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Vector Spaces, Dimension of Subspace

  1. Apr 14, 2007 #1
    Find the dimension of the subspace spanned by the vectors u, v, w in each of the following cases:
    i) u = (1,-1,2)^T v = (0,-1,1)^T w = (3,-2, 5)^T
    ii) u = (0,1,1)^T v = (1,0,1)^T w = (1,1,0)^T

    Right, how do I go about this, do I have to find the subspace first then do the dimension.

    Can someone give me sum clues.

  2. jcsd
  3. Apr 14, 2007 #2

    matt grime

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    Posting it in linear algebra might be more appropriate.

    Anyway, just put the vectors as rows in a matrix, and put it in row echelon form - that tells you the dimension and the space. Just like you ought to have been told in class.

    Of course, clearly those span 2 or 3 dimensional vector subspaces, so you only need to verify linear dependence/independence, which might be easier for you if you're not familiar with row echelon form. The last three are clearly linearly independent by inspection.
  4. Apr 14, 2007 #3
    Sorry, did not know where to put it, an I did a search on vectors, and it showed up in this section, thats why I did it.

    We are not taught anything is class, we are just given work books, and then an assignment, this is one of the final questions in my work book.

    And I can not remember row echelon form as such. So what I find whether they are linearly independant or not and then what?

    Thanks for your help.
  5. Apr 14, 2007 #4


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    I've moved this to the "Linear and Abstract Algebra" forum.

    The dimension of a vector space is the number of vectors in a basis. A basis must both span the space and be independent. Since each of these vectors spaces is spanned by three vectors, the dimension cannot be more than 3. Now determine whether or not these vectors are independent. there are many ways to do that but one, as matt grime said, to construct a matrix having the vectors as rows. The number of independent vectors is the number of non-zero rows you have left after row reducing and therefore the dimension of the space. I strongly recommend you review "row reduction"!
  6. Apr 15, 2007 #5


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    the only way to ,show the first three are dependent would be to kill off the first entry among the two that have a first entry, so the only possible way is to multiply the first one by -3.

    adding that multiple to the last one then does it, since it gives a multiple of the second one.

    so the dimensions are 2,3.
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