- #1
Avatrin
- 245
- 6
Hi
I already know the "usual" Gram-Schmidt Process:
But, then I discovered another Gram-Schmidt process that gives an orthogonal basis (in Matrices and Linear transformations by Charles Cullen):
Essentially, it is the same except you do not divide by <uj,uj>. Cullen calls this the Gram-Schmidt process, yet it is different. Does it give a different basis?
The explanations for the usual Gram-Schmidt process are usually visual while the explanation for this one was algebraic. I cannot make sense of it in any of the visual explanations of the Gram-Schmidt process, yet it makes perfect sense algebraically;
How are these processes different? Or, if they aren't, why?
I already know the "usual" Gram-Schmidt Process:
uk = vk - Ʃj=1k-1projujvk
But, then I discovered another Gram-Schmidt process that gives an orthogonal basis (in Matrices and Linear transformations by Charles Cullen):
uk = vk - Ʃj=1k-1(vk*uj)uj
Essentially, it is the same except you do not divide by <uj,uj>. Cullen calls this the Gram-Schmidt process, yet it is different. Does it give a different basis?
The explanations for the usual Gram-Schmidt process are usually visual while the explanation for this one was algebraic. I cannot make sense of it in any of the visual explanations of the Gram-Schmidt process, yet it makes perfect sense algebraically;
Starting with a basis {u1,u2...,uk-1} and a vector vk that is not in its span to create a vector uk that is orthogonal to all ui:
uk*ui = vk*ui + uiƩj=1k-1ajuj = vk*ui + ai = 0
Thus, ai = -vk*ui for uk to be orthogonal to all ui. This gives us what I wrote above.
How are these processes different? Or, if they aren't, why?
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