How to compute a more general version of Vandermonde / Cauchy double alternant determinant

Consider some variables $$\{X_i\}_{1\le i \le n}$$, $$\{Y_i\}_{1\le i \le n}$$, and $$\{W_i\}_{1\le i \le n}$$. Does anyone know how to compute the following determinant? $$\det ~ \left(\frac{W_j^{i-1}}{X_i+Y_j}\right)_{1\le i,j\le n}.$$

Update: If you could also provide an answer for the case where $$W_j=1$$ for $$j\ge2$$, that would be sufficient for the problem I have encountered in my research.

• It's easy to come up with families of matrices; you had another one at mathoverflow.net/questions/358175/… . Before asking multiple questions of this sort, I think that it is appropriate to give some more information about how this arises, and why it is of particular significance (as opposed to just one of a sequence of close but not identical questions). – LSpice Apr 25 '20 at 1:08
• Thanks for your comment @LSpice. As I mentioned in my question, these matrices are coming up in some polynomial regression that I'm doing for my research. I have been going over Christian Krattenthaler's "Advanced Determinant Calculus", and I realized I can state my question in a more general format as you can see in the new edition of my question. – Ahmadreza Momeni Apr 26 '20 at 22:09

Maybe you can get something out of the technique of displacement equations. It works as follows.

Notice, first, that a matrix $$A$$ is a Cauchy-like matrix if and only if it satisfies the so-called displacement equation $$LA-AR = vu^T$$, where $$L$$ and $$R$$ are diagonal matrices (containing the nodes) and $$vu^T$$ is a generic rank-1 term.

Suppose you are given a matrix $$A$$.

1. Find (if you can!) two matrices $$L$$ and $$R$$ such that $$LA-AR = vu^T$$ has rank 1. If $$A$$ is a Cauchy matrix $$A_{ij} = \frac{1}{X_i + Y_j}$$, then $$L = diag(X_i)$$ and $$R = diag(Y_j)$$ work, while for a Vandermonde matrix $$A = W_j^{i-1}$$ then $$L$$ is a shift matrix and $$R$$ is $$diag(W_j)$$.

2. Diagonalize (if you can do it explicitly) $$L = VD_LV^{-1}$$ and $$R=UD_RU^{-1}$$, and then with some algebra you get $$D_L V^{-1}AU - V^{-1}AUD_R = V^{-1}vu^TU$$.

3. Then, $$V^{-1}AU$$ is a Cauchy matrix with nodes the entries of $$D_L$$ and $$D_R$$, because of that displacement equation. You can compute its determinant, and use it to get the determinant of $$A$$.

My conjecture is that the answer is close to this: $$\frac{\prod_{1\le i < j \le n} (X_j - X_i) (W_jY_j - W_iY_i)} {\prod_{1\le i ,j \le n}(X_i + Y_j)}.$$

• For $n = 2$, the determinant is a fraction whose numerator is an irreducible degree-$3$ polynomial. So I don't think the answer is that easy. – darij grinberg Apr 28 '20 at 14:42
• @darijgrinberg For $n=2$, my conjectured answer has a numerator which is an irreducible degree-3 polynomial too. Am I missing something? – Ahmadreza Momeni Apr 29 '20 at 1:27
• Yours is not irreducible. – darij grinberg Apr 29 '20 at 5:33
• @darijgrinberg I see, Thanks! – Ahmadreza Momeni Apr 30 '20 at 0:27