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The vector space status of all nonconstant functions

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Let V be the set of all nonconstant functions with operations of pointwise addition and scalar multiplication, having the real numbers as their domain. Is V a vectorspace?


    2. Relevant equations
    None.


    3. The attempt at a solution
    My guess is, no. For example
    F(x) = x2
    Among the axioms for vector space, for an arbitrary element of a vector space u there must be a -u. We can put any number in x and it will end up positive. We can multiply it by any negative scalar, but then squaring it will make it positive.

    Am I right? Am I wrong? Am I right for the wrong reasons?

    If I am right, there's another disturbing question that comes up: the explanatory portions of the chapter I'm working on state that "The set of all functions having the real numbers as their domain, with operations of pointwise addition and scalar multiplication, is a vector space." Well, isn't the set of all nonconstant functions a subset of that set?
     
  2. jcsd
  3. Mar 8, 2013 #2

    Dick

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    Right for the wrong reason. Nonconstant functions are a subset (not a subspace) of all functions. But it's not closed. Find two nonconstant functions that you can add to get a function that is non-nonconstant, i.e. constant.
     
  4. Mar 8, 2013 #3
    Oh, okay. So, for example

    F(x)= x+1
    G(x)= -x +1

    Add them together and get a constant, 1. Is that correct?

    If you have a moment, what was wrong with my original reasoning with F(x)=x^2
     
    Last edited: Mar 8, 2013
  5. Mar 8, 2013 #4

    Dick

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    Yes, that's correct. Except if add them you get 2. As for your first reasoning I simply don't understand it. kx^2 isn't the same thing as (kx)^2.
     
  6. Mar 9, 2013 #5

    D H

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    I agree with Dick that you are "right for the wrong reason". The lack of closure is a killer.

    Another killer: What's your zero vector, the additive identity? It can only be f(x)=0, but that's a constant function, so it's not in your set.
     
  7. Mar 9, 2013 #6
    You are correct sir! Thanks.
     
  8. Mar 9, 2013 #7
    Whoops, yeah, two. Thank you for your help.

    Oh, am I just wrong on the order of operations - the exponential would come first in kx^2, then multiplication by the scalar, which could leave us with negatives?
     
    Last edited: Mar 9, 2013
  9. Mar 9, 2013 #8

    Dick

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    Yes. Exactly.
     
  10. Mar 9, 2013 #9
    Thanks for all the help. I usually end up coming here when I'm exhausted and frustrated, and then we see sloppy mistakes on my part. You're a tremendous help.
     
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