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Determining whether or not (a,0,0) and (a,b,0) are subspaces of R3

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data
    "use theorem (below) to determine which of the following are subspaces of R3:

    (a,0,0) and (a,b,0)

    2. Relevant equations

    The theorem: W is a subspace of V iff:

    - u and v are vectors in W, u + v is in W
    - k is a scalar, u is a vector in W, then ku is in W

    3. The attempt at a solution

    I thought that I understood that (a,0,0) was a subspace of R3 since if we take (a,0,0) and (b,0,0) we get (a+b,0,0) which is still in the subspace, the solution says that this is a subspace of R3 but then I thought the same idea would hold for (a,b,0) since if we take (a,b,0) and (c,d,0) we would get (a+b,c+d,0) but the solution sheet says it is not, however my solution sheet doesn't elaborate on why, its just a yes/no answer.

    (PS I also thought multiplying by a scalar would hold too)

    Can anyone help me out? I having a heck of a time truly understanding vector spaces. Thanks in advance
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 7, 2009 #2
    You are correct. The set of all (a,b,0) is a vector subspace of R3. In fact, it is exactly the xy-plane R2 as visualized in R3.

    Likewise the set of all (a,0,0) is simply the x-axis as visualized in R3.

    It's actually a much easier concept that you'd expect. I remember struggling on it somewhat too, but if you just follow the definitions of what a vector subspace or vector space is, and stick to them when you're trying to show something is a vect. space, you'll be fine (as you were above.)
  4. Apr 7, 2009 #3


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    My first reaction was "No, (a, 0, 0) is not a subspace of R3, it is a vector in R3", but what you really mean is:

    Is V= {(a, 0, 0)| a a real number}, with the same definitions of addition and scalar multiplication as in R3, a subspace of R3?

    Use the definition of "subspace"! It is clearly a subset of R3 but does it satify the conditions for a vector space? Fortunately most of the conditions are automatically satisfied because because addition and scalar multiplication already satisfy the necessary condition in R3. All you really need to do is show that the set is closed under those operations. If a and b are real numbers then (a, 0, 0) and (b, 0, 0) are in V. Is (a, 0, 0)+ (b, 0, 0) in V? That is, is it of the form (c, 0, 0) for some real number c? Similarly, if a and x are real numbers so that (a, 0, 0) is in V, is the scalar product, x(a, 0, 0) in V?
  5. Apr 13, 2011 #4
    Right but he doesn't have a problem with (a,0,0), he already knows and proved it as he stated above, what he wants to know is how is it that vectors of the form (a,b,0) are not a subspace? because the ACTUAL Answer in the TEXTBOOK IS it ISN'T a subspace.
  6. Apr 13, 2011 #5
    His book is wrong.
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