Summation of Powers over Product of Differences/Proof 3

Definition
Symbol $\map {\mathbf {Cof } } {M, i, j}$ denotes cofactor $M_{i j}$ of matrix $M$

Lemma 1
Proof of Lemma 1

By Effect of Elementary Row Operations on Determinant, the determinant of $A$ is unchanged by adding a linear combination of the first $n - 1$ columns to the last column.

Let $\map f x$ be the monic polynomial $\map {p_m} x = x^{n - 1} + $ lower power terms.

Then:

The last column on the right has all components zero except $m$th entry $\map {p_m} {x_m}$.

Apply cofactor expansion along column $n$.

Lemma 2
Proof of Lemma 2:

Value of Vandermonde Determinant establishes $(1)$.

To prove (2), let:


 * $\set {y_1, \ldots, y_{n - 1} } = \set {x_1, \ldots, x_n} \setminus \set {x_m}$

then apply $(1)$ with $\set {x_1, \ldots, x_n}$ replaced by $\set {y_1, \ldots, y_{n - 1} }$.

Theorem Details:

The Lemmas change the Theorem's equation to:

Cofactor expansion changes identity $(3)$ to:

The proof is divided into three cases.

Case 1: $0 \mathop \le r \mathop \le n - 2$:

The determinant on the left in $(4)$ is zero by Square Matrix with Duplicate Rows has Zero Determinant.

Case 2: $r = n - 1$:

The determinant on the left in $(4)$ is $\map \det A$.

Case 3: $r = n$:

Expand $\map p x = \prod_{k \mathop = 1}^n \paren {x - x_k}$ by Taylor's Theorem with remainder $R$ of degree $n - 2$:


 * $\map p x = x^n - \paren {x_1 + \cdots + x_n} x^{n - 1} + \map R x$

Let $x = x_k$ for $1 \mathop \le k \mathop \le n$.

Then $x$ is a root of $\map p x = \prod_{k \mathop = 1}^n \paren {x - x_k}$:

Let:

Then:

The left side of $(4)$:

Also see
Sum of Elements in Inverse of Cauchy Matrix

Vandermonde Matrix Identity for Cauchy Matrix