I am solving the exercise from "Linear Algebra" by Georgi E. Shilov. The question asks to prove the following:
\begin{align*} \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots &\vdots \\ x_1^{n-2} & x_2^{n-2} & \cdots & x_n^{n-2}\\ x_1^{n} & x_2^{n} & \cdots & x_n^{n}\\ \end{vmatrix} = \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots &\vdots \\ x_1^{n-2} & x_2^{n-2} & \cdots & x_n^{n-2}\\ x_1^{n-1} & x_2^{n-2} & \cdots & x_n^{n-1}\\ \end{vmatrix} \times \sum_{k=1}^{n}{x_k} \end{align*}
Initially I thought this was a Vandermonde Determinant, but the power of the second last row is $(n - 2)$ after which we skip $(n - 1)$ and go to $n$.
Second, I don't understand the pattern of the last row of the RHS. Is it alternating $(n - 1)$'s and $(n - 2)$'s?
Hence, not sure how to solve this one. Any help would be appreciated.
Preliminaries.
Let $D$ denote the diagonal matrix whose entries are $x_1, x_2, \ldots$. Note that the characteristic polynomial of $D$ is $$ p_D(\lambda ) =\Pi_j (\lambda - x_j) = \lambda^n +\sum_{j=1}^{n} (-1)^{j} \sigma_j \lambda^{n-j}$$ where the $\sigma_j$ are the elementary symmetric polynomials in these values. In particular $\sigma_1=tr(D)=\sum_j x_j$.
The Cayley-Hamilton Theorem asserts that $p_D(D)=0$. Thus one obtains the recurrence relation $$D^n = tr(D) D^{n-1}+\ldots$$ where the suppressed terms are lower powers of $D$. This recurrence relation is the key identity needed below.
Problem solution
Returning to the problem you posted, let $L$ denote the matrix on the left and let $M$ denote the standard vanderMonde matrix that appears on the right.
(1) The matrix $M$ can be generated as follows. Its first row is $v_0=[1,1,\ldots,1]$.
Multiplying the initial row vector $v_0$ by successive powers of $D$ generates the row vectors $v_{k+1}= Dv_k = D^{k} v_0$. That is, $M$ is generated by powers $k=0,1,2 \ldots n-1$.
Write the list of rows of $M$ as $\{ v_0, v_1, \ldots, v_{n-1}\}$
(2) In contrast $L$ consists of the rows $\{v_0, v_1 , \ldots v_{n-2}, v_n\}$
But from the recurrence relation derived above, the row entry $v_{n}$ in $L$ can be replaced by $\sigma_1 v_{n-1}$ plus a linear combination of other rows in $L$ that have no affect when we expand the determinant. After this replacement we obtain a matrix that looks like $M$ except for the extra factor $\sigma_1$. Thus $det(L)= \sigma_1 det (M)$.