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Determine if all vectors of form (a,0,0) are subspace of R3

  1. Jan 21, 2015 #1
    I have the feeling that it is, but Im not really sure how to start the proof. I know I have to prove both closure axioms; u,v ∈ W, u+v ∈ W and k∈ℝ and u∈W then ku ∈ W.
    Do I just pick a vector arbitrarily say a vector v = (x,y,z) and go from there?
     
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  3. Jan 21, 2015 #2

    mathwonk

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    pick vectors that are of the form (a,0,0), and otherwise arbitrary.
     
  4. Jan 21, 2015 #3
    proving u+v
    let u, v ∈ W
    u=(a1,0,0) v= (a2,0,0)
    u+v = (a1+a2, 0,0) ∈ W
    ∴u+v ∈ W

    am i anywhere close to doing that right?
     
  5. Jan 21, 2015 #4

    Fredrik

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    Yes, that's the right way to do it.
     
  6. Jan 21, 2015 #5
    Thank you guys, I really appreciate the help.
     
  7. Jan 24, 2015 #6

    Svein

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    The usual way is to determine the subspace ⊥ (a, 0, 0) - which is the subspace spanned by all vectors (x, y, z) such that (a, 0, 0)⋅(x, y, z) = 0. Since the scalar product is ax, this means that x = 0 and thus the normal subspace is spanned by (0, 1, 0) and (0, 0, 1). Therefore the original subspace has dimension 1 and is spanned by (1, 0, 0).
     
  8. Jan 24, 2015 #7

    Fredrik

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    This is true, but all you have to do to see it is to write ##(a,0,0)=a(1,0,0,)##.
     
  9. Jan 24, 2015 #8

    Svein

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    Yes, in this case it is easy. But if the specification had been more complicated, having a standard recipe is not a bad idea.
     
  10. Jan 24, 2015 #9
    Svein thanks for that, it clears up the concept a little more.
     
  11. Jan 24, 2015 #10

    Fredrik

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    The orthogonal complement of a set ##S## is defined as the set ##S^\perp## of all vectors that are orthogonal to all the vectors in ##S##.

    What Svein described is how to find the orthogonal complement of the orthogonal complement of the set W. This is always a subspace, even if W isn't. If you know this, you can find out if ##W## is a subspace by checking if ##W^{\perp\perp}=W##. This is rarely (never?) the easiest way to do it.
     
  12. Jan 25, 2015 #11

    Svein

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    Well, in the [itex]l_{2}[/itex] space (with the standard scalar product), this algorithm is used in proofs of completeness...
     
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