1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why do we use Gram-Schmidt process?

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data
    I have 3 vectors. I need to find the ortonormal basis by Gram-Schmidt.

    3. The attempt at a solution
    I observed that, without Gram-Schmidt, I get two vectors which I get too by Gram-Schmidt. I have just divided the vectors by their lengths.

    The third vector got without Gram-Schmidt differs however from the vector got by Gram-Schmidt.

    Do you know why?
  2. jcsd
  3. Jan 21, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    How to you 'get' vectors without Gram-Schmidt? If you are dividing the vectors by their length, they will be unchanged if they are unit length. If you apply Gram-Schmidt to list of normalized vectors they will unchanged if they are already orthogonal.
  4. Jan 21, 2009 #3
    My vectors are:
    [tex] v_{1} = [1\ -2\ 0\ 1]^{T}[/tex]
    [tex] v_{2} = [-1\ 0\ 0\ -1]^{T}[/tex]
    [tex] v_{3} = [1\ 1\ 0\ 0]^{T}[/tex]
    in the space of S.

    I get these without G-S:
    [tex] u_{1} = \frac {[1\ -2\ 0\ 1]^{T}} {\sqrt{6}}[/tex]
    [tex] u_{2} = \frac {[-1\ 0\ 0\ -1]^{T}} {\sqrt{2}}[/tex]
    [tex] u_{3} = \frac {[1\ 1\ 0\ 0]^{T}} {\sqrt{2}}[/tex]

    --edit here---
    I get the same vectors with G-S but the the second and the third ones differ:
    [tex] u_{2} = \frac {[-1\ -1\ 0\ -1]^{T}} {\sqrt{3}}[/tex]
    [tex] u_{3} = \frac {[1\ 0\ 0\ -1]^{T}} {\sqrt{2}}[/tex]

    Is the reason that the second and the third ones differ that they are the only vectors which are
    normalised already?
    Last edited: Jan 21, 2009
  5. Jan 21, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper

    The u2 in the second group isn't orthogonal to the u1 in the second group. The final u2 you gave is. That's what's supposed to happen. u3 had better change as well. It's not orthogonal to anything.
  6. Jan 21, 2009 #5
    I agree with you: only the first vector got without G-S is the same as the one got with G-S.

    Is the reason that it is already ortogonal to space?
    I am not sure what the 1st vector is perpendicular to. To space?

    PS: I corrected my previous post: the two last vectors.
    Last edited: Jan 21, 2009
  7. Jan 21, 2009 #6


    User Avatar
    Science Advisor
    Homework Helper

    I'm not sure I get what your problem is. Gram-Schmidt just says to normalize the first vector. You did that. After that it doesn't change. You adjust the other ones to be perpendicular to the first one. You can pick any one of the vectors to remain unchanged just by changing the order of the vectors.
  8. Jan 21, 2009 #7
    Thank you! You cleared my confusion.

    So we need first to normalize the 1st vector. Then, we change the other ones to be perpendicular to the 1st one. To achieve this we use Gram-Schmidt.

    Is there any other method to make vectors perpendicular to the selected vector?

    I am interested in unit circle. Does Gram-Schmidt process allow us to use unit circle?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Why do we use Gram-Schmidt process?
  1. Gram-Schmidt Process (Replies: 1)

  2. Gram-Schmidt Process (Replies: 2)

  3. Gram-Schmidt process (Replies: 10)

  4. Gram-Schmidt Process (Replies: 2)