Isomorphism beetwenn vector space and sub space

1. Nov 30, 2008

Herbststurm

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:

$$U \subseteq V ~ f: U \rightarrow V bijective$$

$$dim ~ U = dim ~ V = \infty$$

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

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

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

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. Nov 30, 2008

morphism

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?

3. Nov 30, 2008

mathwonk

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.

4. Nov 30, 2008

Herbststurm

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

5. Nov 30, 2008