Timeline for Prescribing areas of parallelograms (or 2x2 principal minors)
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Sep 24, 2017 at 2:34 | history | edited | François G. Dorais |
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| May 30, 2012 at 17:05 | vote | accept | Julien Maubon | ||
| May 29, 2012 at 21:02 | answer | added | George Lowther | timeline score: 7 | |
| May 24, 2012 at 14:54 | answer | added | Camilo Sarmiento | timeline score: 0 | |
| May 24, 2012 at 13:18 | history | edited | Julien Maubon | CC BY-SA 3.0 |
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| May 23, 2012 at 23:22 | answer | added | Igor Rivin | timeline score: 1 | |
| May 23, 2012 at 23:06 | history | edited | Julien Maubon | CC BY-SA 3.0 |
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| May 23, 2012 at 22:54 | comment | added | George Lowther | For the case with $a_{ij} > 0$, setting $\lambda_i=\Vert\nu_i\Vert^{-2}$ and $u_i=\nu_i/\Vert\nu_i\Vert$, then $1-\lambda_i\lambda_ja_{ij}=\langle u_iu_j\rangle^2$. So the question becomes whether there exists such $\lambda_i$ such that $1-\lambda_i\lambda_ja_{ij}$ are the elements of the componentwise square of a positive semidefinite matrix. But how do you tell if a matrix is the componentwise square of a positive semidefinite one? I don't know if there is a simple way. This question might be slightly relevant: mathoverflow.net/questions/52642 | |
| May 23, 2012 at 22:43 | comment | added | Pietro Majer | Indeed, one can then take the vectors $v_1,\dots,v_n$ as the columns of $\sqrt{G}$. To find $G$ a possible approach is by induction on $n$: assuming we have found $G_{ij}$ for $i$ and $j$ from 1 to $n-1$ satsfying the equations and defining a semidefinite matrix of order $n-1$ the problem becomes to find $G_{in}$ satisfying the further equations and extending the preceding (n-1)x(n-1) matrix to a semidefinite nxn matrix. | |
| May 23, 2012 at 22:38 | comment | added | Julien Maubon | @Noah Stein: I thought about this way of formulating the problem, and then forgot about it... But after reading your comment I googled something like "prescribed principal minors" and found some papers dealing about prescribing all the principal minors of a symmetric matrix. Maybe I can make something out of it. I will edit my question so that this formulation appears more clearly. Thanks a lot ! | |
| May 23, 2012 at 20:47 | comment | added | Noah Stein | I don't have much to contribute except a slight reformulation. Since you have written everything in terms of the inner products $\langle v_i, v_j\rangle$ you can state the problem in terms of the Gram matrix of the $v_i$, the matrix with entries $G_{ij} = \langle v_i, v_j\rangle$. A square matrix $G$ is the Gram matrix of some list of vectors if and only if it is positive semidefinite, so your question becomes: does there exist a positive semidefinite matrix with prescribed $2\times 2$ principal minors? This seems like something someone must have studied. | |
| May 23, 2012 at 18:31 | comment | added | Victor Protsak | Yes, you are right! I was thinking too quickly, and the diagram didn't work: the projection of the union of two faces is not the remaining face - in your construction, it even has zero area. | |
| May 23, 2012 at 18:00 | comment | added | Julien Maubon | @Victor Protsak: I am not sure I understand what you mean. If $n=3$ the problem always has a solution. Assume for simplicity that $a_{12},a_{13},a_{23}$ are all $>0$. Then taking 3 orthogonal vectors $v_1,v_2,v_3\in{\mathbb R}^3$ such that $\|v_1\|^2=\frac{a_{12}a_{13}}{a_{23}}$, $\|v_2\|^2=\frac{a_{12}a_{23}}{a_{13}}$ and $\|v_3\|^2=\frac{a_{13}a_{23}}{a_{12}}$ does it | |
| May 23, 2012 at 16:40 | comment | added | Victor Protsak | Here is another obvious necessary condition: $a_{ij}+a_{jk}\leq a_{ik}$ (triangle inequality). That this is true in dimension 3 can be seen by projecting two of the three "side" faces of the tetrahedron with edges $v_i, v_j, v_k$ onto the remaining one; it is true in general because any 3 vectors span at most 3-dimensional subspace of $\mathbb{R}^n.$ Moreover, in case of equality $v_i, v_j, v_k$ are coplanar, which imposes a rank condition on the matrix similar to the one you gave for a pair of collinear vectors. | |
| May 23, 2012 at 13:13 | history | edited | Julien Maubon | CC BY-SA 3.0 |
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| May 23, 2012 at 13:03 | history | asked | Julien Maubon | CC BY-SA 3.0 |