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Linear Algebra: Span

  1. Oct 25, 2007 #1
    Hi all,

    I am having a little trouble understanding the concept of span. I do realize the definition of the span of two vectors is all possible linear combinations of the two vectors but I am trying to make the concept understandable with regards to the problems I have been assigned.

    Before I give the problem, let me first ask a question. Since I do understand the concept of column space, can I think of a span of two vectors in a similar manner. For example,
    is the column space of
    [1 1
    1 0
    2 1]
    equal to the span of the column vectors v1 = [1 1 2] and v2 = [1 0 1] ?


    Anyway, here is the problem along with my solution.

    Let column vectors v1 = [1 1 2] and v2 = [1 0 1]
    Find mutually orthogonal vectors u1 and u2 such that the span of {v1, v2} is the same as the span of {u1, u2}.

    First, I check to see if the vectors are multiples of each other. Since they are not, I know that the two vectors make a plane, which is the span of the vectors. I use the cross product to find the equation of the plane which is i + j - k = 0 Since I must find two orthogonal vectors that are in the span of the two given vectors, I am basically looking for two orthogonal vectors that satisfy the above equation of the plane. So if I take v1 = u1, then I can choose u2 = [1 -1 0]

    Is this correct?
     
  2. jcsd
  3. Oct 25, 2007 #2

    Dick

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    Homework Helper

    I think you understand the concept very well. There are other ways to orthogonalize a set of vectors that don't involve the cross product but that will work.
     
  4. Oct 25, 2007 #3
    Could you please elaborate on the method at which you hinted? Does it work in dimensions greater than 3?
     
  5. Oct 25, 2007 #4
  6. Oct 25, 2007 #5

    Dick

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    Science Advisor
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    It's called Gram-Schmidt. And, yes it works in dimensions not equal to 3, where you don't have a cross product. You basically use the dot product to remove parallel components of vectors. I'll let you look it up.
     
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