1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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)?

  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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook