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Determining if sets are in the subspace

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data
    is B in subspace R^2
    B=[x] :x^2+y^2<=1
    [y]


    2. Relevant equations
    1.0∈B
    2.if u,v∈B u+v∈B
    3.if u∈A a∈B then au∈B


    3. The attempt at a solution
    The <= is confusing me. I am not sure if i am suppose to treat it like an equal sign, or is it automatically not in the subspace?

    if i treat it like a equal sign
    0^2+0^2=0
    u^2+uv+v^2 + u^2+uv+v^2 = 0
    0+0=0
    (au)^2+(au)^2=0
    0^2+0^2=0
    so B is in R^2
     
  2. jcsd
  3. Oct 25, 2008 #2

    HallsofIvy

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    Staff Emeritus
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    No, that makes no sense. You are asked to determine whether or not B is a subspace of R2.


    Please do not use "special symbols"; they don't show up on all net-readers. Either use LaTex like [itex]0\in B[/itex] or write
    1. 0 is in B.
    2. if u and v are in B, then u+ v is in V
    3. if a is in R and u is in B then au is in B.
    Notice (3). What you wrote doesn't make sense because you did not say what "A" is. Also, if you have already used "u" to mean a vector, don't use it again to mean a number.
    Finally, you don't need to prove (1). If (3) is true for any number a, take a= 0.

    Which is NOT equal to 1 is it? So (0,0) would not be in this set. If this were only "=" this would immediately NOT be a subspace of R2. But because 0< 1 and the problem DOES say "<= 1", you don't know yet.

    What? Where did you get this? I thought for moment you were doing (u+ v)2+ (u+ v)2 but that would not give you this. And why 0 on the right?

    No! if u is a vector, then u= (x, y) and au= (ax, ay). You want to show that (ax)2+ (ay)2<= 1, not (au)^2+ (au)^2! In a vector space only addition of vectors and multiplication of vectors by numbers is defined, not multiplication of two vectors and so (au)2 makes no sense.
     
    Last edited: Oct 26, 2008
  4. Oct 25, 2008 #3

    Mark44

    Staff: Mentor

    I don't think you have given us the problem's exact wording. I believe that the problem's wording is something like this:
    Let B = {(x, y): x^2 + y^2 <= 1}
    Is B a subspace of R^2?

    Please take a closer look at the problem as defined in your book.

    You are confused about what you are supposed to do. Apparently you are trying to determine whether vectors are "in the subspace." What you need to do is show that the set B, which is a subset of R^2, actually is or is not a subspace of R^2.

    Clearly the vector (0, 0) is in B.
    If you have vectors a = (x1, y1) and b = (x2, y2) that are in B, will a + b also be in B?
    If you have a vector a = (x1, y1) that is in B, will any scalar multiple of it also be in B?
    If the answer to each of these two questions is yes, no matter which vectors a and b and scalar c are chosen, then set B is a subspace of R^2.

    If the answer to at least one question is no, then B is not a subspace of R^2.
     
  5. Oct 25, 2008 #4
    If you have vectors a = (x1, y1) and b = (x2, y2) that are in B, will a + b also be in B?
    a=x1^2+y1^2<=1
    0^2+0^2<=1
    b=x2^2+y2^2=1
    0^2+0^2<=1
    a+b<=1

    If you have a vector a = (x1, y1) that is in B, will any scalar multiple of it also be in B?
    a^2(x^2+y^2)<=1
    if x and y are equal to 0 or a=0, then it would be true.

    so B is in the subspace of R^2?
     
  6. Oct 25, 2008 #5

    Mark44

    Staff: Mentor

    Re: Determining whether a set is a subspace of R^2

    You're still not getting it. The question is not whether B is in the subspace of R^2, but whether B ***is*** a subspace of R^2.

    Suppose a and b are vectors in set B, where a = (x1, y1) and b = (x2, y2).
    a in B ==> x1^2 + y1^2 <= 1, and
    b in B ==> x2^2 + y2^2 <= 1.

    Now, is a + b in B?
    Also, is ca in B, where c is a scalar?

    If the answers to both questions are "yes", then B is a subspace of R^2.

    Think about what B looks like, and what a vector that belongs to B looks like.
     
  7. Oct 26, 2008 #6

    HallsofIvy

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    Why are you writing this? You cannot assume that x1 and y1 are 0.

    Same comment

    This makes no sense. a and b are vectors. You can't compare them to a number.

    But x, y, and a are NOT necessarily 0!

    You seem to be completely misunderstanding this problem.
     
  8. Oct 26, 2008 #7
    I think i am starting to understand it more.
    A(x1^2+x2^2)+B(y1^2+y2^2)<=1 where A and B are scalers
    a+b is not in B

    ca is not in B either since c can be a scaler that makes a not a subspace in B
    so B is not a subspace of R^2
     
  9. Oct 26, 2008 #8

    HallsofIvy

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    Assuming a= (x1,y1) and b= (x2,y2), this is Aa+ Bb. Now how do you conclude it is NOT in B? Do you mean there exist some a, b, A, B such that it is not in B or it is not in B for ALL a, b, A, B?

    (If you are going to use "a" and "b" for vectors it is generally not a good idea to use "A" and "B" for scalars.)

    You can't just say "ca is not in B", you have to show that. Can you give an example of a vector a in B and a number c such that ac is not in B?
     
    Last edited: Oct 26, 2008
  10. Oct 26, 2008 #9
    There exist some a ,b, A, B such that it is not in B.
    a+b is (x1^2+x2^2+y1^2+y2^2)<=1, which is not always true.

    ca is not in B because
    a=(.5,.5) and c=10
    100(.5^2+.5^2) is greater then 1 so ca is not in B.
     
    Last edited: Oct 26, 2008
  11. Oct 26, 2008 #10
    Is this enough proof to show that B is not a subspace of R^2?
     
  12. Oct 27, 2008 #11

    Mark44

    Staff: Mentor

    For the first part, you have to SHOW that if a is in B and b is in B, then a + b isn't in B. You're saying it is not always true. Show us a case where it isn't true.

    Your argument in the second part is more convincing - I truly believe that a is in B, but ca is not in B.
     
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