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Finding an orthonormal basis for a subspace

  1. Oct 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Find an orthonormal basis for the subspace of V4 spanned by the given vectors.
    x1 = (1, 1, 0, 1)
    x2 = (1, 0, 2, 1)
    x3 = (1, 2, -2, 1)
    2. Relevant equations
    Gram-Schmidt Process

    3. The attempt at a solution
    I have used the Gram-Schmidt process but seem to be running into trouble. Here is what I did:
    y1 = x1 = (1, 1, 0, 1)
    y2 = x - y1 = (1-1, 0-1, 2-0, 1-1) = (0, -1, 2, 0)
    y3 = x3 - y1 + y2 = (1-1+0, 2-1-1, -2-0+2, 1-1+0) = (0, 0, 0, 0)

    Now I used these and their norms to find the basis {y1, y2}

    y1 / lly1ll = 1/sqrt(3) (1, 1, 0, 1)
    y2 / lly2ll = 1/sqrt(5) (0, -1, 2, 0)

    Therefore, {1/sqrt(3) (1, 1, 0, 1), 1/sqrt(5) (0, -1, 2, 0)} from my work. However, my book says the answer is {(1/3)(1/sqrt(3)(1, 1, 0, 1), 1/sqrt(42) (1, -2, 6, 1)} which is very different than my answer. Where am I going wrong?
     
  2. jcsd
  3. Oct 7, 2014 #2

    mfb

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    This is not Gram-Schmidt, and randomly subtracting some vectors will not give orthogonal vectors in general. There is an important part missing.
    You can see that your answer is wrong if you check if your y2 and y1 are orthogonal.

    Unrelated to this issue: There is more than one orthonormal basis. Your answer can deviate from the book's answer.
     
  4. Oct 7, 2014 #3

    RUber

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    http://en.wikipedia.org/wiki/Gram–Schmidt_process
    Start with one vector, then from the second vector, subtract off the projection of the second vector onto the first vector. For the third, you have to subtract off the projections onto the two already found orthogonal vectors.
    You can either normalize as you go, or normalize at the end. If you are doing it by hand, waiting until you have the orthogonal vectors is nicer for arithmetic. If you are automating the process, it is usually more efficient to normalize as you go.
     
  5. Oct 7, 2014 #4
    This is the abbreviated formula given to me in my book from the more complex Gram-Schmidt formula. Maybe I will try using the original formulas.
     
  6. Oct 8, 2014 #5

    RUber

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    Your y1 and y2 above are good starting points, since they are helping you to find a set of linearly independent vectors spanning the space. You have already found that the space spanned by the three vectors will only have dimension 2 with y1 and y2 as admissible bases. However, y1 and y2 are neither orthogonal nor normal, bases.

    I was unable to reproduce the answer you had for the book answer, but as mfb said, there are many admissible solutions. It seems like the book answer may be based on conducting the operations in a certain order.
     
  7. Oct 8, 2014 #6

    vela

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    I don't see how those formulas could be an abbreviated form of the Gram-Schmidt process. If they indeed are, you've overlooked important details in their derivation because they don't apply to this problem. I'd guess, however, that you've misunderstood what those particular equations are used for.

    In any case, using the original formulas should get you the correct answer.
     
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