Gram-Schmidt Process: Example 3 in PDF

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Homework Statement


Example 3 in this pdf: http://karin.fq.uh.cu/qct/Tema_04/0...es propios de un sistema/Diagonalization.pdf"

Homework Equations


Gram-schmidt process:
v2 perp = v2 - (u1*v2)u1


The Attempt at a Solution


I don't understand how they got the second orthonormal basis vector. Using the equation above, with u1=1/sqrt(2)[-1 1 0] and v2=[-1 0 1] (these vectors are vertical), I don't get what they got as the second basis vector. Am I doing the Gram-Schmidt process the wrong way?
 
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Any non-zero multiple of an eigenvector is also an eigenvector. The results in the examples have been normalized. Did you normalize yours?
 
The examples state that the normalization has been done through Gram-Schmidt but I don't get the same results when I try to normalize them with Gram-Schmidt. Is there another way of normalizing?
 
Gram-Schmidt doesn't normalize; it orthogonalizes.
 
Okay, so if I normalize [-1 1 0], I get 1/sqrt(2)[-1 1 0] which is what the examples indicate. However, if I normalize [-1 0 1], I get 1/sqrt(2)[-1 0 1], which doesn't match the example. I don't get how they got 1/sqrt(6)[-1 -1 2] from [-1 0 1].
 
Try normalizing [-1 1 2].
 
Where did you get [-1 1 2]? Why can't 1/sqrt(2)[-1 0 1] be a vector in the orthonormal basis?
 
Let's back up a second. Did you calculate the second (unnormalized) vector using the Gram-Schmidt process yet? It's this vector that you want to normalize, not the given vector [-1 0 1].

Also, I meant [-1 -1 2] earlier, not [-1 1 2]. I accidentally dropped a sign. You mentioned [-1 -1 2] in your earlier post.
 
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Using the Gram-Schmidt process, I get [-1/2 -1/2 1] as the unnormalized second vector. So am I allowed to multiply this by 2 and normalize that to get the second vector in the orthonormal basis?
 
  • #10
Yup, though you don't need to multiply by 2 first. You can normalize it as is.
 
  • #11
Okay. Thank you so much for working through this with me!
 

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