Finding an orthonormal basis

  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. Dick

    Dick 25,910
    Science Advisor
    Homework Helper

    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. 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. Dick

    Dick 25,910
    Science Advisor
    Homework Helper

    Why do you need 3 vectors? That's a subspace of R^3. It looks to me like it's two dimensional.
  6. 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. Dick

    Dick 25,910
    Science Advisor
    Homework Helper

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

    HallsofIvy 41,055
    Staff Emeritus
    Science Advisor

    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. 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. 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. Sorry, I meant a orthonormal basis.
  14. Mark44

    Staff: Mentor

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