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Confused on why this subspace spans R^3, matrices fun!

  1. Dec 4, 2005 #1
    I'm studying for my exam on monday and i saw ap roblme that says:
    Determine if the following is a subspace of R^3.
    For it to be a subspace it has to pass
    1) must have the zero vector
    2) closure under addition
    3) closure under multiplaication

    here is my work:
    http://img206.imageshack.us/img206/8067/lastscan1qn.jpg


    now it fails because it r1+r1-2 right? but why? because why wouldn't
    3(s1+s2) also fail? because ur getting a different number if u like say, let s1 and s2 be 1, or r1 and 2 be 1, so did they both fail? Also if there was say

    What if there were only 2 variables, only r and s, and U = [r s r], r & s are in R, now would this automaticallya not be a subpsace in R^3?
     
  2. jcsd
  3. Dec 4, 2005 #2

    LeonhardEuler

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    Actually, it doesn't fail. The problem said that r,s, and t were real numbers. So, if r1 and r2 are real numbers, then (r1+r2-1) is a real number. Call it R. Then r1+r2-2=R-1, which means that this component of the vector is ok.
    No, it would still be a subspace. Being a subspace of a vector space means being a subset that is also a vector space (with the same rules of addition and scalar multiplication as the original vector space). So the x-y plane is a subspace of R^3, as is the x-axis.
     
  4. Dec 4, 2005 #3
    LeonhardEuler, thanks for the reply!! My professor did this problem,and we are having an exam on monday, and he told us it fails, because r1+r2-1 isn't equal to r1+r2-2, he said u cna't get the same number so it fails, is he wrong?
     
  5. Dec 4, 2005 #4

    LeonhardEuler

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    Well, before I go saying that the professor is wrong, I would need to know the exact wording of the problem. As I understand it, r is allowed to be any real number and vectors in the subspace are allowed to have the first component equal to (r-1). Is that the case?
     
  6. Dec 4, 2005 #5
    He really didn't expand on what it said, all it says is:
    Determine if the follwing is a subpsace of R^3. then the problem says
    U = {
    r-1
    3s
    t,

    r,s,t in R }

    So they are all real numbers.
     
  7. Dec 4, 2005 #6

    matt grime

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    I think the question is badly phrased (either by you or your teacher). As a set the vectors given are all of R^3, and so that subset is certainly a subspace of R^3. Have you failed to define something properly?

    In any case, what he's trying to get at is that whilst it is a vector space it is not a subspace since we have to define a different addition on the first component, that is the addition of x+y is different if done in R^3 or done in the set itself. Ie the operation u@v=u+v-1 on the underyling set R is an isomorphism of additive groups but does not identify R with a subgroup of itself under addition.
     
    Last edited: Dec 4, 2005
  8. Dec 4, 2005 #7
    How is it suppose to be phrased if you want it to fail in the case of r1-r2 -2, because thats how he thought it failed, so what would u have to add to the directions for that case to occur? (failure under closure of addition)
     
  9. Dec 4, 2005 #8

    matt grime

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    Unless you define the operations on the vectors it fails to be well phrased since as a set that is clearly all of R^3! You need to explain how the description (parametrization) of the vectors in terms of r,s,t interacts with the operations.

    It is sort of clear what he's getting at...
     
  10. Dec 4, 2005 #9

    HallsofIvy

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    If, for example your vectors were of the form (r-1, s, r) where the first component was always 1 less than the third, then it would not be a subspace because (1, 1, 2) is in that set and so is (2, 5, 3). But their sum, (3, 6, 5) is not in the set: 3= 5-2 not 5-1. Of course, here the zero vector (0, 0, 0) is also not in the set.

    In the example you first gave, (r-1, 3s, t), if r, s, and t can be any numbers, let x= r-1, y= 3s, z= t and you have (x, y, z) where x, y, and z can be any numbers: that's all of R3. The set you gave spans all of R3, which, interestingly, is what you title this thread.
    If the problem was to show that this set was NOT a subspace, why did you title the thread "confused on why this set spans R3"?
     
  11. Dec 4, 2005 #10
    Thanks Ivey, cleary my professor is wrong, becuase no the problem wasn't to show that this set was NOT a subspace the question says:
    Determine if the following is a subspace of R^3. and yes r,x, and t can be any real number. Because the problem says: r,s,t in R. Also he said the set of vectors (r-1,3s,t) does contain the zero vector shows that he didn't set any pre-conditions that says, the first component is less than the thrid. So as a general rule, can you say, as long as the addition makes sense, it passes under closure of addition? like as you showed,
    3 = 5-2 not 5-1.

    Note: the reason i asked, "so what would u have to add to the directions for that case to occur? (failure under closure of addition)" was becuase i was trying to see if my professor just forgot to tell us somthing, but that wasn't the case.
     
  12. Dec 4, 2005 #11

    HallsofIvy

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    "can you say, as long as the addition makes sense, it passes under closure of addition?"

    I'm not sure what you mean by "the addition makes sense". In the example I gave, (r-1, s, r), didn't the addition "make sense"?

    By the where, where did you get "Ivey"?
     
  13. Dec 4, 2005 #12
    hah Sorry, I don't know why I put an 'e' in there. When i said, "the adddition make sense" i ment, like 5-2 = 3, but 5-1 is not equal to 3, so if u made 2 vector sets like I did above to prove closure under addition works, how does this prove it?
    (r1-1, 3s1, t1) + (r2-1,3s2,t2);
    (r1+r2-2 ,3(s1+s2), t1+t2);

    So since its in the same form as
    (r-1,3s,t) it works? but how is it in the same form, if its not longer
    r-1, but insteed, R-2? if R = r1+r2?

    THe more i look into this the more i get confused!

    Like.
    let r = 0, u get -1 for the case where (r-1,3s,t).
    for case 2:
    r1+r2-2
    r1 = 0, r2 = 0, u get -2
    not -1!
     
    Last edited: Dec 4, 2005
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