Timeline for Unique matrix satisfying a system of equations
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 3, 2011 at 1:36 | vote | accept | Kap | ||
| Nov 3, 2011 at 0:49 | comment | added | Kap | The entries in the matrix are from the real numbers. One could view this as an equation $Ax = 1$, where the distinct elements from $G$ are in the vector $x$ and the $j:th$ row in $A$ contains the distinct elements from the matrix $2v_jv_j^T - diag(v_jv_j^T)$. Because of the rank constraint, the elements in $x$ are dependent. We can write them as $\sqrt{g_{ii}g_{jj}}\cos(\alpha_{ij})$ where some $\alpha_{ij} = \alpha_i - \alpha_j$,i.e., some depend on the other angles. So the question is whether any $np - p(p-1)/2$ equations are enough to either get a finite number solutions or no solutions. | |
| Nov 2, 2011 at 23:32 | answer | added | Robert Israel | timeline score: 4 | |
| Nov 2, 2011 at 17:35 | comment | added | Igor Rivin | AH, I have missed the rank constraint. | |
| Nov 2, 2011 at 16:29 | comment | added | Gerhard Paseman | Also, isn't it the case that there are potentially infinitely many solutions in the case that the entries come from a skew field? The poster has not made it clear from where the matrix entries come. Gerhard "Ask Me About System Design" Paseman, 2011.11.02 | |
| Nov 2, 2011 at 14:26 | comment | added | Denis Serre | @Igor. Just try the case $n=2$ and $p=1$. You'll find two solutions. | |
| Nov 2, 2011 at 14:21 | comment | added | Denis Serre | @Igor. This is not a linear system, unless $p=0$. There is the nonlinear constraint that $G$ has rank $p$. | |
| Nov 2, 2011 at 13:40 | comment | added | Igor Rivin | You are asking whether a non degenerate system of $n$ linear equations in $n$ unknowns has a finite number of solutions? The answer is "yes", and one can sharpen the result to "a unique solution". | |
| Nov 2, 2011 at 13:16 | history | asked | Kap | CC BY-SA 3.0 |