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Column Space problem

  1. Aug 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Obtain the column space of the following matrix

    B = 2 -3 -1

    2 -3 -1

    -3 3 2


    2. Relevant equations


    Linear independence test

    c1V1 + c2V2 + ...... + cnVn = 0

    c1=c2=...cn=0

    3. The attempt at a solution

    Please see attachment

    I cannot get the answer which is CS(B) = span{ (2,2,-3)^T, (1,1,-1)^T } where T denotes transpose


    The (1,1,-1)^T can be realised by adding a(2,2,-3)^T and c(-1,-1,2)^T together (where a=c) and setting a=1

    but the problem is that it destroys the linear independence of (1,1,-1)^T since it depends on (2,2,-3)^T to generate the result
     

    Attached Files:

  2. jcsd
  3. Aug 21, 2011 #2
    So if you want to find the column space, you have to find the column vectors. To do this, you need to identify which columns are pivot columns by row reducing your matrix B. once this is done, the column space will simply be the span of these vectors. you can check the linear independence of these vectors pretty simply.

    also, when you're worrying about the vector (1,1,-1)^T depending on the other one, you have to take into account that you're using a second vector to get to (1,1,-1)^T (namely the vector (-1, -1, 2)^T). You would need to worry if you could somehow form one vector purely from one of your other ones. make sense?
     
  4. Aug 22, 2011 #3
    I tried using Gauss-Jordan elimination method, but it doesn't put it in the form that the answer gives

    This is why I used the linear cominbation method


    Okay a vector in a set is linearly independent if it cannot be formed by the linear combination of any other vectors in that same set. The (1,1,-1)^T must depend on the
    (-1, -1, 2)^T, which is outside the spanning set given in the answers.

    This is why (1,1,-1)^T is linearly independent of (2,2,-3)^T

    right ?
     
  5. Aug 22, 2011 #4

    HallsofIvy

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    Two vectors are linearlydependent if and only if one is a multiple of the other. (2, 2, 3) clearly is not a multiple of (1, 1, -1).
     
  6. Aug 22, 2011 #5
    The whole point of the row reduction is to find which columns are pivot columns. Once you identify those, you have to go back to the original matrix for the actual vectors that will form the basis. The row reduction will tell you which column, but the actual matrix tells you the vector corresponding to the pivot column.

    Your reasoning is pretty sound for linear independence. Just be careful about which vectors you're actually trying to form. Also keep in mind that there are multiple bases that are possible for this kind of problem...
     
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