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Vector Spaces

  1. Feb 15, 2005 #1
    I had this question in my book asking me to show these things in detail, but it seems easy yet i dont understand why teacher said it was a little difficult:

    1) Prove that R^2(with the rules of addition and scalar multiplication) is a vector space and find (zero vector)?

    2)Deduce from the rules of addition and scalar multiplication that
    0v=0(vector) for all v in V and (-1)v=-v? (v is a vector)

    3) Show that a vector space U has another just 1 element or infinitely many.

    I just need to see what you guys comment on this, as to me it seems like i just simply have to rewrite the rules and thats pretty much it, but again i dont think thats what its asking for, i just dont understand i guess.
  2. jcsd
  3. Feb 16, 2005 #2


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    Start with the definition of a vector space. Show that R^2 with the given rules of addition and scalar multiplication satisfies all the requirements of a vector space.

    So rewriting the rules, that's pretty much it. You should only be careful to prove the required rules only from properties of operations in R and not R^2.

    For b) and c) you must prove it from the defining properties of a vector space.
  4. Feb 16, 2005 #3

    matt grime

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    3) is false, but not if we mean "over R".
  5. Feb 16, 2005 #4


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    I had to think about that for a moment- it's false if the vector space is over a finite field!
  6. Feb 16, 2005 #5
    2) just seems a simple application of the scalar multiple property. Start by defining the vector in some arbitrary way (v_1, v_2, ... , v_n) and use the property of real numbers, etc. to deduce what you need.
  7. Feb 16, 2005 #6
    yea for 1) is this enough to prove that R^2 is a vector space:
    The vector space of R^2 consists of the set of ordered real numbers
    x = (x1, x2). Addition of these real numbers is defined by adding the components:
    x + y = (x1 + y1, x2 + y2) = y + x.
    And multiplication by a scalar k in R^2 is defined as
    kx = (kx1, kx2).
    (now what does it mean to find a zero vector?)
    and for 3) i came up with, the zero vector as having one element and everything else besides zero vector in U, will have infinetly many elements. right?
  8. Feb 16, 2005 #7
    Find a vector in R2 that behaves like the zero vector in the vector space axioms. You may think the choice is obvious, but you must supply the rigor.
    Have you rigorously shown why they will have infinitely many elements (over the field R) ?
  9. Feb 16, 2005 #8
    The set {0} is a subspace with just one element.
    Every other subspace of U is infinite, so it has infinite number of elements, since 0 =/= v in U, then for any real number k, kv in U, since U is closed under scalar multiplication and for k ≠ d, kv ≠ dv. Proof of this is as follows:
    If kv=dv, then (k – d)v = 0, so k – d ≠ 0, then
    v = 1/(k – d) x (k – d)v = 1/(k – d) x 0 = 0.

    Does this show everything or not?
  10. Feb 17, 2005 #9

    matt grime

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    No, you've not proven that any (real) vector space must have infinitely many elements - there are vector spaces that do not have infinitely many elements, and it'd be good for you to see why.

    If it doesn't have just one element it has at least 2, 0 and v, say, but it must also have v, 2v, 3v,... which is an infinite list as kv-0 if and only if... now do you see why you had to prove it?
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