1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Infinite dimensions

  1. Apr 6, 2006 #1
    Let V consists of all sequences [x0,x1x,...) of numbers and define vectors operations
    [tex] [x_{0},x_{1},...) + [y_{0},y_{1},...) = (x_{0}+y_{0},...) [/tex]
    [tex] r[x_{0},x_{1},...) = [rx_{0},...} [/tex]

    SHow taht V is a vector space of infinite dimension

    Well for some linear transformation T: V- >V

    dim V = dim(ker T) + dim(im T)
    ker T = {T(v) = 0, v in V}
    i dont see how i can find the dimension of the ker or image for the matter...

    Any ideas/suggestions?
  2. jcsd
  3. Apr 6, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    Why are you looking at linear transformations? And I'm pretty sure that formula you're using only works for finite dimensional V, otherwise it would give equations of the form [itex]\infty = X + Y[/itex]. Either way, it's irrelevant to the question you've asked. First, just show that V is a vector space, i.e. that V under the given operations satisfies the vector space axioms. Then, show that V is not finite dimensional, by showing that no finite collection of vectors in V spans V. To do this, take an arbitrary finite collection of vectors, and construct a new vector that is not in the span of this collection.
  4. Apr 7, 2006 #3
    this arbitrary collection of vectors, should be in V? Since V is collection of all sequences of numbers, it shouldnt matter?

    So form the question isnt
    [tex] V = span[x_{0},x_{1},...) [/tex]
    our finite collection is
    [tex] W = span\{v_{1},v_{2},...,v_{N}\} = r_{1} v_{1} + r_{2} v_{2} + ... r_{N} v_{N}[/tex]
    No matter what the N is, W cannot span V because V doesn't haev a finite N. WE are trying to span an infinite span with a finite span. So it isnt possible? Unless N - > infinity?
  5. Apr 7, 2006 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    I think you are confusing the index denoting which sequence in a finite collection of sequences with the index denoting the term in a single sequence.
  6. Apr 7, 2006 #5
    so each of the v's are supposed to individual collections of numbers themselves??
  7. Apr 7, 2006 #6


    User Avatar
    Science Advisor

    A somewhat more straightforward way to show that V is not finite dimensional is to show that for any natural number n, you can find n + 1 independent vectors in V, so V cannot be n-dimensional for any n.
  8. Apr 7, 2006 #7
    well ok then suppose
    [tex] V = span\{v_{1},v_{2},...,v_{N}\} [/tex]
    [tex] r_{1} v_{1} + r_{2} v_{2} + ... r_{N} v_{N}= 0 [/tex]
    and the r i are zero

    so far so good?
    since V is a collection of all possible numbers there must be exist an [itex] v_{n +1} [/tex] st vn+1 is in V
    but how do i prove that v n+1 is not in the span??
  9. Apr 7, 2006 #8


    User Avatar
    Science Advisor
    Homework Helper

    No. What do you mean the ri are zero? Do you just mean to say that the vi are linearly independent? V is not a collection of all possible numbers, V is a collection of sequences. And what do you mean that "there must exist a vn+1 such that vn+1 is in V"? That sentence there is essentially meaningless, it says nothing more than that V is nonempty. Here are three sequences:

    v = [1, 2, 3, 5, 0, -1, 2, 1, 2, 1, 2, 1, 2, 1, ....)
    w = [0, 0, 1, 2, 1, 1, 1, 2.3, 8.99, 8.99, 8.99, 8.99, 8.99, ...)
    u = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ...)

    Let a = 1, b = 1, c = 2. What is av + bw + cu? Just tell me the first 4 entries in the answer. Now, suppose I tell you that I have some constants a', b', and c' such that a'v + b'w + c'u has the same first three entries as av + bw + cu (which is just v + w + 2u). What can you say about the fourth entry in a'v + b'w + c'u?
  10. Apr 7, 2006 #9
    v + w + 2u = (5,8,14,21,...)

    so for a'v + b' w + c'u the fourth entry may or may not be different?
  11. Apr 8, 2006 #10


    User Avatar
    Science Advisor
    Homework Helper

    It may not be different? How so? How about this: I've told you that a'v + b'w + c'u also has first three entries 5, 8, and 14. Here, the v, w, and u are the same as before, but my a', b', and c' might be different. See if you can solve for a', b', and c'. See if they really can be different. If not, then the fourth entry must be the same.
  12. Apr 8, 2006 #11
    it can't be different because the a',b,',c' are all unique. I dunno what i was thinkin when i was saying taht th3ey may not be different
  13. Apr 8, 2006 #12


    User Avatar
    Science Advisor
    Homework Helper

    What do you mean, "the a', b', c' are all unique"? The point is that if av + bw + cu and a'v + b'w + c'u have the same first three entries, then ALL their entries are the same, and a' = a, b' = b, and c' = c, assuming that {v,w,u} is a linearly independent basis. Can you prove this? If so, what can you say about the possibility of v, w, and u spanning V? Generalize this.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Infinite dimensions
  1. Dimension of a surface (Replies: 2)

  2. Dimension of matrix (Replies: 5)

  3. Finding Dimensions (Replies: 18)

  4. Span and dimension (Replies: 1)