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Show the set of all functions f(x) = ax+b, a,b real numbers forms a vector space, but not for a<2

  1. Oct 12, 2014 #1
    Question: Show that the set of all functions of the form f(x) = ax+b, with a and b real numbers forms a vector space, but that the same set of functions with a > 2 does not.

    Equations: the axioms for vector spaces

    Attempt:
    I think that the axiom about the zero vector is the one I need to use, but I can't figure out how to show this, or how it only affects a>2 functions. Think I may not be visualising the vector space correctly.
     
  2. jcsd
  3. Oct 12, 2014 #2

    Mark44

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    There are several vector space axioms that aren't satisfied if a > 2, not just the one about the zero vector.
     
  4. Oct 12, 2014 #3

    vela

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    And you need to use all the axioms for this problem because you're first asked to verify that the original set is a vector space.
     
  5. Oct 12, 2014 #4

    RUber

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    Take the advice above. Work through the axioms one at a time, and you will see where it breaks. As far as visualizing the space, the function space of linear functions ax+b is the set of lines with slope a and y-intercept b. The zero function in this space is the horizontal line at zero. That is, f(x) = 0x+0. So I think your intuition is correct that you can show that the zero is not in the space. That is not the same as saying there is no function in the space that can touch the origin, but instead that there is no one function in the space which is identically zero for all x .
     
  6. Oct 13, 2014 #5
    I've worked through the axioms, and have found that all of them hold apart from two in the case of a<2.

    These two are:
    There exists a zero vector such that f(x) + 0(x) = f(x) which can't work for a>2, since no zero function can exists as f(x) = (+ve integer)x + b

    And that for any f(x), there exists a function -f(x), such that f(x) + (-f(x)) = 0(x).
    This can't be true for a>2 since the x-coefficient of (-f(x)) couldn't possibly cancel the positive coefficient of f(x), since both can only be positive, so the zero function (which has an x coefficient of 0) could never be achieved.

    Is this correct? All of the other rules seem to hold, with or without the constraint on a.
     
  7. Oct 13, 2014 #6

    Mark44

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    Did you mean a > 2?
    I get your reasoning, but this isn't the cleanest way to say it. You need to show what the "zero" function has to look like, and show why f(x) + 0(x) can't be equal to f(x).
    Let g(x) = 3x + b, which is clearly a member of the set. For any choice of a constant k, is kg also a member of the set?
     
  8. Oct 13, 2014 #7

    vela

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    I don't know if this was just a typo, but the coefficient of x doesn't have to be an integer.
     
  9. Oct 13, 2014 #8

    Fredrik

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    Since that second set is a subset of the first, you only need to check the three usual things to determine if it's a subspace.

    But you still need to know how to check if the vector axioms are satisfied, so please continue along that path. It's only a little harder.
     
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