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Equality of spans

  1. Feb 8, 2005 #1
    Hi,
    I have a T/F which I need to prove.

    X1, X2, X3 belong to vector-space V.
    Y1 = X1 + X2, Y2 = X3.
    Span{Y1, Y2} is contained in but not equal to span{X1, X2, X3}.

    I am not sure which one it is:
    since y-span can be represented as span{X1 + X2, X3} it may be false, but then if all spans are subspaces, these two subspaces are not of the same dimension, i.e. they are not equal, then the statement is true. Obviously one of my reasonings is wrong. :uhh:
    Could someone clear up this for me?
    Thank you in advance.
     
  2. jcsd
  3. Feb 8, 2005 #2

    It's false. Let X_i=0. The span of both is {0}.
     
  4. Feb 9, 2005 #3
    But what if X_i is not lin. dep? In which case, {0} would not work...
     
  5. Feb 9, 2005 #4

    Galileo

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    If X_1 and X_2 are linearly dependent then clearly span(X_1+X_2)=span(X_1,X_2),
    so the assertion is false in general.

    If X_1 and X_2 are lin. indep., X_3 could still be dep. on X_1 + X_2.
     
  6. Feb 9, 2005 #5
    Actually, that is what I was not sure about, because this:
    span{X1, X2} = span{X1, X2, c1X1 + c2X2} I see.
    But this:
    span{X1, X2} = span{X1 + X2}
    I don't quite. Could you outline the proof?
    If I look at it from the stand-point of dimension, first one has dim. of 2, second -- 1, which I take to mean that they are not equal, even if one is lin. comb. of the other.
    Am I totally off?
    Thank you very much.
     
  7. Feb 9, 2005 #6

    Galileo

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    Of X_1 and X_2 are dependent, then they span a line, one is simply a scalar multiple of the other.

    Use the definition of linear dependence to prove this:
    X_1 and X_2 are linearly dependent means X_1=cX_2 for some c.
     
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