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Test if these 4 vectors span R^3

  1. Jul 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine if the vectors v1=(3,1,4), v2=(2,-3,5), v3=(5,-2,9), v4=(1,4,-1) span ℝ3

    2. Relevant equations

    3. The attempt at a solution

    So I first arranged it as a matrix,


    \begin{bmatrix}
    \begin{array}{cccc|c}
    3&2&5&1&b_1\\
    1&-3&-2&4&b_2\\
    4&5&9&-1&b_3
    \end{array}
    \end{bmatrix}

    Now I know what to do if it's a square matrix. I just have to see if the coefficient matrix is invertible (det ≠0). If yes that would mean that any vector b can be expressed as a linear combination. Since this is not a square matrix, I thought I'd have to row reduce it.

    Row reduced:

    \begin{bmatrix}
    \begin{array}{cccc}
    1&0&1&1\\
    0&1&1&1\\
    0&0&0&0
    \end{array}
    \end{bmatrix}

    Now what do I do? It seems to me that the system has infinitely many solutions and therefore the vectors span ℝ3. But the solution manual says that it doesn't. What am I doing wrong? Thanks.
     
  2. jcsd
  3. Jul 21, 2012 #2
    Hey jey1234 and welcome to the forums.

    Think about the rank of your matrix: how many non-zero rows do you have?

    If you have one non-zero row, then everything is a multiple of the first row which means that everything is a multiple of some vector.

    Building on this idea, what is the difference between having two non-zero rows and no non-zero rows in this matrix in terms of the dimension of the space (for one non-zero row we have one dimension since everything is a scalar multiple of that vector)?
     
  4. Jul 22, 2012 #3

    HallsofIvy

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    The question was whether the vector span the space, not whether or not the form a basis. The fact that the system "has infinitely many solutions" means it has solutions- and so the vectors do span the space. The fact there there is not a unique solution means they are not independent and do not form a basis for R3. But you already knew that- no set of four vectors can be a basis for a three dimensional vector space.
     
  5. Jul 22, 2012 #4

    LCKurtz

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    But the OP's row reduction just shows the homogeneous system has infinitely many solutions. If you try to write [1,1,1] as a linear combination of the given vectors you get an inconsistent system.
     
  6. Jul 22, 2012 #5

    SammyS

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    jey1234,

    Have you noticed that v3 = v1 + v2

    and v4 = v1 - v2 ?
     
    Last edited: Jul 22, 2012
  7. Jul 22, 2012 #6

    LCKurtz

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    Who, me? Halls??
     
  8. Jul 22, 2012 #7

    SammyS

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    Neither you nor Halls ...

    The OP, jey1234 .

    (I will edit that !)
     
  9. Jul 22, 2012 #8
    Do you understand the consequences of your line of zeros in your row reduced matrix? That means one of the 3 vectors can be obtained by a linear combination of the other 2. What does that tell you? Well, the 3 vectors lie in a plane.

    Can you see why they do not span ℝ3?

    EDIT: Sorry, I thought you had 3 vectors in R4, but you actually have 4 vectors in R3.
    Since you wrote the vectors vertically, then you actually have to column-reduce them. If I were you, I would transpose the matrix and complete the row reduction. You will see that 2 of your vectors can be obtained by a linear combination of the other 2. In other words, you will have 2 lines of zeros.
     
    Last edited: Jul 22, 2012
  10. Jul 22, 2012 #9
    OP's matrix is confusing because he wrote his vectors as columns instead of rows. 2 of the vectors are actually linear combinations of the other 2, which means they do not span R3.
     
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