Timeline for Eigenvalues of a matrix with entries involving combinatorics
Current License: CC BY-SA 3.0
15 events
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| Nov 29, 2016 at 12:58 | comment | added | No_way | @AlexeyUstinov Yes, I think they bear some links to algebraic topology. As I pointed out in an answer to your another MO question on the eigenvectors of these matrices, these polynomials are related to the generating function $\displaystyle\left(\frac{t}{\sinh t}\right)^y$. While I do not know precisely the link, but the similar function $\displaystyle \frac{\frac{\sqrt{t}}{2}}{\sinh \frac{\sqrt{t}}{2}}$ is used to define the $\widehat{A}$-genus of manifolds. See en.wikipedia.org/wiki/Genus_of_a_multiplicative_sequence | |
| Nov 29, 2016 at 10:47 | comment | added | Alexey Ustinov | @No_way This matrices map any polynomial vector to the polynomial vector of the same (or lower) degree. If matrix is degenerate then we can find some polynomial in its kernel. It is very nice that you are interested in this polynomials. Will they have any meaning in algebraic topology? | |
| Nov 29, 2016 at 9:39 | comment | added | No_way | @AlexeyUstinov Eigenvectors do not depend on $l$ because those matrices (for a fixed $n$) commute with each other and thus are simultaneously diagonalizable. Thank you for your answer, and I am very interested in the appearance of the special numerical sequence in a particular entries of an eigenvector. Could you please explain why this sequence shows up? By the way I have just given another set of eigenvectors (not normalised to integers as you did here) in an answer to your another question in MO. | |
| Nov 29, 2016 at 4:23 | comment | added | Alexey Ustinov | @Lev Borisov Yes, integer values which are mutually coprime. For examle for $v_1$ we need a factor $1/(n,2)$ | |
| Nov 28, 2016 at 16:19 | comment | added | Lev Borisov | @Alexey Ustinov I presume that this means that there are formulas for the eigenvectors that involve sums of $1/k$. The LCM in A025529 must be an artifact of trying to get integer values for the eigenvectors. | |
| Nov 28, 2016 at 14:07 | comment | added | Pat Devlin | Oh! My apologies! | |
| Nov 28, 2016 at 14:06 | comment | added | Alexey Ustinov | They are rows in the matrix $V$ for $n=4$. You can check it for $M$ above or for $M$ with $l=3$: $\left( \begin{array}{cccc} 10 & 16 & 1 & 0 \\ 4 & 19 & 4 & 0 \\ 1 & 16 & 10 & 0 \\ 0 & 10 & 16 & 1 \\ \end{array} \right)$ | |
| Nov 28, 2016 at 14:01 | comment | added | Pat Devlin | Neat. Would you mind listing the eigenvectors for say $n=4$? This would likely give some insight. | |
| Nov 28, 2016 at 13:59 | comment | added | Alexey Ustinov | @Pat Devlin I was also surprised. I've checked this fact for $n$ and $l$ large enough. | |
| Nov 28, 2016 at 13:16 | comment | added | Pat Devlin | Well, I admit I was thinking of general $n,$ but I thought even for $n=4$ or so there was dependence on $l$. Of course everything in sight should be a polynomial. | |
| Nov 28, 2016 at 13:13 | comment | added | Alexey Ustinov | @Pat Devlin No, you can easely see it in the case $n=2$. I don't see any reasons to ignore the last row and column of $M$ so far. | |
| Nov 28, 2016 at 12:50 | comment | added | Pat Devlin | Eigenvectors do seem to depend on $l$ in the last coordinate, no? Which is why I think we should ignore the last row and column of $M.$ | |
| Nov 28, 2016 at 6:50 | history | edited | Alexey Ustinov | CC BY-SA 3.0 |
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| Nov 28, 2016 at 6:39 | history | edited | Alexey Ustinov | CC BY-SA 3.0 |
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| Nov 28, 2016 at 5:35 | history | answered | Alexey Ustinov | CC BY-SA 3.0 |