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Confused with Vector Spaces

  1. Aug 8, 2012 #1
    Hi, I have this problem that is solved, but I don't understand the theory behind it.

    It says: Which of the following sets, with the natural definitions of addition and scalar multiplication, form real vector spaces?

    A) The set of all differentiable functions [itex]f:(0,1)\rightarrow\Re[/itex] such that [itex]f+f'=0[/itex].
    B) The set of all differentiable functions [itex]f:(-1,1)\rightarrow\Re[/itex] such that [itex]f+f'=0[/itex] and [itex]f(0)=1[/itex].

    The answer says the first one is a vector space, but that the second one is not because zero does not belong to the set... However, I don't see the reasoning behind it. Maybe they are supposed to be the opposite (A is not a vector space and B is) and the answer is wrong? No one has complained about it and there is nothing on the course forum, so I think the answers should be correct :/

    Thanks a lot!
     
  2. jcsd
  3. Aug 8, 2012 #2
    Does the second set contain the zero function? Can a set of functions be a vector space over the reals if they lack an additive identity (the zero function)? What does the definition of "vector space" say?
     
  4. Aug 8, 2012 #3
    Let's call the sets defined in A) and B), [itex]A[/itex] and [itex]B[/itex]. Then try to think of a zero element in set [itex]A[/itex] and in set [itex]B[/itex].

    A zero element in set [itex]A[/itex] would be an element [itex]g \in A[/itex] such that for all [itex]f \in A: f + g = f[/itex].

    Can you find such an element in [itex]A[/itex]?
     
  5. Aug 8, 2012 #4

    Bacle2

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    Another general result that answers your questions is:

    Given a vector space V, then a subset S of V is a subspace if V is closed under

    scaling and under linear combinations (you can clearly describe the first condition

    as part of the second.). It is a nice exercise to show that if S satisfies these,

    then S is a subspace of V .
     
  6. Aug 8, 2012 #5
    Remember, when they say "zero" in this context they do not mean the number zero. They mean the zero function. This is an ambiguous usage that you have to keep in mind when dealing with sets of functions.

    What is the zero function? It's the function that takes the value zero for every value of its domain. It's the function we'd denote as f(x) = 0. Its graph is the x-axis.

    It's differentiable on the entire real line. Its derivative is also the zero function.

    Now, what is the value of the zero function at the point x = 0? Well, (x) = 0 for all values of x, so f(0) = 0. But in part (b) we are asked to consider functions such that f(x) = 1. The zero function doesn't have that property, so it's not in that set.
     
    Last edited: Aug 8, 2012
  7. Aug 8, 2012 #6

    jbunniii

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    It's easy to see that B is not closed under addition. If f(0) = 1 and g(0) = 1, then what is (f+g)(0)?
     
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