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The dimension of the span of three linearly independent R^3 vectors

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data


    [tex]x_{1}= \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}, x_{2}= \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}, x_{3}= \begin{pmatrix} 2 \\ 6 \\ 4 \end{pmatrix}[/tex]

    Find the dimension of [tex]Span(x_{1}, x_{2}, x_{3})[/tex]

    2. Relevant equations

    If V is a vector space of dimension n > 0 then any set of n linearly independent vectors spans V.

    If the x vectors are combined into a matrix and the matrix is nonsingular then the vectors are linearly independent.

    [tex]dim R^{3} = 3[/tex]

    3. The attempt at a solution

    [tex]det \begin{vmatrix} 2 & 3 & 2 \\ 1 & -1 & 6 \\ 3 & 4 & 4 \end{vmatrix} = -28.[/tex]

    This shows that the three vectors are linearly independent. This should mean that any vector in [tex]R^{3}[/tex] should be able to be represented as a linear combination of the three given vectors. Therefore, shouldn't [tex]Span(x_{1}, x_{2}, x{3})[/tex] have a dimension of 3? The answers section in the back of my book says it has 2 dimensions. I can't quite figure out why.
  2. jcsd
  3. Feb 28, 2008 #2
    Heh, I'm being silly. I miscalculated the determinant. It should be 0. Looks like we can represent any one of the vectors with any of the other 2.
  4. Feb 28, 2008 #3

    what's wrong is that your calc is wrong----> det=0 in this case,

    do it another time, pay attection to the signs.

  5. Feb 28, 2008 #4


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    Staff Emeritus
    Science Advisor

    Be careful. Knowing that the determinant is 0 only tells you that the dimension of the span is less than 3. Either one of the vectors can be written as a combination of the other two (and so the span has dimension 2) or two can be written as multiples of the other one (in which case the span would have dimension 1). I would use row reduction rather than finding the determinant to distinguish those possibilities.
  6. Feb 28, 2008 #5
    Yes you are right, in general is like you say; but with three vector if the determinant is zero you can be sure that two are independent just looking at the numbers, i.e. choose two that are not multiple of each others. In this case is easy.

  7. Feb 28, 2009 #6
    I'm actually working a problem similar to this, could you please explain this more? Does writing something as a combination involves multiplying the first vector by a scalar to get the second vector? So if I can multiply a vector by a scalar and get the other two then the dimension is 2?

    I'm pretty lost.
  8. Mar 1, 2009 #7


    Staff: Mentor

    With a 3 x 3 matrix whose determinant is zero, the set of three vectors is linearly dependent. This means that one of the following must be true:
    1. One vector is a scalar multiple of another vector, such that there are two vectors that are linearly independent.
    2. One vector is a linear combination of the other two, such that there are two vectors that are linearly independent.
    3. Two vectors are (possibly different) scalar multiples of the third.
    4. All three vectors are zero vectors.
    The dimensions of the span{v1, v2, v3} in these situations would be, respectively:
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