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Positive Quadrant Vector Space

  1. Feb 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Show that the positive quadrant
    [tex]Q = ( (x,y) | x,y > 0 ) \in \mathbb{R}^2[/tex]
    is a vector space.

    2. Relevant equations
    Addition is redefined by
    [tex](x_1,y_1) + (x_2,y_2) = (x_1 x_2, y_1 y_2)[/tex]
    and scalar multiplication by
    [tex] c(x,y) = (x^c , y^c)[/tex]

    3. The attempt at a solution
    There are two properties I am having trouble with - the additive identity and additive inverse.

    Additive identity
    [tex](x,y) + (0,0) = (0,0)[/tex]
    which violates the definition that v + 0 = v

    Additive inverse
    [tex](x,y) + (-x,-y) = (-x^2, -y^2) \not \in \mathbb{V}[/tex]
    also violates that v + (-v) = 0

    For the identity I might just be thinking about the zero element in the wrong way, but I really have no idea how I could have messed up the additive inverse.
     
    Last edited: Feb 11, 2007
  2. jcsd
  3. Feb 11, 2007 #2

    radou

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    The zero element could be (1, 1), according to your definition of addition.
     
  4. Feb 11, 2007 #3
    I suspected that the zero element could be (1, 1), but I am not really sure why, other than that it satisfies the additive identity problem.

    What about the additive inverse? The inverse of (x,y) would have to be (0,0), which is probably what it is, but again I don't really know the justification other than satisfying the property.

    I never payed much attention to vector spaces because they never seemed all that applicable or important, so things like zero elements that aren't actually zero seem to elude me.
     
  5. Feb 12, 2007 #4

    radou

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    If I'm not mistaken, for every element you can take (0, 0) as the additive inverse, since, by your definition of addition, you have (x, y) + (0, 0) = (x*0, y*0) = (0, 0).
     
  6. Feb 12, 2007 #5

    HallsofIvy

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    What do you think an additive identity is? Any thing that satisfies the additive identity property: v+ e= v.


    WHAT!!!! Vector spaces are everything! Everything "linear" at any rate- and almost all methods of solving "non-linear" problems involve reducing to a linear problem.

    Absolutely not! Each element of a vector space must have its own additive inverse: when the two are added the result is the additive identity.

    Here, are you forgetting that (0,0) is not the additive identity? (1,1) is. Given any (x,y) in this space, its additive inverse is (a,b) such that (x,y)+ (a,b)= (xa, yb)= (1, 1). What is a in terms of x and b in terms of y? Do they always exist?
     
  7. Feb 12, 2007 #6

    radou

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    Yes, I am. :yuck: Excuzes moi.
     
  8. Feb 12, 2007 #7

    matt grime

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    The point of axiom is that if something satisfies the axioms to be a FOO, then it is a FOO. You have used the wrong axioms in the second part. There is absolutely no reason why (0,0) MUST be the additive identity. I think you're confused because in every other example where you have to verify V is a SUBspace of R^n, (0,...,0) is the additive identity, but that is because of the prefix SUB in there and the additive identity of R^n is (0,0,..,0).


    Here we just have a set, with operations. We just need to verify these things satisfy the axioms of a vector space.
     
  9. Feb 12, 2007 #8
    Oh, I see now. Since the inverse must go back to the additive identity then it would have to be (1/x, 1/y), which would always exist since (x, y) > 0.

    Now if I wanted to find a subspace the subspace must satisfy the altered axioms for closure, scalar multiplication, and addition.
     
  10. Feb 12, 2007 #9

    HallsofIvy

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    Yes, that is correct. You said in your original post that you were only having problems with the existance of an additive identity and additive inverses. Can we assume that you have proved that the operations satisfy the require properties: associative, scalar multiplication distributes over addition, etc.? Of course if you have proven them for the space, you don't have to do that again for a subspace- just showing that the subspace is closed under addition and scalar multiplication is sufficient- the other properties follow from the fact that they are true for the entire space.

    By the way, it is an easy theorem that the scalar product of 0 with any vector is the additive identity. Do you see how that works here?
     
  11. Feb 12, 2007 #10
    Yes, I was able to prove the other axioms fairly easily. The scalar product of zero with any vector would have to be the additive identity because it is identically zero for all vector functions.

    So, here is another question then.
    [tex]\mathbb{V} = C^0 \mathbb{(R)}[/tex] with [tex]f:\mathbb{R} \rightarrow \mathbb{R}[/tex]
    where the vector space is the set of all continuous functions.

    I know that f(1) = 0 would be a subspace because the zero element is identically zero for all zero functions. Also, it is trivial from calculus that continuous functions added, and multiplied by a constant, remain continuous. On the other hand, f(0) = 1 would not be a subspace because the zero element is not identically zero. Considering the more general form of f(a)=b, what values of a and b would make vector spaces? It seems to be that a can be any real number and b must be zero.
     
  12. Feb 13, 2007 #11

    matt grime

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    I'm going to be picky.

    f(1)=0 is not a subspace. It is an equation.

    {f : f(1)=0}

    is a subspace.
     
  13. Feb 13, 2007 #12

    HallsofIvy

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    You are saying the same thing in 3 different ways here!

    It's not at all clear what you are saying here! Yes, the set of all continuous functions from R to R forms a vector space over R because we can add continuous functions and multiply continuous functions by real numbers and still get continuous functions. It is a subspace of the vector space of all functions.

    But, as matt grime said, "f(1)= 0 would be a supspace" makes no sense. I think you meant it as shorthand for "the set of all continuous functions f, such that f(1)= 0" form a subspace of the space of all continuous functions". Shorthand is not always a good thing! Yes, that statement is true since if f and g are two such functions, and a a real number, then f+ g is continuous with f+g(1)= f(1)+ g(1)= 0+ 0= 0, and af is a continuous function with af(1)= a(0)= 0. That would be true with any number in place of the "1": For any number a, "the set of all continuous functions such that f(a)= 0 forms a subspace of the space of all continuous functions" is true.

    On the other hand, "the set of all continuous functions f, such that f(a)= b, forms a subspace" only if b= 0. If f and g are two continuous functions such that f(a)= b and g(a)= b then f+ g(a)= f(a)+ g(a)= 2b which is equal to b only if b= 0. More simply, for any number p, pf(a)= pb which is equal to b only if b= 0.
     
  14. Feb 13, 2007 #13
    Yeah, sorry about saying the same thing three times. I don't know notation all that well and figured one of them had to get the point across. I meant {f: f(1) = 0}, I guess I should be more explicit (I've never really cared about formalities doing applied physics). I think I have a better handle on the more abstract vector spaces (ones that aren't R^n), thanks guys.
     
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