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MathGroup Archive 2005

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Re: GramSchmidt problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57076] Re: GramSchmidt problem
  • From: Heath Gerhardt <heathgerhardt at hotmail.com>
  • Date: Sat, 14 May 2005 04:58:16 -0400 (EDT)
  • Organization: Department of Computer Science, University of Calgary
  • References: <d5uvda$9at$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Wow, wasn't expecting such useful responses. Thanks for the excellent 
tips guys! I'm sure they will come in handy on many future projects.

Heath

Heath Gerhardt wrote:
> Does anyone know if there are any know problems with GramSchmidt
> crashing Mathematica? When I run it on these 4 25-dimensional vectors it
> crashes on my system:
> 
> \!\({1\/2\ \((3 + \@5)\), 1\/2\ \((\(-3\) - \@5)\), 1\/2\ \((1 + \@5)\), 0, 
>     1\/2\ \((\(-1\) - \@5)\), 1\/2\ \((\(-3\) - \@5)\), 1\/2\ \((3 + \@5)\), 
>     1\/2\ \((\(-1\) - \@5)\), 0, 1\/2\ \((1 + \@5)\), 1\/2\ \((1 + \@5)\), 
>     1\/2\ \((\(-1\) - \@5)\), 1, 0, \(-1\), 0, 0, 0, 0, 0, 
>     1\/2\ \((\(-1\) - \@5)\), 1\/2\ \((1 + \@5)\), \(-1\), 0, 1}\)
> 
> \!\({1\/2\ \((1 + \@5)\), 1\/2\ \((\(-3\) - \@5)\), 1\/2\ \((3 + \@5)\), 
>     1\/2\ \((\(-1\) - \@5)\), 0, 1\/2\ \((\(-1\) - \@5)\), 
>     1\/2\ \((3 + \@5)\), 1\/2\ \((\(-3\) - \@5)\), 1\/2\ \((1 + \@5)\), 0,
> 1, 
>     1\/2\ \((\(-1\) - \@5)\), 1\/2\ \((1 + \@5)\), \(-1\), 0, 0, 0, 0, 0, 
>     0, \(-1\), 1\/2\ \((1 + \@5)\), 1\/2\ \((\(-1\) - \@5)\), 1, 0}\)
> 
> \!\({1\/2\ \((1 + \@5)\), 1\/2\ \((\(-1\) - \@5)\), 1, 0, \(-1\), 
>     1\/2\ \((\(-3\) - \@5)\), 1\/2\ \((3 + \@5)\), 1\/2\ \((\(-1\) - \@5)\), 
>     0, 1\/2\ \((1 + \@5)\), 1\/2\ \((3 + \@5)\), 1\/2\ \((\(-3\) - \@5)\), 
>     1\/2\ \((1 + \@5)\), 0, 1\/2\ \((\(-1\) - \@5)\), 
>     1\/2\ \((\(-1\) - \@5)\), 1\/2\ \((1 + \@5)\), \(-1\), 0, 1, 0, 0, 0, 0, 
>     0}\)
> 
> \!\({1, 1\/2\ \((\(-1\) - \@5)\), 1\/2\ \((1 + \@5)\), \(-1\), 0, 
>     1\/2\ \((\(-1\) - \@5)\), 1\/2\ \((3 + \@5)\), 1\/2\ \((\(-3\) - \@5)\), 
>     1\/2\ \((1 + \@5)\), 0, 1\/2\ \((1 + \@5)\), 1\/2\ \((\(-3\) - \@5)\), 
>     1\/2\ \((3 + \@5)\), 1\/2\ \((\(-1\) - \@5)\), 0, \(-1\), 
>     1\/2\ \((1 + \@5)\), 1\/2\ \((\(-1\) - \@5)\), 1, 0, 0, 0, 0, 0, 0}\)
> 
> 
> thanks in advance
> Heath
> 


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