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.

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    $\begingroup$ 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). $\endgroup$ – LSpice Apr 25 '20 at 1:08
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    $\begingroup$ 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. $\endgroup$ – 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: $$ \frac{\prod_{1\le i < j \le n} (X_j - X_i) (W_jY_j - W_iY_i)} {\prod_{1\le i ,j \le n}(X_i + Y_j)}. $$

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

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