Finding an orthonormal basis

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Find an orthonormal basis for the subspace of R^3 consisting of all vectors (a, b, c) such that a + b + c = 0.


    2. Relevant equations



    3. The attempt at a solution
    I know how to find an orthonormal basis just for R^3 by taking the standard basis vectors (1, 0, 0), (0, 1, 0), and (0, 0, 1) and applying the Gram-Schmidt process to make them orthonormal. The condition that a + b + c must equal 0 is throwing me off, however. Can anyone give me any suggestions for how to approach problems like these? thanks!
     
  2. jcsd
  3. Nov 13, 2008 #2

    Dick

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    If you know Gram-Schmidt, that's the hard part. What's so hard about about finding two vectors (a,b,c) such that a+b+c=0 to do Gram-Schmidt with? (1,-1,0) sounds like a good start. Give me another one.
     
  4. Nov 13, 2008 #3
    I think I need 3 vectors, right? such that a + b + c = 0. Can I start with any vector? Or is there a reason why (1, -1, 0) is a good choice? If it can be any 3 vectors that add to 0 it should be simpler, but then I can't be sure they are a basis?
     
  5. Nov 13, 2008 #4

    Dick

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    Why do you need 3 vectors? That's a subspace of R^3. It looks to me like it's two dimensional.
     
  6. Nov 13, 2008 #5
    Ohh.. sorry I was being confused, I thought a, b, and c referred to vectors and not the components.

    So I can choose (1, -1, 0) like you said and also (1, 0, -1)? I would still have to check if the 2 vectors I choose are actually a basis (span R^3 and are linearly independent) right?
     
  7. Nov 13, 2008 #6

    Dick

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    Sure, exactly. That's a good choice. There aren't too many dumb choices. (1,-1,0) and (-1,1,0) would have been a dumb choice because they're linearly dependent. But you didn't make the dumb choice. Now do Gram-Schmidt.
     
  8. Nov 13, 2008 #7

    Mark44

    Staff: Mentor

    If you apply Gram-Schmidt to these vectors, you'll just get the same ones. They are already orthogonal and have length 1.
     
  9. Nov 14, 2008 #8

    Mark44

    Staff: Mentor

    I'm not sure it's clear to you where Dick got the vectors he did. Start with your equation:
    a + b + c = 0

    If you solve for a, you get
    Code (Text):

    a = -b - c, where b and c are arbitrary
    b =  b
    c =      c
     
    That last two equations are obviously true.
    To make things even more explicit,
    Code (Text):

    a = -1b - 1c
    b =  1b + 0c
    c =  0b  +1 c
     
    or,
    [tex]
    \left[
    \begin{array}{ c }
    a \\
    b \\
    c
    \end{array} \right] = b\left[
    \begin{array}{ c }
    -1 \\
    1 \\
    0
    \end{array} \right] + c\left[
    \begin{array}{ c }
    -1 \\
    0 \\
    1
    \end{array} \right][/tex]
    Voila, there's your basis.
     
  10. Nov 14, 2008 #9

    HallsofIvy

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    You need to review basic definitions. Those two vectors CAN'T be a basis for R^3 and can't span R^3 because that requires THREE vectors. You must have misread the original problem. As Dick said, the set of all (a, b, c) such that a+ b+ c= 0 is a two dimensional subspace of R^3. NO set of such vectors can be a basis for R^3. (1, -1, 0) and (1, 0, -1) form a basis for that two dimensional subspace.
     
  11. Jul 27, 2010 #10
    I'm currently doing the same problem so I figured I'd revive this old thread.

    Once I have my two vectors,
    u1=<-1,1,0> and u2=<-1,0,1>

    I then go through the Gram-Schmidt process to find the normalized basis?
     
  12. Jul 27, 2010 #11

    Mark44

    Staff: Mentor

    If you already have a basis, you don't need Gram-Schmidt to find a normalized basis.

    Do you know what the term "normalized" means?
     
  13. Jul 27, 2010 #12
    Sorry, I meant a orthonormal basis.
     
  14. Jul 27, 2010 #13

    Mark44

    Staff: Mentor

    OK, in that case you need to use G-S.
     
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