# Closed-form solution for the determinant of a Vandermonde-like matrix

I'm trying to find a closed-form solution $$\forall$$ odd integer $$n\ge 3$$ for the determinant of a matrix with some structure on it. After some manipulation, I've reduced it to the following matrix:

$$\small\begin{bmatrix}\boldsymbol{t_{1}^{n}-t_{a}^{n}} & \boldsymbol{t_{2}^{n}-t_{a}^{n}} & \boldsymbol{\cdots} & \boldsymbol{t_{a-1}^{n}-t_{a}^{n}} & nt_{1}^{n-1} & \cdots & nt_{a-1}^{n-1} & nt_{a}^{n-1}\\ \boldsymbol{t_{1}^{n-1}-t_{a}^{n-1}} & \boldsymbol{t_{2}^{n-1}-t_{a}^{n-1}} & \boldsymbol{\cdots} & \boldsymbol{t_{a-1}^{n-1}-t_{a}^{n-1}} & (n-1)t_{1}^{n-2} & \cdots & (n-1)t_{a-1}^{n-2} & (n-1)t_{a}^{n-2}\\ \boldsymbol{\vdots} & \boldsymbol{\vdots} & \boldsymbol{\ddots} & \boldsymbol{\vdots} & \vdots & \ddots & \vdots & \vdots\\ \boldsymbol{t_{1}^{2}-t_{a}^{2}} & \boldsymbol{t_{2}^{2}-t_{a}^{2}} & \boldsymbol{\cdots} & \boldsymbol{t_{a-1}^{2}-t_{a}^{2}} & 2t_{1} & \cdots & 2t_{a-1} & 2t_{a}\\ \boldsymbol{t_{1}-t_{a}} & \boldsymbol{t_{2}-t_{a}} & \boldsymbol{\cdots} & \boldsymbol{t_{a-1}-t_{a}} & 1 & \cdots & 1 & 1 \end{bmatrix}_{n\times n}$$

where $$a:=\frac{n+1}{2}$$, the bold block is $$n\times(\frac{n+1}{2}-1)$$, and the non-bold block is $$n \times \frac{n+1}{2}$$.

Although it has some similarities with the Vandermonde Matrix or some generalizations, it's not the same. Using some values of n, its determinant looks pretty simple, which leads me to think that there should be a closed-form solution:

$$n=3$$: $$det\left( \left[\begin{array}{ccc} {t_{1}}^3-{t_{2}}^3 & 3\,{t_{1}}^2 & 3\,{t_{2}}^2\\ {t_{1}}^2-{t_{2}}^2 & 2\,t_{1} & 2\,t_{2}\\ t_{1}-t_{2} & 1 & 1 \end{array}\right] \right)= -{\left(t_{1}-t_{2}\right)}^4$$

$$n=5$$: $$det\left( \left[\begin{array}{ccccc} {t_{1}}^5-{t_{3}}^5 & {t_{2}}^5-{t_{3}}^5 & 5\,{t_{1}}^4 & 5\,{t_{2}}^4 & 5\,{t_{3}}^4\\ {t_{1}}^4-{t_{3}}^4 & {t_{2}}^4-{t_{3}}^4 & 4\,{t_{1}}^3 & 4\,{t_{2}}^3 & 4\,{t_{3}}^3\\ {t_{1}}^3-{t_{3}}^3 & {t_{2}}^3-{t_{3}}^3 & 3\,{t_{1}}^2 & 3\,{t_{2}}^2 & 3\,{t_{3}}^2\\ {t_{1}}^2-{t_{3}}^2 & {t_{2}}^2-{t_{3}}^2 & 2\,t_{1} & 2\,t_{2} & 2\,t_{3}\\ t_{1}-t_{3} & t_{2}-t_{3} & 1 & 1 & 1 \end{array}\right] \right)= -{\left(t_{1}-t_{2}\right)}^4\,{\left(t_{1}-t_{3}\right)}^4\,{\left(t_{2}-t_{3}\right)}^4$$

I was wondering if there is a known closed-form solution for this determinant, or if it could be found using the determinant of a generalized Vandermonde matrix

Thanks!

These seem to be a direct consequence of Schendel's 1891 theorem about "confluent Vandermonde matrices", as explained in (say) "On a Recursion Formula Related to Confluent Vandermonde", Shui-Hung Hou and Edwin Hou, The American Mathematical Monthly, Vol. 122, No. 8 (October 2015), pp. 766-772, or here or here. An example of a confluent Vandermonde determinant is $$\begin{vmatrix}x^3&3x^2&y^3&3y^2\\x^2&2x&y^2&2y\\x&1&y&1\\1&0&1&0\end{vmatrix}$$ where certain columns of an ordinary Vandermonde determinant have been replaced by derivatives of others. Schendel's formula is that this determinant is equal to $$(x-y)^4$$. If you subtract the third column from the first, and then expand by minors, you end up with the identities $$\begin{vmatrix}x^3&3x^2&y^3&3y^2\\x^2&2x&y^2&2y\\x&1&y&1\\1&0&1&0\end{vmatrix} = \begin{vmatrix}x^3-y^3&3x^2&y^3&3y^2\\x^2-y^2&2x&y^2&2y\\x-y&1&y&1\\0&0&1&0\end{vmatrix} = -\begin{vmatrix}x^3-y^3&3x^2&3y^2\\x^2-y^2&2x&2y\\x-y&1&1\end{vmatrix}.$$ This is the OPs $$n=3$$ example.

Schendel's general formula for confluent Vandermonde determinants is $$\prod_{i, where the value $$x_i$$ is used to form $$n_i$$ columns consisting of the first $$n_i$$ derivatives (the $$0$$-th up through the $$n_i-1$$-th derivative) of the usual Vandermonde column $$(1,x_i, x_i^2,\ldots)'$$. In the OP's case, all of the $$n_i=2$$. The ordinary Vdm formula has all $$n_i=1$$.

• Thanks! Yeah, I think this is the answer, I've posted below the example for n=5. Just a minor thing: I think the element (3,2) of the matrix of the last determinant should be 1. But otherwise, this is correct, thanks!
– Lab
Commented Jul 3, 2020 at 13:23
• Yes, it should be a $1$. Thanks for the catch. Commented Jul 3, 2020 at 13:26

For the case $$n=5$$, and using the suggestion from the author of the correct answer:

$$det\left(\begin{bmatrix}t_{1}^{5} & 5t_{1}^{4} & t_{2}^{5} & 5t_{2}^{4} & t_{3}^{5} & 5t_{3}^{4}\\ t_{1}^{4} & 4t_{1}^{3} & t_{2}^{4} & 4t_{2}^{3} & t_{3}^{4} & 4t_{3}^{3}\\ t_{1}^{3} & 3t_{1}^{2} & t_{2}^{3} & 3t_{2}^{2} & t_{3}^{3} & 3t_{3}^{2}\\ t_{1}^{2} & 2t_{1} & t_{2}^{2} & 2t_{2} & t_{3}^{2} & 2t_{3}\\ t_{1} & 1 & t_{2} & 1 & t_{3} & 1\\ 1 & 0 & 1 & 0 & 1 & 0 \end{bmatrix}\right)=\left(t_{1}-t_{2}\right)^{4}\left(t_{1}-t_{3}\right)^{4}\left(t_{2}-t_{3}\right)^{4}=$$

$$=det\left(\begin{bmatrix}t_{1}^{5}-t_{3}^{5} & 5t_{1}^{4} & t_{2}^{5}-t_{3}^{5} & 5t_{2}^{4} & t_{3}^{5} & 5t_{3}^{4}\\ t_{1}^{4}-t_{3}^{4} & 4t_{1}^{3} & t_{2}^{4}-t_{3}^{4} & 4t_{2}^{3} & t_{3}^{4} & 4t_{3}^{3}\\ t_{1}^{3}-t_{3}^{3} & 3t_{1}^{2} & t_{2}^{3}-t_{3}^{3} & 3t_{2}^{2} & t_{3}^{3} & 3t_{3}^{2}\\ t_{1}^{2}-t_{3}^{2} & 2t_{1} & t_{2}^{2}-t_{3}^{2} & 2t_{2} & t_{3}^{2} & 2t_{3}\\ t_{1}-t_{3} & 1 & t_{2}-t_{3} & 1 & t_{3} & 1\\ 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}\right)=-det\left(\begin{bmatrix}t_{1}^{5}-t_{3}^{5} & 5t_{1}^{4} & t_{2}^{5}-t_{3}^{5} & 5t_{2}^{4} & 5t_{3}^{4}\\ t_{1}^{4}-t_{3}^{4} & 4t_{1}^{3} & t_{2}^{4}-t_{3}^{4} & 4t_{2}^{3} & 4t_{3}^{3}\\ t_{1}^{3}-t_{3}^{3} & 3t_{1}^{2} & t_{2}^{3}-t_{3}^{3} & 3t_{2}^{2} & 3t_{3}^{2}\\ t_{1}^{2}-t_{3}^{2} & 2t_{1} & t_{2}^{2}-t_{3}^{2} & 2t_{2} & 2t_{3}\\ t_{1}-t_{3} & 1 & t_{2}-t_{3} & 1 & 1 \end{bmatrix}\right)=$$

$$=-det\left(\begin{bmatrix}t_{1}^{5}-t_{3}^{5} & t_{2}^{5}-t_{3}^{5} & 5t_{1}^{4} & 5t_{2}^{4} & 5t_{3}^{4}\\ t_{1}^{4}-t_{3}^{4} & t_{2}^{4}-t_{3}^{4} & 4t_{1}^{3} & 4t_{2}^{3} & 4t_{3}^{3}\\ t_{1}^{3}-t_{3}^{3} & t_{2}^{3}-t_{3}^{3} & 3t_{1}^{2} & 3t_{2}^{2} & 3t_{3}^{2}\\ t_{1}^{2}-t_{3}^{2} & t_{2}^{2}-t_{3}^{2} & 2t_{1} & 2t_{2} & 2t_{3}\\ t_{1}-t_{3} & t_{2}-t_{3} & 1 & 1 & 1 \end{bmatrix}\right)$$