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Discuss the following assertion:if two rational vector spaces have the same cardinal number(i.e., if there is some one-to-one correspondence between them), then they are isomorphic(i.e., there is linearity-preserving one-to-one correspondence between them).

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Is this homework? The imperative wording looks like it. (Also: consider $\mathbb Q$ versus $\mathbb Q^2$). – Henning Makholm Feb 6 '12 at 18:38
No Bud i am myself working through Paul Halmos's Finite Dimensional vector spaces book and this is a question in there. Since i was not sure about the answer. I thought of seeking help from those who know better. Yes you are right Q versus \mathbb Q^2Q2 also form infinite dimensional vector spaces and should have the same cardinal number, although i am unsure how we would show such a one-to-one correspondence of a vector space perhaps we could use tuples ?? – Comic Book Guy Feb 6 '12 at 18:51
@Hardy: Please provide context in the post, not in comments; some of us consider posts in the imperative to be rather rude, so perhaps you may try, in the future, not to simply post as if you were issuing orders to the readers. – Arturo Magidin Feb 6 '12 at 19:10
Definitely not true if the two spaces are countable. $\mathbb Q^k$ is countable for any finite $k$. – Thomas Andrews Feb 6 '12 at 19:27
@ArturoMagidin Thanks very much for the tip my goal was n't to be rude. I mearly copied the question as was in the book. I would try rephrase in future with this in mind. – Comic Book Guy Feb 6 '12 at 20:15
up vote 3 down vote accepted

In general assuming the axiom of choice then there is a unique cardinality for all maximal linearly independent sets, i.e. bases.

This means that the idea of "dimension" is well defined, and two vector spaces whose dimension is the same are isomorphic. Infinities are quite varied and there are more than one size of infinity.

So there can be two vector spaces over a field $F$, both have an infinite dimension, but alas they are not isomorphic.

However, do note that any nonzero vector space contains a copy of the field, $F$. If this $F$ is infinite (such examples are $\mathbb Q$, $\mathbb R$, and other fields) then the vector space is infinite, as a set. The dimension, however, may still be finite.

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Thank you very much for that. I have n't studied axiom of choice or about different sizes of infinity. Will do ASAP. That was very insightful. – Comic Book Guy Feb 6 '12 at 19:11
@Hardy: The axiom of choice is not needed for developing the theory of finite dimensional linear algebra, but is indeed used in infinitely dimensional linear algebra. As for different sizes of infinity, this is an important fact. For example $\mathbb Q$ and $\mathbb R$ do not have the same cardinality. – Asaf Karagila Feb 6 '12 at 19:28
thanks for the clarification actually i was aware of Q and \mathbb RR not having the same cardinality fact, but not sure if i have come across yet what they are or how they are represented in mathematical literature. – Comic Book Guy Feb 6 '12 at 20:18

No. Any two finite-dimensional vector spaces over $\mathbb{Q}$ have the same cardinality, but two such vector spaces are isomorphic if and only if they have the same dimension (which in general they don't).

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I was under the impression that any vector space over the field q has infinite dimensions, is n't that true ? If it is so then would n't two such vec space over the field Q both have infinite dimension ? – Comic Book Guy Feb 6 '12 at 19:05
I just re-read your post and was thinking of what finite dimensional vector space exists with field Q ? Can you think of an example possibly ? Cheers – Comic Book Guy Feb 6 '12 at 19:06
@Hardy: $\mathbb{Q}^n$ for every positive integer $n$. – Brian M. Scott Feb 6 '12 at 19:12

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