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A basis

  1. Jan 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Find a basis for the subspace

    S = {(a+2b,b-a+b,a+3b) | a,b [tex]\in[/tex] R } [tex]\subseteq[/tex] R^4

    What is the dimension of S?

    2. Relevant equations



    3. The attempt at a solution

    a(1,0,-1,1) + b(2,1,1,3) , a,b [tex]\in[/tex] R

    span { (1,0,-1,1) , (2,1,1,3) }

    So I put (1,0,-1,1) as V1 and (2,1,1,3) as V2 and then formed a matrix with V1 and V2 in columns. Then i reduced it to row echelon form. The column corresponding to the leading entry form a basis which is essentially just (1,0,-1,1) and (2,1,1,3) . Thus the dimension is just 2. Am I doing the right thing because I dont have an answer to refer to.
     
  2. jcsd
  3. Jan 22, 2009 #2

    Mark44

    Staff: Mentor

    There has to be an error in your definition for this set.
    Your vector has only three coordinates, which is too few for it to be in R^4. Are you missing a comma in there?
     
  4. Jan 23, 2009 #3
    Hi Jeffrey,
    The first step you have done is shown that S = span {(1,0,-1,1), (2,1,1,3)}. Then if (1,0,-1,1) and (2,1,1,3) are linearly independent, they form a basis (recall the definition) for S.

    It is correct that n vectors are linearly independent if and only if the matrix formed with them as its columns has rank n, but in this case it is probably simpler just to recognise that two non-zero vectors are linearly independent iff they're not scalar multiples of each other.
     
  5. Jan 25, 2009 #4
    Oops, my bad. Yes, there is a comma in there. Should be, S = {(a+2b,b,-a+b,a+3b) | a,b R } R^4
     
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