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So I understand that the effective formula for the orthogonal basis of a matrix is the same in both modified and classical Gram Schmidt algorithm. Can someone explain whats the numerical instability that arises with classical gram schmidt? and how does modified solve it.

flag	asked Jun 18 2011 at 10:53 dasa 31●2

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this sounds like classic numeric linear algebra textbook stuff. please have a look at: maths.manchester.ac.uk/~higham/asna/index.php – S. Sra Jun 18 2011 at 13:29

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Type "numerical instability gram-schmidt" into Google and you get plenty of well-informed responses. Even YouTube video demonstrations. – Ryan Budney Jun 18 2011 at 22:03

Nikola-M · Answer 1 · 2011-06-18 18:58:34Z UTC

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Here's the link:

http://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process

answered Jun 18 2011 at 18:58

Nikola-M
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Nikola-M · Answer 2 · 2011-06-18 18:54:15Z UTC

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As I remember, because of rounding off the numbers - vectors loose their orthoginality in non-modified version. Wikipedia article about Gram-Schmidt explains the modification.

answered Jun 18 2011 at 18:54

Nikola-M
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modified gram schmidt…

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