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Finding a basis of a space

  1. Mar 8, 2015 #1
    Let u = [1, 2, 3, -1, 2]T, v = [2, 4, 7, 2, -1]T in 5.
    Find a basis of a space W such that wu and wv for all wW.

    I think the question is quite easy. Given this vector w in the space W is orthogonal to both u and v. I can only think of w being a zero vector. But would this be too trivial?

    wTu = wTv = 0

    wT(u - v) = 0

    I believe there is something wrong. Things above are what I can compute...
    Is there any other way to solve this question?

    -----------------------------------------------------------------------------------------------------

    Oh, I've thought of another way.

    Set a augmented matrix
    [1 2 3 -1 2 | 0]
    [2 4 7 2 -1 | 0]

    ~

    [1 2 3 -1 2 | 0]
    [0 2 4 3 -3 | 0]

    Let x5 = t, x4 = s, x3 = z where t, s, z ∈ ℝ.
    ∴ x2 = 1.5t - 1.5s - 2z
    x1 = -5t + 4s + z

    Therefore, the solution set is {t[-5 1.5 2 0 0 1]T + s[4 -1.5 0 1 0]T + z[1 -2 1 0 0]T}.
    The basis is {[-5 1.5 2 0 0 1]T, [4 -1.5 0 1 0]T, [1 -2 1 0 0]T}.


    Is this correct?
     
    Last edited: Mar 8, 2015
  2. jcsd
  3. Mar 8, 2015 #2

    Svein

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    You have started OK, but then you have hurried along too fast. Let w = [w1, w2, w3, w4, w5]. Now apply your conditions (written as scalar products: [itex]\left\langle w, u \right\rangle = 0 [/itex] and [itex]\left\langle w, v \right\rangle = 0 [/itex] ). You will not find a complete solution (after all, there are 5 variables and two equations), but you will have some constraints on the coordinates.
     
  4. Mar 8, 2015 #3
    So my second try is correct?
    Better to let w1, w2, w3, w4, w5 instead of x...?
     
  5. Mar 8, 2015 #4

    Mark44

    Staff: Mentor

    No.
    The first vector in your set has too many components (typo?). Each vector in your set has to be perpendicular to u and v, which you can easily check by doing dot products.
     
  6. Mar 9, 2015 #5
    Yeah typo...sorry
    the 2 is redundant
     
  7. Mar 9, 2015 #6

    WWGD

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    Seems like W would be the intersection of the ortho complements of u,v .
     
  8. Mar 13, 2015 #7

    HallsofIvy

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    I would do it this way- write the general vector in your subspace as (a, b, c, d, e). Such a vector is perpendicular to (1, 2, 3, -1, 1) if and only if a+ 2b+ 3c- d+ e= 0. It is perpendicular to (2, 4, 7, 2, -1) if and only if 2a+ 4b+ 7c+ 2d- e= 0. Adding those two equations together, 3a+ 6b+ 10c- d= 0 so that d= 3a+ 6b+ 10c. Putting that into either or the first equations lets you solve for, say, e in terms of a, b, c, and d. Replace d and e in (a, b, c, d, e) by those and then write the result as a( ...)+ b(...)+ c(...).
     
  9. Mar 13, 2015 #8
    for your first answer that w=0, you should understand that a set of vectors with the zero vector is never linearly independent. Think about it for a bit. You can easily prove it yourself. And if you dont have a linearly independent set then you dont have a basis. A basis is defined to be a linearly independent spanning list afterall.
     
  10. Mar 13, 2015 #9

    WWGD

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    A 1-line proof: take any non-zero c , then c.0=0 is a non-trivial combination that gives you 0.
     
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