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A question

  1. Nov 16, 2007 #1
    there is a space called V
    wich is defined by f:R>>R

    (R-real numbers)

    does V1={f|f(0)=f(1)} a part of V space??

    i tried to prove it by cheking if 0 vector exists in V1
    and checking multiplication by a constant

    but i cant figure out a way to do it wright because
    i dont have actual (a1,a2,a3,a4) vectors to play with

    ???
     
    Last edited: Nov 16, 2007
  2. jcsd
  3. Nov 16, 2007 #2
    there is a space called V
    wich is defined by f:R>>R

    (R-real numbers)

    does V1={f|f(0)=f(1)} a part of V space??

    i tried to prove it by cheking if 0 vector exists in V1
    and checking multiplication by a constant

    but i cant figure out a way to do it wright because
    i dont have actual (a1,a2,a3,a4) vectors to play with
    ???
     
  4. Nov 16, 2007 #3

    quasar987

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    indeed you can't because the space of all functions from R to R is infinite dimensional, which means there is no base {e1,...,en} that you can use to express the elements of V as an ordered n-tuple.

    but you don't need that; the definition of the 0 vector is simply the element of your space such that v+0=f for any vector v your space. so you must first find what this 0 vector is in the case of your space of function, and then to determine whether V1 is a subspace or not you must asses whether or not this 0 vector is also in V1. That is to say, whether or not it satisfies f(0)=f(1).

    If it is not, then V1 is not a vector space, because it does not satisfy the axiom of existence of a 0 element. But if it is, then you can't conclude anything and you have to actually verify whether or not V1 satisfies the axioms of a subspace.
     
  5. Nov 16, 2007 #4

    quasar987

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  6. Nov 16, 2007 #5

    matt grime

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    You do have vectors to play with. The functions are the 'vectors' - techincally we can't call them vectors yet as we've not shown they form a vector space.

    What is the zero vector? It is the function f such that f(x)=0 for all x. Does this lie in V1? If f and g are functions in V1 is f+g? If r is a real number and f is in V1, is rf in V1?
     
  7. Nov 16, 2007 #6

    HallsofIvy

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    In other words, V is the set of all functions from R to R? (I would write "that R->R".)

    Surely that's not what the question really said! "a part of" or "a subspace of"?

    What do you think vectors are? A vector space over the real numbers is any "algebraic structure" in which we can add vectors and multiply vectors by real numbers. We can certainly add functions (If f and g are functions, then f+ g is the function such that (f+g)(x)= f(x)+ g(x)) and multiply functions by numbers (if f is a function and r is a real number, then rf is the function such that (rf)(x)= r*f(x)).
    Now to decide whether V1 is a subspace of V you need only determine whether it is "closed" under addition and multiplication by numbers. (If it is closed under multiplication by real numbers, then the 0 vector is just the number 0 times any vector in V1 so you don't need to check that separately.)

    Suppose f and g are two functions such that f(0)= f(1) and g(0)= g(1) (in other words, f and g are in V1). What can you say about f+ g? In particular, what can you say about (f+g)(0) and (f+g)(1)?
     
    Last edited: Nov 17, 2007
  8. Nov 16, 2007 #7

    HallsofIvy

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    And don't post it multiple times!! I'm going to merge these and move to the Homework area.
     
  9. Nov 16, 2007 #8
    i know that we need to proove that (f+g)(0) and (f+g)(1) are R type of fuction

    but i cant figure out a reason why they R type too??
     
  10. Nov 16, 2007 #9

    HallsofIvy

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    I have no idea what you mean by that! (f+b)(0) and (f+g)(1) are NOT functions, they are specific values of functions- at x= 0 and x= 1. "R"= "are"? Surely a three letter word is not that difficult to type and would save some confusion!

    The "type" of function you are trying to show they are is the type that has the same value at 0 and 1. What is (f+g)(0)? What is (f+g)(1)? Now use the fact that f(0)= f(1) and g(0)= g(1).
     
  11. Nov 16, 2007 #10
    R is for real numbers

    i need to proove that (f+g)(0) is of Real numbers type
    and that (f+g)(1) is of Real numbers type.

    does (f+g)(0) equals to f(0)+g(0) ??
    then the only thing i can conclude is that(f+g)(1)=(f+g)(0) because f(0)= f(1) g(0)= g(1).
    which prooves that our function i closed under addition.
    am i correct??

    now i want to proove that if i mulitply a vector from V1 by a constact
    i will get a vector from the same group ("R"=real numbers).

    now i am just guessing ,i dont know if its ok
    because this is not the normal vector form:

    if f(0)=f(1)
    then k*f(0)=k*f(1)
    ??
    now i need to proove that the vector 0 exists in V1

    here i dont know what to do
    ??
     
    Last edited: Nov 16, 2007
  12. Nov 16, 2007 #11

    matt grime

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    I think that your lack of command of the English language is holding you back. I suggest you find someone to help you in your own language, as this is proving to be a major barrier.

    "does (f+g)(0) eqauls f(0)+g(0)?"

    Those two things are _defined_ to be equal, thus I also suggest you review the definitions of everything in the question.
     
  13. Nov 17, 2007 #12

    HallsofIvy

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    Yes, (f+g)(x)= f(x)+ g(x). That is exactly the definition of "f+g". You should have learned that long before Linear Algebra.

    Exactly HOW does (f+g)(1)= (f+g)(0) follow from f(0)= f(1) and g(0)= g(1)? It's just one line of equations but you should write it out.

    No, we don't have one function, we have a set of functions. Once you have added the clarification, you will have proved that this set of functions is closed under addition.

    You are missing a line. You need to show that the function (kf) satisfies (kf)(0)= (kf)(1). (kf)(0)= k*f(0)= k*f(1)= (kf)(1) is sufficient.

    What is the 0 vector in this vector space? If we call it [itex]\vec{0}(x)[/itex], what is [itex]\vec{0}(0)[/itex]? What is [itex]\vec{0}(1)[/itex]? Are they the same?

    Strictly speaking, in order to prove a (non-empty) subset of a vector space is a subspace, you don't need to prove that the 0 vector is in the subset, nor that "whenever v is in the subset so is -v". Just prove closure under addition and scalar multiplication. Then you automatically have [itex]\vec{0}= 0*v[/itex] (where v is any vector in the subset) and, if v is in the subset then -v= -1*v is also.
     
    Last edited: Nov 17, 2007
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