Timeline for Is there a reasonable way to check intersection of these set of vectors?
Current License: CC BY-SA 3.0
15 events
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| Oct 1, 2017 at 18:39 | history | edited | Peter Heinig | CC BY-SA 3.0 |
Many corrections of obvious typos. Stylistic improvements. The 'placement' of '$a$' in front of the 'such that' was unfortunate. The new version is logically exactly the same as before, modulo the corrected errors.
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| Oct 1, 2017 at 17:51 | history | edited | Turbo | CC BY-SA 3.0 |
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| Oct 1, 2017 at 17:45 | history | edited | Turbo | CC BY-SA 3.0 |
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| Oct 1, 2017 at 17:14 | history | edited | Turbo | CC BY-SA 3.0 |
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| Oct 1, 2017 at 17:05 | comment | added | Peter Heinig | One more attempt (starting with this comment MO is exhorting me not to): what I mean is that $a$ can be eliminated in the sense that your condition ' $\forall i\in \{1,..,n\}\quad\langle v_i,w\rangle = a$ ' is equivalent to ' $\color{blue}{\forall i,j\in\{1,..,n\}\quad \langle v_i-v_j,w \rangle = 0}$ ' , the latter being the form I would strongly recommend to write your condition in, removing the '$a$' completely from the OP. | |
| Oct 1, 2017 at 16:40 | comment | added | Turbo | @PeterHeinig inner product all $v_i$ with $w$ is $a$. | |
| Oct 1, 2017 at 16:33 | comment | added | Peter Heinig | Re "is this trivial": it isn't 'trivial' in any sense of trivial I know, I think, but for the time being, please let me continue critizsing the formulation of the OP: it seems to me that the 'parameter' " $a$ " is totally redundant and (to me) confusing. The given data are $v_1,\dotsc, v_n,w\in\mathbb{Z}^n$, and $b\in\mathbb{Z}$, subject to the 'axiom' that the $v_i$ be linearly independent. The equation containing '$a$' is a tautology/does-not-impose-any-condition. It seems to me that the '$a$' can be eliminated from the context for good, right? | |
| Oct 1, 2017 at 16:22 | comment | added | Turbo | @PeterHeinig clarified but is this trivial? | |
| Oct 1, 2017 at 16:21 | history | edited | Turbo | CC BY-SA 3.0 |
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| Oct 1, 2017 at 16:10 | history | edited | Peter Heinig | CC BY-SA 3.0 |
The exponent at the symbol naming the type of $a$ and $b$ was evidently wrong: an inner product between vectors returns an integer, not a vector. Corrected. Small stylistic improvements. 'full rank system' is less clear than my formulation. Post respected.
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| Sep 30, 2017 at 20:49 | history | edited | Federico Poloni |
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| Sep 30, 2017 at 20:31 | history | edited | Turbo | CC BY-SA 3.0 |
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| Sep 30, 2017 at 20:30 | comment | added | Peter Heinig | The last line is puzzling: the condition ' $v=(\langle v_1,v\rangle,\langle v_2,v\rangle,\dots,\langle v_n,v\rangle)\in\Bbb Z^n$ ' is equivalent to $Mv=v$, where $M$ is the (as per your hypothesis: full-rank) matrix which has the $v_i$ for it rows; but row-rank equals column-rank, also for linear algebra over $\mathbb{Z}$, so this 'matrix equation' is non-contradictory only if $M$ is the identity matrix, i.e., $v_i=e_i$=the $i$-th standard basis vector, whereupon your last line becomes tautologous. Would you please clarify what you mean? | |
| Sep 30, 2017 at 20:17 | history | edited | Turbo | CC BY-SA 3.0 |
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| Sep 30, 2017 at 20:10 | history | asked | Turbo | CC BY-SA 3.0 |