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Linear algebra: vector spaces

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Hi, I am really having trouble with questions regarding proving whether a given set is a vector space or not.

    So one of the questions is [ x ε R2|x12=x23 ]

    So I have to prove whether the following set is a vector space
    2. Relevant equations



    3. The attempt at a solution

    Having a peek at the textbook, it tells me that there are ten axioms that satisfy whether a set belongs to a vector space.

    What I did was go x12 - x23 = 0

    u+ v = v + u -For this one, I did the following: I let x1 and x2 have a value each and named it a vector u, then I let x1 and x2 have some other values and named it a vector v.

    So x1 = 1 and x2 = 2 and u = 12 - 23
    And x1 = 4 and x2 = 5 and v = 42 - 53

    u + v = v + u should work out and I used the same method for the rest of the axioms....I need to know two things

    (a) Is the method I am doing correct? If not, could you point me in the right direction?
    (b) How do I go about checking the first axiom for the above set u + v lies in the set. If I do the same old substitution, all i get is a number, whatever it is!

    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 23, 2012 #2

    jbunniii

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    Many of the axioms will be satisfied automatically because this set (let's give it a name: [itex]S[/itex]) is a subset of a vector space.

    For example, if [itex]u[/itex] and [itex]v[/itex] are in [itex]S[/itex], then they are also in [itex]\mathbb{R}^2[/itex], and since [itex]\mathbb{R}^2[/itex] is a vector space, addition is commutative. Therefore [itex]u + v = v + u[/itex].

    The main things to check are whether [itex]S[/itex] is closed under vector addition and scalar multiplication. The fact that [itex]S[/itex] is defined in terms of squares and cubes of the vector coordinates is a pretty good indication that [itex]S[/itex] is not a vector space. This is because squaring and cubing are both nonlinear operations, and vector spaces are all about linearity.

    Try picking a specific element of [itex]S[/itex], for example [itex]v = (1,1)[/itex]. Is there a scalar multiple of [itex]v[/itex] that is not in [itex]S[/itex]? Is [itex]v + v[/itex] in [itex]S[/itex]?
     
  4. Oct 23, 2012 #3
    Yea, that is really the part (b) of the problem that I faced. I don't think I can take v = (1,1) though since both numbers have to be different x12 - x23 Anyway, even if I take two different numbers, what I get is another number. How do I know whether that number is part of set S or not?

    For instance, if I take v = (1,2), I will get 1-8 = -7. How do I know -7 is in S?

    I did check all the rules using the method that you have mentioned, taking numbers and figuring out whether the axioms are satisfied...but I wasn't sure whether that was a reasonable method...but since you argue in the same way, I guess it is.

    Thanks for your input.
     
  5. Oct 23, 2012 #4

    jbunniii

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    I think you're misunderstanding the condition [itex]x_1^2 = x_2^3[/itex]. The vector [itex]v[/itex] is in [itex]S[/itex] if and only if its components satisfy [itex]x_1^2 = x_2^3[/itex]. In your example, [itex]x_1 = 1[/itex] and [itex]x_2 = 2[/itex], so [itex]x_1^2 = 1[/itex] and [itex]x_2^3 = 8[/itex]. Since [itex]1 \neq 8[/itex], this means that [itex]v[/itex] is not in [itex]S[/itex].

    Now consider my example again: [itex]v = (1,1)[/itex]. In this case, [itex]x_1 = 1[/itex] and [itex]x_2 = 1[/itex], so [itex]x_1^2 = 1[/itex] and [itex]x_2^3 = 1[/itex]. Therefore [itex]x_1^2 = x_2^3[/itex], so [itex]v \in S[/itex].

    So now that we have established that [itex]v = (1,1)[/itex] is in [itex]S[/itex], you can easily use this to show that [itex]S[/itex] is not closed under addition and scalar multiplication, so [itex]S[/itex] is not a vector space. It doesn't matter that it satisfies the other axioms. If it isn't closed under addition and scalar multiplication, it cannot be a vector space.

    To show that [itex]S[/itex] is not closed under addition and scalar multiplication, consider the vector [itex]v + v = (2,2)[/itex]. If [itex]S[/itex] is a vector space, then [itex](2,2)[/itex] must be in [itex]S[/itex]. Is it?
     
    Last edited: Oct 23, 2012
  6. Oct 23, 2012 #5
    (2,2) is not in S because 22 = 4 and 23 = 8 and 4≠8.

    I am still confused about showing that S is not closed under addition and scalar multiplication.

    With addition, I know that the axiom is u + v = v + u

    So if I take v = (2,2), u = (1,2), do I just do it the normal vector addition way; adding the x1s together and the x2s together?

    So v+ u = [ (22 + 12) , ( 23 + 23) ] = (5, 16)

    u + v= [ (12 + 22) , (23 + 23) ] = (5,16)

    Seems like it is closed under vector addition or I am probably doing something wrong?
     
  7. Oct 23, 2012 #6

    jbunniii

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    Addition is defined in the usual way: [itex](2,2) + (1,2) = (2+1, 2+2) = (3,4)[/itex]. The [itex]x_1^2 = x_2^3[/itex] relation is just a constraint which tells you which vectors are in [itex]S[/itex]. It has nothing to do with how you add vectors or multiply by scalars.
     
  8. Oct 23, 2012 #7
    Oh, I think I get it. What we have done so far is find vectors that belong or do not belong to S. Example: v = (1,2) does not belong to S and v= (1,1) belongs to S.

    But v = (1,1), even if it belongs to S, that does not necessarily make it a vector space, I need to do do the addition and scalar multiplication to figure it out. Is my understanding correct so far?

    v = (2,2) is not in S so S is not a vector space? What I know about vector spaces is that, take R2 for an example. R2 is a vector space because you can take any linear combinations of vectors (x1, x2) and they all lie in R2.
     
  9. Oct 23, 2012 #8

    jbunniii

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    Yes, that's right.

    Right. Now, we already established that v = (1,1) is in S. Now let's take a linear combination of elements of S: (1,1) + (1,1) = (2,2) is a linear combination, namely 1v + 1v. If S is a vector space, then S must contain all linear combinations of its elements. Therefore if (2,2) is not in S, then S cannot be a vector space.
     
  10. Oct 23, 2012 #9
    Yes, I finally get it now. Using the same method, I should be able to get through all the problems.

    Thanks a lot for your help!
     
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