7
$\begingroup$

As we know, a Cauchy determinant of size n admits the following explicit formula: $$\det \left(\frac{1}{x _i+y _j}\right) _{1\le i,j \le n}=\frac{\prod _{1\le i < j\le n} (x _j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i+y _j)}.$$

Is there something known about the following generalized Cauchy determinant? $$\det \left(\frac{A_i+B_j}{x _i+y _j}\right) _{1\le i,j \le n}.$$

More specially, how does it go for $$\det \left(\frac{A_i+A_j}{x _i+x _j}\right) _{1\le i,j \le n}.$$

A simple case is for $x_i=i$.

I wonder if there are some references about them. Thank you.

Improve this question
$\endgroup$
1
  • $\begingroup$ I don't know about the determinant, but there is some theory on how to compute inverses (and sums and products) of generalized Cauchy matrices. An important quantity is the rank of the numerator (displacement rank), in your case 2. For instance, the inverse of such a matrix (when it exists) will have the same displacement rank. Or there are algorithms to compute their LU decomposition in $O(n^2\cdot \text{(displacement_rank)})$ $\endgroup$
    – Federico Poloni
    Commented Mar 14, 2015 at 7:49

5 Answers 5

Reset to default
6
$\begingroup$

See (3.1) of Okada, S. "Generalizations of Cauchy’s Determinant Identity and Schur’s Pfaffian Identity"

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you. Maybe it is useful. But It seems not a "nice explicit" formula. $\endgroup$
    – cd14
    Commented Mar 14, 2015 at 0:38
  • $\begingroup$ do you have a result? "See (3.1) of Okada, S. "Generalizations of Cauchy’s Determinant Identity and Schur’s Pfaffian Identity" I do not find a nice result $\endgroup$
    – mathworker
    Commented Dec 31, 2020 at 16:12
3
$\begingroup$

A determinant of this type is discussed in these two publications on the two-dimensional square lattice Ising model on the rectangle:

Hucht, Alfred, The square lattice Ising model on the rectangle. I: Finite systems, J. Phys. A, Math. Theor. 50, No. 6, Article ID 065201, 23 p. (2017). ZBL1357.81154. https://arxiv.org/abs/1609.01963

Hucht, Alfred, The square lattice Ising model on the rectangle. II: Finite-size scaling limit, J. Phys. A, Math. Theor. 50, No. 26, Article ID 265205, 23 p. (2017). ZBL1369.82005. https://arxiv.org/abs/1701.08722

Improve this answer
$\endgroup$
3
$\begingroup$

There certainly is a pattern, although I have trouble finalizing it. Denominators are $\prod_{i,j}(x_i+y_j)$ while the numerators go like this:

$n=2$:

$$ \begin{aligned}\left(A_1 A_2+B_1 B_2\right) \left(x_1-x_2\right) \left(y_1-y_2\right)\\+\left(A_2 B_1+A_1 B_2\right) \left(x_2+y_1\right) \left(x_1+y_2\right)\\-\left(A_1 B_1+A_2 B_2\right) \left(x_1+y_1\right) \left(x_2+y_2\right)\end{aligned} $$

$n=3$:

$$ \begin{aligned} &\left(A_1 A_2 A_3+B_1 B_2 B_3\right) \left(x_1-x_2\right) \left(x_1-x_3\right) \left(x_2-x_3\right) \left(y_1-y_2\right) \left(y_1-y_3\right) \left(y_2-y_3\right)\\ \quad\\ +&\left(A_1 A_2 B_3+A_3 B_1 B_2\right) \left(x_1-x_2\right) \left(y_1-y_2\right)\left(x_3+y_1\right) \left(x_1+y_3\right)\left(x_3+y_2\right) \left(x_2+y_3\right)\\ +&\left(A_1 A_3 B_2+A_2 B_1 B_3\right) \left(x_1-x_3\right)\left(y_1-y_3\right) \left(x_2+y_1\right) \left(x_1+y_2\right) \left(x_3+y_2\right) \left(x_2+y_3\right)\\ +&\left(A_2 A_3 B_1+A_1 B_2 B_3\right) \left(x_2-x_3\right) \left(y_2-y_3\right) \left(x_2+y_1\right)\left(x_1+y_2\right) \left(x_3+y_1\right) \left(x_1+y_3\right)\\ \quad\\ -&\left(A_1 A_2 B_2+A_3 B_1 B_3\right) \left(x_1-x_2\right)\left(y_1-y_3\right) \left(x_2+y_2\right) \left(x_3+y_3\right) \left(x_1+y_2\right) \left(x_3+y_1\right)\\ -&\left(A_1 A_3 B_3+A_2 B_1 B_2\right) \left(x_1-x_3\right)\left(y_1-y_2\right) \left(x_2+y_2\right) \left(x_3+y_3\right) \left(x_1+y_3\right) \left(x_2+y_1\right)\\ +&\left(A_1 A_2 B_1+A_3 B_2 B_3\right) \left(x_1-x_2\right)\left(y_2-y_3\right) \left(x_1+y_1\right) \left(x_3+y_3\right) \left(x_2+y_1\right) \left(x_3+y_2\right)\\ +&\left(A_2 A_3 B_3+A_1 B_1 B_2\right) \left(x_2-x_3\right) \left(y_1-y_2\right) \left(x_1+y_1\right) \left(x_3+y_3\right) \left(x_2+y_3\right) \left(x_1+y_2\right)\\ -&\left(A_1 A_3 B_1+A_2 B_2 B_3\right) \left(x_1-x_3\right)\left(y_2-y_3\right) \left(x_1+y_1\right) \left(x_2+y_2\right) \left(x_3+y_1\right) \left(x_2+y_3\right)\\ -&\left(A_2 A_3 B_2+A_1 B_1 B_3\right) \left(x_2-x_3\right) \left(y_1-y_3\right) \left(x_1+y_1\right)\left(x_2+y_2\right)\left(x_3+y_2\right) \left(x_1+y_3\right) \end{aligned} $$

$n=4$:

$$\scriptstyle{ \begin{aligned} \ &(A_1A_2A_3A_4+B_1B_2B_3B_4)(x_1-x_2)(x_1-x_3)(x_2-x_3)(x_1-x_4)(x_2-x_4)(x_3-x_4)(y_1-y_2)(y_1-y_3)(y_2-y_3)(y_1-y_4)(y_2-y_4)(y_3-y_4)\\ \,\\ +&(A_1A_2B_1B_2+A_3A_4B_3B_4)(x_1-x_2)(x_3-x_4)(y_1-y_2)(y_3-y_4)(x_1+y_1)(x_2+y_2)(x_3+y_3)(x_4+y_4)(x_2+y_1)(x_1+y_2)(x_4+y_3)(x_3+y_4)\\ +&(A_1A_3B_1B_3+A_2A_4B_2B_4)(x_1-x_3)(x_2-x_4)(y_1-y_3)(y_2-y_4)(x_1+y_1)(x_2+y_2)(x_3+y_3)(x_4+y_4)(x_3+y_1)(x_1+y_3)(x_4+y_2)(x_2+y_4)\\ +&(A_1A_4B_1B_4 +A_2A_3B_2B_3)(x_1-x_4)(x_2-x_3)(y_1-y_4)(y_2-y_3)(x_1+y_1)(x_2+y_2)(x_3+y_3)(x_4+y_4)(x_1+y_4)(x_4+y_1)(x_2+y_3)(x_3+y_2)\\ +&(A_1A_2B_3B_4 +A_3A_4B_1B_2)(x_1-x_2)(x_3-x_4)(y_1-y_2)(y_3-y_4)(x_1+y_3)(x_3+y_1)(x_2+y_4)(x_4+y_2)(x_1+y_4)(x_4+y_1)(x_2+y_3)(x_3+y_2)\\ +&(A_1A_3B_2B_4+A_2A_4B_1B_3)(x_1-x_3)(x_2-x_4)(y_1-y_3)(y_2-y_4)(x_1+y_2)(x_2+y_1)(x_3+y_4)(x_4+y_3)(x_1+y_4)(x_4+y_1)(x_2+y_3)(x_3+y_2)\\ +&(A_1A_4B_2B_3+A_2A_3B_1B_4)(x_1-x_4)(x_2-x_3)(y_1-y_4)(y_2-y_3)(x_1+y_2)(x_2+y_1)(x_3+y_4)(x_4+y_3)(x_1+y_3)(x_3+y_1)(x_2+y_4)(x_4+y_2)\\ \,\\ +&(A_1A_2B_1B_4 +A_3A_4B_2B_3)(x_1-x_2)(x_3-x_4)(y_1-y_4)(y_2-y_3)(x_1+y_1)(x_3+y_3)(x_2+y_4)(x_4+y_2)(x_2+y_1)(x_4+y_3)(x_1+y_4)(x_3+y_2)\\ +&(A_1A_2B_2B_3+A_3A_4B_1B_4)(x_1-x_2)(x_3-x_4)(y_1-y_4)(y_2-y_3)(x_2+y_2)(x_4+y_4)(x_1+y_3)(x_3+y_1)(x_1+y_2)(x_3+y_4)(x_4+y_1)(x_2+y_3)\\ +&(A_1A_4B_1B_2+A_2A_3B_3B_4)(x_1-x_4)(x_2-x_3)(y_1-y_2)(y_3-y_4)(x_1+y_1)(x_3+y_3)(x_2+y_4)(x_4+y_2)(x_1+y_2)(x_3+y_4)(x_4+y_1)(x_2+y_3)\\ +&(A_2A_3B_1B_2+A_1A_4B_3B_4)(x_1-x_4)(x_2-x_3)(y_1-y_2)(y_3-y_4)(x_2+y_2)(x_4+y_4)(x_1+y_3)(x_3+y_1)(x_2+y_1)(x_4+y_3)(x_1+y_4)(x_3+y_2)\\ -&(A_1A_2B_1B_3+A_3A_4B_2B_4)(x_1-x_2)(x_3-x_4)(y_1-y_3)(y_2-y_4)(x_1+y_1)(x_4+y_4)(x_2+y_3)(x_3+y_2)(x_2+y_1)(x_3+y_4)(x_1+y_3)(x_4+y_2)\\ -&(A_3A_4B_1B_3+A_1A_2B_2B_4)(x_1-x_2)(x_3-x_4)(y_1-y_3)(y_2-y_4)(x_2+y_2)(x_3+y_3)(x_4+y_1)(x_1+y_4)(x_1+y_2)(x_4+y_3)(x_3+y_1)(x_2+y_4)\\ -&(A_2A_4B_2B_3+A_1A_3B_1B_4)(x_1-x_3)(x_2-x_4)(y_1-y_4)(y_2-y_3)(x_1+y_1)(x_2+y_2)(x_3+y_4)(x_4+y_3)(x_3+y_1)(x_4+y_2)(x_1+y_4)(x_2+y_3)\\ -&(A_1A_3B_2B_3+A_2A_4B_1B_4)(x_1-x_3)(x_2-x_4)(y_1-y_4)(y_2-y_3)(x_3+y_3)(x_4+y_4)(x_1+y_2)(x_2+y_1)(x_1+y_3)(x_2+y_4)(x_4+y_1)(x_3+y_2)\\ -&(A_1A_3B_1B_2+A_2A_4B_3B_4)(x_1-x_3)(x_2-x_4)(y_1-y_2)(y_3-y_4)(x_1+y_1)(x_4+y_4)(x_3+y_2)(x_2+y_3)(x_1+y_2)(x_4+y_3)(x_3+y_1)(x_2+y_4)\\ -&(A_2A_4B_1B_2+A_1A_3B_3B_4)(x_1-x_3)(x_2-x_4)(y_1-y_2)(y_3-y_4)(x_2+y_2)(x_3+y_3)(x_4+y_1)(x_1+y_4)(x_2+y_1)(x_3+y_4)(x_1+y_3)(x_4+y_2)\\ -&(A_2A_3B_1B_3+A_1A_4B_2B_4)(x_1-x_4)(x_2-x_3)(y_1-y_3)(y_2-y_4)(x_3+y_3)(x_4+y_4)(x_1+y_2)(x_2+y_1)(x_3+y_1)(x_4+y_2)(x_1+y_4)(x_2+y_3)\\ -&(A_1A_4B_1B_3+A_2A_3B_2B_4)(x_1-x_4)(x_2-x_3)(y_1-y_3)(y_2-y_4)(x_1+y_1)(x_2+y_2)(x_4+y_3)(x_3+y_4)(x_1+y_3)(x_2+y_4)(x_4+y_1)(x_3+y_2)\\ \,\\ +&(A_1A_2A_3B_4+A_4B_1B_2B_3)(x_1-x_2)(x_1-x_3)(x_2-x_3)(y_1-y_2)(y_1-y_3)(y_2-y_3)(x_1+y_4)(x_4+y_1)(x_2+y_4)(x_4+y_2)(x_3+y_4)(x_4+y_3)\\ +&(A_1A_2A_3B_2+A_4B_1B_3B_4)(x_1-x_2)(x_1-x_3)(x_2-x_3)(y_1-y_3)(y_1-y_4)(y_3-y_4)(x_2+y_2)(x_4+y_4)(x_1+y_2)(x_4+y_3)(x_4+y_1)(x_3+y_2)\\ +&(A_1A_2A_4B_3+A_3B_1B_2B_4)(x_1-x_2)(x_1-x_4)(x_2-x_4)(y_1-y_2)(y_1-y_4)(y_2-y_4)(x_1+y_3)(x_3+y_1)(x_2+y_3)(x_3+y_2)(x_3+y_4)(x_4+y_3)\\ +&(A_1A_2A_4B_1+A_3B_2B_3B_4)(x_1-x_2)(x_1-x_4)(x_2-x_4)(y_2-y_3)(y_2-y_4)(y_3-y_4)(x_1+y_1)(x_3+y_3)(x_2+y_1)(x_3+y_4)(x_4+y_1)(x_3+y_2)\\ +&(A_1A_3A_4B_2+A_2B_1B_3B_4)(x_1-x_3)(x_1-x_4)(x_3-x_4)(y_1-y_3)(y_1-y_4)(y_3-y_4)(x_1+y_2)(x_2+y_1)(x_2+y_3)(x_3+y_2)(x_2+y_4)(x_4+y_2)\\ +&(A_1A_3A_4B_4+A_2B_1B_2B_3)(x_1-x_3)(x_1-x_4)(x_3-x_4)(y_1-y_2)(y_1-y_3)(y_2-y_3)(x_2+y_2)(x_4+y_4)(x_2+y_1)(x_3+y_4)(x_1+y_4)(x_2+y_3)\\ +&(A_2A_3A_4B_1+A_1B_2B_3B_4)(x_2-x_3)(x_2-x_4)(x_3-x_4)(y_2-y_3)(y_2-y_4)(y_3-y_4)(x_1+y_2)(x_2+y_1)(x_1+y_3)(x_3+y_1)(x_1+y_4)(x_4+y_1)\\ +&(A_2A_3A_4B_3+A_1B_1B_2B_4)(x_2-x_3)(x_2-x_4)(x_3-x_4)(y_1-y_2)(y_1-y_4)(y_2-y_4)(x_1+y_1)(x_3+y_3)(x_1+y_2)(x_4+y_3)(x_1+y_4)(x_2+y_3)\\ -&(A_1A_2A_3B_1+A_4B_2B_3B_4)(x_1-x_2)(x_1-x_3)(x_2-x_3)(y_2-y_3)(y_2-y_4)(y_3-y_4)(x_1+y_1)(x_4+y_4)(x_3+y_1)(x_4+y_2)(x_2+y_1)(x_4+y_3)\\ -&(A_1A_2A_3B_3+A_4B_1B_2B_4)(x_1-x_2)(x_1-x_3)(x_2-x_3)(y_1-y_4)(y_2-y_4)(y_1-y_2)(x_3+y_3)(x_4+y_4)(x_1+y_3)(x_4+y_2)(x_4+y_1)(x_2+y_3)\\ -&(A_1A_2A_4B_4+A_3B_1B_2B_3)(x_1-x_2)(x_1-x_4)(x_2-x_4)(y_1-y_2)(y_1-y_3)(y_2-y_3)(x_3+y_3)(x_4+y_4)(x_3+y_1)(x_2+y_4)(x_1+y_4)(x_3+y_2)\\ -&(A_1A_2A_4B_2+A_3B_1B_3B_4)(x_1-x_2)(x_1-x_4)(x_2-x_4)(y_1-y_3)(y_1-y_4)(y_3-y_4)(x_2+y_2)(x_3+y_3)(x_1+y_2)(x_3+y_4)(x_3+y_1)(x_4+y_2)\\ -&(A_1A_3A_4B_3+A_2B_1B_2B_4)(x_1-x_3)(x_1-x_4)(x_3-x_4)(y_1-y_2)(y_1-y_4)(y_2-y_4)(x_2+y_2)(x_3+y_3)(x_2+y_1)(x_4+y_3)(x_1+y_3)(x_2+y_4)\\ -&(A_1A_3A_4B_1+A_2B_2B_3B_4)(x_1-x_3)(x_1-x_4)(x_3-x_4)(y_2-y_3)(y_2-y_4)(y_3-y_4)(x_1+y_1)(x_2+y_2)(x_2+y_4)(x_3+y_1)(x_4+y_1)(x_2+y_3)\\ -&(A_2A_3A_4B_2+A_1B_1B_3B_4)(x_2-x_3)(x_2-x_4)(x_3-x_4)(y_1-y_3)(y_1-y_4)(y_3-y_4)(x_1+y_1)(x_2+y_2)(x_4+y_2)(x_1+y_3)(x_1+y_4)(x_3+y_2)\\ -&(A_2A_3A_4B_4+A_1B_1B_2B_3)(x_2-x_3)(x_2-x_4)(x_3-x_4)(y_1-y_2)(y_1-y_3)(y_2-y_3)(x_1+y_1)(x_4+y_4)(x_1+y_3)(x_2+y_4)(x_1+y_2)(x_3+y_4)\\ \end{aligned}} $$ and so on...

Improve this answer
$\endgroup$
5
  • 2
    $\begingroup$ Each second product (with x's and y's) is a determinant of a Vandermonde matrix. The first product (with A's and B's) shows which x's and y's stand in Vandermonde matrix. $A_2B_1B_2B_3$ means that we must construct Vandermonde using $x_2$, $y_1$, $y_2$ and $y_3$. $\endgroup$
    – Alexey Ustinov
    Commented Jun 14, 2017 at 6:45
  • $\begingroup$ @AlexeyUstinov I certainly understand, and you are certainly right, it is just that I still cannot bring it to the end. I made it community wiki, feel free to do it (if you want). $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 14, 2017 at 7:46
  • 1
    $\begingroup$ The pattern is easier to prove than to write down, sadly. The matrix $\left( \dfrac{A_i+B_j}{x_i+y_j} \right)_{1\leq i\leq n, \ 1\leq j\leq n}$ can be written as the sum $P + Q$, where $P = \left( \dfrac{B_j}{x_i+y_j} \right)_{1\leq i\leq n, \ 1\leq j\leq n}$ and $Q = \left( \dfrac{A_i}{x_i+y_j} \right)_{1\leq i\leq n, \ 1\leq j\leq n}$. There is a known (tedious) formula for $\det\left(P+Q\right)$ as an alternating sum of products of a minor of $P$ with the complementary minor of $Q$. Finally, ... $\endgroup$
    – darij grinberg
    Commented Jun 17, 2017 at 16:27
  • $\begingroup$ ... computing minors of $P$ is easy, because after factoring out the $B_j$'s (which are uniform per column and thus factor out easily) each such minor becomes a little Cauchy matrix. Similarly for minors of $Q$. This gives the exact huge sums you have found, except that there are some terms that can be combined (e.g., you can combine $A_1 A_2 \left(x_1-x_2\right) \left(y_1-y_2\right)$ with $B_1 B_2 \left(x_1-x_2\right) \left(y_1-y_2\right)$ to get $\left(A_1 A_2 + B_1 B_2\right) \left(x_1-x_2\right) \left(y_1-y_2\right)$). There are no cancellations, since each choice of minor of $P$ gives ... $\endgroup$
    – darij grinberg
    Commented Jun 17, 2017 at 16:27
  • 1
    $\begingroup$ ... rise to one term, and no two such terms share the same $A_? A_? \cdots A_? B_? B_? \cdots B_?$ monomial. As for which terms can be combined, if my quick combinatorics isn't wrong, the answer is that each term corresponding to some minor $P^{I}_{J}$ and the complementary minor $Q^{\overline{I}}_{\overline{J}}$ can be combined with the term corresponding to $P^{\overline{I}}_{\overline{J}}$ and $Q^{I}_{J}$, and these are the only possible combinations (i.e., the terms get combined into pairs). $\endgroup$
    – darij grinberg
    Commented Jun 17, 2017 at 16:30
1
$\begingroup$

Not an answer, but: experiment seems to show no nice pattern for the numerator of the determinant, but the denominator seems to be the product of all the $x$ variables and all the squares of the sums of pairs of $x$-es.

Improve this answer
$\endgroup$
3
  • 4
    $\begingroup$ It is immediate from the expansion of the determinant (in the special case $y_i=x_i$) that it can be written in the form suggested by Igor Rivin. $\endgroup$
    – Richard Stanley
    Commented Mar 14, 2015 at 4:54
  • $\begingroup$ Thanks. The numerator has no nice formula? If we impose on the restriction $A_i+A_{i+1}=\frac 1 i$, what will happen? $\endgroup$
    – cd14
    Commented Mar 14, 2015 at 7:30
  • $\begingroup$ @cd14 I did not try the special case, i will look... $\endgroup$
    – Igor Rivin
    Commented Mar 14, 2015 at 16:04
1
$\begingroup$

As already noted by others, it does not seem feasible a general formula shapes up here. Having said that, the special case $A_j=Ay_j$ and $B_i=Bx_i$ finds a closed form involving (unsurprisingly) the Vandermonde determinant; see page 2 of the paper

Tewodros Amdeberhan and Doron Zeilberger, "Trivializing" generalizations of some Izergin-Korepin-type determinants, Discrete Mathematics and Theoretical Computer Science, DMTCS, 2007, 9 (1), pp.203–206, hal-00966504 (pdf)

To wit, $$ \det\left(\frac{Ay_j+Bx_i}{y_j+x_i}\right)_{i,j}^{1,n} =(A-B)^{n-1}\frac{(A\pmb{y}+(-1)^{n-1}B\pmb{x})\cdot\prod_{i<j}(x_i-x_j)(y_i-y_j)}{2\prod_{i,j}^{1,n}(x_i+y_j)}; $$ where $\pmb{x}=\prod_{i=1}^nx_i$ and $\pmb{y}=\prod_{j=1}^ny_j$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .