Find mod ( det A )

**FernandoRevilla** · December 17th 2011, 01:19 AM

Challenge problem

Find $\mod (\det A)$ being $A\in\textrm{Mat}_{n\times n}(\mathbb{C})$ defined by

$A=\begin{bmatrix} 1 & 1 & 1 & \ldots & 1\\ 1 & \epsilon & \epsilon^2 & \ldots & \epsilon^{n-1} \\ 1 & \epsilon^2 & \epsilon^4 & \ldots & \epsilon^{2n-2}\\ \vdots&&&&\vdots \\ 1 & \epsilon^{n-1} & \epsilon^{2(n-1)}&\ldots & \epsilon^{(n-1)^2}\end{bmatrix}\quad (\epsilon=\cos(2\pi/n)+i\sin (2\pi/n))$

**NonCommAlg** · December 17th 2011, 06:04 PM

Originally Posted by FernandoRevilla

Challenge problem

Find $\mod (\det A)$ being $A\in\textrm{Mat}_{n\times n}(\mathbb{C})$ defined by

$A=\begin{bmatrix} 1 & 1 & 1 & \ldots & 1\\ 1 & \epsilon & \epsilon^2 & \ldots & \epsilon^{n-1} \\ 1 & \epsilon^2 & \epsilon^4 & \ldots & \epsilon^{2n-2}\\ \vdots&&&&\vdots \\ 1 & \epsilon^{n-1} & \epsilon^{2(n-1)}&\ldots & \epsilon^{(n-1)^2}\end{bmatrix}\quad (\epsilon=\cos(2\pi/n)+i\sin (2\pi/n))$

what does $\mod (\det A)$ mean? obviously $\det A=\prod_{r < s}(\epsilon^s - \epsilon^r).$ maybe you want a more simplified answer?

**FernandoRevilla** · December 17th 2011, 11:06 PM

Originally Posted by NonCommAlg

what does $\mod (\det A)$ mean?

Modulus of the complex number $\det (A)$ .

obviously $\det A=\prod_{r < s}(\epsilon^s - \epsilon^r).$

Right, $A$ is a Vandermonde matrix.

maybe you want a more simplified answer?

Being this the Math Challenge Problem forum, I know the solution (and its resolution)

. Another version is: prove that $\mod (\det A)=n^{n/2}$ .

**NonCommAlg** · December 18th 2011, 01:17 AM

Originally Posted by FernandoRevilla

Modulus of the complex number $\det (A)$ .

Right, $A$ is a Vandermonde matrix.

Being this the Math Challenge Problem forum, I know the solution (and its resolution)

. Another version is: prove that $\mod (\det A)=n^{n/2}$ .

so the question is to find $|\det(A)|.$ so far i got $|\det(A)| = 2^{\binom{n}{2}}a^n,$ where $a = \prod_{k=1}^m \sin(k \theta)$ and $\theta = \frac{\pi}{n}$ and $m = \lfloor \frac{n-1}{2} \rfloor.$

so the problem now is to evaluate $a.$

**FernandoRevilla** · December 18th 2011, 04:28 AM

A better approach is to find previously $A^2$ .

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