Timeline for Null space of random $(0,1)$ binary matrix
Current License: CC BY-SA 3.0
13 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Dec 25, 2013 at 22:01 | comment | added | user117230 | Yes zero coordinates are allowed. However in my example matrix above there is no such vector I believe even with zero coordinates. The problem comes originally from a version of group testing I have become interested in. | |
| Dec 25, 2013 at 20:29 | comment | added | Andy Novocin | In your new problem do you also want to allow zero coordinates? Is there a place where your restrictions are coming from? | |
| Dec 25, 2013 at 20:01 | comment | added | user117230 | I didn't properly express the requirements of my problem in this problem so have posted a follow up question. In particular, the elements of the vector all have to have the same absolute value. | |
| Dec 25, 2013 at 19:56 | comment | added | Andy Novocin | $\begin{bmatrix} -2 \\ -1 \\ 1 \\ 1 \end{bmatrix}$ | |
| Dec 25, 2013 at 19:55 | vote | accept | user117230 | ||
| Dec 25, 2013 at 19:54 | comment | added | user117230 | I see what is going on now. (The answer to my question is (4,2,-2,-2)). Thank you. | |
| Dec 25, 2013 at 19:51 | comment | added | Andy Novocin | Lattices are more restrictive than vector spaces (less scalars to work with), further $\mathbb{Q}^n$ contains $\mathbb{Z}^n$, so if there is no answer in $\mathbb{Q}^n$ then there is no answer in $\mathbb{Z}^n$. If there is an answer in $\mathbb{Q}^n$ then there is an answer in $\mathbb{Z}^n$. | |
| Dec 25, 2013 at 19:49 | review | First posts | |||
| Dec 25, 2013 at 19:57 | |||||
| Dec 25, 2013 at 19:46 | comment | added | user117230 | That's very interesting, thank you. Just for my interest, which non-zero integer vector is in the null space of $ M = \begin{pmatrix} 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1\\ 0 & 1 & 0 & 1\ \end{pmatrix}.$ ? | |
| Dec 25, 2013 at 19:32 | comment | added | Andy Novocin | If you have a rational vector in your null space you can just multiply by the lcm of the denominators and have an integer vector in the null space. A scalar times a null space vector is still in the null space. | |
| Dec 25, 2013 at 19:31 | comment | added | user117230 | I am not sure it is the same. I am asking if there is a non-zero vector with only integer coordinates in the null space. In my case all operations are over $\mathbb{Z}$ and I don't see the mapping to your examples. | |
| Dec 25, 2013 at 19:30 | history | answered | Andy Novocin | CC BY-SA 3.0 |