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Linear Algebra: 4 Fundamental Subspaces

  1. Mar 3, 2008 #1
    1. The problem statement, all variables and given/known data

    Without computing A, find the bases for the 4 fundamental subspaces.

    [1 0 0][1 2 3 4]
    [6 1 0][0 1 2 3]=A=LU
    [9 8 1][0 0 1 2]

    2. Relevant equations

    N/A

    3. The attempt at a solution

    There was an "example" in the book. It just showed the answers.
    It was:

    [1 0 0][1 3 0 5]
    [2 1 0][0 0 1 6]=A
    [5 0 1][0 0 0 0]
    Where
    Row Space: Basis (1,3,0,5) and (0,0,1,6)
    Column Space: Basis (1,2,5) and (0,1,0)

    In the problem we have to do, I take it that the row space's basis is just the 3 rows of U similar the the example. But, I'm unsure of the column space. Is it the 3 columns of L?

    And, also for the nullspace, I put U in its row reduced echelon form, and solved for the nullspace as one would normally do. Is this correct?

    [1 2 3 4]
    [0 1 2 3] -->
    [0 0 1 2]

    [1 0 0 0]
    [0 1 0 -1]
    [0 0 1 2]
    So, the nullspace basis is (0,1, -2, 1)

    And, since there isn't a zero row, there also isn't a left nullspace, correct?

    Sorry if that's confusing. And, thank you in advance!
     
  2. jcsd
  3. Mar 4, 2008 #2
    Bump?
     
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