I'm working on some problems for fun and came across this problem I'm stuck with. Here's the question: Use row operations to show that $ x + \omega y + \omega^2 z $ is a factor of $ \Delta $, where $ \Delta $ is defined as: $$ \Delta \equiv \begin{vmatrix} x & y & z \\ z & x & y \\ y & z & x \\ \end{vmatrix} = x^3 + y^3 + z^3 - 3xyz $$
and $ \omega $ is a cubic root of unity.
What I've tried:
I first applied the row operations $ R_1 \to R_1 + R_2 $ and then $ R_1 \to R_1 + R_3 $. This allows us to rewrite the determinant as:
$$ \Delta = (x + y + z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ y & z & x \\ \end{vmatrix} $$
Then I can transpose since the determinant is unaffected by transposing to get:
$$ \Delta = (x + y + z) \begin{vmatrix} 1 & z & y \\ 1 & x & z \\ 1 & y & x \\ \end{vmatrix} $$
I then wanted to get zeros in the (2,1) element. I know $ 1 + \omega + \omega^2 = 0 $ since $ \omega $ is a cubic root of unity, so I applied the row operations $ R_2 \to R_2 + \omega R_3 $ and $ R_2 \to R_2 + \omega^2 R_1 $ which gives:
$$ \Delta = (x + y + z) \begin{vmatrix} 1 & z & y \\ 0 & x + \omega y + \omega^2 z & z + \omega x + \omega^2 y \\ 1 & y & x \\ \end{vmatrix} $$
Then to set the (3,1) element to zero, I performed the row operation $ R_3 \to R_3 - R_1 $ where I got to my final point.
$$ \Delta = \begin{vmatrix} 1 & z & y \\ 0 & x + \omega y + \omega^2 z & z + \omega x + \omega^2 y \\ 0 & y - z & x - y \\ \end{vmatrix} $$
I was hoping to get a hint on how to proceed from here. I should mention that I know how to factor $ x + \omega y + \omega^2 z $ from $ x^3 + y^3 + z^3 - 3xyz $, but I don't know how to obtain $ x + \omega y + \omega^2 z $ from row operations alone. Thanks!
This is actually a really interesting puzzle!
Firstly the matrix you are tackling is in the family of circulant matrices, which is interesting and useful in applied math.
A property of these matrices is that it's diagonalized by the DFT matrix. In other words, each of its eigenvector is a geometric series of the $n$th unit root. $$ C=\begin{bmatrix} c_0 & c_{n-1} & c_{n-2} & ... & c_2 & c_1\\ c_1 & c_{0} & c_{n-1} & ...& c_3 & c_2\\ &&...&&\\ c_{n-2} & c_{n-3} & c_{n-4} & ...& c_0& c_{n-1}\\ c_{n-1} & c_{n-2} & c_{n-3} & ... & c_1&c_0\\ \end{bmatrix} $$ $$ v_j=[1,\omega^j,\omega^{2j},...\omega^{(n-1)j}]^T\\ \lambda_j = c_0+c_1 \omega^j+c_2 \omega^{2j}+ ... + c_{n-1}\omega^{(n-1)j} $$ Then given $$ \Delta =\prod_i\lambda_i $$ I think you can easily prove your target with $n=3$