1. Not finding help here? Sign up for a free 30min 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!

Vector space

  1. Oct 11, 2006 #1
    I am supposed to determine whether or not the following two sets constitute a vector space.

    1) The set of all polynomials degree two.
    2) The set of all diagonal 2 x 2 matrices.

    For the first one, it will not be a vector space because it does not satisfy the closure property. Also the distributive property would be broken because a(x+y)^2 would not be (ax+ay)^2, or did I do that wrong?

    The second one would not satisfy additivity properties, and not be a vector space.

    Right? I don't know the set notation in LaTeX, so I'm not really sure how to put up much of my work.
     
  2. jcsd
  3. Oct 11, 2006 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Note that A vector space is a set V (and a field F) together with a binary operation [itex]\phi : V\times V\rightarrow V[/itex]. It does not ask the question "is V a vector space?" if you don't specify under whch binary operation you consider V. For instance, for the set of polynomials, the closure property is certainly satisfied under regular addition, but it is not under regular multiplication.
     
  4. Oct 11, 2006 #3

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    A piece of my post seems to be missing. Make that second sentence

    "It does not make sense to ask the question [...]"

    And make

    "For instance, for the set of degree 2 polynomials, the closure property is certainly satisfied under regular addition, but it is not under regular multiplication."

    Because the set of all polynomials is certainly closed under regular multiplication.
     
    Last edited: Oct 11, 2006
  5. Oct 12, 2006 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    The second one, whilst true, in general, has nothing to do with why the set of degree polys does not form a vector space in the natural way.
     
  6. Oct 12, 2006 #5

    matt grime

    User Avatar
    Science Advisor
    Homework Helper


    That should be FxV to V.
     
  7. Oct 12, 2006 #6
    Okay, I was thinking about these a bit more. Now I have come to the conclusion that I was thinking about the question of degree two polynomials wrong. I thought it meant everything of the for ax^2, bz^2, etc. Which would mean that since there is no zero vector, that they are not a vector space. But since I think it is really asking for, ax^2 + bx + c, so as far as I can tell it would be a vector space.

    Then the 2 by 2 matrices should also be a vector space because all the properties are fulfilled.

    I don't really know what you mean by a field though. By the field do you mean the function, like polynomials of degree two would be the field?

    Do you guys know of a place that would go through a bunch of examples in vector spaces? They aren't hard at all, but they are definitely something I need to get used to, and I think having some examples would help.
     
  8. Oct 12, 2006 #7

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Roughly, the field is the set that you take your scalars from for the properties that involve scalar multiplication. Usually, the field is the real numbers, but it could also be the complex numbers, or something else. Less roughly, a field is any set satisfying the "field axioms" (which is a set of conditions similar to the ~8 axioms defining a vector space).


    Matt: no, no I really meant V x V --> V. But so we're both happy, let me restate the definition by taking your remark into account:

    A vector space is a set V (together with a field [itex]\mathbb{F}[/itex]) and two binary operations [itex]\phi_1 : V\times V\rightarrow V[/itex], [itex]\phi_2 : \mathbb{F} \times V\rightarrow V[/itex], satisfying the vector space axioms.

    Do you agree with that?
     
    Last edited: Oct 12, 2006
  9. Oct 12, 2006 #8
    Ahh, okay, our fields are elements of real numbers for now. I understand the [tex]\phi_1 : \mathbb{F} \times V\rightarrow V[/tex] now that I know what the field is, but I don't see how two binary operations can form a vector space without the field. It seems like that first definition is making an undefined space out of an undefined space, no?
     
  10. Oct 12, 2006 #9

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Ehh, I don't understand your objection (if you're making any)
     
  11. Oct 13, 2006 #10

    matt grime

    User Avatar
    Science Advisor
    Homework Helper


    No, it is not. The zero vector will be the zero poly f(x)=0, and that is not degree two. If you add x^2 and -x^2 you get zero, so it is not closed under addition. The set of polynomials of degree two spefically requires a=/=0. The set of polys of degree less than or equal to 2 (or any other natural number) is a vector space.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Vector space
  1. Vector Space (Replies: 1)

  2. Vector space (Replies: 8)

  3. Vector Space (Replies: 3)

  4. Not a Vector Space (Replies: 3)

  5. Vector Space (Replies: 10)

Loading...