Test if these 4 vectors span R^3

  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. 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. HallsofIvy

    HallsofIvy 40,674
    Staff Emeritus
    Science Advisor

    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. LCKurtz

    LCKurtz 8,386
    Homework Helper
    Gold Member

    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. SammyS

    SammyS 8,740
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    jey1234,

    Have you noticed that v3 = v1 + v2

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

    LCKurtz 8,386
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    Gold Member

    Who, me? Halls??
     
  8. SammyS

    SammyS 8,740
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Neither you nor Halls ...

    The OP, jey1234 .

    (I will edit that !)
     
  9. 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. 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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