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Determine whether the given vectors form a basis

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data

    w1 = 2 1 2
    w2 = 1 -2 -3
    w3 = 5 0 1

    for
    R3


    2. Relevant equations



    3. The attempt at a solution

    The books says the above is not a basis, why not? There are no free variables, none of the vectors are multiples of the other, they are linearly independent and the number of unknowns equals the number of equations. That check list to me signals a basis
     
  2. jcsd
  3. May 12, 2012 #2
    What makes you think they are lineally independent?
     
  4. May 12, 2012 #3
    As I already said: There are no free variables, none of the vectors are multiples of the other

    This is what I get in Reduced Ech form

    1 -2 -3
    0 5 -4
    0 0 -6
     
  5. May 12, 2012 #4
    Actually is:
    [tex]w_3=2w_1+w_2[/tex]
     
  6. May 12, 2012 #5
    In echelon form I get
    [itex]\left( \begin{matrix}
    1 & -2 & -3 \\
    0 & 5 & 8 \\
    0 & 0 & 0
    \end{matrix} \right)[/itex]

    Also be careful with difference between echelon form and reduced echelon form. The matrix you wrote (and the one above) is in echelon form, as every pivot has only zeros below it. Whereas in reduced echelon form the piviot is the only non zero entry in the column i.e. the above matrix becomes:

    [itex]\left( \begin{matrix}
    1 & 0 & \frac{1}{5} \\
    0 & 1 & \frac{8}{5} \\
    0 & 0 & 0
    \end{matrix} \right)[/itex]
     
  7. May 12, 2012 #6
    Ok, I understand now.

    Also thanks for the tip about echelon form and reduced echelon form, I thought they were the same.
     
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