Linear Algebra: Orthonormal Basis


by tylerc1991
Tags: algebra, basis, linear, orthonormal
tylerc1991
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#1
Apr14-11, 09:16 PM
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1. The problem statement, all variables and given/known data

Find an orthonormal basis for the subspace of R^4 that is spanned by the vectors: (1,0,1,0), (1,1,1,0), (1,-1,0,1), (3,4,4,-1)

3. The attempt at a solution

When I try to use the Gram-Schmidt process, I am getting (before normalization): (1,0,1,0), (0,1,0,0), (1,0,-1,2), (0,0,0,0). So obviously there is some mistake that I am making but I have checked this at least 3 times. Can someone help me and let me know if it is something on my end or the problem. Thank you.
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Dick
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Apr14-11, 09:29 PM
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Without have actually worked it out fully, why do you think you made a mistake? The dimension of the subspace is less than 4. You are only going to get a number of orthonormal vectors equal to the dimension of the subspace.
tylerc1991
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Apr14-11, 09:33 PM
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Quote Quote by Dick View Post
Without have actually worked it out fully, why do you think you made a mistake? The dimension of the subspace is less than 4. You are only going to get a number of orthonormal vectors equal to the dimension of the subspace.
I thought that the basis had to span R^4? And since of of the elements was (0,0,0,0) the basis can't span R^4.

Dick
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#4
Apr14-11, 09:35 PM
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Linear Algebra: Orthonormal Basis


Quote Quote by tylerc1991 View Post
I thought that the basis had to span R^4? And since of of the elements was (0,0,0,0) the basis can't span R^4.
It doesn't span R^4. It spans the subspace of R^4 spanned by the given vectors.
tylerc1991
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#5
Apr14-11, 09:37 PM
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Quote Quote by Dick View Post
It doesn't span R^4. It spans the subspace of R^4 spanned by the given vectors.
I see. So will (0,0,0,0) be included in the orthonormal basis?
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Apr14-11, 09:41 PM
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Quote Quote by tylerc1991 View Post
I see. So will (0,0,0,0) be included in the orthonormal basis?
(0,0,0,0) is in EVERY subspace. You throw that away. It's never part of a basis. A basis is a set of linearly independent vectors. And it certainly isn't normal.


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