Isomorphism beetwenn vector space and sub space

**Herbststurm** · November 30th 2008, 09:04 AM

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

**clic-clac** · November 30th 2008, 10:05 AM

Hi.

I think I understand what you mean, but take care, the sums $\sum\limits_{i,n=1}^{\infty}x_{i}e_{n}\ \ \& \sum\limits_{j,m=1}^{\infty}x_{j}e_{m}$ are exactly the same, changing indices' letters doesn't change anything.

But indeed if you want to have a vector space in bijection with one of his proper subspaces, then they must have the same infinite dimension.

Why don't you take an example, like $\mathbb{R}[X]$ , whose dimension in infinite (countable), and its subspace $\{P(X) \in \mathbb{R}[X]\ ;\ \deg (P(X)) \in 2\mathbb{N} \}$ .

Is there an isomorphism between them?

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