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How many dimensions is this vector space?

  1. Feb 28, 2013 #1
    So I just got out of my linear algebra midterm, and this question is confusing the hell out of me. Basically, it's a subspace of R^4, such that the coordinates satisfy the following qualifications:

    (a, b - a, b, 2(b - a))

    So basically, a and b can range over the xz plane, and y and w sort of follow suit. I said that this space had the basis given by the three vectors:

    (a,0,0,0)
    (0,0,b,0)
    (0,b-a,0,2(b-a))

    But now that I look back at it, the more it seems like this is a 2-dimensional subspace, since you're essentially mapping from the xz plane to a line in yw, which means that it's basically isomorphic to a plane in three space.

    Any thoughts?
     
  2. jcsd
  3. Feb 28, 2013 #2

    jbunniii

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    Note that you can write ##(a, b-a, b, 2(b-a))## as the linear combination of two vectors: $$(a, b-a, b, 2(b-a)) = a(1, -1, 0, -2) + b(0, 1, 1, 2)$$
    What does this tell you about the dimension of the subspace?
     
  4. Feb 28, 2013 #3

    pasmith

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    It is indeed 2-dimensional: you can re-write it as [itex]a(1, -1, 0, -2) + b(0,1,1,2)[/itex].

    None of the vectors you give as basis vectors are actually in that subspace for arbitrary a and b: to get the second component to vanish you must take a = b, so either the first and third components are both non-zero or they are both zero, in which case all components are zero. To make the first and third components vanish you must take a = b = 0, in which case all components are zero.
     
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