3
$\begingroup$

The ubiquitous Vandermonde matrix, of entries $(x_i^{j-1})_{i,j}^{1,n}$, and its determinant $$\prod_{i<j}^{1,n}(x_j-x_i)$$ have found many utilities in Combinatorics and Physics, among other places.

At present, I got interested in a slight variation given by the matrix entiries $$\pmb{M}_n:=(x_i^j-x_j^i)_{i,j}^{1,n}.$$ Experimental calculation prompted me to ask:

QUESTION 1. Is this true? The determinant $\det\pmb{M}_{2n}$ is the square of a multi-variable polynomial.

QUESTION 2. Is this true? The determinant $\det\pmb{M}_{2n+1}$ vanishes.

Example. If $2n=2$ then $\det\pmb{M}_2=(x_1^2-x_2)^2$. If $2n=4$ then \begin{align*} &\det\begin{pmatrix} 0&x_1^2-x_2&x_1^3-x_3&x_1^4-x_4 \\ -x_1^2+x_2&0&x_2^3-x_3^2&x_2^4-x_4^2 \\ -x_1^3+x_3&-x_2^3+x_3^2&0&x_3^4-x_4^3 \\ -x_1^4+x_4&-x_2^4+x_4^2&-x_3^4+x_4^3&0 \end{pmatrix} \\ &=(x_1^4x_2^3 - x_1^3x_2^4 - x_1^4x_3^2 + x_1^2x_3^4 + x_1^3x_4^2 - x_1^2x_4^3 + x_2^4x_3 - x_2x_3^4 - x_2^3x_4 + x_2x_4^3 + x_3^2x_4 - x_3x_4^2)^2. \end{align*}

Remark. The numerical matrix, with $x_j=j$ so that $\pmb{M}_n=(i^j-j^i)_{i,j}^{1,n}$, appears in OEIS but without further details.

Improve this question
$\endgroup$
2
  • 9
    $\begingroup$ Your matrix is anti-symmetric. The determinant of any odd anti-symmetric matrix is 0, and the determinant of any even anti-symmetric matrix is a square (known as the Pfaffian). The first part is completely elementary and follows from applying the determinant to the identity $A=-A^T$, while the second part is due to Cayley and has several proofs. $\endgroup$
    – Ofir Gorodetsky
    Commented Feb 2, 2022 at 16:58
  • 1
    $\begingroup$ Thank you, indeed. $\endgroup$
    – T. Amdeberhan
    Commented Feb 2, 2022 at 17:18

0

Reset to default

You must log in to answer this question.