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Homework Help: Orthogonal Base

  1. Jul 25, 2012 #1
    1. The problem statement, all variables and given/known data


    I'm trying to solve these problems

    given the vectors u=(3,-2,1) and v=(2,-3,1)

    1. Find an orthogonal base for the space H generated by {u,v}

    2. Find the orthogonal projection of w=(3,0,1) on H

    2. Relevant equations

    3. The attempt at a solution

    Im not sure how to get an orthogonal base.
    Im thinking about finding a vector by inspection in which the dot product of v and u of that vector would be zero. Is there e method to find an orthogonal base for the space H ? Can I use gram-schmidt to solve the first problem and if so, how?

    Thanks a lot for your comments.
  2. jcsd
  3. Jul 25, 2012 #2

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    Hi Jimmy84! :smile:

    The Gram-Schmidt algorithm is exactly what you need.
    Can you apply its steps?

    When you have found an orthogonal base, it becomes easy to find an orthogonal projection.
    Let's first try to apply Gram-Schmidt though...

    What do you have about Gram-Schmidt and how much of it can you apply?
  4. Jul 25, 2012 #3

    My problems is that Im not sure how to apply Gram-Schmidt

    given the vectors u=3,-2,1 and v=2 -3 1 I try to apply gram schmidt to them in this way

    the algorithm says that

    w1 = u


    w2 = v - (vu)/(uu) u

    the result is w2 = -23/8, 1/4 ,-5/8 which is not orthogonal to u nor to v. I'm wondering what am I doing wrong?
  5. Jul 25, 2012 #4

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    You're on the right track! :)

    But when I calculate w2 I get a different result.

    What did you get for (vu)? And for (uu)?
  6. Jul 25, 2012 #5
    for vu I got 13 for uu 14 I think I found a mistake let me check the calculations again
  7. Jul 25, 2012 #6
    I found the mistake :) it is -11/14, -8/7 , 1/14 now can I say that w2 is an orthogonal basis of H? and how can I project w=(3,0,1) on H?
  8. Jul 25, 2012 #7

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    Actually, a basis is a set.
    So {w1, w2} is an orthogonal basis of H.

    Do you have a formula for a projection?
    (That formula is rather closely related to Gram-Schmidt.)
  9. Jul 25, 2012 #8
    I think that the formula for the projection of v into u is Proj u (v) = uv/(uu) u but that is for two vectors, I have no idea how to find the projection of a vector into a space.
  10. Jul 25, 2012 #9

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    It's the summation of the projections on the individual vectors that form the orthogonal basis.

    In your case:
    Proj H (x) = w1x/(w1w1) w1 + w2x/(w2w2) w2
  11. Jul 25, 2012 #10
    Thanks a lot, I was just wondering about a similar problem I'm trying to solve.

    If there is a plane W with bases b1, b2 and an orthogonal base n , then

    Proj W (b1) = b1 , Proj W (b2) = b2 , and Proj W (n) = 0

    Im curious how can I compute Proj W (b1) = b1 ? Should I use the formula you just wrote for the plane W?

  12. Jul 25, 2012 #11

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    Yes, you can use that formula... can you apply it?

    Also, you can take a look at the geometric meaning of it.
    If you start with a vector in the plane you're projecting onto... you'll get that same vector!
  13. Jul 25, 2012 #12
    I'm have been for weeks trying to see the geometric meaning of it the problem is that this material is not covered in many books that I have consulted when it comes to reflection and orthogonal projection in planes. Can you recommend me a book that might touch the subject?

    I'm reading anton howard`s elementary linear algebra.

    I think the formula would be Proj W (b1) = (w1b1)/(w1w1) w1 + (w2b1)/(w2w2) w2

    where w1 and w2 are an orthogonal basis for b1 and b2 isnt it?
  14. Jul 25, 2012 #13

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    Sorry, the books that I studied from are not available anymore.

    Try w1=b1 and w2=b2.
    And since b1 is orthogonal to b2, what is (b1b2)?
  15. Jul 25, 2012 #14
    so Proj W (b1) = b1b1/b1b1 b1 ? which is b1

    Proj W (b2) = b2b2/b2b2 b2 = b2

    I'm confused about the Proj W (n) how would it look like in order to compute it? my guess

    is that it could be Proj W (n) = b1n/b1b1 b1 + b2n/ b2b2 b2
  16. Jul 25, 2012 #15

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    The projection using a normal vector n is:

    Proj W (x) = x - (xn)/(nn) n
  17. Jul 25, 2012 #16
    Thanks a lot, have a good night.
  18. Jul 26, 2012 #17
    well I was checking some details about the problem we were discussing yesterday in the example Im given b1 and b2 which are not orthogonal. Can the formula work if b1 and b2 are just non orthogonal bases of the plane W ?
  19. Jul 26, 2012 #18

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    Sorry, but no. b1 and b2 have to be orthogonal for the projection formula to work.
    There's a reason why it's useful for a base to be orthogonal (or even better: orthonormal)!
  20. Jul 26, 2012 #19
    The problem is this one

    1. Given a linear operator which can be an orthogonal projection or a reflection (passing through the origin) calculate the eigenvalues and eigenspaces associated which each eigenvalue.

    The matrix is

    2 1 -1
    -1 0 1
    1 1 0

    2. based on that, determine if the operator is a reflection or an orthogonal projection.

    3. Describe an orthogonal base of the plane, and complete it to a base of R^2. The matrix of the operator with respect to the calculated base must have the form




    I got the eigenvalues 1 and 0 therefore I assume that it is an orthogonal projection.
    the eigenvectors for the eigenvalue 1 are B { b1 = -1,1,0 and b2 = 1,0,1}

    which were the ones I was thinking about applying in the formula that we were just discussing.

    "Proj W (x) = w1x/(w1w1) w1 + w2x/(w2w2) w2 "

    In order to get the orthogonal projection matrix which vectors should I try in the formula of the Proj of W ?

    im considering
    Proj W (w1) = w1 , Proj W (w2) = w2 , where w1 and w2 are elements of the orthogonal basis of b1 and b2 however I'm not sure how to get the next column of zeros of the orthogonal matrix that I'm ask to find.
  21. Jul 26, 2012 #20

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    However, the projection formula only works for an orthogonal base.
    If you want to use it, you will first have to make the base orthogonal.
  22. Jul 26, 2012 #21
    Im a bit confused,

    Should I use Proj W (w3) = w1w3/w1w1 w1 + w2w3/w2w2 w2

    for the last column of the orthogonal projection matrix ?

    The first member is zero, but since w2 and w3 are not orthogonal it seems Proj W (w3) won't be zero?
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