeta}\ze$ for $\exp(2\pi i/n)$. The determinant is $$\newcommand{\la}{\lambda}D=\prod_{r=0}^{n-1}\sum_{k=0}^{n-1}\ze^{jk}(a+k\la)$$ by the usual formula. The $r=0$ summand is $$\sum_{k=0}^{n-1}(a+k\la)=na+\frac{n(n-1)}2\la.$$ The summand for $0<r\le n-1$ is $$\sum_{k=0}^{n-1}\ze^{rk}(a+k\la)=\la\sum_{k=0}^{n-1}k\ze^{rk}.$$ This is an arithmetic-geometric progression: $$(1-\ze^r)^2\sum_{k=0}^{n-1}k\ze^{rk}=\ze^r-n\ze^{rn}+(n-1)\ze^{r(n+1)} =-n(1-\ze^r).$$ Then $$\sum_{k=0}^{n-1}\ze^{rk}(a+k\la)=-\frac{n\la}{1-\ze^r}$$ and so $$\prod_{r=1}^{n-1} \sum_{k=0}^{n-1}\ze^{rk}(a+k\la)=(-n\la)^{n-1}\prod_{r=1}^n \frac1{1-\ze^r}=\frac{(-n\la)^{n-1}}{n}.$$ If you multiply this by the $r=0$ summand, you should get your formula.

Answer 2

The determinant of circulant matrix $M$ is given by the formula: $$\det M = \prod_{\ell=0}^{n-1} f(\omega^\ell)$$ where $\omega = e^{\frac{2\pi i}{n}}$ is a primitive $n^{th}$ root of unity and

$$\begin{align} f(x) &= \sum_{k=0}^{n-1} a_{k+1} x^k = \sum_{k=0}^{n-1} (a_1 + \lambda k)x^k\\ &= \left(a_1 + \lambda x\frac{d}{dx}\right)\sum_{k=0}^{n-1} x^k = \left(a_1 + \lambda x\frac{d}{dx}\right)\frac{1 - x^n}{1-x}\\ &= a_1 \frac{1 - x^n}{1-x} + \lambda\left[\frac{-n x^n}{1 -x} + x\frac{1-x^n}{(1-x)^2}\right] \end{align} $$ Notice $\omega,\omega^2,\cdots,\omega^{n-1}$ are all the roots of $g(x) = \sum_{k=0}^{n-1} x^k = \frac{1 - x^n}{1-x}$. We have $x^n = 1$ at these points and

$$f(\omega^\ell) = \frac{-\lambda n}{1-\omega^\ell}\quad\text{ for }\quad 1 \le \ell \le n-1$$

As a result, $$\prod_{\ell=1}^{n-1} f(\omega^\ell) = \prod_{\ell=1}^{n-1}\frac{-\lambda n}{1 - \omega^\ell} = \frac{(-\lambda n)^{n-1}}{g(1)} = \frac{(-\lambda n)^{n-1}}{n}$$

Together with $$f(1) = \sum_{k=0}^{n-1} (a_1 + \lambda k) = a_1 n + \lambda \frac{n(n-1)}{2} = n\frac{a_1 + a_n}{2}$$

We obtain $$\det M = f(1)\prod_{\ell=1}^{n-1} f(\omega^\ell) = \frac{a_1 + a_n}{2}(-\lambda n)^{n-1}$$