Timeline for Determinant of walk matrix for a skew-symmetric matrix of even order
Current License: CC BY-SA 4.0
20 events
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| Apr 29, 2021 at 6:42 | history | edited | W. Wang | CC BY-SA 4.0 |
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| Apr 27, 2021 at 19:05 | comment | added | Ilya Bogdanov | @darij As for criterion: Apply your argument to each irreducible factor separately. As for reasoms to put an alternative proof: this is kind of technique which is widely applicable, and it replaces some art with more routine actions; I think this may be quite jnstructive. | |
| Apr 27, 2021 at 18:38 | comment | added | darij grinberg | @IlyaBogdanov: Your answer is beautiful; I don't see much of a point in adding my own, particularly as I'd have to prove or source some commutative algebra (I'm not sure how easy your criterion for no multiple factors is to prove). | |
| Apr 27, 2021 at 15:20 | history | became hot network question | |||
| Apr 27, 2021 at 12:28 | comment | added | Ilya Bogdanov | (Sorry for multiple comments) the Pfaffian has no multiple factors merely because it is linear in each variable. | |
| Apr 27, 2021 at 12:21 | comment | added | Ilya Bogdanov | @darij BTW you don't need irreducibility; you obly beed that the Pfaffian has no multiple irreducible factors (still I don't know whether it is easier to prove) | |
| Apr 27, 2021 at 12:18 | comment | added | Ilya Bogdanov | @darij Thanks a lot! Could you put this as another answer? | |
| Apr 27, 2021 at 11:33 | comment | added | W. Wang | @darij:Now, I understand your approach. It is also very cool, and is more conceptual( borrowing the language of IIya Bogdanov.) | |
| Apr 27, 2021 at 11:14 | comment | added | darij grinberg | ... by Hilbert's Nullstellensatz (and using the irreducibility of the Pfaffian, which I believe is well-known), this entails that the Pfaffian of $S$ divides the determinant of $W$ as a polynomial over the prime field, which we can take to be $\mathbb{Q}$. Finally, this results in a divisibility over $\mathbb{Z}$ by Gauss's lemma. | |
| Apr 27, 2021 at 11:13 | comment | added | darij grinberg | @W.Wang: Ah, I see. You can actually argue a bit simpler: Over a field, $\det S = 0$ is equivalent to $\operatorname{Pf} S = 0$ and implies that $S$ has rank $\leq n-2$ (since the rank of an alternating matrix over a field is always even). Hence, for any vector $w$, the $n-1$ vectors $S^1 w, S^2 w, \ldots, S^{n-1} w$ are $n-1$ vectors in a space of dimension $\leq n-2$ (viz., the column space of $S$), and therefore are linearly independent. Hence, the columns of $W$ are linearly dependent, so that $\det W = 0$. Now, ... | |
| Apr 27, 2021 at 10:48 | comment | added | W. Wang | @darij, det S=0 implies S has 0 as a multiple eigenvalue. Since rank W is at most the number of distinct eigenvalues of S (using the spectral decomposition theorem for normal matrices) . This fact may imply the division relation as you comment. I know little about commutative algebra, and I will try your approach. I recall similar argument can be used to give an interesting (unusual) proof for Vandermonde determinant. | |
| Apr 27, 2021 at 9:55 | comment | added | Ilya Bogdanov | @darij Ah, I see. Anyway, in fact, I would be glad to see something more conceptual and connected with the Pfaffian than my answer... | |
| Apr 27, 2021 at 9:51 | comment | added | darij grinberg | @IlyaBogdanov: Note that I wrote $\operatorname{Pf}$, not $\det$. But this is moot anyway, as your answer appears to be much better (will read it later tonight). | |
| Apr 27, 2021 at 9:36 | vote | accept | W. Wang | ||
| Apr 27, 2021 at 9:07 | comment | added | Ilya Bogdanov | @darij The commutative algebra approach from your first comment won’t work, otherwise you would prove $\det S\mid \det W$ which may fail (as in the OP’s example). | |
| Apr 27, 2021 at 9:02 | comment | added | darij grinberg | Actually, if I am understanding things right, $e$ can be any vector, not necessarily the all-ones vector. | |
| Apr 27, 2021 at 9:01 | answer | added | Ilya Bogdanov | timeline score: 7 | |
| Apr 27, 2021 at 8:48 | comment | added | darij grinberg | Actually, it is sufficient to prove that $\operatorname{Pf} S = 0$ implies $\det W = 0$ over any field, since the Pfaffian is a primitive irreducible polynomial over the integers (at least I believe so). Maybe it is even sufficient to restrict oneself to characteristic-$0$ fields. | |
| Apr 27, 2021 at 8:14 | comment | added | darij grinberg | "In particular, when $\det(S)=0$ then $\det(W)=0$. This particular case is not hard to show." If you can prove that $\operatorname{Pf} S = 0$ implies $\det W = 0$ over an arbitrary commutative ring, then you have solved your problem. Do you use the field property somewhere? | |
| Apr 27, 2021 at 7:20 | history | asked | W. Wang | CC BY-SA 4.0 |