Isomorphism beetwenn vector space and sub space

In summary, the exercise is asking to find a vector space V with a real subspace U and a bijective linear map f from U to V. The linear map must be an isomorphism, meaning that the dimensions of U and V must be the same. To satisfy this constraint, the dimension of U must be infinite. Possible solutions include choosing V as the space of all sequences of real numbers and U as an infinite-dimensional subspace of V, or choosing V as the space of all real sequences with finitely many nonzero terms and U as a proper subspace of V.
  • #1
Herbststurm
30
0
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
 
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  • #2
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
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
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
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:

1. What is the definition of isomorphism between a vector space and a subspace?

An isomorphism between a vector space and a subspace is a bijective linear transformation that preserves the vector space structure, meaning that it maps addition and scalar multiplication in the vector space to addition and scalar multiplication in the subspace.

2. How can we prove that a vector space and a subspace are isomorphic?

To prove isomorphism between a vector space and a subspace, we need to show that there exists a linear transformation that is both injective and surjective. This means that the transformation must be one-to-one and onto, preserving the structure of the vector space.

3. What are the advantages of using an isomorphism between a vector space and a subspace?

Using an isomorphism between a vector space and a subspace allows us to simplify the study of the subspace by using the structure and properties of the vector space. This can make it easier to understand and work with the subspace.

4. Can a vector space and a subspace be isomorphic if they have different dimensions?

No, a vector space and a subspace cannot be isomorphic if they have different dimensions. Isomorphism preserves the dimensions of the vector space, so if the dimensions are different, the two spaces cannot be isomorphic.

5. Are all subspaces of a vector space isomorphic to the vector space itself?

No, not all subspaces of a vector space are isomorphic to the vector space itself. Isomorphism is a specific type of relationship between two spaces, and not all subspaces will have this relationship with the vector space.

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