1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    x1=x2=x3=(1,0), x4=(0,1). I'm really starting to wonder if you got a) and b) right.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook