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Calculate dim(U) and company - linear algebra

  1. Mar 27, 2007 #1
    1. The problem statement, all variables and given/known data
    Before I start, I should mention that I'm taking linear algebra in French so I hope I translated all the terms correctly. I'm studying for a test and here is a question i'm having a little trouble with...hope you can help!

    U = span{[1 1 1 1]T, [-2 -2 1 1]T, [3 3 3 1]T, [0 0 1 0]T}

    a) Calculate dim(U) and give a base of U.
    b) Complete the base found in a) to obtain a base of R4.

    2. Relevant equations

    3. The attempt at a solution

    a) I get a base of (-5/6, -1/3, -1/2, 1) so dim(U) has dim(U) = 1

    b) This is where I don't know what to do. How to you complete a base? I don't need the answer, I just want to know how to complete a base. The only thought I had was to use the identity matrix but then again, I really don't know what to do or how to approach this. Any help is appreciated!
  2. jcsd
  3. Mar 27, 2007 #2


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    You seem to be misunderstanding the question since your answer to a) is not at all right. What you want to do is take the 4 vectors in your set and find a minimal set of linearly independent vectors with the same span. The usual technique to do this is called 'row reduction' (sorry I don't know the french!).
  4. Mar 27, 2007 #3
    The dimension is determined by the smallest amount of linearly independent vectors that you can find, which describe entirely U. To do so, reduction is the way out.

  5. Mar 27, 2007 #4
    hmm I thought thats what I did...

    I reduced the matrix from:
    1 -2 3 0
    1 -2 3 0
    1 1 3 1
    1 1 1 0

    to get..

    1 0 0 5/6
    0 1 0 1/3
    0 0 1 1/2
    0 0 0 0 <--- isn't this the "row reduction" you were talking about?
  6. Mar 27, 2007 #5


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    That's sort of it. Though I usually write the vectors as rows. Since you've written them as columns and then (I think) row reduced, it's a little hard for me to interpret the results exactly - but I can clearly see that U has dimension 3. (How many obvious independent columns in your set?). Try the same exercise writing the vectors as rows.
  7. Mar 27, 2007 #6


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    Yes, that is the "row reduction" but that does NOT mean that the last column gives you the only basis vector. In fact, since you have 3 non-zero rows left, that tells you that the dimension of this subspace is 3.

    Use the given vectors as rows, not columns, and row-reduce. The remaining non-zero rows will be the basis vectors.
  8. Mar 27, 2007 #7


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    The "row reduction" result doesn't look correct, unless I missed something. Go through it again.

    Edit: actually, it's almost correct, you missed a sign (figure out which one!) So, you have shown that the vector (0, 0, 1, 0) can be represented as a linear combination of the other vectors, so the set without the vector is a span too, and linearly independent, which makes it a basis for U.
    Last edited: Mar 27, 2007
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