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Let $s_k = x_1^k + \cdots + x_n^k$. Compute

$$\begin{vmatrix} s_0 & s_1 & \cdots & s_{n-1} & 1\\ s_1 & s_2 & \cdots & s_{n} & y\\ \vdots & \vdots & \vdots & \ddots &\vdots \\ s_n & s_{n+1} & \cdots & s_{2n-1} & y^n \end{vmatrix}$$

The proof "observes" that the determinant can be written as the product of the two determinants:

$$\begin{vmatrix} 1 & \cdots & 1 & 1\\ x_1 & \cdots & x_{n} & y\\ x_1^2 & \cdots & x_{n}^2 & y^2 \\ \vdots & \cdots & \vdots &\vdots \\ x_1^n & \cdots & x_{n}^n & y^n \end{vmatrix} \cdot \begin{vmatrix} 1 & x_1 & \dots & x_1^{n-1} & 0\\ 1 & x_2 & \dots & x_2^{n-1} & 0 \\ \vdots & \vdots & \cdots & \vdots &\vdots \\ 1 & x_n & \dots & x_n^{n-1} & 0 \\ 0 & 0 & \dots & 0 & 1 \end{vmatrix}$$

The answer then being $\prod (y-x_i) \prod_{i > j}(x_i -x_j)^2$

My question is: where does the observation come from?

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Notice that we can write:

$$\begin{bmatrix} 1 & \cdots & 1 & 1\\ x_1 & \cdots & x_{n} & y\\ x_1^2 & \cdots & x_{n}^2 & y^2 \\ \vdots & \cdots & \vdots &\vdots \\ x_1^n & \cdots & x_{n}^n & y^n \end{bmatrix} \cdot \begin{bmatrix} 1 & x_1 & \dots & x_1^{n-1} & 0\\ 1 & x_2 & \dots & x_2^{n-1} & 0 \\ \vdots & \vdots & \cdots & \vdots &\vdots \\ 1 & x_n & \dots & x_n^{n-1} & 0 \\ 0 & 0 & \dots & 0 & 1 \end{bmatrix} = \begin{bmatrix} s_0 & s_1 & \cdots & s_{n-1} & 1\\ s_1 & s_2 & \cdots & s_{n} & y\\ \vdots & \vdots & \vdots & \ddots &\vdots \\ s_n & s_{n+1} & \cdots & s_{2n-1} & y^n \end{bmatrix}$$

To see why this is true, notice that the dot product of the $i$-th row of the left matrix and the $j$-th column of the right matrix is $x_1^{i-1} \cdot x_1^{j-1}+\cdots+x_n^{i-1}\cdot x_n^{j-1}+y^{i-1}\cdot 0 = s_{i+j-2}$ for $j \neq n$, and $x_1^{i-1}\cdot 0 + \cdots x_n^{i-1} \cdot 0 + y^{i-1} \cdot 1 = y^{i-1}$ for $j = n$.

The result then follows from using the identity $\det(AB) = \det(A)\det(B)$.

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  • $\begingroup$ Thanks. I want to ask, given the two matrices we multiply them to get the original. But first we have to find/stumble upon them somehow in the proof. How does one see the reverse of what you have done? $\endgroup$ – T2S Dec 14 '19 at 3:23
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    $\begingroup$ @T2S: When you see a matrix whose entries are sums of products, doesn't it sound rather natural to try and factor it as a product of matrices? $\endgroup$ – darij grinberg Dec 14 '19 at 5:24

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