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Shankar Exercise 1.1.3 (Vector Spaces)

  1. Apr 13, 2017 #1
    <Mentor's note: moved from a technical forum, therefore no template.>

    I'm long out of college and trying to teach myself QM out of Shankar's.
    I'm trying to understand the reasoning here because I think that I am missing something...

    1.1.3
    1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
    2) How about periodic functions? obeying f(0)=f(L) ?
    3) How about functions that obey f(0)=4 ?

    If the functions do not qualify, list what go wrong.

    The first 2 questions seem straight forward. X and L go to zero, and outside those boundaries you would get the null vector. If the function is periodic, you are certain to return a vector in the f(x).

    #3 is tripping me up. I thought of an example function... lets say f(x)=x+4, so f(0)=4.
    How is this not a vector space?

    I was told that if g(x) and h(x) were in the set of f(x), g(0)+h(0)=8 which is not 4.

    I do understand how this proves anything? What am I misunderstanding?
     
  2. jcsd
  3. Apr 13, 2017 #2
    1. The problem statement, all variables and given/known data
    I'm long out of college and trying to teach myself QM out of Shankar's.
    I'm trying to understand the reasoning here because I think that I am missing something...

    1.1.3
    1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
    2) How about periodic functions? obeying f(0)=f(L) ?
    3) How about functions that obey f(0)=4 ?

    If the functions do not qualify, list what go wrong.

    The first 2 questions seem straight forward. X and L go to zero, and outside those boundaries you would get the null vector. If the function is periodic, you are certain to return a vector in the f(x).

    #3 is tripping me up. I thought of an example function... lets say f(x)=x+4, so f(0)=4.
    How is this not a vector space?

    I was told that if g(x) and h(x) were in the set of f(x), g(0)+h(0)=8 which is not 4.

    I do understand how this proves anything? What am I misunderstanding?

    2. Relevant equations
    The first 2 questions seem straight forward. X and L go to zero, and outside those boundaries you would get the null vector. If the function is periodic, you are certain to return a vector in the f(x).
    3. The attempt at a solution

    #3 is tripping me up. I thought of an example function... lets say f(x)=x+4, so f(0)=4.
    How is this not a vector space?

    I was told that if g(x) and h(x) were in the set of f(x), g(0)+h(0)=8 which is not 4.

    I do understand how this proves anything? What am I misunderstanding? It said f(0)= 4, not f(x)=4. In my head 8 would be an acceptable answer.
     
  4. Apr 13, 2017 #3

    PeroK

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    Your working shows that you perhaps lack a good enough grasp of linear algebra and formal mathematics. You seem to be confused between an individual function (such as ##f(x)##) and a vector space, which is an infinite set of functions. Normally, you would denote a vector space by ##V## and have something like ##V## is the set of functions that vanish on the endpoints.
     
    Last edited: Apr 13, 2017
  5. Apr 13, 2017 #4

    MathematicalPhysicist

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    Well for 3) it's not closed under addition, take two functions with f(0)=g(0)=4, does h(x)=f(x)+g(x) satisfy this condition?
     
  6. Apr 15, 2017 #5
    3 is a collection of functions all satisfying ##f(0)=4 ##. These functions can't form a vector space over ##\mathbb R ## (or ##\mathbb C## for that matter), because ##2f(0)=8\neq 4##. There is no homogeneity.
     
  7. Apr 15, 2017 #6

    vela

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    The way the question is worded might be confusing you. In each case, you have a set of functions, and the question is asking you if this set along with the usual definitions of adding functions and multiplying by a scalar satisfy the requirements of being a vector space. So first, you have to know all of the conditions required of a vector space. There's probably a list in the book, or if not, you can check Wikipedia.

    What you wrote here doesn't make sense. For the first question, let ##V = \{f\ |\ f(0)=0\text{ and }f(L)=0\}##, and let ##f, g \in V##. The fact that ##f## is an element of ##V## means you know that ##f(0)=0## and ##f(L)=0##. Likewise for ##g##. One of the conditions of a vector space is that of closure under addition. In other words, the function ##f+g## should also be in ##V##. So you can check to see if ##(f+g)(0)=0## and if ##(f+g)(L)=0##. If this is the case, then the function ##(f+g) \in V##. It turns out that's true because ##(f+g)(0) = f(0)+g(0) = 0+0 = 0## and ##(f+g)(L) = f(L)+g(L) = 0+0 = 0##. So ##V## is closed under addition.

    You have to verify all of the other requirements before concluding ##V## is a vector space.

    Given the explanation above, do you now understand why set 3 isn't closed under addition?
     
  8. Apr 15, 2017 #7
    For each of the cases 1), 2) and 3) you will have a set of functions (probably infinite number, but it doesn't effect the problem). To show that they form a vector space ##V##, you have to show that for all members of the set, the conditions in Definition 1 are satisfied.
     
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