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Problem:

Let $\delta \in \mathbb{R}^+$ and $n\in \mathbb{N}$. The matrix $A_n = (a_{i,j}) \in \mathbb{R}^{n\times n}$ is defined as

$$ a_{i,j} = \prod_{k=0}^{i-2}\left((j-1)\delta +n-k\right) $$ Prove that $$\det A_n = \delta ^{\frac{1}{2}n(n-1)}\prod^{n-1}_{k=0}k!$$

So there is

$$ A_1 = \pmatrix{ 1\\ } $$

$$ A_2 = \pmatrix{ 1&1\\ 2&\delta+2\\ } $$

$$ A_3 = \pmatrix{ 1&1&1\\ 3&\delta+3&2\delta+3\\ 6&(\delta+3)(\delta+2)&(2\delta+3)(2\delta+2)\\ } $$

$$\vdots$$

I eventually managed to prove it by converting the matrix to upper triangular using elementary row operations, but the proof is just too complicated, involves things like $(k−2)\delta+n−(i−(k−1))+1)$-th multiples of certain rows (for that matter it is quite long, so I am not fully including it here). So it somehow feels like not the best possible way to do this.

What are some another ways to prove this?

Sil
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1 Answers1

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Finally found this in a literature, namely Calculation of some determinants using the s-shifted factorial by Jean-Marie Normand. In Lemma $1$ he proves equation $(3.5)$:

For complex numbers $z_j, s, b_i$ we have$$\tag{1}\det [(b_i+z_j)_{s;i}]_{i,j=0,\dots,n-1}=\prod_{0\leq i \leq j \leq n-1}(z_j-z_i)$$ where $(z)_{s;i}=z(z+s)\dots(z+(n-1)s)$ is called $s$-shifted factorial and $\prod_{0\leq i\leq j\leq n-1}(z_j-z_i)$ is a Vandermonde determinant.

Furthermore Appendix B shows in equation (B$.5$) that for special case $b_i=0, z_j=b+aj$ one has

$$ \det[(b_i+z_j)_{s;i}]_{i,j=0,\dots,n-1}=\det[(b+aj)_{s;i}]_{i,j=0,\dots,n-1}=a^{n(n-1)/2}\prod_{j=0}^{n-1}j!.\tag{2} $$

Before applying to our problem, we first need to re-index the range to $i,j=0,\dots,n-1$, then we have $a_{i,j}=(j\delta+n)_{1;i}$. Now choosing $b_i=0, a=\delta, b=n, s=1$ in $(2)$ gives exactly the statement we wanted to prove.

Sil
  • 16,612