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Find a Basis B for the subspace

  1. Apr 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Let V be the subspace of R3 defined by V={(x,y,z)l2x-3y+6z=0}
    Find a basis B for the subspace.

    2. Relevant equations



    3. The attempt at a solution

    First I broke apart the equation such that:

    [[x,y,z]] = [[3/2s-3t, s, t]] = s [[3/2, 1 ,0]] +t[[-3, 0, 1]]



    So, a basis for V =

    [[3/2, 1, 0]
    -3, 0, 1]]
    Does this sound correct?
     
  2. jcsd
  3. Apr 9, 2014 #2

    Dick

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    Sounds correct.
     
  4. Apr 9, 2014 #3
    so it would be acceptable to have a two dimensional basis for a subspace of R3?

    Its because its a plane in R3?
     
  5. Apr 9, 2014 #4

    Dick

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    Exactly. It's a two dimensional subspace, so you should get two basis vectors.
     
  6. Apr 9, 2014 #5
    Awesome, thank you for the conformation.
     
  7. Apr 9, 2014 #6

    HallsofIvy

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    But you don't say where these, and in particular, "s" and "t", came from.
    I would have been inclined to say that 2x- 3y+ 6z= 0 is the same as x= (3/2)y- 3z so that
    (x, y, z)= ((3/2)y- 3z, y, z)= ((3/2)y, y, 0)+ (-3z, 0, z)= y(3/2, 1, 0)+ z(-3, 1, 0)

    Since every vector can be written as a linear combination of (3/2, 1, 0) and (-3, 1, 0), yes, they form a basis.


    Personally, I would have solved 2x- 3y+ 6z= 0 for z giving z= (-1/3)x+ (1/2)y so that
    (x, y, z)= (x, y,(-1/3)x+ (1/2)y)= x(1, 0, -1/3)+ y(0, 1, 1/2).

    Of course, there are an infinite number of bases for this space.
     
  8. Apr 9, 2014 #7


    I see what you are talking about.
     
  9. Apr 9, 2014 #8

    LCKurtz

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    Yeah, Dick is good at all the angles :smile:
     
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