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Linear subspaces and dimensions Proof

  1. Sep 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Let Pn denote the linear space of all real polynomials of degree </= n, where n is fixed. Let S denote the set of all polynomials f in Pn satisfying the condition given. Determine whether or not S is a subspace of Pn. If S is a subspace, compute dim S.

    The given condition if f(0)=f(1)

    2. Relevant equations
    Closure axioms:
    (1)Closure under addition: For every pair of elements x and y in V there corresponds a unique element in V called the sum of x and y denoted by x+y.
    (2) Closure under multiplication: For every x and y in V and every real number a there corresponds an element in V called the product of a and x, denoted by ax.


    3. The attempt at a solution
    I
    know that S is a subspace of Pn if S is a subset of Pn and it satisfies the closure axioms. So I need to prove these three things, but they seem trivial to me.

    S c Pn because S: f(0)=f(1) has all elements in Pn. Is this enough of a proof to draw the conclusion?

    For the closure axioms, I know that I have to show that f(x)+g(x) is an element in Pn, but again this seems trivial, I am unsure how to do this. Same for the closure axiom of multiplication.
     
    Last edited: Sep 26, 2014
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  3. Sep 26, 2014 #2

    vela

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    I take is S is defined as ##S = \{ f \in P_n\ \vert\ f(0)=f(1) \}##. If so, why would you think every element of ##P_n## is in S? What about f(x)=x? It's in ##P_n## (if n>0) but ##f(0) \ne f(1)##.

    Or is your reasoning that all elements of S are in ##P_n##, so closure is automatically satisfied? It's true that closure is satisfied for ##P_n##, e.g., f+g will be in ##P_n##, but you can't say the same thing about S, i.e., f+g isn't necessarily in S even if f and g are both in S.
     
  4. Sep 26, 2014 #3

    LCKurtz

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    Is it a secret or are you going to tell us what the "condition given" is?

    Is f(0) = f(1) the secret condition?

    You haven't shown closure, If ##f,g\in S## what does that give you? Then show ##cf## and ##f+g## are in ##S## too.
     
  5. Sep 26, 2014 #4
    I'm sorry I forgot to write the given condition. I just edited the post to add it. Yes, the condition if f(0)=f(1). I am very confused as how to prove all of this. I am trying to prove that S is a subset of Pn and prove the 2 closure axioms for S.
     
  6. Sep 26, 2014 #5

    LCKurtz

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    Isn't ##S## given to be a subset satisfying f(0)=f(1)? You don't have to prove it is a subset.

    In post #4 at the end I asked:

    If f,g∈S what does that give you? Once you answer that you can talk about whether cf and f+g are in S.
     
  7. Sep 26, 2014 #6
    So would I say, let f and g be in S such that they satisfy the given condition. Now f+g exist in S because f(0)+g(0)=f(1)+g(1) therefore the first axiom stands. And for the second cf exists in S because cf(0)=cf(1), therefore the second axiom stands. Is this a valid argument?
     
  8. Sep 26, 2014 #7

    LCKurtz

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    That's the idea, but you could certainly phrase it better. Instead of saying "let f and g be in S such that they satisfy the given condition" you could say: Suppose ##f,g\in S##. Then f(0)=f(1) and g(0) = g(1). Adding these equations gives f(0)+g(0) = f(1)+g(1) so ##f+g\in S##. It is very simple but needs to be stated properly. Similarly for the ##cf## case.
     
  9. Sep 26, 2014 #8
    Thank you, this helps. I knew what the conclusion needed to be but had a hard time phrasing it. Thank you!
     
  10. Sep 27, 2014 #9

    Fredrik

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    Seems like a strange thing to say when all elements of S satisfy the given condition.

    This is how I usually phrase these things: Let ##f,g\in S## be arbitrary. We have
    $$(f+g)(0)=f(0)+g(0)=f(1)+g(1)=(f+g)(1).$$ This implies that ##f+g\in S##.

    This version makes it very clear that we're using the definition of f+g. (If we don't use it, we can't claim to have proved a statement about f+g, so it's pretty essential to make it clear that we're using it). Similarly, you should make it clear that you're using the definition of cf when you prove the other one.
     
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