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Subspace Proof

  1. Aug 17, 2009 #1
    1. The problem statement, all variables and given/known data

    For the following subset W of R3 determine whether or not W is a subspace of R3. If the subset is not a subspace give a specific example to indicate why it is not a subspace.

    ii.) W = {(x,y,z): 2x + y + 3z = 0

    3. The attempt at a solution

    I know how to do this mostly, but there's two bits that I don't understand.

    For the 'closed under addition' test, I said if (x1, y1, z1) and (x2, y2, z2) are in W then 2x1 + y1 + 3z1 = 0 and 2x2 + y2 + 3z2 = 0

    Thus, 2(x1+x2) + y1 + y2 + 3(z1 + z2) = 0

    But how can you just add them together and say they equal zero? Surely you'd have to subtract one from the other? Like x = 0, y = 0 therefore x = y and x - y = 0?

    Then, if that is true, (x1 + x2, y1+y2, z1+z2) is in W.

    This is the bit I really don't understand. How can you jump from 2(x1+x2) + y1 + y2 + 3(z1 + z2) = 0 to (x1 + x2, y1+y2, z1+z2)?

    Thanks!
     
  2. jcsd
  3. Aug 17, 2009 #2
    0 = 0 + 0 = (2x1 + y1 + 3z1) + (2x2 + y2 + 3z2) = 2(x1+x2) + (y1 + y2) + 3(z1 + z2)

    If you write x3=x1+x2, y3=y1+y2, z3=z1+z2, then it will become clear that this vector (x3,y3,z3) is also in W.
     
  4. Aug 17, 2009 #3
    Also it's non-empty, since 2.0+0+3.0=0, [tex]0 \in W[/tex] hence [tex]W \neq \emptyset[/tex]

    And you also need to show it's closed under scalar multipication. For x,y,z in W and for [tex]\lambda \in R[/tex]

    [tex]\lambda (2x+y+3z) = \lambda 2x+ \lambda y +\lambda 3z[/tex][tex]= (\lambda . 2)x+ (\lambda) y + (\lambda . 3)z[/tex]
     
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