Let $\Bbb K$ be a field. Let $$f := t^n - \sum\limits_{i = 0}^{n-1} a_i t^i \in {\Bbb K} [t]$$ be a polynomial. I construct a $2n \times 2n$ matrix $A$ in the following way:
The upper-left quarter is the $n \times n$ identity matrix.
The lower-left quarter is constructed iteratively the following way:
- The first row is $(a_0, \dots, a_{n-1})$.
- If the $i$-th row is $(b_0 \dots b_{n-1})$, then the $i+1$-th row is $$ b_{n-1} \cdot (a_0, \dots, a_{n-1}) + (0, b_0, \dots, b_{n-2}) $$
The $i$-th row of the right side is $(i-1)$ times the $(i-1)$-row of the left side.
For example, for $n = 2$, the matrix is
$$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ a_0 & a_1 & 0 & 2 \\ a_0a_1 & a_0 + a_1^2 & 3 a_0 & 3 a_1\end{pmatrix}$$
I have checked with a computer (up to $n = 11$) that $\det(A)$ is always $\pm \operatorname{discriminant}(f)$.
Question: How can one show that $\det(A) = \pm \operatorname{discriminant}(f)$ in general?
The discriminant is defined as the determinant of the Sylvester-matrix of $f$ and $f'$. This is a $(2n-1) \times (2n-1)$ matrix, but since $A$ has a nice first line, we can do Laplace-expansion there, and get that the Determinant of $A$ is equal to a determinant of a $(2n-1) \times (2n-1)$ matrix. I then tried to do row-and column operations to get the Sylvester-matrix, but could not find a way. Maybe there is another property of the discriminant one can use?
Motivation
Maybe some words on where this matrix comes from: Let $R := {\Bbb K} [x',y'] / \langle f(x'),y'^2 \rangle$ and $x = \pi(x'), y = \pi(y')$ (with $\pi$ the canonical epimorphism). I want to check if ${\Bbb K} [x+y] = R$. It is $A \cdot (1,x,\dots,x^{n-1},y,xy,x^{n-1}y)^T = (1,(x+y)^1,\dots,(x+y)^{2n-1})^T \in {\Bbb K}[x+y]^{2n}$. If $A$ is invertible, then in particular $x,y \in {\Bbb K}[x+y]$, and thus $R = {\Bbb K}[x+y]$ (with $\supseteq$ being clear anyway).
