8
$\begingroup$

One of the more utilized determinant is that of Vandermonde's

$$\begin{vmatrix} 1&x_1&x_1^2&\dots&x_1^{n-1}\\ 1&x_2&x_2^2&\dots&x_2^{n-1}\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ 1&x_n&x_n^2&\dots&x_n^{n-1}\\ \end{vmatrix}=\prod\limits_{1\leq i<j\leq n}(x_j-x_i).$$ In short, we write $\det(x_i^{j-1})$. I became curious of a Hankel-type (not quite) formulation of these monomial entries as $\det(x_{i+j}^{j-1})$. A quick check reveal that the latter has no neat evaluation.

On the other hand, if we specialize $x_{i+j}$ to the numerical values $i+j$ then it appears that $$\det((i+j)^{j-1})=\prod_{k=1}^{n-1}k!\tag1$$

Question. Why is true?

Remark. The above evaluation equals the familiar $\det(i^{j-1})$ which OEIS lists as A000178 with many interpretations, including other determinants but (1) is not one of them.

$\endgroup$
13
$\begingroup$

Here is a slightly more general version: If $P_n(x)$ are some polynomials of degree $n-1$ with leading term $a_nx^{n-1}$ then $$\det(P_j(x_i))=\prod_{j=1}^n a_n\prod\limits_{1\leq i<j\leq n}(x_j-x_i).$$ The proof is a simple reduction to Vandermonde's determinant. The first column is all constant, and every column after that can be written as the vector $a_j(x_1^{j-1},\dots,x_n^{j-1})^{T}$ minus a linear combination of previous columns. Therefore a collection of elementary column operations bring our matrix to the form $(a_jx_i^{j-1})$.

Your determinant corresponds to the case $P_n(x)=(x+n)^{n-1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.