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Subspaces and Orthogonality

  1. Apr 11, 2010 #1
    Ok so I've been working on this problem and I'm really having some struggles grasping it. Here it is:

    Let W be some subspace of Rn, let WW consist of those vectors in Rn that are orthognoal to all vectors in W.

    1) Show that WW is a subspace of Rn?

    So for this part I'm thinking that because WW is a linear combination of W (maybe) then therefore it forms a subspace of Rn

    2) If {v1, v2,...vt} is a basis for W, show that a vector X in Rn lies in WW if and only if x is orthogonal to each of the vectors v1, v2,...vt?

    And for this one I'm really at a lose for where to start. Any help would be appreciated.

    Thanks
     
  2. jcsd
  3. Apr 11, 2010 #2
    What have you tried, so far.?

    Basically, in order to show that any subset S of a vector space V is a subspace,

    is to show that elements of S are closed under scaling and under linear combinations;

    in this case, show that if w^ and w'^ are in W^ =(" W Perp" ) , so is their sum,

    and for any w^, cw^ (c a scalar) is also in W^. The key here is the properties

    of the inner-product.



    For 2, one side follows by definition. For the other side ( w^ is perpendicular

    to each of v1,..,vt ) , think of writing any x in W using the elements of v1,..,vt,

    and, again, use the properties of the inner-product (multilinearity).

    Good Luck.
     
  4. Apr 11, 2010 #3
    1) Well I understand the definition of a subspace, i think im just finding it difficult to fully understand the proof. Would it be correct to say then that:

    WW is a subspace of Rn because
    i) It contains the zero vector
    ii) WW= {v1', v2',...vt'}

    Let x, y span (WW)
    x=av1' + av2' + avt'
    y=bv1' + bv2' + bvt'

    Therefore (x+y)=(a1+b1)v1' + (a2+b2)v2' + ...(at + bt)vt' and is closed under vector addition.

    iii) Let x span (WW)

    Then x= a1v1' + a2v2' + atvt' for some a1, a2,...at
    Then kx=ka1v1' + ka2v2' + katvt'

    Therefore it is closed under scalar multiplication. Since it satisfies all three, it is a subspace of Rn.

    So that's what I have for part 1

    And for part 2 I'm still lost lol

    2)
     
  5. Apr 11, 2010 #4
    "" 1) Well I understand the definition of a subspace, i think im just finding it difficult to fully understand the proof. Would it be correct to say then that:

    WW is a subspace of Rn because
    i) It contains the zero vector
    ii) WW= {v1', v2',...vt'}

    Let x, y span (WW)
    x=av1' + av2' + avt'
    y=bv1' + bv2' + bvt'

    Therefore (x+y)=(a1+b1)v1' + (a2+b2)v2' + ...(at + bt)vt' and is closed under vector addition. ""


    I don't see how you have shown it is closed. How do you know that x+y is in W^.?



    iii) Let x span (WW)

    What do you mean here.?. How do you know that a single vector x spans WW.?.
    I think (reading below ) you mean: let x be a vector in WW. Right.?


    Then x= a1v1' + a2v2' + atvt' for some a1, a2,...at
    Then kx=ka1v1' + ka2v2' + katvt'

    Therefore it is closed under scalar multiplication.

    Not clear to me. How do you know kx is in WW.? You need to check this, or give a good
    argument to that effect.


    Since it satisfies all three, it is a subspace of Rn.

    So that's what I have for part 1

    And for part 2 I'm still lost lol.

    Well, assume that a vector x in R^n is orthogonal to each of the basis vectors

    {v1,..,vt} of W . You then want to show that x is orthogonal to any vector v in W.

    How do you show any two vectors are orthogonal.?.

    Assume x is orthogonal to each of v1,..,vt (what does this mean.?) and let w be in W.

    What do you need to do to show that x and v are orthogonal.?. How can you write w in

    order to show this.?
     
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