Page 1 of 2 12 LastLast
Results 1 to 15 of 25

Math Help - Determining a set is a vector space

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    54

    Determining a set is a vector space

    Hey everyone,

    I'm a little confused. I am confused on how to determine if a set is a vector space. I've look at some examples and I am still a bit confused. How do you go about determining if a set is vector space?

    Here is a problem I am confused about:
    The set of all pairs of real numbers of the form (x,0) with the standard operations on R^2

    Any help is appreciated.
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    The set, together with the scalar multiplication and vector addition operations, has to satisfy the axioms of a vector space. Now, in your case, you know already (I assume?) that R^2 is a vector space with "the standard operations". When, therefore, you're trying to determine if a set is a vector subspace, your job is a little easier. All you have to do is determine three things:

    1. The set is nonempty.
    2. The set is closed under scalar multiplication.
    3. The set is closed under vector addition.

    If those three conditions are satisfied, you've got yourself a subspace. Do those three conditions hold for your set?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    54
    What is meant when it is "closed" by addition and multiplication?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Ok, let's get into the notation. Let V be a vector space over the scalar field F. (You could think of \mathbb{R}^{2} as the vector space, over the field of real numbers \mathbb{R}.) The word "closed", in this context, means the following:

    Scalar multiplication: for any vector \mathbf{v}\in V and scalar f\in F, it follows that f\mathbf{v}\in V. That is, multiplying a vector in V by a scalar in F keeps you inside the vector space V.

    Vector addition: for any two vectors \mathbf{u},\mathbf{v}\in V, it follows that \mathbf{u}+\mathbf{v}\in V. Again, the addition of two vectors in V is still inside V.

    That's what closure means. Does that make sense?

    Here's an example. Let

    V=\{\mathbf{v}\in\mathbb{R}^{2}|v_{1}+v_{2}=0\}.

    Take the reals as the field. We check if it's closed under the two operations.

    Let r\in\mathbb{R}, and let \mathbf{v}\in V be given by \mathbf{v}=\langle v_{1},v_{2}\rangle. By assumption, we have that v_{1}+v_{2}=0. Then r\mathbf{v}=\langle r v_{1},r v_{2}\rangle. Does the property still hold? We check to see if rv_{1}+rv_{2}=0, or r(v_{1}+v_{2})=0, which we can see is true, since v_{1}+v_{2}=0. Thus, we can see that we're closed under scalar multiplication, since the scalar and vector I chose was completely arbitrary.

    The vector addition case will be very similar. Finally, you note that the zero vector is, indeed, in our space, and hence the set is nonempty.

    Make sense?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2010
    Posts
    54
    I think I understand what you are explaining. If you add two vectors or multiply by a scalar the new vector is still in the space?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Right. If that condition holds for all vectors and scalars, you've got yourself a vector subspace.

    Now, I should point out that if you're trying to see if a given set is a vector space, and it's not a subset of a known vector space, then you've got to verify every single one of the axioms to which I linked in post # 2. Only then, a full vector space, will you have.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2010
    Posts
    54
    So with the problem I posted:
    The set of all pairs of real numbers of the form (x,0) with the standard operations on R^2

    So (x,0) considered "u"
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jan 2010
    Posts
    54
    I understand your example, but how do I see if (x,0) is a vector space of R^2 if there is no condition like in your example such as v1+v2=0.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  9. #9
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    If there's no "condition", your job is actually easier.

    1. Is the zero vector in your set? (Really just need nonempty, but the zero vector is often the easiest to check).

    2. Closed under scalar multiplication?

    3. Closed under vector addition?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Jan 2010
    Posts
    54
    So would this set no be a vector space because of the axiom (c+k)u = cu + ku

    (cx + kx, 0) is not equal to (cx + kx)?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    No, no, no. Just check what I mentioned in post # 9. Is the zero vector in your set?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Jan 2010
    Posts
    54
    So 0 + (x,0) = (x,0)? I think I am making this harder than it really is.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  13. #13
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Is the vector \mathbf{0}=\langle 0,0\rangle in your set? That is, does \langle 0,0\rangle look like a vector of the form \langle x,0\rangle, for some x?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Jan 2010
    Posts
    54
    If x is 0 then if would be (0,0)
    Follow Math Help Forum on Facebook and Google+

  15. #15
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Right. So that tells you that the zero vector is in the set, and hence the set is nonempty. Next question:

    If you take an arbitrary scalar r\in\mathbb{R}, and multiply it times an arbitrary vector \mathbf{x}=\langle x,0\rangle in your set, is the result still in your set?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Dual Space of a Vector Space Question
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 16th 2011, 04:02 AM
  2. Replies: 2
    Last Post: April 1st 2011, 03:40 AM
  3. Banach space with infinite vector space basis?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 24th 2011, 07:23 PM
  4. Replies: 15
    Last Post: July 23rd 2010, 12:46 PM
  5. Isomorphism beetwenn vector space and sub space
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 30th 2008, 11:05 AM

Search Tags


/mathhelpforum @mathhelpforum