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Finding the basis of a subspace

  1. Dec 7, 2014 #1
    1. The problem statement, all variables and given/known data
    How do I find a basis for:
    the subspace of R^3 consisting of all vectors x such that x ⋅ (1,2,3) = 0.

    2. Relevant equations
    I believe this is performed through setting x = x,y,z, setting each parameter sequentially equal to 1 while the others are set to o, putting into a matrix where a^i --> v(subscript i), and transforming into row reduced echelon form.

    3. The attempt at a solution
    I believe I have done this correctly, but please tell me if I have not.

    x ⋅ (1,2,3) = 0.
    (x,y,z) ⋅ (1,2,3) = 0.
    x + 2y + 3z = o
    x = -2y -3z
    that is,
    x = -2s - 3t
    y = s
    z = t

    Giving: (-2s-3t,s,t)

    Set s = 1, t = 0
    (-2,1,0)
    Set s = 0, t = 1
    (-3,0,1)

    -2 -3
    1 0
    0 1

    ROW REDUCTION
    yields:

    1 0
    0 1
    0 0

    Transforming v1 and v2 into the elementary columns. Thus,
    {(-2,1,0),(-3,0,1)} form a basis for the subspace of R^3 consisting of all vectors x such that x ⋅ (1,2,3) = 0.
    DimS = 2

    _____

    If it is of any assistance, the answer in the back of the textbook is:
    {(-2,1,0),(-3,0,1)} with dimS = 2
     
    Last edited: Dec 7, 2014
  2. jcsd
  3. Dec 7, 2014 #2
    I apologize for the redundancy setting y = s and z = t. That was clearly not needed.
     
  4. Dec 7, 2014 #3

    BiGyElLoWhAt

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    Gold Member

    Your subspace is going to be a plane in r3. That plane will consist of all your vectors x, which are orthogonal to 1,2,3. Your method looks alright, if you doubt the answer, find a way to check it. *cough* cross some vectors *uncough*

    Also, if this were any more complex, using x for a vector equal to <x,y,z> could be confusing. I wouldn't recommend making a habit of it.
     
  5. Dec 7, 2014 #4
    Thank you for the reply!

    Do you mean to say I would be better saying x = <x1,x2,x3>?
    Perhaps I've misunderstood you completely.
     
  6. Dec 7, 2014 #5

    BiGyElLoWhAt

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    Gold Member

    Something along those lines, yea. or x = <e1,e2,e3>

    Just setting x = <x,*,*> can be confusing, you know, since its just x on the left side, and it's just x in one of the components on the right side. However you want to differentiate between those is up to you; but I suggest you do, otherwise you'll get lost in your own work no doubt about it.
     
  7. Dec 7, 2014 #6
    Thank you!
    It's been a while since I've done this.
     
  8. Dec 7, 2014 #7

    BiGyElLoWhAt

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    Gold Member

    No problem, glad to help =]
     
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