# 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