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Intersection of subspaces

  1. Sep 5, 2005 #1
    I have 2 subspaces U and V of R^3 which
    U = {(a1, a2, a3) in R^3: a1 = 3(a2) and a3 = -a2}
    V = {(a1, a2, a3) in R^3: a1 - 4(a2) - a3 = 0}

    I used the information in U and substituted it into the equation in V and I got 0 = 0. So, does it mean that the intersection of U and V is the whole R^3 which has no restrictions on a1, a2 and a3 (they are free)? Or do the original restrictions on both the original subspaces still being applied to the intersection?
  2. jcsd
  3. Sep 5, 2005 #2


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    The intersection of U and V cannot possibly be all of R³. How could the intersection of two sets be bigger than both of the sets? Both subspaces are 1-dimensional, so the intersection is either 1-dimensional or 0-dimensional. Can you find a non-zero point that is in both U and V? If so, then the intersection of U and V is U and is also V (i.e. U = V). A point in U takes the form (x, x/3, -x/3). Would such a point be in V?

    x - 4(x/3) - (-x/3) = x - (4/3)x + (1/3)x = 0

    so the answer is "yes."
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