On a determinant

Let $p$ be a prime number and let $\omega$ be a primitive $p$-th root of unity. Define:

$$\mathcal{V} = \det \begin{pmatrix}
1 &1  &1  &\cdots  &1 \\
 1& \omega &\omega^2  &\cdots  &\omega^{p-1} \\
1 &\omega^2  &\left ( \omega^2 \right )^2  &\cdots  &\left ( \omega^2 \right )^{p-1} \\
 \vdots&\vdots  &\vdots  &\ddots  & \vdots\\
 1& \omega^{p-1} &\left ( \omega^{p-1} \right )^2  &\cdots  & \left ( \omega^{p-1} \right )^{p-1}
\end{pmatrix}$$

Evaluate the rational number $\mathcal{V}^2$.

Solution



The $ij$ -th entry of $\mathcal{V}$ is  $\omega^{(i-1)(j-1)}$. Thus the $ij$-th entry of $\mathcal{V}^2$ is equal to:

$$\sum_{\ell} \omega^{(i-1)(\ell-1)} \omega^{(\ell-1) (j-1)} = \sum_{\ell} \omega^{(i-1+j-1)(\ell-1)} = \left\{\begin{matrix}
0 &\text{if}  & (i-1) + (j-1)  \neq 0 \mod p\\
 p& \text{if}  & (i-1) + (j-1) = 0 \mod p
\end{matrix}\right.$$

since it is known that $\sum \limits_{0 \leq \ell <p} \omega^{\ell}=0$ for any $p$-th root of unity $\omega$ rather than $1$. Thus:

$$\mathcal{V}^2 = \begin{pmatrix}
p &0  & 0 &\cdots  &0  &0 \\
0 & 0 & 0 & \cdots & 0 & p\\
 0& 0 & 0 &\cdots & p &0 \\
 \vdots& \vdots &\vdots  &\ddots  &\vdots  &\vdots \\
 0& 0 & p&  \cdots&0  & 0\\
 0& p & 0 &\cdots  &0  &0
\end{pmatrix}$$

Thus is there is a $p$ at the upper left corner and $p$'s along the anti diagonal in the lower right $(n-1)$ by $(n-1)$ block. Thus:

$$\left ( \det \mathcal{V} \right )^2 = \det \left ( \mathcal{V}^2 \right )= (-1)^{(p-1)(p-2)/2} \; p^p $$