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I tried, in vain, to prove the following formula for the determinant of a cyclic matrix: $$ \begin{vmatrix}a_1&a_2&a_3&\cdots&a_n\\a_n&a_1&a_2&\cdots&a_{n-1}\\a_{n-1}&a_n&a_1&\cdots&a_{n-2}\\\vdots&\vdots&\vdots&&\vdots\\a_2&a_3&a_4&\cdots&a_1\end{vmatrix}=f(\epsilon_0)f(\epsilon_1){\cdots}f(\epsilon_{n-1}), $$ where $\epsilon_0,\epsilon_1,\dots,\epsilon_{n-1}$ are the $n$-th roots of unity, and $f(x)=a_1+a_2x+a_3x^2+\cdots+a_nx^{n-1}$.

I tried to add all columns to the first one and factor $a_1+a_2+\cdots+a_n=f(\epsilon_0)$ but I got nowhere. On the internet I found only one proof which uses eigenvectors but I wonder whether this identity could be proven using only column/row operations.

user26857
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S. Stefan
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1 Answers1

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Proof without the use of eigenvalues :

Let $$A=\begin{pmatrix}a_1&a_2&a_3&\cdots&a_n\\a_n&a_1&a_2&\cdots&a_{n-1}\\a_{n-1}&a_n&a_1&\cdots&a_{n-2}\\\vdots&\vdots&\vdots&&\vdots\\a_2&a_3&a_4&\cdots&a_1\end{pmatrix}$$

and $\Omega$ be the matrix with entries $\Omega= (\omega^{(i-1)(j-1)})_{1 \leq i,j \leq n} \in \mathcal{M}_n(\mathbb{C})$, where $\omega = e^{2i\pi/n}$.

Then it is easy to see that the $(i,j)$-entry of the product $A\Omega$ is $\omega^{(i-1)(j-1)}f(\omega^{j-1})$, where $f(x)=a_1+a_2x+a_3x^2+\cdots+a_nx^{n-1}$.

Hence $$A\Omega=\begin{pmatrix}f(1)&f(\omega)&\cdots&f(\omega^{n-1})\\f(1)&\omega f(\omega)&\cdots&\omega^{n-1}f(\omega^{n-1})\\\vdots&\vdots&&\vdots\\f(1)&\omega^{n-1}f(\omega)&\cdots&\omega^{(n-1)(n-1)}f(\omega^{n-1})\end{pmatrix}$$

so $$\det(A\Omega)=f(1)f(\omega)\cdots f(\omega^{n-1})\begin{vmatrix}1&1&\cdots&1\\1&\omega &\cdots&\omega^{n-1}\\\vdots&\vdots&&\vdots\\1&\omega^{n-1}&\cdots&\omega^{(n-1)(n-1)}\end{vmatrix} $$ $$=\det(A)\det(\Omega)=f(1)f(\omega)\cdots f(\omega^{n-1})\det(\Omega).$$

Because $\det(\Omega) \neq 0$ (it is a Vandermonde), you deduce that $$\boxed{\det(A)=f(1)f(\omega)\cdots f(\omega^{n-1})}.$$

user26857
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TheSilverDoe
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