How to construct a vector orthogonal to all but one?

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To construct a vector orthogonal to all but one of a set of n linearly independent vectors, both the Gram-Schmidt process and a brute-force method of solving equations can be utilized. The Gram-Schmidt process is seen as a way to orthogonalize a complete set, but it can be modified to focus solely on the vector of interest. The brute-force approach involves solving n-1 equations to ensure the desired orthogonality. The discussion highlights that while both methods are valid, they do not inherently create orthogonal vectors among the other vectors. A proposed expression for the orthogonal vector is w = [v1; v2; ... ; vn] - 1(1, 0, 0, ..., 0).
td21
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Given n linearly independent vectors, v1, v2, v3, ...vn.
How to find construct a vector that is orthogonal to v2, v3, ..., vn (all v but not v1)?
Is Gram Schmitt process the way to do this? or just by brute force?
 
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I take it this is in Rn ? Then: yes !
Gram Schidt is brute force as far as I am concerned.
 
BvU said:
I take it this is in Rn ? Then: yes !
Gram Schidt is brute force as far as I am concerned.
Thanks for the reply. But why is Gram Schmidt process needed? My original brute force idea is to solve n-1 equations.
 
I also wonder if there is a neat expression for such vector?:oldsmile:
 
Ah, sorry, I didn't read carefully enough. You have no prior knowledge of the vv ... vn to exploit, so orthogonalizing seems to me the only way to get rid of the components of v1 that are in the subspace spanned by vv ... vn ... But by now I'm not all that certain any more...
 
Gram-Schmidt is not necessary. People usually think of Gram-Schmidt as making an entire set of n orthogonal vectors. Is that what you mean?

Since you only care about finding the orthogonal part of v1, I think you can modify the Gram-Schmidt process to work on only v1. Just keep subtracting the projection of the next vi on what remains of v1.

Your "brute-force" method of solving equations also works. Just like the modified Gram-Schmidt, it does nothing to make the v2, ..., vn orthodonal. I don't know which approach would be less "brute-force".

PS. Don't you mean n equations? You want the dot product of x with v1 to be nonzero (set it = 1) and all the other dot products to be =0.
 
td21 said:
I also wonder if there is a neat expression for such vector?:oldsmile:
I think that would be w = [v1; v2; ... ; vn]-1(1, 0, 0, ..., 0)
 
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