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Vector S, dimension of subspace Span(S)?

  1. Apr 14, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider the set of vectors S= {a1,a2,a3,a4}
    where
    a1= (6,4,1,-1,2)
    a2 = (1,0,2,3,-4)
    a3= (1,4,-9,-16,22)
    a4= (7,1,0,-1,3)

    Find the dimension of the subspace Span(S)?

    Find a set of vectors in S that forms basis of Span(S)?

    2. Relevant equations
    dimension of V = n in Rn?




    3. The attempt at a solution

    - Part one: Dimension

    If equation is true and Span(S) is in fact basis of V then is the dimension 4? Is it that easy?

    - Part two: Basis
    Is this asking for the k1,k2,k3,k4 that make Span(S) = V?
     
    Last edited: Apr 14, 2014
  2. jcsd
  3. Apr 14, 2014 #2

    vela

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    What equation?

    Span(S) is a subspace, not a basis. To say that Span(S) is a basis of V doesn't make any sense.

    What are the k's supposed to be?

    It seems you're not clear about basic definitions of terms like span and basis. You should get those down first before you attempt to do this problem.
     
  4. Apr 14, 2014 #3
    Sorry my reply was confusing I mixed up a few terms here and there.
    So what I have determined thus far:
    1. the dimension is 3
    I used row eliminations in a matrix with [ a1 a2 a3 a4] and got two rows of zeroes out of the total five rows. Thus 5-2 = 3 and the dimension is 3.

    2. what I need to figure out now is which of the vectors is a basis for the Span(S). How do I do this?

    the options are:
    {a2,a3}
    {a1,a2,a3}
    {a1,a2}
    {a1,a2,a3,a4}
    {a1,a2,a4}
     
  5. Apr 15, 2014 #4

    HallsofIvy

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    Since you have already determined that the dimension is three, those sets of two or four vectors are immediately eliminated. To determine whether {a1, a2, a3} or {a1, a2, a4} is a basis, do the same thing you did with the entire set. See if they are linearly independent.
     
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