Timeline for How to calculate one Cauchy type determinant
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 31, 2020 at 16:39 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 4.0 |
tried to make mathjax formatting more robust
|
| Oct 12, 2018 at 20:47 | history | edited | Wolfgang | CC BY-SA 4.0 |
restructured n=3
|
| Jun 20, 2017 at 16:44 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
did not fit in my browser window otherwise
|
| Jun 18, 2017 at 8:52 | history | edited | Wolfgang | CC BY-SA 3.0 |
still some more rearrangements
|
| Jun 17, 2017 at 21:20 | history | edited | Wolfgang | CC BY-SA 3.0 |
some re-arranging of terms to better show the patterns
|
| Jun 17, 2017 at 16:30 | comment | added | darij grinberg | ... 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). | |
| Jun 17, 2017 at 16:27 | comment | added | darij grinberg | ... 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 ... | |
| Jun 17, 2017 at 16:27 | comment | added | darij grinberg | 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, ... | |
| Jun 17, 2017 at 14:20 | history | edited | Wolfgang | CC BY-SA 3.0 |
deleted 2 characters in body
|
| Jun 17, 2017 at 14:06 | history | edited | Wolfgang | CC BY-SA 3.0 |
edited body
|
| Jun 17, 2017 at 13:53 | history | edited | Wolfgang | CC BY-SA 3.0 |
some re-arranging of terms to better show the patterns (not done in all lines)
|
| Jun 15, 2017 at 7:42 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
mathjax fitting
|
| Jun 14, 2017 at 7:46 | comment | added | მამუკა ჯიბლაძე | @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). | |
| Jun 14, 2017 at 7:44 | history | made wiki | Post Made Community Wiki by მამუკა ჯიბლაძე | ||
| Jun 14, 2017 at 6:45 | comment | added | Alexey Ustinov | 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 $y_3$. | |
| Jun 14, 2017 at 5:42 | history | answered | მამუკა ჯიბლაძე | CC BY-SA 3.0 |