# Determinant of matrix $(x_j^{n-i}- x_j^{2n-i})_{i,j=1}^{n}$

Good evening all, I am determined to determine this determinant:

$$D = \det{\left[x_j^{n-i} - x_j^{2n-i}\right]_{i,j=1}^{n}}$$

Looking at the smaller cases, leads me to believe that

$$D = \prod_{1 \leq i < j \leq n}\left(x_i-x_j\right)\prod_{i=1}^n \left(1-{x_i}^n\right)$$

although I am having trouble showing this. I know that, since the determinant is an alternating function in the variables $x_1,\dots x_n$ it follows that

$$\frac{D}{\displaystyle\prod_{1 \leq i < j \leq n}\left(x_i-x_j\right)}$$

is a symmetric polynomial of degree $n^2$ (the degree of D minus the degree of the Vandermonde part).

How can I show that this symmetric polynomial is exactly $\prod_{i=1}^n \left(1-{x_i}^n\right)$ ?

Your help is, as always, much appreciated.

• "I am determined to determine this determinant" -Nice – user38268 Feb 16 '12 at 1:24

We would like to know the determinant $\det M$ of the matrix $n\times n$-matrix $$M_{ij} = x_j^{n-i} - x_j^{2n-i}.$$

Note that $$M_{ij} = (1- x_j^n) x_j^{n-i} = (1-x_j^n) V_{ij}.$$ It is easy to see (column operation) that $$\det M = \det V \,\prod_{j=1}^n (1-x_j^n).$$

Now $V_{ij} = x^{n-i}_j$ is a Vandermonde matrix whose determinant is known (and can be proven by induction) to be $$\det V= \prod_{1 \leq i < j \leq n}\left(x_i-x_j\right).$$

• Nice answer Fabian. It turns out it was not so "easy" for me to see, and indeed I was unaware that scaling a column by a constant also scales the determinant by the same magnitude. +1 – Daniel Pietrobon Feb 15 '12 at 15:30
• @DanielPietrobon: maybe you were unaware of the fact but I'm sure once told it is also kind of obvious (e.g., via Cramer's rule). "Easy" was only indicating that no proof is needed. – Fabian Feb 15 '12 at 18:10

The $(i,j)$-entry can be expressed as: $x_j^{n-i}-x_j^{2n-i}=x_j^{n-i}(1-x_j^n)$.

So column $j$ has the common factor $(1-x_j^n)$. Using multilinearity of the determinant pull these factors out (from each column). Then you have

$$D = \prod_{j=1}^n (1-x_j^n) \; \mathrm{det} \begin{bmatrix} x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \\ x_1^{n-2} & x_2^{n-2} & \cdots & x_n^{n-2} \\ \vdots & \vdots & & \vdots \\ x_1 & x_2 & \cdots & x_n \\ 1 & 1 & \cdots & 1 \end{bmatrix}$$

This an upside down Vandermonde. Interchanging rows $1$ and $n$, $2$ and $n-1$, etc. gives us the sign $(-1)^{\ell}$ where $\ell$ is the floor of $n/2$. Then using the Vandermonde determinant formula finishes the job.

So I think your result is off by sign. As Fabian pointed out, the OP's product is $x_i-x_j$ for $i<j$ instead of $x_j-x_i$ (that appear in the standard Vandermonde formula), so the sign cancels back out and the result follows (no adjustment needed).

• I think his result is correct. The usual form of the Vandermonde matrix has factors $x_j - x_i$ in the product. In the question the factors are written as $x_i - x_j$ which cancels the factor $(-1)^\ell$ (in a sense everything depends on how exactly one defines the Vandermonde matrix) – Fabian Feb 15 '12 at 15:03
• @Fabian Thanks! I didn't catch that. – Bill Cook Feb 15 '12 at 15:14