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Finding a basis of a subspace

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Find a basis for each of these subspaces of R4

    All vectors that are perpendicular to (1,1,0,0) and (1,0,1,1)

    2. The attempt at a solution
    I'm not sure how to approach this question. The only thing I can think of is that a vector that would be perpendicular to both would be where the dot product would equal zero aye?

    So then that would give me

    1x1 + 1x2 = 1x1 + 1x3 + 1x4 = 0.
    So in which case, I'd do RREF
    [[1,1,0,0]
    [1,0,1,1]]

    to

    [[1,0,1,1]
    [0,1,-1,-1]]

    I get stuck here because I'm not sure how to solve for the basis at this point.
    I'm also unsure if what I did for rref was correct because vectors are usually denoted by column spaces, rather than what row spaces such as what I have done.
     
    Last edited: Oct 11, 2009
  2. jcsd
  3. Oct 11, 2009 #2

    Dick

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    That's the right start. But now you have to solve the linear equations. You've got 2 equations in 4 unknowns. You should be able to express the vector (a,b,c,d) in terms of two parameters by eliminating two unknowns. Can you do that?
     
  4. Oct 11, 2009 #3
    I'm not sure if this is what I'm supposed to do but

    Given that I have the RREF form, that gives me
    x1 + x3 + x4 = 0
    x2 - x3 - x4 = 0

    Given that x1 and x2 are pivots, this gives me
    x1 = -x3 - x4
    x2 = x3 + x4

    So, if I substitute x3 and x4 with a and b, this makes my basis the span of <-1,1,1,0> and <-1,1,0,1> ?
     
  5. Oct 11, 2009 #4

    Dick

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    Well, both of those vectors are perpendicular to the given vectors. And they are linearly independent, so sure, they are a basis. The subspace they span is the perpendicular subspace, the span itself isn't a 'basis'. Those two vectors are a basis.
     
  6. Oct 11, 2009 #5
    Ah, your help was much appreciated.
     
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