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Question regarding n-space and inner product

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data

    I have posted simular questions a couple of times but now I feel I have a better understanding(hopefully).

    Given a Vectorspace M which is defined as a sequence of realnumber [tex]\{r_n\}[/tex] and where [tex]\sum_{r=1}^{\infty} r_n < \infty[/tex]

    Show that M has an innerproduct given by

    [tex]\langle \{r_n\}, \{p_n\}\rangle = \sum_{n=1}^{\infty} r_n \cdot p_n[/tex]

    There is also a question regarding that one needs to show that M is complete with respect to the induced norm [tex]\|\{r_n\}\|_2 = \sqrt{(\sum r_n^2)}[/tex]


    3. The attempt at a solution

    First I will try to verify that M is indeed a Vector Space. From the definition Our space M is defined as a sequences of real numbers [tex]\{r_n\}[/tex]. Thus that leads me to think of it as a Euclidian n-space definition. Which if I remember my linear algebra correct leads to the conclusion that our space M satisfies the axioms of a the Vector Space and hence M is indeed a Vector Space.
    How is that?

    If this is true (if I understand this correctly) then this will lead me to that if [tex]r_n[/tex] is a vector in both M and [tex]\mathbb{R}^n[/tex], and I can choose a scalar called lets call it and choose S, and then I choose a second and third vector in [tex]\mathbb{R}^n[/tex]
    and then prove that the axioms of the inner product can be applied to M successfully.

    How is that?

    Have I understood this correctly?

    Cheers
    Cauchy
     
  2. jcsd
  3. Feb 16, 2009 #2

    Mark44

    Staff: Mentor

    The vectors (sequences) in M are not elements of R^n; the vectors in M have an infinite number of elements, and their components sum to a finite number.

    What you need to show is that the inner product that you have described satisfies the three axioms of an inner product on an inner product space:
    symmetry: <{rn}, {pn}> = <{pn}, {rn}>
    linearity in the first component: <{a*rn}, {pn}> = a<{pn}, {rn}>
    positive-definiteness: <{rn}, {rn}> > 0
     
  4. Feb 16, 2009 #3
    Hi Mark,

    Thank You for your reply. (Maybe this is a stupid question) But how can M be seen as a Vector Space if it doesn't belong to R^n?

    Would you say its enough to show that the three axioms of inner product true and then conclude "Hence that the inner product between (r_n, p_n) exist?

    In Question 2) I have been told to look closely at the sequence from Q1 and then this should allow me deduce the completness?

    Best Regards
    Cauchy
     
  5. Feb 16, 2009 #4

    quasar987

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    The question as you have written it doesn't make sense to me.

    Look at the question again. How is M defined? I'm certain it's not defined as "a sequence of real numbers {r_n} and where [itex]\sum r_n <\infty[/itex]".
     
  6. Feb 16, 2009 #5
    The question is as follows:

    Is says that Let M = [tex]\{\{r_n\}_{n=1}^{\infty}| r_n \in \mathbb{R}, \sum_{n=1}^{\infty} r_n^2 <\infty\}[/tex] Show that M is a Vectorspace with the inner product

    [tex]\langle \{r_n\}_{n=1}^{\infty} \{s_n\}_{n=1}^{\infty} \rangle = \sum_{n=1}^{\infty} r_n \cdot s_n [/tex]

    and show that M is complete with respect to L2 norm.

    How else can one do this than first verify that M is Vector Space and then next verify that the axioms of the inner product applies here ad finally use this to deduce the completeness of M?

    Best Regards
    Cauchy
     
  7. Feb 16, 2009 #6

    quasar987

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    Alright, now the question makes sense. This complete inner product space is usually denoted [tex]\ell^2[/tex].

    I do not think there is any shortcut to the solution. You must verify that all the axioms of a vector space are verified. Then that all the axioms of the inner product are verified.
     
  8. Feb 16, 2009 #7
    My space M isn't that an Euclidian N-Space? Because that deals also with a sequence of real valued vectors. Thusly making into an Vector Space in R^n ?
     
  9. Feb 16, 2009 #8

    quasar987

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    Please re-read yourself before posting because most of your sentences do not make sense.
     
  10. Feb 16, 2009 #9

    Mark44

    Staff: Mentor

    M is NOT a finite-dimensional Euclidean space. As I said in post 2, a vector/sequence in M has an infinite number of components, whereas any vector in R^n has n components.
     
  11. Feb 17, 2009 #10
    Hi Mark what I don't get here. You say M is not a Vector Space? How is that possible since its the premise for the whole assignment?
     
  12. Feb 17, 2009 #11

    HallsofIvy

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    Please read what you are being told more carefully. Mark did NOT say M is not a vector space! He said it is not "finite-dimensional Euclidean space".

    A vector space is any set and operations that satisfy the definition of "vector space". "Euclidean space" is specifically Rn with the usual component wise operations but it can be shown tnat any n dimensional vector space is isomorphic to that. The crucial point is that M is infinite dimensional.

    By the way: you have an error, or mistype, right at the start of your first post. You say that M is the space of all sequences [tex]{r_n}[/tex] of real numbers such that [tex]\sum_{n=1}^\infty r_n< \infty[/tex]. That is wrong. You have the correct requirement later in your post: [tex]\sum_{n=1}^\infty r_n^2< \infty[/tex]. Those two requirements give very different sequences. For example, the sequence [tex]\{\frac{1}{n}\}[/tex] satisfies the second but no the first.
     
    Last edited: Feb 17, 2009
  13. Feb 17, 2009 #12

    Anyway I just thought about something after re-reading my linear algebra book and talked to my professor he said something like this:

    An infinite dimensional Vector Space is defined as

    [tex]\mathbb{R}^\infty = \{(v_1, v_2, \ldots)| v_n \in \mathbb{R}, \forall n\}[/tex]

    Then if Our M is of subspace of [tex]\mathbb{R}^{\infty}[/tex] s.t.

    [tex]M = \{(r_1, r_2, \ldots)| r_n \in \mathbb{R} \mathrm{\ and \ that} \sum_{n=1}^\infty r_n^2 < \infty\}[/tex]

    Then in order to show that M is a subspace of [tex]\mathbb{R}^\infty[/tex] Isn't it then sufficiant to show that for all [tex]r_1, r_2 \in M[/tex]

    1) [tex]s \cdot r_1 \in M[/tex] for some real valued scalar.
    2) [tex]r_1 + r_2 \in M[/tex]

    And then conclude that M is a subspace of [tex]\mathbb{R}^\infty[/tex] and an also a Vectorspace?

    1) This must mean that [tex]s \cdot \sum r_1^2 < \infty[/tex]? Hence since s in an element of [tex]\mathbb{R}[/tex] and [tex]r_n \in \mathbb{R}[/tex], then if we multiply by this scalar then the result must also be in M (Condidition (1) for being a subspace is forfilled).

    2) Any idears on howto show this?

    If we choose two arbitrary vectors in M calling them [tex]r_n^' \in M[/tex] and if

    [tex]\sum(r_n')^2 < \infty \Rightarrow \sum(r_n' + r_n)^2 < \infty [/tex]

    If I can show that sum(1) results in sum(2) then condidation 2 for being a subspace of the infinite dimensional vector space be true!

    If the above is surficiant way of verifying that M is a vector space could someone then please give me a hint on how this will lead me to conclude that M has an inner product?

    Best regards
    Cauchy
     
    Last edited: Feb 17, 2009
  14. Feb 17, 2009 #13

    Mark44

    Staff: Mentor

    No. By your problem description, r1 and r2 are merely coordinates or elements in a sequence, and are real numbers. You have to show that the set M, together with the operations of addition and scalar multiplication, is a vector space. When you have done that, you have to show that the product of two sequences as defined in this problem satisfies the three axioms for an inner product on a vector space.

    Also, item 1 in your list, quoted above, has to hold for any scalars, not just some scalar.
     
  15. Feb 17, 2009 #14
    Just lets take it from the top. My understand that M can considered at a subspace of R_infinity. That hopefully correct?

    So what You are simply saying is simply to show that this is infact the case then I say for any real scalar S in M then [tex]S \cdot \sum_{n=1}^\infty} r_n^2 < \infty[/tex] ???

    and here since both any scalar S is in M and therefore i R(Infinity) then this operation exists in M? Yes?

    and next I say for any vector r' in M then their addition between r exists

    since [tex]\sum_{n=1}^\infty} r_n^2 + \sum_{n=1}^\infty} r_n'^2 < \infty[/tex]

    also exists since both r' and r are seen to be in M? Yes?

    Cauchy
     
  16. Feb 17, 2009 #15

    Mark44

    Staff: Mentor

    Not according to your post #5. "Show that M is a Vectorspace with the inner product ..."
    I assume that this means you have to show that 1) M is a vector space, and 2) that the inner product you have described satisfies the usual axioms.
    M is not a set of scalars; it is a set of sequences.
    Yes, but there are 10 axioms you need to verify if you are going to show that M is indeed a vector space.
     
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