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Spans and linear independance

  1. Mar 8, 2008 #1
    [SOLVED] Spans and linear independance

    1. The problem statement, all variables and given/known data

    Let V be a vector over a field F.

    a.) Let x1,...,xn[tex]\in[/tex]V and y1,...,ym[tex]\in[/tex]V. Show that

    Span(x1,...,xn,y1,...,ym) = Span(x1,...,xn) + Span(y1,...,ym)

    B.) Let x1, x2, x3, x4 be four linearly independent vectors in V. Show hat

    Span(x1, x2,x3) [tex]\cap[/tex] Span(x2, x3, x4) = Span(x2,x3)

    c.) Show that the equality in part b.) does not hold if we drop the assumption that x1, x2, x3, x4 are linearly independent.

    3. The attempt at a solution

    a.) Does it suffice to show:
    For a,b[tex]\in[/tex]R,

    (a1x1+...+anxn+b1y1+...+bmym) = (a1x1+...+anxn)+(b1y1+...+bmym) ?

    b.)
    Does it suffice to show:
    For a,b[tex]\in[/tex]R,

    Span(x1, x2,x3) [tex]\cap[/tex] Span(x2, x3, x4)= (a1x2(1)+...+anx2n+b1x3(1)+...+bmx3m) = Span(x2, x3) ?

    c.) If what i have doen so far is reasonably correct, the only part i'm unsure about is part c.) I will do some reading about it if i can, but any hints would be greatly appreciated. Thanks
     
  2. jcsd
  3. Mar 8, 2008 #2

    HallsofIvy

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    Yes, that is exactly right.

    By "x2(1)", "x2n", etc, you mean the components of vector x1? You shouldn't need to look at components. Just use the definition of "span". You will need to show why vectors x1 and x4 are not needed- and that has to do with the fact that the four vectors are independent.

    To prove that something is NOT true, you just need a counter-example. Take 4 vectors in, in R2, say, that are NOT idependent and see shy you can't do that. In fact, taking x1= x2= x3= x4 should work!
     
  4. Mar 9, 2008 #3
    Oh yes, so for b.)

    Span(x1, x2,x3)[tex]\cap[/tex] Span(x2, x3, x4)= (ax2+bx3) = Span(x2, x3)
    (cx1, dx4) [tex]\neq[/tex] (ax2, bx3) and so x1 and x4 are not of Span(x2, x3)

    I doubt this is enough to show what you are asking, how do i show x1 and x4 are not needed.
     
  5. Mar 9, 2008 #4

    HallsofIvy

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    Don't talk in generalities- be specific and write specific equations:

    If v is in span{x1, x2, x3}, what does that mean? v= what?

    If v is in span{x2, x3, x4}, what does that mean? v= what?

    If v is in the intersection of the two spans, then both of those are true. Set them equal, shift everything to one side of the equation and use the fact that the vectors are independent.
     
    Last edited: Mar 9, 2008
  6. Mar 9, 2008 #5
    So for v[tex]\in[/tex]R,

    Span(x1, x2,x3) [tex]\cap[/tex]Span(x2, x3, x4)= (v1x1+v2x2+v3x3)[tex]\cap[/tex](v2x2+v3x3+v4x4)= v2x2+v3x3= Span(x2,x3)

    Because the vectors are linearlry independent there is only one trivial solution to the equation:

    v1x1+v2x2+v3x3=v2x2+v3x3+v4x4=0

    Which is v1=v2=v3=v4=0.Therefore x1 and x4 are not in the given intersection asked for in the question.

    Is this right?? Thanks for all you help
     
  7. Mar 9, 2008 #6
    For part c.)

    If x1, x2, x3, x4 are not linearly independent then

    v1x1+v2x2+v3x3=v2x2+v3x3+v4x4=0

    has a solution other than the trivial solution, that is v1, v2, v3, v4 are not all zero.

    Therefore,

    Span(x1, x2,x3)[tex]\in[/tex] Span(x2, x3, x4)= (v1x1+v2x2+v3x3)[tex]\cap[/tex](v2x2+v3x3+v4x4)=v1x1+v2x2+v3x3+v4x4= Span(x1, x2, x3, x4)

    If v1, v2, v3, v4 [tex]\neq[/tex]0

    Is this right? Im not sure if i need to say that all the vectors aren't zero or if just not all of them are zero?

    Edit: Oh yes, i can do this one by counterexample,

    So if x1, x2, x3, x4 are not linearly independent,

    Span(x1, x2,x3)[tex]\in[/tex] Span(x2, x3, x4)[tex]\neq[/tex]Span(x2,x3)

    By counterexample:

    If x1=x2=x3=x4 and For v [tex]\in[/tex]R,

    Span(x1, x2,x3)[tex]\in[/tex] Span(x2, x3, x4)= (v1x1+v2x2+v3x3)[tex]\cap[/tex](v2x2+v3x3+v4x4)=v1x1+v2x2+v3x3+v4x4= Span(x1, x2, x3, x4)[tex]\neq[/tex]Span(x2,x3)

    Is this okay? Also was my other answer sufficient, if not it would be helpful to know why. Thanks
     
    Last edited: Mar 9, 2008
  8. Mar 10, 2008 #7

    HallsofIvy

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    You are assuming here that the coefficents of x2 and x3, in the two linear combinations, are the same! That is not necessarily true.

    I find the use of v1, etc. confusing, since "v" is often used to represent a vector. If v is in both spans then
    v= ax1+ bx2+ cx3= px2+ qx3+ rx4 for some numbers, a, b, c, p, q, r.

    If all four are 0, then NONE of the vectors are in the intersection!

    Hard to tell since you haven't said that what "v1", etc. are! They must be members of the field in order that "v1x1", etc. make sense. In that case, "Span(x1, x2,x3) [tex]\cap[/tex]Span(x2, x3, x4)= (v1x1+v2x2+v3x3)[tex]\cap[/tex](v2x2+v3x3+v4x4)= v2x2+v3x3= Span(x2,x3)" makes no sense because "Span(x1, x2,x3) [tex]\cap[/tex]Span(x2, x3, x4)" is a set of vectors while (v1x1+v2x2+v3x3) is a single vector.

    Again, you are writing generalities. Answer the questions I asked: if v is in Span(x1, x2, x3, x4), v= what? If v is in Span(x2, x3, x4), v= what? Write done the equations for v.
     
    Last edited: Mar 10, 2008
  9. Mar 10, 2008 #8

    HallsofIvy

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    Again, you are assuming that the one combination has the same coefficients as the other.

    I'm not sure I like this counter example. If x1= x2= x3= x4, the saying you need all of them is really the same as saying you only need x2 and x3.

    Consider this: x1= <1, 1, 0>, x2= <0, 1, 0>, x3= <0, 0, 1>, x4= <1, 0, 1>.

    What does a vector in their intersection look like? Can you write it using only x2 and x3?
     
  10. Mar 10, 2008 #9
    Thats great hallsofivy, i think i see why the counterexample wasnt good and the other example is better. I think i've got it now. Thanks
     
  11. Mar 22, 2009 #10
    Re: [SOLVED] Spans and linear independance

    Suppose that u, v and w are vectors in Rn. Show that if w E span{u,v} then span{u,v} = span{u,v,w}
     
  12. Mar 22, 2009 #11
    Re: [SOLVED] Spans and linear independance

    How do I tackle this problem?
     
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