# Determinant of the vandermonde matrix times the sum of its base terms?

I've been following along Georgi Shilov's dense but pleasant Linear Algebra. To my chagrin, I'm stuck on problem 10 of the first chapter, which has withstood a few hours' toil.

The problem is to prove that the equation below holds, where the vertical bars signify taking the determinant: $$\begin{vmatrix} 1 & 1 & ...& 1 \\ x_1 & x_2 & ... & x_n \\ x_1^2 & x_2^2 & ... & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-2} & x_2^{n-2} & ...& x_n^{n-2} \\ x_1^n & x_2^n & ...& x_n^n \\ \end{vmatrix} = \begin{vmatrix} 1 & 1 & ...& 1 \\ x_1 & x_2 & ... & x_n \\ x_1^2 & x_2^2 & ... & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-2} & x_2^{n-2} & ...& x_n^{n-2} \\ x_1^{n-1} & x_2^{n-1} & ...& x_n^{n-1} \\ \end{vmatrix} \times \sum_{k=1}^nx_k$$ Let us call the lefthand matrix $A$ and the righthand matrix $B$.

Now, here's the rub. Based on the book's presentation (and hint at the back), there exists a solution using the below facts/operations:

(1) $\det(A)$ is a polynomial in $x_n$ of degree $n$, with factors the $\{x_1, x_2, ..., x_{n-1}\}$. That is:

$$\det(A) = (\alpha + \beta x_n)\prod_{k=1}^{n-1}(x_n-x_k)$$ (2) Cofactor expansion on the rightmost column of $A$. That is:

$$\det(A) = C_n^1 + ... + x_n^{n-2}C_n^{n-2} + x_n^nC_n^n$$ where the $C_n^j$ are the corresponding cofactors of the $j$th element in the rightmost column.

## My observations:

1. The polynomial (1) has the below leading and next-leading terms (see Wolfram):

$$\det(A) = \beta x_n^n - \beta(\sum_{k=1}^{n-1}x_i)x_n^{n-1} + \alpha x_n^{n-1}...$$

1. The polynomial (2) has leading coefficient $C_n^n$. Therefore $\beta = C_n^n$.

2. The polynomial (2) does not have any $x_n^{n-1}$ term. Therefore, equating the coefficient from Observation 1:

$$\beta(\sum_{k=1}^{n-1}x_i) = C_n^n(\sum_{k=1}^{n-1}x_i) = \alpha$$

1. $\alpha$ has the exact form of the righthand side of our equation, were it one column and row smaller!

Unfortunately, I can't see past this observation. Any pointers?

• No idea for your question but the ratio $A/B$ is a special case of something called Schur polynomial. look at wiki entry for similar decomposition. – achille hui Apr 14 '18 at 7:57

You've shown $(\beta x_n +\alpha)=C^{n}_{n} (x_n+\sum_{1}^{n-1}x_j)$ so $$\det A= C^{n}_{n} (x_n+\sum_{1}^{n-1}x_j)\prod^{n-1}_{1}(x_n-x_j)= C^{n}_{n} \sum_{1}^{n}x_j\prod^{n-1}_{1}(x_n-x_j).$$

A similar argument on $B$ shows that it is $$\det B= \hat{C}^{n}_{n} \prod^{n-1}_{1}(x_n-x_j).$$ where $\hat{C}$ are its cofactors.

Now note that the $(n,n)$ cofactors are the same.

I've been struggling with this problem for a considerable amount of time.

I've not been able to really grasp the whole way of proving the equation even with the hint in the book and some explanation above.

My lack of some knowledge turned out to be the main reason why I was stuck.

Despite the fact that the question was asked quite a long time ago, I suppose my experience would help linear algebra newbies like myself.

So, let me share what I got.

1. Let's set $$x_n$$ as a variable and all other $$x_i$$ where $$1\leqslant i \leqslant (n-1)$$ as constants. Then the determinant on the left is a polynomial of degree $$n$$ in $$x_n$$.

$$\begin{vmatrix} 1 & 1 & ...& 1 \\ x_1 & x_2 & ... & x_n \\ x_1^2 & x_2^2 & ... & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-2} & x_2^{n-2} & ...& x_n^{n-2} \\ x_1^n & x_2^n & ...& x_n^n \\ \end{vmatrix} = a_nx_n^n+a_{n-2}x_n^{n-2}+...+a_2x_n^2+a_1x_n+a_0$$ The polynomial coefficients $$a_i$$ where $$0\leqslant i \leqslant n \ \text{and} \ i \neq (n-1)$$ depends only on the constants from $$x_1$$ up to $$x_{n-1}$$.

Good explanation of the connection between polynomial interpolation problem and Vandermonde determinant

1. The polynomial (1) has no term with $$x_n^{n-1}$$. It means that the coefficient $$a_{n-1}$$ of the term $$a_{n-1}x_n^{n-1}$$ equals to zero.

2. If $$x_n$$ is equal to any of the constants in the set $$\{x_1, x_2, ..., x_{n-1}\}$$, then the determinant on the left has two identical columns. It's known that a determinant with two identical columns is equal to zero. Hence, all the constants $$\{x_1, x_2, ..., x_{n-1}\}$$ are roots of the polynomial (1). Therefore, according to Polynomial remainder theorem, the polynomial (1) is divisible by the product $$\prod_{k=1}^{n-1}(x_n-x_k)$$

3. According to Vieta's formulas $$\prod_{k=1}^{n-1}(x_n-x_k)=x_n^{n-1}-x_n^{n-2}\sum_{k=1}^{n-1}x_k+...$$ Good explanation how to create polynomial coefficients from its roots.

4. So the polynomial (1) can be represented like this $$a_nx_n^n+a_{n-2}x_n^{n-2}+...+a_2x_n^2+a_1x_n+a_0 = (A+Bx_n)\prod_{k=1}^{n-1}(x_n-x_k)= (A+Bx_n)(x_n^{n-1}-x_n^{n-2}\sum_{k=1}^{n-1}x_k+...)=A(x_n^{n-1}-x_n^{n-2}\sum_{k=1}^{n-1}x_k+...)+Bx_n(x_n^{n-1}-x_n^{n-2}\sum_{k=1}^{n-1}x_k+...)=Ax_n^{n-1}-Ax_n^{n-2}\sum_{k=1}^{n-1}x_k+...+Bx_n^n-Bx_n^{n-1}\sum_{k=1}^{n-1}x_k+...=Bx_n^n+(A-B\sum_{k=1}^{n-1}x_k)x_n^{n-1}+...$$

5. On the other hand, the determinant on the left can be expanded with respect to its last column. $$\begin{vmatrix} 1 & 1 & ...& 1 \\ x_1 & x_2 & ... & x_n \\ x_1^2 & x_2^2 & ... & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-2} & x_2^{n-2} & ...& x_n^{n-2} \\ x_1^n & x_2^n & ...& x_n^n \\ \end{vmatrix} = \begin{vmatrix} 1 & 1 & ...& 1 \\ x_1 & x_2 & ... & x_{n-1} \\ x_1^2 & x_2^2 & ... & x_{n-1}^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-3} & x_2^{n-3} & ...& x_{n-1}^{n-3} \\ x_1^{n-2} & x_2^{n-2} & ...& x_{n-1}^{n-2} \\ \end{vmatrix}x_n^n+\begin{vmatrix} 1 & 1 & ...& 1 \\ x_1 & x_2 & ... & x_{n-1} \\ x_1^2 & x_2^2 & ... & x_{n-1}^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-3} & x_2^{n-3} & ...& x_{n-1}^{n-3} \\ x_1^n & x_2^n & ...& x_{n-1}^n \\ \end{vmatrix}x_n^{n-2}+...$$ Therefore, $$B = \begin{vmatrix} 1 & 1 & ...& 1 \\ x_1 & x_2 & ... & x_{n-1} \\ x_1^2 & x_2^2 & ... & x_{n-1}^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-3} & x_2^{n-3} & ...& x_{n-1}^{n-3} \\ x_1^{n-2} & x_2^{n-2} & ...& x_{n-1}^{n-2} \\ \end{vmatrix}$$ $$B$$ happens to be Vandermonde determinant of order (n-1), so $$B = \prod_{1 \leqslant i \leqslant j \leqslant (n-1)} (x_j-x_i)$$

6. It's been shown in (2) that the coefficient $$a_{n-1}$$ in the polynomial (1) is equal to zero, so $$A-B\sum_{k=1}^{n-1}x_k=0$$ $$A=B\sum_{k=1}^{n-1}x_k=\prod_{1 \leqslant i \leqslant j \leqslant (n-1)} (x_j-x_i)\sum_{k=1}^{n-1}x_k$$

7. So the determinant on the left is $$(A+Bx_n)\prod_{k=1}^{n-1}(x_n-x_k)=(\prod_{1 \leqslant i \leqslant j \leqslant (n-1)} (x_j-x_i)\sum_{k=1}^{n-1}x_k + \prod_{1 \leqslant i \leqslant j \leqslant (n-1)} (x_j-x_i)x_n)\prod_{k=1}^{n-1}(x_n-x_k)=\prod_{1 \leqslant i \leqslant j \leqslant (n-1)} (x_j-x_i)\prod_{k=1}^{n-1}(x_n-x_k)(\sum_{k=1}^{n-1}x_k + x_n)=\prod_{1 \leqslant i \leqslant j \leqslant n} (x_j-x_i)\sum_{k=1}^nx_k$$

$$\prod_{1 \leqslant i \leqslant j \leqslant n} (x_j-x_i)$$ is Vandermonde determinant of order $$n$$. Therefore, the left expression is equal to the the right one, which is $$\begin{vmatrix} 1 & 1 & ...& 1 \\ x_1 & x_2 & ... & x_n \\ x_1^2 & x_2^2 & ... & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-2} & x_2^{n-2} & ...& x_n^{n-2} \\ x_1^{n-1} & x_2^{n-1} & ...& x_n^{n-1} \\ \end{vmatrix}\sum_{k=1}^nx_k$$