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Homework Help: Condition for equality between subspaces.

  1. Dec 3, 2012 #1
    1. The problem statement, all variables and given/known data

    What would be the/a condition on vectors in K so that V=W, where V is a vector space which K={v1,v2,v3,v4} spans, and W is a subspace of V defined thus:

    2. Relevant equations

    3. The attempt at a solution

    I believe V would be equal to W if W were linearly independent, but by writing that mathematically I get a condition for the scalars, not the vectors in K themselves.

    I hope one of you could assist. Thanks in advance!
  2. jcsd
  3. Dec 3, 2012 #2
    I know what you mean, but your terminology is wrong. You can't say that W is linearly independent because it is not true. What you mean is that the four vectors


    are linearly independent. That would indeed be the correct condition.

    What did you get when you wrote that mathematically??
  4. Dec 3, 2012 #3
    I have tried to find conditions so that:
    a1v1 + a2v2 + a3v3 + a4v4 = v1(b1+b4) +
    v2(b2+b1) + v3(b3+b2) + v4(b4+b3).
    But that yielded conditions on the scalars, not the vectors. Can conditions on the vectors themselves be found?
  5. Dec 3, 2012 #4
    How did you get that? In order for [itex]\{v_1+v_2,v_2+v_3,v_3+v_4,v_4+v_1\}[/itex] to be a basis, you must prove that any linear combination of the form

    [itex]\alpha(v_1+v_2)+\beta(v_2+v_3)+\gamma (v_3+v_4)+\delta(v_4+v_1)=0[/itex]

    only if [itex]\alpha=\beta=\gamma=\delta=0[/itex].

    Now, try to use that [itex]\{v_1,v_2,v_3,v_4\}[/itex] is a basis.
  6. Dec 3, 2012 #5
    These yielded alpha=-delta=-beta=gamma.
    But how does this affect the vectors in K themselves? I mean, what is then the condition on v1,v2,v3,v4?
  7. Dec 3, 2012 #6
    OK, so what if you take the equation

    [tex]\alpha(v_1+v_2)+\beta(v_2+v_3)+\gamma(v_3+v_4)+ \delta(v_4+v_1)=0[/tex]

    and if you substitute [itex]\alpha[/itex] for [itex]\gamma[/itex] and [itex]-\alpha[/itex] for [itex]\delta[/itex] and [itex]\beta[/itex]?
  8. Dec 3, 2012 #7
    You get alpha*0=0. How does that help?
  9. Dec 3, 2012 #8
    It shows that there is always a nontrivial linear combination that ends up in zero. Doesn't that show that your set [itex]\{v_1+v_2,v_2+v_3,v_3+v_4,v_4+v_1\}[/itex] is never linearly independent?
  10. Dec 3, 2012 #9
    Let us go back a bit, momentarily.
    I am slightly confused. Why is it that for V to be equal to W, the elements in W must be linearly independent? Is it because dimV is equal to or less than the number of elements in K, i.e. 4?
    Furthermore, I know that if the elements in K are linearly independent, then V is not equal to W. Does that mean that for any K whose elements are linearly dependent, V would be equal to W?
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