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Isomorphism beetwenn vector space and sub space

  1. Nov 30, 2008 #1
    Hi,

    I have to find a vector space V with a real sub space U and a bijective linear map.

    Here my Ideas and my questions:

    If the linear map is bijective, than dim V = dim U

    Because U is a real sub space the only way to valid this constraint is if the dimension is infinity. I wrote:

    [tex]U \subseteq V ~ f: U \rightarrow V bijective[/tex]

    [tex]dim ~ U = dim ~ V = \infty[/tex]

    [tex]U = x_{1}e_{1} + x_{2}e_{2} + x_{i}e_{n} = \sum\limits_{i,n=1}^{\infty} x_{i}e_{n} \ x_{i} \in k, ~ e_{n} \in U, ~ i,n \in \mathbb{N}[/tex]

    [tex]V = x_{1}e_{1} + x_{2}e_{2} + x_{j}e_{m} = \sum\limits_{j,m=1}^{\infty} x_{j}e_{m} \ x_{j} \in k, ~ e_{m} \in U, ~ j,m \in \mathbb{N}[/tex]

    [tex]\sum\limits_{i,n=1}^{\infty} x_{i}e_{n} = \sum\limits_{j,m=1}^{\infty} x_{j}e_{m} ~ \Leftrightarrow ~ f: U \ \rightarrow V ~ isomorphism[/tex]

    1.) Are my minds up to now correct?

    2.) How to go on? Maybe a complete induction? But I have different indices.

    Thank you
    all the best
     
  2. jcsd
  3. Nov 30, 2008 #2

    morphism

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    I'm assuming you're looking for a proper subspace U, then? (V is a subspace of itself, and there's an obvious isomorphism from V onto V!)

    Your idea of going to infinite dimensions is of course correct. But I don't quite understand what you're doing after that. What is your choice of V? Of U? Of f?
     
  4. Nov 30, 2008 #3

    mathwonk

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    since linear maps of vector spaces are equivalent to functions on their bases, and every set is a basis of some vector space, it suffices to find an injection from a set to itself which is not surjective.
     
  5. Nov 30, 2008 #4
    Hi,

    thanks for help. Maybe it is a good idea if I quote the exercise:

    Find a vector space V and a real sub vector space U in V such that a linear map f from U to V is isomorph. Specify the isomorphism and proof you statement.

    That are all informations in the exercise.

    greetings
     
  6. Nov 30, 2008 #5
    Technically, the exercise doesn't say that U is a proper subspace of V, so you could use the identity map as your isomorphism. But that's probably not what they're asking for ;)

    What's the simplest infinite vector space you can think of? For example, let V be the space of all sequences of real numbers, and look at any infinite-dimensional subspace. (Unfortunately in this case V has no basis (it has no spanning set) so you have to define the isomorphism explicitly, but if you pick your subspace properly then it will still be easy. Alternatively, let V be the space of all real sequences with finitely many nonzero terms; this space has a basis.)
     
    Last edited: Nov 30, 2008
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