Timeline for Irreducibility of determinant of symmetric matrix
Current License: CC BY-SA 2.5
21 events
| when toggle format | what | by | license | comment | |
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| Aug 20, 2016 at 19:14 | comment | added | Emolga | @zroslav, I don't understand the proof you gave as an update in the question. I will be grateful for a clarification of it here: math.stackexchange.com/questions/1893344/… | |
| Jul 24, 2014 at 5:21 | answer | added | Luke Oeding | timeline score: 1 | |
| Dec 29, 2010 at 21:10 | comment | added | zroslav | @Petya: I don't understand how to see it. | |
| Dec 28, 2010 at 18:12 | comment | added | Petya | There is a well-known connection between algebraic geometry and geometry of polytopes. Concerning your original question - it is more or less easy to see that a set of non-singular points of the discriminant hypersurface is connected. It implies the irreducibility of the hypersurface. | |
| Dec 28, 2010 at 16:55 | comment | added | zroslav | @Petya: no, it's still algebraic in my point of view. I meant the proof in terms of algebraic geometry (such proofs usually can be generalized). | |
| Dec 27, 2010 at 20:23 | comment | added | Petya | The question on polytopes looks (for me) as geometric as it possible. | |
| Dec 27, 2010 at 20:13 | history | edited | Denis Serre | CC BY-SA 2.5 |
added 1 characters in body
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| Dec 27, 2010 at 8:20 | comment | added | Aaron Meyerowitz | Aha.. given that the determinant of a general n by n matrix $A$ with $n^2$ variables is irreducible let $B=A A^{T}$ then $B$ is symmetric etc. etc. that is pretty cute. | |
| Dec 27, 2010 at 7:31 | comment | added | zroslav | @Davidac897: the construction of resultants is given in the third chapter | |
| Dec 27, 2010 at 7:28 | vote | accept | zroslav | ||
| Dec 27, 2010 at 7:24 | comment | added | zroslav | @Davidac897: It follows from the irreducibility of resultant which follows from irreducibility of variety dual to irreducible (the first chapter). @Aaron: $A=(a_{i,j})$, $\det(A)\in\mathbb C[a_{i,j}, 1\leq i,j\leq n]$, $\det(A A^T)\in\mathbb C[a_{i,j}, 1\leq i,j\leq n]$ ($a_{i,j}\ne a_{j,i}$). Do you understand what I meant? @Petya: This is a still algebraic proof. | |
| Dec 27, 2010 at 4:37 | comment | added | Aaron Meyerowitz | I did not understand your algebraic proof. It is true for diagonal matrices (for example) that $\det (A A^{T})=det(A)^2$ and yet the determinant there is neither irreducible nor the square of an irreducible. What is the ingredient I am missing? | |
| Dec 26, 2010 at 15:48 | vote | accept | zroslav | ||
| Dec 27, 2010 at 7:16 | |||||
| Dec 26, 2010 at 7:48 | history | edited | zroslav | CC BY-SA 2.5 |
algebraic proof
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| Dec 26, 2010 at 3:25 | answer | added | algori | timeline score: 8 | |
| Dec 26, 2010 at 3:22 | comment | added | Petya | Is it possible to decompose Newton polytope of the determinant of symmetric matrix into Minkowski sum of two polytopes? Negative answer implies irreducibility. | |
| Dec 26, 2010 at 1:43 | comment | added | David Corwin | Could you please provide a reference to the exact place inside the book? | |
| Dec 26, 2010 at 0:05 | answer | added | Luis H Gallardo | timeline score: 0 | |
| Dec 25, 2010 at 20:52 | history | edited | zroslav | CC BY-SA 2.5 |
corrected formulating
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| Dec 25, 2010 at 18:20 | answer | added | Dick Palais | timeline score: 0 | |
| Dec 25, 2010 at 18:13 | history | asked | zroslav | CC BY-SA 2.5 |