Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving a set to be a vector space,

  1. Aug 12, 2011 #1
    I know that a set (let's call it V) of all functions which map (R -> R) is a vector space under the usual multiplication and addition of real numbers, but i am having trouble proving it, i understand that the zero vector is f(x)=0, do i just have to prove that each element of V remains in V under additon and scalar multiplication? If that's the case then i cant work out how to go about proving it, if it isnt the case, then what exactly am i proving?
  2. jcsd
  3. Aug 12, 2011 #2
    You have to notice that when you want to see how a set of functions forms a vector space, you are *not * taking the addition operation of real numbers.
    f ( c ) may be a real number for any c, but the images of a function are not the function "object" itself. You need to think about how you can take two functions, add them, and have a resulting function that is still from R to R
  4. Aug 12, 2011 #3
    No, that is not true. You don't have the correct definition of addition and scalar multiplication of functions. The usual multiplication of real numbers doesn't apply, since we are dealing with a space of functions.

    The correct definitions are that V is a vector space under pointwise addition and scalar multiplication. What does that mean? If f and g are elements of V, then we define f+g to be the function defined by

    (f+g)(x) = f(x) + g(x).

    Now we have to verify that f+g, as defined, is in fact an element of V. You should prove that.

    Scalar multiplication of a function f by a scalar a is defined as

    (af)(x) = a * f(x).

    Again, we need to show that if f is in V, then so is af for any scalar a.

    Having shown that V is closed under addition and scalar multiplication as defined, you then need to verify each of the vector space axioms.

    Yes, you need to prove that; then you need to prove that V satisfies the vector space axioms. I think you'll find it easier once you understand how addition and scalar multiplication of functions are defined. Did you understand the definitions I gave?
  5. Aug 13, 2011 #4
    @ SteveL27, How would i go about proving those? perhaps an example would help considerably,
    What i was thinking was that i could set V := {v1, v2, ... , vn} where each V is a function which maps R -> R and thus an element of V, but i cant quite get a rigorous proof using this notation..
  6. Aug 13, 2011 #5
    I hope you can clear something up for me.

    I'm not quite sure what you mean here with [itex]V=\{v_{1} ,v_{2} ,\dots ,v_{n} \}[/itex]. This would imply that [itex]V[/itex] has [itex]n[/itex] elements, which is not true if [itex]V[/itex] is a vector space. This can be seen that if [itex]V[/itex] is a vector space over the field [itex]\mathbb R[/itex] and [itex]f\in V[/itex], then [itex]af\in V,\,\forall a\in \mathbb R[/itex]. However, there are uncountably many [itex]a\in \mathbb R[/itex], so [itex]V[/itex] has much more than only [itex]n[/itex] elements.

    Does that make sense? I hope it helps.

    Do you mean for [itex]\{v_{1} ,v_{2} ,\dots ,v_{n} \}[/itex] to be a basis of [itex]V[/itex]? Because if so, then I want you to ask yourself if we know that [itex]V[/itex] is finite-dimensional.
  7. Aug 13, 2011 #6
    We need to get the definition of addition and scalar multiplication clear first. Do you understand what the definitions of f+g and af for a scalar a? Your idea is unfortunately not in the right direction, but before discussing that, it's important to get the definition of V nailed down.

    Are you taking a class or working from a book? Have you worked through other examples yet of being given a set with operations, and showing that it is or isn't a vector space?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook