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RE: Re: Are points co-planar in (numDimensions-1)?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg43287] RE: [mg43240] Re: Are points co-planar in (numDimensions-1)?
  • From: "Ingolf Dahl" <ingolf.dahl at telia.com>
  • Date: Sat, 23 Aug 2003 08:09:18 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Use NullSpace!
The following vectors are evidently coplanar:

vectors = {{1., 0., 0.}, {0., 1., 0.}, {1., 1., 0.}};

If you take NullSpace of this, you get the orthogonal complement 
(a set of all orthogonal vectors) to this set.

NullSpace[vectors]

gives

Out[2]= {{0., 0., 1.}}

which is orthogonal to all the vectors.

If you take 

NullSpace[NullSpace[vectors]]

you get back a set of vectors that span your original set.
This method works for any number of dimensions.

Ingolf Dahl
Sweden

>-----Original Message-----
>From: AngleWyrm [mailto:no_spam_anglewyrm at hotmail.com]
To: mathgroup at smc.vnet.net
>Sent: Thursday, August 21, 2003 04:26
>To: mathgroup at smc.vnet.net
>Subject: [mg43287] [mg43240] Re: Are points co-planar in (numDimensions-1)?
>
>
>My last post on this subject lacked depth, so here's more info.
>
>Given some n-dimensional vectors, are they coplanar in n-1? Let 
>a1, a2, ..., an be vectors. If they
>are coplanar, then there exists a set of coefficients {k1, k2, 
>..., kn}, not all zero, which satisfy
>the equation:
>
>k1 a1 + k2 a2 + ... + kn an = 0.
>
>I have a dataset I wish to test for this property in various dimensions:
>
>numDimensions = 2;
>dataSet = ReadList["data.txt", Number ];
>dataSet = Partition[ dataSet, numDimensions ];
>
>At this point I have a set of 2D vectors, and if I take any two of 
>them they HAVE to be coplanar,
>right?
>
>sample = Take[ dataSet, numDimensions];
>Sum[ k\_i sample[[i]], {i, numDimensions} ]    (I've used \_ to 
>indicate subscript here)
>
>Now my problem is in solving for zero on the last equation. I've 
>tried like so:
>Solve[ Sum[ k\_i sample[[i]], {i, numDimensions} ] ==0, {k\_i} ]
>
>but it gives an empty set every time. Thanks for any help.
>-Jonathan
>


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