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LA - spanning R^3

  1. Mar 22, 2007 #1

    daniel_i_l

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    1. The problem statement, all variables and given/known data
    you have 4 vectors:
    v1=(2,1,-1), v2=(-m,-1,3), v3=(-3,2,m+1), v4=(1,2,1)
    a) for which values of m do the vectors span R^3
    b) the vector w=(m+1, m-1, 1). for which m is there a solution to:
    x1v1 + x2v2 + x3v2 + x4v4 = w ?

    2. Relevant equations
    the definition of spanning - every vector in R^3 can be made of those 4 vectors.


    3. The attempt at a solution
    a) i made the following matrix where the first few coloums are v4,v1,v3,v4 and the last one is some vector (a,b,c):
    Code (Text):

    [1  2   -3 -m  a]
    [2  1    2 -1  b]
    [1 -1 (m+1) 3  c]
    and reduced it to:
    [1  2   -3    -m   a]
    [0 -3    8  (2m-1) b-2a]
    [0  0 (m-4) (4-m) c-b+a]  
     
    now the only way that there isn't a solution is if there's a row with the form:
    [0 0 0 0 alpha] where alpha is any scalar. since the only row that can possibly look like this is the third one where m=4 then the answer is that they span R^3 for every m except 4.
    Is that right?
    b) here's where i got confused, if m doesn't equal 4 then there's a solution because according to a) the vectors v1,v2,v3 and v4 span R^3 which includes w. and if m=4 then there's no solution because for w:
    c-b+a = 3 (not 0)
    so the answer for b should be the same as a). is that true? it doesn't seem likely that there isn't some trick or something.
    Thanks.
     
    Last edited: Mar 22, 2007
  2. jcsd
  3. Mar 22, 2007 #2
    Woah. If you have 3 independent vectors they automatically span R^3. I can give you four vectors (1,0,0), (2,0,0), (3,0,0) and (4,0,0) but they do not span R^3.

    Your solution is right. If m is not 4 then you can reduce the matrix to find all solutions (pick such an m if you want to see). Note how these 4 vectors 'overspan' R^3 so that you would get infinitely many solutions.

    We know that for m not equal to four there is a solution to w (as then the 4 vectors span R^3). What happens when you try to solve the system when m=4?
     
  4. Mar 22, 2007 #3

    daniel_i_l

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    When m=4 then there's no solution because you can reduce the matrix to a matrix that has the row:
    [0 0 0 0 3]
    is that right? if so why would they give two questions with the same answer?
     
  5. Mar 22, 2007 #4

    Dick

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    I think you are reducing the transpose of the matrix you want to reduce. Reduce this one (and I would say forget about the a,b,c). You are missing one of the values of m that makes this group of vectors fail to span.

    [ 1 2 1 ]
    [ - m - 1 3 ]
    [ - 3 2 m + 1 ]
    [ 1 2 1 ]
     
  6. Mar 22, 2007 #5

    daniel_i_l

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    i made the matrix that i posted above to check when an arbitrary vector is a linear combination of the four given vectors, to do that you'd put in the vectors as coloums right?
     
  7. Mar 22, 2007 #6

    Dick

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    If you are going to do row reduction I think you should put in the vectors as rows. When the reduction is finished you can tell by inspection how many of the vectors are linearly independent. If there are three of them, then they span R^3.
     
  8. Mar 22, 2007 #7
    Yes, you can have four vectors, as long as one is linearly dependent. The range has to be 3 dimensional.
     
  9. Mar 22, 2007 #8

    Dick

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    If he row reduces, linear dependence will show up as zero vectors in the list.
     
  10. Mar 23, 2007 #9

    daniel_i_l

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    yes, you get the same answer regardless of whether you put the vectors in as rows and coloums as long as you know how to interpert the final matrix. my question is if my answer for b is correct.
    thanks
     
  11. Mar 23, 2007 #10

    Dick

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    I can't argue with that. Today I think it's correct. Yesterday, I was getting a quadratic for m. I think I was wrong yesterday.
     
  12. Mar 23, 2007 #11

    radou

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    If you know how to add ordered triples, then I don't see why the issue about putting the vectors into the matrix as rows or columns would arise.
     
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