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Prove sum of two subspaces is R^3

  1. Jan 31, 2010 #1
    How do you prove that the sum of the following subspaces is R^3?
    U = {(x,y,z) : x - y = z}
    W = {(t,-t,-t) : t∈R}

    I guess I need to show that any vector (x,y,z)∈R^3 can be written as the sum of a vector from U and a vector from W, but I'm not sure how to do that. I know intuitively that U+W=R^3 because U is a plane and W is a line not contained in U, but I don't know how to show that mathematically.

    Any hints?
  2. jcsd
  3. Jan 31, 2010 #2


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    How does that work?
  4. Jan 31, 2010 #3
    x-y-z=0 is the equation of a plane and n=(1,-1,-1) is a normal vector, so (1,-1,-1)t is a line perpendicular to the plane, and the span of the basis elements is R^3.
  5. Jan 31, 2010 #4


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    That sounds like a sketch for a mathematical proof to me. Can you justify each part of it?

    (p.s. nobody said anything about a basis....)
  6. Feb 1, 2010 #5


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    Any vector in U can be written as <x, y, x- y>= <x, 0, x>+ <0, y, -y>= x<1, 0, 1>+ y<0, 1, -1>. What is a basis for U?

    Any vector in W can be written as <t, -t, -t>= t<1, -1, -1>. What is a basis for U?

    Suppose you put those two bases together? Are they still independent?
  7. Feb 1, 2010 #6
    Moreover, notice that Sp(B1)+Sp(B2)=Sp(B1uB2), where B1,B2 are sets or bases (in our case).
  8. Feb 3, 2010 #7
    You can also use the dimension theorem dim(U+V)=dim(U)+dim(V)-dim(U^V)

    And the fact that if U is a subspace of V and they have the same dimension then U=V.
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