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P is not a subspace of R3. Why?

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data
    P={(x,y,z)|x+2y+z=6}, a plane in R3. P is not a subspace of R3. Why?
    2. Relevant equations
    See below.

    3. The attempt at a solution
    I am really quite confused here.

    My text says:
    "A subset W of a vector space V is called a subspace of V if W is itself a vector space under the addition and scalar multiplication defined on V"
    and goes on to say that the only axioms that need verification are:
    (a) If u and v are vectors in W, then u+v is in W.
    (b) If k is any scalar and u is any vector in W, then ku is in W.

    So from here...I'm a bit confused. Where does this x+2y+z=6 come in to play? I'm really quite lost and cannot find any relevant examples in my text.
     
  2. jcsd
  3. Mar 23, 2010 #2
    Re: Subspaces


    Hi dcramps :smile:

    I'm learning linear algebra by myself , and i'll try to help you to the best of my ability.
    If P is a subspace of [tex]R^{3}[/tex] what should be satisfied?
    [Hint: It must passes through what?]
     
  4. Mar 23, 2010 #3
    Re: Subspaces

    This is a bit of a guess here, since I am a bit confused on the whole thing still, but I believe it must pass through the origin, and since it states x+2y+z=6, that is not satisfied...?
     
  5. Mar 23, 2010 #4

    Mark44

    Staff: Mentor

    Re: Subspaces

    Right, the origin is not in the plane, which means that set P does not include (0, 0, 0). There are a couple of other requirements for P to be a subspace of R3, but since this one failed, there's no need to check the others.
     
  6. Mar 23, 2010 #5
    Re: Subspaces

    Just to make a clearer picture for dcramps , when we say it must pass through the origin it must satisfy the equation of having x,y,z=0 , hence the equation of your plane would be 0+0+0=6 => 0=6 which is absurd. Therefore , we could have the plane passing through the origin and fail to satisfy the criteria of a subspace. :biggrin:
     
  7. Mar 23, 2010 #6

    Mark44

    Staff: Mentor

    Re: Subspaces

    I'm not sure this is a clearer picture...
     
  8. Mar 24, 2010 #7
    Re: Subspaces

    Both answers were helpful. I do have another question though, since the solution given did not mention the origin. The solution to the question I asked above is:

    (x+x') + 2(y+y') + (z+z') = (x+2y+z) + (x'+2y'+z') = 6 + 6 = 12
    thus
    (x+x') + 2(y+y') + (z+z') is not in P, and so P is not a subspace of R3

    I don't understand why it isn't. Because it's 12? Huh?
     
  9. Mar 24, 2010 #8

    statdad

    User Avatar
    Homework Helper

    Re: Subspaces

    The text's solution shows that if you take two points that are on the plane and "add them" the result is not in the plane (because of the 12) - the set of points in the plane is not closed under addition.
     
  10. Mar 24, 2010 #9

    Mark44

    Staff: Mentor

    Re: Subspaces

    If (x, y, z) is in P, then x + 2y + z = 6.
    If (x', y', z') is in P, then x' + 2y' + z' = 6.

    Now let's check whether (x, y, z) + (x', y', z') is in P.

    If so, then (x + x', y + y', z + z') is in P, since (x, y, z) + (x', y', z') = (x + x', y + y', z + z').

    (x + x', y + y', z + z') is in P provided that x + x' + 2y + 2y' + z + z' = 6. But x + x' + 2y + 2y' + z + z' = x + 2y + z + x' + 2y' + z' = 6 + 6 = 12, from previous work.
    So x + x' + 2y + 2y' + z + z' [itex]\neq[/itex] 6, hence (x + x', y + y', z + z') is NOT in P.

    If (x, y, z) + (x', y', z') = (x + x', y + y', z + z') is in P. If so, then x + x' + 2(y + y') + z + z' = 6.

    But we've already seen that

    (x, y, z) + (x', y', z')
     
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