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Span question with Restrictions

  1. Jun 29, 2011 #1
    1. The problem statement, all variables and given/known data

    Let U be the span of
    [ 0 ] [ 2 ]
    [ 1 ] and [ 0 ]
    [ 2 ] [ -1]
    [ 4 ] [ 1 ]

    Give conditions on a, b, c and d so that [a b c d] (transposed) is in U (as in, give restrictions that are equations of only a, b, c and d.

    2. Relevant equations



    3. The attempt at a solution

    I've set up the problem to set the two given vectors multiplied by scalar x and y, yielding four equations:
    2y = a,
    x = b,
    2x - y = c, and
    4x + y = d.

    Is this the right first step to take? After that, I take x or b as a free variable, and try to rewrite the other equations in terms of b, yielding:

    2y = a, b = b, 2b - y = c, and 4b + y = d.

    After this, however, I am stuck. Any pointers? Did I set up the problem the right way?

    Edit: I also know that [a b c d] must be a linear combination of the two given vectors if it is in the span. However, if it is a linearly independent set, the trivial zero vector could be matched with [a b c d], although I am pretty sure the answer isn't that simple.
     
    Last edited: Jun 29, 2011
  2. jcsd
  3. Jun 29, 2011 #2

    micromass

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    Hi jberg074! :smile:

    You also know that 2y=a, so [itex]y=\frac{a}{2}[/itex]. Substitute that into the equations

    2b - y = c
    4b + y = d.

    and you'll find your conditions on a,b,c and d.
     
  4. Jun 29, 2011 #3
    Is the answer as simple as this?
    a=6b-c
    b=b
    c=2b-a/2
    d=4b+a/2
     
  5. Jun 29, 2011 #4

    micromass

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    I'm not sure where you got the a=6b-c, actually. But the two last equations are what you need. These give necessary and sufficient conditions for something to be in the span. In general, the span is given by

    [tex]\{(a,b,2b-a/2,4b+a/2)~\vert~a,b\in \mathbb{R}\}[/tex]
     
  6. Jun 29, 2011 #5
    [tex]\{(a,b,2b-a/2,4b+a/2)~\vert~a,b\in \mathbb{R}\}[/tex][/QUOTE]

    I used other questions to eliminate variables and to try and get a in terms of the other. However, in the text above, shouldn't "a" be replaced by "a/2"?
     
  7. Jun 29, 2011 #6

    micromass

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    I used other questions to eliminate variables and to try and get a in terms of the other. However, in the text above, shouldn't "a" be replaced by "a/2"?[/QUOTE]

    No, we have concluded that

    a=a
    b=b
    c=2b-a/2
    d=4b+a/2

    So, we will have

    [tex](a,b,c,d)=(a,b,2b-a/2,4b+a/2)[/tex]
     
  8. Jun 29, 2011 #7
    Ah, I understand now. I wasn't exactly sure what the question was asking for at first; thank you very much for the help!
     
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