Summation of Powers over Product of Differences/Proof 3

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

Lemma 1
Proof of Lemma 1

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

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:

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 $\det \paren {A}$.

Case 3: $r = n$:

Expand $\map p x = \prod_{k = 1}^n \paren { x - x_k }$ 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 $x = x_k$ ($1 \mathop \le k \mathop \le n$), 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

Historical Note
The Example according to Knuth (1997) is a discovery of fictitious character Dr. Matrix.