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Homework Help: Vector Spans

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Let V be a vector over a field F.

    a.) Let x1,...,xn∈V and y1,...,ym∈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) ∩ 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.

    2. Relevant equations

    3. The attempt at a solution

    I have done a and b just not sure about c:

    If x1=x2=x3=x4 and For a ∈ in R,

    Span(x1, x2,x3) ∩ Span(x2, x3, x4)= (a1x1+a2x2+a3x3) ∩ (a2x2+a3x3+a4x4)=a1x1+a2x2+a3x3+a4x4= Span(x1, x2, x3, x4)≠ Span(x2,x3)

    Is this okay?

    or is it ok to just do a counter example using vectors,

    if so could somebody show me an example counter example

    thanks in advance
  2. jcsd
  3. Mar 13, 2010 #2


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    No, whatever it is you think you did is not ok. Picking x1=x2=x3=x4 is not going to give you a counter example. Suppose you pick x1=x2=x3 and x4 to be 'something else'. Use numerical vectors to make it super clear.
  4. Mar 13, 2010 #3
    How about x1=x4

    then let v∈Span(x1, x2,x3) and v∈Span(x2, x3,x4)


    v = a1x1 + a2x2 + a3x3

    and v = b1x2 + b2x3 + b3x4

    so we have: a1x1 + a2x2 + a3x3 = b1x2 + b2x3 + b3x4

    equating coefficents we get


    and as x1 = x4:

    a1 = b3

    Therefore let:

    a1 = b3 = c1
    a2 = b1 = c2
    a3 = b2 = c3

    then v = c1x1 + c2x2 + c3x3

    so v = (c1/2)x1 + (c1/2)x1 + c2x2 + c3x3

    so v = (c1/2)x1 + (c1/2)x4 + c2x2 + c3x3

    so v∈Span(x1, x2,x3,x4)

    hows that?
  5. Mar 13, 2010 #4
    ok ill try the numerical method tomorow, need some sleep now!

    any comments by anyone throughout the night will be massivley appreciated
  6. Mar 13, 2010 #5


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    x1=x2=x3=(1,0), x4=(0,1). I'm really starting to wonder if you got a) and b) right.
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