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Cardinality and Dimention

by rhobymic
Tags: cardinality, dimention
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rhobymic
#1
Nov19-12, 05:55 PM
P: 5
This is not a homework question .....

If two vector spaces, say V and W, have equal cardinality |V|=|W| .... do they then have the same dimension? That is dim(V)=dim(W)?

I am struggling with making this call one way or the other. This is no area of expertise for me by any means so I know I am missing something important but here are my thoughts:

-> No it does not mean they have the same dim. Dimension is the value of the cardinality of the BASIS vectors of a vector space not the cardinality of the full vector space.


-> Yes it does because if |V|=|W| is true then there is a bijection between V and W and therefor an isomorphic linear transformation T between V and W. This would imply that T carries a basis from V into W and so V and W would have the same cardinality of basis vectors er go the same dimension....


I am still leaning towards "No" because I think the assumption that if V and W are bijective then there is an isomorphic linear transformation is probably not possible...

Thanks for any help!
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micromass
#2
Nov19-12, 07:20 PM
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P: 18,345
Quote Quote by rhobymic View Post
This is not a homework question .....

If two vector spaces, say V and W, have equal cardinality |V|=|W| .... do they then have the same dimension? That is dim(V)=dim(W)?

I am struggling with making this call one way or the other. This is no area of expertise for me by any means so I know I am missing something important but here are my thoughts:

-> No it does not mean they have the same dim. Dimension is the value of the cardinality of the BASIS vectors of a vector space not the cardinality of the full vector space.
This is a nice observation, but it is not a proof. To answer the question as "no", you just need to come up with two vector spaces that have equal cardinality but not equal basis.

-> Yes it does because if |V|=|W| is true then there is a bijection between V and W and therefor an isomorphic linear transformation T between V and W. This would imply that T carries a basis from V into W and so V and W would have the same cardinality of basis vectors er go the same dimension....
I don't really see why an arbitrary bijection would be an isomorphism...
Erland
#3
Nov21-12, 07:41 PM
P: 353
Quote Quote by rhobymic View Post
-> Yes it does because if |V|=|W| is true then there is a bijection between V and W and therefor an isomorphic linear transformation T between V and W. This would imply that T carries a basis from V into W and so V and W would have the same cardinality of basis vectors er go the same dimension....
Thanks for any help!
A bijection needs not be linear. In fact, Cantor proved that R and R2 have the same cardinality, and hence that all Rn (n>0) have the same cardinality. For a proof, see Theorem 2 here. (The proof is actually incomplete, since the author has forgotten the 0.99999.....=1 problem. But it can be fixed.)


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