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Question regarding the subspace status of a vector in R^2

  1. Mar 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Is the following set a subspace of R2?


    2. Relevant equations

    (a,b2)

    3. The attempt at a solution

    I'm exhausted and stumped.
     
  2. jcsd
  3. Mar 7, 2013 #2

    Zondrina

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    (a, b2) is a set? Not a vector?
     
  4. Mar 7, 2013 #3

    Mark44

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    No, this represents a set of vectors of this form.
     
  5. Mar 7, 2013 #4
    Sorry, sorry - I miswrote. The question was "Are the following sets subspaces of R2? The set of all vectors of the form...

    and then (a,b2)

    I'm trying to think of every reason. There's a zero vector for sure...I don't think scalar multiplication is a problem...but I don't know.
     
  6. Mar 7, 2013 #5

    Mark44

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    You still need to show what you have tried. This problem is similar to the other one you posted today - you need to show that for vectors in the set, addition is closed, and scalar multiplication is closed.
     
  7. Mar 7, 2013 #6
    I'll be more explicit in the future. I did things, and I didn't know "showing attempt" was such a hard and fast rule.

    a and b could both be zero, so there is a zero vector.

    2(a,b2) I do not see any problem with.

    (a,b2) + (c,d2) = (a+c, b2+d2)

    I'm not sure if I'm dealing with a subspace here or not. I suspect not, and that it's not closed under addition.
     
    Last edited: Mar 7, 2013
  8. Mar 7, 2013 #7

    Mark44

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    Are there any restrictions on a and b? IOW, are they restricted to integers or can they be any real numbers?

    Clearly (1, 4) is in the set, since it could be written as (1, 22). What about (2, -9)?
     
  9. Mar 7, 2013 #8
    They can be any real numbers.
     
  10. Mar 7, 2013 #9
    I had thought that we had to begin by assuming a and b are within R2? I mean, I see where you're going - the square root of negative nine is not within R2, so the vector is not a subspace. There just weren't any ground rules laid down in this class for how we could go about this. Do you, perhaps, have a link to a webpage that really goes over vector spaces and subspaces?
     
  11. Mar 7, 2013 #10

    Mark44

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    No, a and b are real numbers. A vector <a, b> with these as its components would be in R2.
    More accurately, it's not in R (i.e., it's not real).
    So the set of vectors does not form a subspace (of R2).
    I don't have a link, but I know there's an article on wikipedia, and no doubt many other sites. The basic idea is that a vector space is a set of things with two operations: addition and scalar multiplication. A vector space must satisfy ~10 vector space axioms, most of which deal with vector addition and scalar multiplication.

    A subspace of a vector space is a vector space in its own right. To show that a subset of a vector space is in fact a subspace of that vector space, you need to establish three things:
    1. The 0 vector is contained in the set.
    2. If u and v are arbitrary vectors in the set, then u + v is also in the set.
    3. If u is an arbitrary vector in the set and k is a scalar, then ku is in the set.

    That's really about it.
     
  12. Mar 7, 2013 #11
    But your suggestion for proving that (a,b2) are not a set of vectors involves making b a non-real number, correct? So we're not actually beginning with a subset of R2 - or do I misunderstand? Thank you for your help thus far.
     
  13. Mar 7, 2013 #12

    vela

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    You're given the set S=##\{(a,b^2)\in \mathbb{R}^2: a, b \in \mathbb{R}\}##. It's a collection of ordered pairs, right? So any element in S is also in ##\mathbb{R}^2##, so it's a subset of ##\mathbb{R}^2##. The vector (1,4) = (1,22) is in the set. The vector (2,-9), however, isn't, because you can't express it in the form (a,b2) where a and b are real numbers. S is a set of vectors, but it doesn't contain every vector in ##\mathbb{R}^2##.

    To prove S is a subspace, you have to show the three things Mark44 listed in his post.

    You've already shown that (0,0) is an element of the set because a=0, b=0 works.

    Now take two elements of S, (a,b2) and (c,d2). When you add them together, you get (a+c, b2+d2). Can this be written in the form (e, f2) for some real numbers e and f? You need to explain why you know there are such e and f, or you need to find a counterexample to prove you can't.

    Now let k be a real number. Is k(a,b2) = (ka, kb2) in S? In other words, can you always find real numbers e and f such that e=ka and f=kb2? Remember k can be any real number.
     
    Last edited: Mar 7, 2013
  14. Mar 8, 2013 #13
    As far as I can tell, yes, it's closed under addition and scalar multiplication, but the answer is that it's not a subspace. This is why I am stumped.
     
  15. Mar 8, 2013 #14

    mfb

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    It is not closed under scalar multiplication. Try a factor of -1 for any vector with non-zero second component.
     
  16. Mar 8, 2013 #15
    Like, -1(a,b2)? So we get (-a,-b2)? Is this a problem - am I viewing the order of operations wrong or something?
     
  17. Mar 8, 2013 #16

    mfb

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    (1,4) is element of your set of vectors (coming from a=1, b=2), but -1(1,4)=(-1,-4) is not an element of your set (as there is no real b where b^2=-4).
     
  18. Mar 8, 2013 #17
    It may take me a little longer than the other kids, but I get there. Thank you, sir.
     
  19. Mar 8, 2013 #18

    Mark44

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    MoreDrinks, now do you see where I was going with my question?
     
    Last edited: Mar 8, 2013
  20. Mar 8, 2013 #19
    Surely, Mark. Follow-up question: when you're checking to see if a subset of vectors is closed under scalar multiplication, must a scalar be within the original vector space from which the subset is derived?
     
    Last edited by a moderator: Mar 8, 2013
  21. Mar 8, 2013 #20

    vela

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    It would help if, instead of simply claiming you get it, you demonstrate your understanding by explaining your solution. For example, you said S is closed under vector addition. How do you know? I wonder if you truly understand the concepts because you later claimed S was closed under scalar multiplication.
     
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