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Prove that $$\det\begin{pmatrix} x_0 & x_1 &x_2&\ldots&\ldots&x_{n-1} \\ x_{n-1} & x_0 &x_1&\ldots&\ldots&x_{n-2} \\ \vdots & \ddots&&&&\vdots \\ \vdots & &\ddots&&&\vdots \\ x_1 & x_2 &x_3&\ldots&\ldots&x_0 \\ \end{pmatrix} = \displaystyle \prod_{\alpha^n=1} (x_o+\alpha x_1+\alpha^2x_2+\ldots+\alpha ^{n-1}x_{n-1})$$

For each$j(2\leq j\leq n)$ I added all the rows times $\alpha^{j-1}$ to the first row, then the elements of $i$th column first row is now $$\alpha^{i-1}(x_0+\alpha x_1+\alpha^2 x_2+\ldots+\alpha^{n-1}x_{n-1})$$

It's quite similar to right hand side, but I could not find out how to proceed from here.

ohisamadaigaku
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    Well, assuming that $$x_0+\alpha x_1+\alpha^2 x_2+\ldots+\alpha^{n-1}x_{n-1} = 0$$ (for some $\alpha^n = 1$), your matrix resulting from applying the row operations has a zero row, hence is singular.

    Since rank is preserved by row operations, this shows your original matrix is singular, and hence has determinant 0.

    I think this means that this polynomial divides the determinant (as a homogeneous polynomial in the $x_i$).

     – Good Boy Sep 02 '20 at 16:44
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    See https://math.stackexchange.com/questions/770117/determinant-of-circulant-matrix – Angina Seng Sep 02 '20 at 16:53

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Hint multiply this matrix by $$\begin{vmatrix} 1 & 1 & \cdots & 1 \\ \zeta_1 & \zeta_2 & \cdots & \zeta_n \\ \vdots & \vdots & \ddots & \vdots \\ \zeta_1^{n-1} & \zeta_2^{n-1} & \cdots & \zeta_n^{n-1} \end{vmatrix}$$

Then you will have:

$$\begin{vmatrix} f(\zeta_1) & f(\zeta_2) & \cdots & f(\zeta_n) \\ \zeta_1f(\zeta_1) & \zeta_2f(\zeta_2) & \cdots & \zeta_nf(\zeta_n) \\ \vdots & \vdots & \ddots & \vdots \\ \zeta_1^{n-1}f(\zeta_1) & \zeta_2^{n-1}f(\zeta_2) & \cdots & \zeta_n^{n-1}f(\zeta_n) \end{vmatrix}$$ where $f(\zeta) = \sum x_i \zeta^i$

openspace
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