Isomorphism beetwenn vector space and sub space

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SUMMARY

The discussion centers on finding a vector space V with a real subspace U and a bijective linear map f: U → V that demonstrates isomorphism. Participants confirm that if the linear map is bijective, then the dimensions of V and U must be equal, leading to the conclusion that both dimensions must be infinite. Suggestions include using the space of all sequences of real numbers as V and identifying an appropriate infinite-dimensional subspace U. The identity map is also mentioned as a potential isomorphism, although it may not align with the exercise's intent.

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  • Understanding of vector spaces and subspaces
  • Knowledge of bijective linear maps and isomorphisms
  • Familiarity with infinite-dimensional spaces
  • Basic concepts of linear algebra, including bases and dimensions
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  • Study the space of all sequences of real numbers and its subspaces
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Herbststurm
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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
 
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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?
 
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.
 
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
 
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.)
 
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